y--^':im: r:-.;':',^;tp-Y<:i- '■1^/' r: fv^„:; > \, ',, '■j"^. .'.,,'.'' '•• w{ i^M^r''- :^h<,^'^}i^:6'- -'''i/!' '■■''''^'' ■j;-.r;: 'iv:'-' :::■:'■] "f. :./.,' ':}!V' r, ■.:-■;;■"■■■ V Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofanalytOOsmitrich THE ELEMENTS OF ANALYTIC GEOMETRY BY PERCEY F. SMITH, Ph.D. Professor of Mathematics in the Sheffield Scientific School Yale University ARTHUR SULLIVAN GALE, Ph.D. Fayerweather Professor of Mathematics in The University of Rochester 1^ GINN & COMPANY BOSTON . NEW YORK • CHICAGO • LONDON ^'>^'?-°' SCNEUl Copyright, 1904, by ARTHUR SULLIVAN GALE ALL RIGHTS RESERVED 99.1 ^fae iatbtnaum l^vtest GINN & COMPANY . PRO- PRIETORS . BOSTON • U.S.A. PREFACE In preparing this volume the authors have endeavored to write a drill book for beginners which presents the elements of the subject in a manner conforming with modern ideas. The scope of the book is limited only by the assumption that a knowledge of Algebra through quadratics must suffice for any investigation. This does not mean a treatise on conic sections. In fact, the authors have intentionally avoided giving the book this form. Conic sections naturally appear, but chiefly as illustrative of general analytic methods. A chapter is devoted to their study, but the numerous properties of these curves are developed inci- dentally as applications of methods of general importance. The subject-matter is rather more than is necessary for the usual course of sixty exercises. It has been made so intentionally, to permit of choice on the part of the teacher, and also in order to include all topics strictly elementary in the sense defined above. The table of contents will show topics not usually treated. For example, in discussing the nature of the locus of the general equation of the second degree (Chapter XII), invariants are introduced. Again, three chapters are devoted to the simple transformations in the plane. After mastering the entire book, the student is assured of an acquaintance with all that is funda- mental in modern Analytic Euclidean Geometry. Attention is called to the method of treatment. The subject is developed after the Euclidean method of definition and theorem. 196499 iv PREFACE without, however, adhering to formal presentation. The advan- tage is obvious, for the student is made sure of the exact nature of each acquisition. Again, each method is summarized in a rule stated in consecutive steps. This is a gain in clearness. Many illustrative examples are worked out in the text. Emphasis has everywhere been put upon the analytic side, that is, the student is taught to start from the equation. He is shown how to work with the figure as a guide, but is warned not to use it in any other way. Chapter III may be referred to in this connection. The same methods have been used uniformly for the plane and for space. In this way the extension to three dimensions is made easy and profitable. Acknowledgments are due to Dr. W. A. Granville for many helpful suggestions, to Professor E. H. Lockwood for suggestions regarding some of the drawings, and to Mr. L. C. Weeks for assistance in proof reading. New Haven, Connecticut December, 1904 CONTENTS CHAPTER I REVIEW OF ALGEBRA AND TRIGONOMETRY SECTION PAGE 1. Numbers 1 2. Constants 1 3. The quadratic. Typical form 2 4. Special quadratics 4 5. Cases when the roots of a quadratic are not independent . .6 6. Variables 10 7. Variation in sign of a quadratic 10 8. Infinite roots . . . 14 9. Equations in several variables 16 10. Functions of an angle in a right triangle 18 11. Angles in general 18 12. Formulas and theorems from Trigonometry .... 19 13. Natural values of trigonometric functions . ' . . . .21 14. Rules for signs 22 15. Greek alphabet . . 22 CHAPTER II CARTESIAN COORDINATES 16. Directed line 17. Cartesian coordinates 18. Rectangular coordinates 19. Angles .... 20. Orthogonal projection . 21. Lengths . 22. Inclination and slope . 23. Point of division 24. Areas .... 25. Second theorem of projection 23 24 25 28 29 31 34 38 42 47 Vi CONTENTS CHAPTER III THE CURVE AND THE EQUATION SECTION PAGE 26. Locus of a point satisfying a given condition 51 27. Equation of the locus of a point satisfying a given condition . 61 28. First fundamental problem 53 29. General equations of the straight line and circle ... 57 30. Locus of an equation 59 31. Second fundamental problem 60 32. Principle of comparison 62 33. Third fundamental problem. Discussion of an equation . . 67 34. Symmetry . .72 35. Further discussion . 73 36. Directions for discussing an equation 74 37. Points of intersection 76 38. Transcendental curves ^ . .79 39. Graphical representation in general . 83 CHAPTER IV THE STRAIGHT LINE AND THE GENERAL EQUATION OF THE FIRST DEGREE 40. Introduction . . .85 41. The degree of the equation of a straight line .... 85 42. The general equation of the first degree, Ax + By -]- C = . . 86 43. Geometric interpretation of the solution of two equations of the first degree 89 44. Straight lines determined by two conditions 92 45. The equation of the straight line in terms of its slope and the coordi- nates of any point on the line 95 46. The equation of the straight line in terms of its intercepts . . 96 47. The equation of the straight line passing through two given points 97 48. The normal form of the equation of the straight line . . , 101 49. The distance from a line to a point 105 50. The angle which a line makes with a second line .... 109 51. Systems of straight lines 113 52. The system of lines parallel to a given line li6 53. The system of lines perpendicular to a given line . . . 117 54. The system of lines passing through the intersection of two given lines . 119 55. The parametric equations of the straight line .... 123 CONTENTS vii CHAPTER V THE CIRCLE AND THE EQUATION ac^ + y"- + I)x + Ey + F = SECTION PAGE 56. The general equation of the circle 130 57. Circles determined by three conditions 132 58. Systems of circles 136 59. The length of the tangent 144 CHAPTER VI POLAR COORDINATES 60. Polar coordinates 149 61. Locus of an equation 150 62. Transformation from rectangular to polar coordinates . . . 154 63. Applications 156 64. Equation of a locus • 157 CHAPTER VII TRANSFORMATION OF COORDINATES 65. Introduction 160 66. Translation of the axes 160 67. Rotation of the axes , . 162 68. General transformation of coordinates 163 69. Classification of loci ' 164 70. Simplification of equations by transformation of coordinates . 165 71. Application to equations of the first and second degrees . . 168 CHAPTER VIII CONIC SECTIONS AND EQUATIONS OF THE SECOND DEGREE 72. Equation in polar coordinates 173 73. Transformation to rectangular coordinates .... 178 74. Simplification and discussion of the equation in rectangular coordi- nates. The parabola, e = \ 178 75. Simplification and discussion of the equation in rectangular coordi- nates. Central conies, e ^ 1 182 76. Conjugate hyperbolas and asymptotes -. 189 77. The equilateral hyperbola referred to its asymptotes . . . 191 viii CONTENTS SECTION PAGE 78. Focal property of central conies 192 79. Mechanical construction of conies 192 80. Types of loci of equations of the second degree . . . .194 81. Construction of the locus of an equation of the second degree . 197 82. Systems of conies 200 CHAPTER IX TANGENTS AND NORMALS 83. The slope of the tangent 207 84. Equations of tangent and normal 210 85. Equations of tangents and normals to the conic sections . . 212 86. Tangents to a curve from a point not on the curve . . . 215 87. Properties of tangents and normals to conies . . . . 217 88. Tangent to a curve at the origin . . . . . . .221 89. Second method of finding the equation of a tangent . . 223 CHAPTER X RELATIONS BETWEEN A LINE AND A CONIC. APPLICATIONS OF THE THEORY OF QUADRATICS 90. Relative positions of a line and conic 226 91. Relative positions of lines of a system and a conic, and of a line and conies of a system 228 92. Tangents to a conic 2.30 93. Tangent in terms of its slope 233 94. The equation in p 235 95. Tangents 237 96. Asymptotic directions and asymptotes . • 238 97. Centers 240 98. Diameters 241 99. Conjugate diameters of central conies 244 CHAPTER XI LOCI. PARAMETRIC EQUATIONS 100. Introduction 248 101. Loci defined by a construction and a given curve . . . 249 102. Parametric equations of a curve 253 103. Loci defined by the points of intersection of systems of curves . 259 CONTENTS ix CHAPTER XII THE GENERAL EQUATION OF THE SECOND DEGREE SECTION PAGE 104. Introduction 204 105. Condition for a degenerate conic 264 106. Degenerate conies of a system 267 107. Invariants under a rotation of the axes 269 108. Invariants under a translation of the axes 273 109. Nature of the locus of an equation of the second degree . . 275 110. Equal conies 278 111. Conies determined by five conditions 279 CHAPTER XIII EUCLIDEAN TRANSFORMATIONS WITH AN APPLICATION TO SIMILAR CONICS 112. Introduction 281 113. Equal figures 281 114. Translations 282 115. Rotations 282 116. Displacements . 284 117. The reflection in a line 287 118. Symmetry transformations 287 119. Congruent and symmetrical conies . . . . . . 291 120. nomothetic transformations 291 121. Similitude transformations 292 122. Similar conies 293 CHAPTER XIV INVERSION 123. Definition 297 124. Equations of an inversion 297 125. Inversion of conic sections 299 126. Angle formed by two circles 303 127. Angles invariant under inversion 304 128. Inversion of systems of straight lines 306 129. Inversion of a system of concentric circles .... 307 130. Orthogonal systems of circles 308 xii CONTENTS CHAPTER XXII QUADRIC SURFACES AND EQUATIONS OF THE SECOND DEGREE IN THREE VARIABLES SECTION PAGE 176. Quadric surfaces ' 397 177. Simplification of the general equation of the second degree in three variables 398 178. The ellipsoid ^ + ^ + ^ = 1 400 a2 62 c2 a^2 y1 ^1 179. The hyperboloid of one sheet ~ -^ ~ = 1 . . . . 401 a^ h: c^ ^•2 yl ^1 180. The hyperboloid of two sheets = 1 . . . .402 181. The elliptic paraboloid ^ + ^ = 202 . . . . . 405 182. The hyperbolic paraboloid ---- =r 2 cz . . > . . 406 183. Rectilinear generators 408 CHAPTER XXIII RELATIONS BETWEEN A LINE AND QUADRIC. APPLICATIONS OF THE THEORY OF QUADRATICS 184. The equation in p. Relative positions of a line and quadric . 410 185. Tangent planes 411 186. Polar planes 412 187. Circumscribed cones 412 188. Asymptotic directions and cones ....... 415 189. Centers 419 190. Diametral planes 419 Index 423 i AINTALYTIC GEOMETRY CHAPTER I REVIEW OF ALGEBRA AND TRIGONOMETRY 1. Numbers. The numbers arising in carrying out the opera- tions of Algebra are of two kinds, real and imaginary. A real number is a number whose square is a positive number. Zero also is a real number. A pure imaginary number is a number whose square is a nega- tive number. Every such number reduces to the square root of a negative number, and hence has the form 6 V— 1, where ^ is a real number, and (V— 1)^ = — 1. An imaginary or complex number is a number which may be written in the form a-\-b V— 1, where a and b are real numbers, and b is not zero. Evidently the square of an imaginary number is in general also an imaginary number, since (a^b V^)2 = a''-b''-\-2ab V^^, which is imaginary if a is not equal to zero. 2. Constants. A quantity whose value remains unchanged is called a constant. Numerical or absolute constants retain the same values in all problems, as 2, — 3, Vl, tt, etc. Arbitrary constants, or parameters, are constants to which any one of an unlimited set df numerical values may be assigned, and these assigned values are retained throughout the investigation. Arbitrary constants are denoted by letters, usually by letters from the first part of the alphabet. In order to -increase the number of symbols at our 1 2 . ANALYTIC GEOMETRY disposal, it is convenient to use primes (accents) or subscripts or both. For example : Using primes, a' (read "a prime or a first ")j a'' (read "a double prime or a second"), a'''(read "a third"), are all different constants. Using subscripts, &i (read " b one "), b^ (read " b two "), are different constants. Using both, c/ (read "c one prime"), Cg'' (read "c three double prime"), are different constants. 3. The quadratic. Typical form. Any quadratic equation may by transposing and collecting the terms be written in the Typical Form (1) Ax'^ + Bx + C = 0, in which the unknown is denoted by x. The coefficients A, B, C are arbitrary constants, and may have any values whatever, except that A cannot equal zero, since in that case the equation would be no longer of the second degree. C is called the con- stant term. The left-hand member (2) Ax"" -\-Bx-\-C is called a quadratic, and any quadratic may be written in this Typical Form, in which the letter x represents the unknown. The quantity B^ — AAC is called the discriminant of either (1) or (2), and is denoted by A. That is, the discriminant A of a quadratic or quadratic equa- tion in the Typical Form is equal to the square of the coefficient of the first power of the unknown diminished by four times the product of the coefficient of the second power of the unknown by the constant term. The roots of a quadratic are those numbers which make the quadratic equal to zero when substituted for the unknown. The roots of the quadratic (2) are also said to be roots of the quadratic equation (1). A root of a quadratic equation is said t() satisfy that equation. l-^--^l^r^^c. REVIEW OF ALGEBRA AND TRIGONOMETRY 3 In Algebra it is shown that (2) or (1) has two roots, x^ and x^, obtained by solving (1), namely, (3) Adding these values, we have (4) x^^-x^=--' Multiplying gives (5) ^ -2, a;<5, are written in the more compact form -2 5 or x = 5 ,, are abbreviated to a; < — 2 and a; > 5. 7. Variation in sign of a quadratic. In many problems it is important to determine the algebraic signs of the results obtained by substituting in a quadratic different values for the variable unknown, that is, to determine the algebraic signs of the values of a quadratic for given values of the variable. The discussion of this question depends upon the definitions of greater and less already given, the precise point necessary being the statement : If a is a given real constant and x a real variable, then J when xa, x — a is a positive number. By the aid of this statement and the identities (7), p. 4, we easily prove * The meaning of greater and less for real numbers (§ 1) is defined as follows : a is greater than b when a-bisa, positive number, and a is less than b when a -fe is negative. Hence any negative number is less than any positive number ; and if a and b are both negative, then a is greater than b when the numerical value of a is less than the numer- ical value of b. Thus 3<5, but -3>-5. Therefore changing signs throughout an inequality reverses the inequality sign. REVIEW OF ALGEBRA AND TRIGONOMETRY 11 Theorem III. If the disGriminant of a quadratic is positive, the value of the quadratic* and the coefficient of the second power differ in sign for all values of the variable lying between the roots, and agree i7i sign for all other values. If the discriminant is zero or negative, the value of the quadratic and the coefficient of the second power always agree in sign. Proof Denoting the variable by x, and writing the quadratic in the Typical Form, (1), p. 2, we have, by (7), p. 4, Case I. Ax^ -[- Bx + C = A (x — x-^ (x — x^ if A is positive. Case II. Ax'^ -{- Bx -\- C = A{x — x-^'^ if A is zero. (a) + — 1 + ^ ^, if A i« negciiive. ^ ^ Consider these cases in turn. Case I. Since the roots are unequal, let x^ < x^. Then, by (1), we have at once {x — iCi) (x — X2) is negative when x^<.x<. x^, since a: — Xi is positive, and x — x^\s> negative ; {x — £Ci) {x — CC2) is positive when xKx^oy x> x^, since x — x^ and x — x^ are both negative or both positive. Therefore the quadratic has the sign of — ^ in one case, and of A in the other. Case II. Since {x — cci)^ is positive (p. 1), the sign of the quadratic agrees with that of A. Case III. Since A is negative, ^ AC — B'^ = — ^ \^ positive ; hence the expression within the brackets is always positive, and the sign is the same as that of A. q.e.i>. For example, consider the quadratic 2 ^2 _ 3 ^ + 1. Here A = 9-8=+l, vl = 2, and the roots are \ and 1. ' ... 2r2-3< + l = 2(i- 0(«-l). *It is assumed that all the numbers involved are real. Also, since the value of the quadratic is zero for a value of the variable equal to a root, any such value of the variable is excluded. 12 ANALYTIC GEOMETRY If now any real number be substituted for t in the quadratic, it will be found that when l 1, the quadratic 2 <2 _ 3 ^ 4. 1 > 0. Again, consider the quadratic in r, 3 y2 + 4 r + 9. Here A = 16 - 108 == - 92, and A = Z. Hence, by Theorem III, if any real number whatever be substituted for r, the result will always be a posi- tive number. Applications of Theorem III. The following examples illustrate appli- cations of Theorem III. Ex. 1. Determine all real values of the variable for which the following radicals are real. (a) V3- 2x-x2; (b) V2 ^2 + 3 ?/ + 9. Solution. Consider the quadratic under the radical. In (a), A = 4 + 12 =: 16, ^ = — 1, and the roots are 1 and — 3. Applying Theorem III, when — 3 < X < 1, the quadratic 3 — 2x — x2>0; when x< — 3orx>l, the quadratic 3 — 2 x — x^ < 0. Since under the condition of the problem the given quadratic must be either positive or zero, we have — 3 ; (b) the roots are imaginary if - 8 {k^ + A; - 2) < 0. REVIEW OF ALGEBRA AND TRIGONOMETRY 13 Applying Theorem III to the quadratic -S{k^ + k-2), we have, since A = 64 + 512 = 576, J. = — 8, and the roots are — 2 and 1, when - 2 < A; < 1, the quadratic - 8 (A:^ + fc - 2) > ; when A; < - 2 or A; > 1, the quadratic - 8 (A;2 + A; - 2) < 0. Hence (a) the roots of (2) are real and unequal if — 2 < A; < 1 ; (b) the roots of (2) are imaginary ifA;< — 2orA:>l. -4ns. Ex. 3. Show that the simultaneous equations (4) y = mx + 3 (5) ^ 4x2 + 2/2 4-6x -16 = have two real and distinct common solutions for every real value of m. Solution. Substituting the value of y from (4) in (5), and arranging the result in the Typical Form, we get (6) (4 + m2) x2 + (6 771 + 6) X - 7 = 0. Calculating the discriminant of (6), we find, neglecting the positive factor 4, (7) 16 m2 + 18 m + 37. Applying Theorem III, p. 11, to the quadratic (7), A = 324 - 64 • 37 is negative, A = 16. Therefore the quadratic (7) has a positive value for every real value of ?n, and hence the roots of (6) are, by Theorem II, p. 3, always real and unequal. That is, (6) always has two real roots, Xi and X2, and from (4) we find the corresponding real values of y, namely, i/i and 2/2, so that the equations (4) and (5) have two real and distinct common solutions, (Xi, yi), (X2, 2/2), for every value of m. q.e.d. PROBLEMS 1. Write inequalities to express that the values of the variable named are limited as stated. (a) X has any value from to 5 inclusive. (b) y has any positive value. (c) t has any negative value. (d) X has any value less than — 2 or greater than — 1. (e) r has any value from — 3 to 8 inclusive. (f) z has any negative value, or any positive value not less than 3. (g) X has any value not less than — 8 nor greater than 2. 14 ANALYTIC GEOMETRY 2. Determine the sign of each df the quadratics of the first problem on p. 8 for all values of the variable. 3. Determine all real values of the variable for which the square root of the quadratics of problem 1, p. 8, are real. 4. Determine all real values of the parameter for which the roots of each equation of problem 4, p. 8, are (a) real and unequal ; (b) imaginary. 5. In problem 6, p. 9, find all real values of the parameter in each case such that the two common solutions are (a) real and unequal ; (b) imaginary , 6. Determine the algebraic sign of the value of the cubic 2(x + l)(x-2)(x-4) for any value of the variable. Hint. In this case the roots are -1, 2, 4 in the order of magnitude. Hence, when x<—l, each factor is negative [(1), p. 10] and the cubic is negative, etc. Ans. For x < — 1, cubic <0; — l j 2 < x < 4, cubic < ; 4 < X, cubic > 0. 7. Determine the sign of the value of each of the following quantics for any value of the variable. Hint. From Algebra we know that any quantic with real coefficients may be resolved into real factors of the first and second degrees. The sign of each factor for any value of the variable may then be determined by (1), p. 10, and Theorem III, p. 11. It is well first to arrange the real roots of the quantic in the order of magnitude, and then it is necessary to consider only values of the variable less than any root, lying between each successive pair, and greater than any root, as in problem 6. (a) (X + 1) (2 x2- 4 X + 7). (f) (x2 - 9) (x2 - 16) (x2 - 25). (b) (x2-2x-3)(x3-4x2). (g) (3x2-12) (2-x)(3-2x)(5x+4). (c) (3x + 8)(x2-4x + 4)(x3-l). (h) (x - 1)2(3 + 2x)(4 - 5x) (6 -x)^. (d) (2 x2 -f 3) (x2 - 4) (x4 - 1). (i) 7 (x2 - 4) (9 - x2) (16 - x2). (e) (2x + 3)(x-l)(x + 2)(x-3). (j) (x2 - 8) (2x2 - 8) (3x2 - 27). (k) (2x + 8)2(9-3x)(7-6x)(12-llx). 8. Infinite roots. Consider the quadratic equation (1) Ax"^ -{-Bx+C = 0, whose roots are x^ and x^ [(3), p. 3]. Then the equation (2) Cx'^-{-Bx-{-A=0, REVIEW OF ALGEBRA AND TRIGONOMETRY 15 obtained from (1) by reversing tbe order of the coefficients, has the roots* — and — j that is, the reciprocals of the roots of (1). Let us now fix the values f of ^ and C, but allow A to dimin- ish indefinitely in numerical value, that is, allow A to approach zero. Then, in (2), since — . (Theorem I, p. 3) is the product of the roots, this product must also approach zero. Therefore one root of (2) must approach zero ; and hence its reciprocal, that is, one root of (1), must increase indefinitely. Again, let us in (1) and (2) fix the value t of C only, and assume that both B and A approach zero. Then, in (2), both the B A sum, ) and the product, — ? of the roots approach zero, and c c hence both roots also approach zero. Hence their reciprocals, the roots of (1), must increase indefinitely. This reasoning establishes Theorem IV. If the coefficient of the second power in a quadratic equation is variable and approaches zero as a limit, then one root of the equation becomes infinite, t If the coefficient of the first power is also variable and approaches zero as a limit, then both roots become infinite. Ex. 1. What value must the variable k approach as a limit in order that a root of the equation 3x2 + 2A;x - A;2x2 - 3 - 2A:x2 = may become infinite ? Solution. Arranging the equation in the Typical Form, we have (A;2 + 2 fc - 3) x2 - 2 A:x + 3 = 0. If fc2 + 2 fc — 3 = 0, then one root must become infinite. Hence k must approach 1 or — 3. Ans. * This theorem is demonstrated in Algebra and may be easily verified thus : The equation whose roots are — and — \s(x ^ (x ^ = 0. a?, X2 \ Xi/ \ xj Multiplying out and reducing, this becomes x^x^ • a;' - (aT^ + a^gV a: + 1 = 0. O R By Theorem I, p. 3, XyX^ = - . x^ + x^ = , and substitution of these values and multiplication by A gives (2). t We give C a value diif erent from zero. X A variable whose numerical value becomes greater than any assigned number is said to " become infinite." 16 ANALYTIC GEOMETRY Ex. 2. What values must k and m approach in order to make both roots of the equation (62 _ a2?n2) x2 _ 2 a^kmx - a^k^ - aW = become infinite ? Solution. By Theorem IV we must have 52 _ ^2^2 ^0, or fw = ± - , and 2 a^km = 0, or k = 0. Hence m must approach + - or , and k must approach zero. Ans. PROBLEMS 1. What real value must the parameter approach as a limit in each of the following equations in order to make a root become infinite ? (a) kz^-Sx-\-5 = 0. (d) (m^ - i)x^ - 3x + 8 = 0. (b) (A:2 - l)x2 + 6x - 5 = 0. (e) (c2 - 3)y^ + 2cy - 6 = 0. (c) 2 x2 - 3 X + A;2x2 + 5 = kx^. (f ) 2 62^2 - 3 ?/ - 3 5?/2 + 2 = - 2 2/2. 2. What real values must the parameters k and m approach in order that both roots of each of the following equations may become infinite ? (a) m2x2 ^(2k-m + l)x + 6 = 0. (b) (m2 - 3 m + 2) 2/2 + (3 fc - 2 m) y + 2 = 0. (c) (m2 ^ k'^ - 25)P + {m - 7 k + 2b)t + 8 = 0. (d) m2x2 4- 3A;x + A:2x2 - 4mx + 25x - 25x2 = 2. (e) (m2 + 3)x2 + (2 fc - 5)x + 8 = 0. 8. Equations in several variables. In Analytic Geometry we are concerned chiefly with equations in two or more variables. An equation is said to be satisfied by any given set of values of the variables if the equation reduces to a numerical equality when these values are substituted for the variables. For example, x = 2,y=—S salasfy the equation «;2a;2 + 3?/2=35, since 2(2)2 + 3 ( - 3)2 = 35. Siitti\^rly, x = — 1, j/ = 0,- z = — 4 satisfy the equation 2a;2_3?y2_|.22_i8 = o, since 2(- 1)2 _ 3.O + (- 4)2 - 18 = 0. REVIEW OF ALGEBRA AND TRIGONOMETRY 17 An equation is said to be algebraic in any number of variables, for example x, y, «, if it can be transformed into an equation each of whose members is a sum of terms of the form ax^^y^z^, where a is a constant and m, n,p are positive integers or zero. Thus the equations x* + xhf'^ — z^ -\-2x — h = Q, x^y + 2 x'hf^ =r_?/3 4.5a;2 + 2-a; are algebraic. The equation x^ -\-y^ — a^ is algebraic. For, squaring, we get a; + 2 x^y^ -{-y — a. Transposing, 2 xhj^ — a — x — y. Squaring, 4iXy = a^ + x^ -\- y'^ — 2 ax — 2 ay ■}- 2xy. Transposing, x^ -{- y^ — 2xy — 2 ax — 2 ay -{- a^ ~ 0. q.k.d. The degree of an algebraic equation is equal to the highest degree of any of its terms.''^ An algebraic equation is said to be arranged with respect to the variables when all its terms are transposed to the left-hand side and written in the order of descending degrees. For example, to arrange the equation 2x'^ + 3y'-{-6x'— 2x'y' — 2 + x'^= x'^y'—y'^ with respect to the variables x', y\ we transpose and rewrite the terms in the order x'^ — x'Y + 2aj'2 _ 2xY + y'2 + ex' + 3?/' — 2 = 0. This equation is of the third degree. An equation which is not algebraic is said to be transcendental. Examples of transcendental equations are y = sin x, y— 2^, log y = 3 x. PROBLEMS 1. Show that each of the following equations is algebraic; arrange the terms according to the variables x, y, or x, jsfeg, and determine the degree. (a) x2 + Vy-5 + 2 X = 0. (b) x^ + y + 3 X = 0. (c) xy + 3 X* + 6 x2?/ - 7 xy8 -f- (d) X + y + z 4- x^^; - 3xy - 2g2 (e) 2/ = 2 + Vx2 - 2 X - 5. The degree of any term is the sum of the exponents of the variables in that term. 18 ANALYTIC GEOMETRY (f)2/ = (g) ^ = X + 5 + V2 x2 - -6x + 3. -Ey- -2/2. (h) y = ^x + -B + Vix2 + ifx + iV". 2. Show that the homogeneous quadratic * ^x2 + 5X2/ + C?/2 may be written in one of the three forms below analogous to (7), p. 4, if the discriminant A = JB^ — 4 ^ (7 satisfies the condition given : Case I. Ax'^ + Bxy + Gy'^ = A{x- hy) (x - hy), if A > ; Case II. Ax^ + Bxy + Cy^=A{x- hyf, if A = ; r/ B \2 4AC-B^ Case III. Ax^ + Bxy + Cy^ = A\ (x y) + ],. A<0. 2A / 4^2 10. Functions of an angle in a right triangle. In any right triangle one of whose acute angles is A, the functions of A are defined as follows : sin A = opposite side hypotenuse adiacent side cos A = -T^ — 7 : hypotenuse opposite side adjacent side tan^ csc^ = sec A = cot A hypotenuse opposite side hypotenuse adjacent side adjacent side opposite side From the above the theorem is easily derived : ^ In a right triangle a side is equal to the product of the hypotenuse and the sine of the angle opposite to that side, or of the hypote- a nuse and the cosine of the angle adjacent to that side. A 6 (7 11. Angles in general an angle XOA is considered as gen- erated by the line OA rotating from an initial position OX. The angle is positive when OA rotates from OX counter-clockwise, and negative when the direction of rotation of OA is clockwise. In Trigonometry * The coefficients A, B, C and the numbers l^, Z,, are supposed real. REVIEW OF ALGEBRA AND TRIGONOMETRY 19^ The fixed line OX is called the initial line, the line OA the terminal line. Measurement of angles. There are two important methods of measuring angular magnitude, that is, there are two^unit angles. Degree measure. The unit angle is ^^^^ of a complete revolu- tion, and is called a degree. Circular measure. The unit angle is an angle whose subtend- ing arc is equal to the radius of that arc, and is called a radian. The fundamental relation between the unit angles' is given by the equation 180 degrees = ir radians (tt = 3.14159 • • •). Or also, by solving this, 1 degree = j|^ = .0174 . • • radians. 180 1 radian = ^^ = 57.29 • TT • • degrees. These equations enable us to change from one measurement to another. In the higher mathematics circular measure is always used, and will be adopted in this book. The generating line is conceived of as rotating around through as many revolutions as we choose. Hence the important result : Any real number is the circular measure of some angle, and conversely, any angle is measured by a real number. 12. Formulas and theorems from Trigonometry. Ill 1. cotx = ; secx = ; cscx = -; tan X cos x sin x sin a; , cosx 2. tanx = ; cotx = , cosx sinx 3. sin2x + cos2x = 1 ; 1 + tan2x = sec^x ; 1 + cot2x = csc2x. 4. sin ( — x) = — sin x ; esc ( — x) = — esc x ; cos ( — x) = cos X ; sec ( — x) = sec x ; tan (— x) = — tanx ; cot (— x) = — cot x. 20 ANALYTIC GEOMETRY 5. sin (tt — x) = sin x ; sin {tt -\- x) = — sin x ; cos (tt - x) = — cos X ; cos (tt + ic) =: — cos x ; tan (tt — x) = — tan x ; tan (tt + x) = tan x ; 6. sin ( X j = cos X ; sin ( -- + x j == cos x ; cos ( X ) = sin X ; cos ( — j- a^ ) = — sin x ; tan( X )= cotx; tan(~- + X j = — cotx. 7. sin (2 TT - x) = sin (- x) = - sin x, etc. 8. sin [x -\- y) = sin x cos y + cos x sin y. 9. sin (x — y) = sin x cos y — cos x sin y. 10. cos (x + ?/) = cos X cos ?/ — sin x sin ?/. ' , ; 1 1 . cos (x — y) = cos X cos 2/ + sin x sin ^. tan X + tan y 10*/ \ tan x - tan y 12. tan {x + y)= --^- 13. tan {x - y) - 1 - tan X tan 2/ 1 + tan x tan y 2tanx 14. sin 2 X = 2 sin x cos x ; cos 2 x = cos^ x — sin^ x ; tan 2 x 1 - tan2x X ll — cosx X /l + cosx , X /I — 16. sin-=±i^-^— ;cos-=±^-^— ;tan-=±^^ 2\22 \22 \l + cosx 16. Theorem. Law of sines. In any triangle the sides are proportional to the sines of the opposite angles ; that is, sin A sin B sin C 17. Theorem. Law of cosines. In any triangle the square of a side equals the sum of the squares of the two other sides diminished by twice the product of those sides by the cosine of their included angle ; that is, a2 _ 52 _f. c2 _ 2 6c cos^. 18. Theorem. Area of a triangle. The area of any triangle equals one half the product of two sides by the sine of their included angle ; that is, area = ^ a6 sin C = i 6c sin -4 = ^ ca sin B. REVIEW OF ALGEBRA AND TRIGONOMETRY 21 13. Natural values of trigonometric functions. Angle in Radians Angle in Degrees Sin Cos Tan Cot .0000 0° .0000 1.0000 .0000 00 90° 1.5708 .0873 5° .0872 .9962 .0875 11.430 85° 1.4835 .1745 10° .1736 .9848 .1763 5.671 80° 1.3963 .2618 15° .2588 .9659 .2679 3.732 75° 1.3090 .3491 20° .3420 .9397 .3640 2.747 70° 1.2217 .4.363 25° .4226 .9063 .4663 2.145 65° 1.1345 .5236 30° .5000 .8660 .5774 1.732 60° 1.0472 .6109 35° .5736 .8192 .7002 1.428 55° .9599 .6981 40° .6428 .7660 .8391 1.192 50° .8727 .7854 45° .7071 .7071 . 1.0000 1.000 45° .7854 Cos Sin Cot Tan Angle in Degrees Angle in Radians Angle in Radians Angle in Degrees Sin Cos Tan Cot Sec Csc 0° 1 00 1 00 It 2 90° 1 GO 00 1 It 180° • -1 00 -1 00 2 270° -1 CO ° 00 -1 2Tt 360° 1 00 1 00 22 ANALYTIC GEOMETRY Angle in Radians Angle in Degrees Sin Cos Tan Cot Sec Csc 0° 1 CO 1 CO Tt 6 30° 1 2 V3 2 V3 3 V3 2V3 3 2 It 4 45° 2 2 1 1 V2 V2 It 3 60° V3 2 1 2 V3 V3 3 2 2V3 3 Tt 2 90° 1 CO CO 1 14. Rules for signs.. Quadrant Sin Cos Tan Cot Sec Csc First . " . . . + + + + + + Second . . . + - - - - + Third. . . . - - + + - - Fourth . . . - + - - + - 9 15. Greek alphabet. Letters Names Letters Names A a Alpha I L Iota / B^ Beta K K Kappa r 7 Gamma A X Lambda A 5 Delta M/i Mil E e Epsilon N V Nu z f Zeta S 1 Xi H^ Eta Omicron e d Theta n IT Pi \r JU-t ^ ^? Mf^ ■ Letters Names P^ Rho S (T S Sigma T T Tau T V Upsilon 4. Phi Xx Chi ^ 1// Psi ft U) Omega CHAPTER II CARTESIAN COORDINATES 16, Directed line. Let X'X be an indefinite straight line, and let a point 0, which we shall call the origin be chosen upon it. Let a unit of length be adopted and assume that lengths measured from to the right are positive, and to the left negative. -5-4-3-2-1 O-hl-t-2 +3 + 4+5 unit 1 Then any real number (p. 1), if taken as the measure of the length of a line OP, will determine a point P on the line. Conversely, to each point P on the line will correspond a real number, namely, the measure of the length OP, with a positive or negative sign according as P is to the right or left of the origin. The direction established upon X^X by passing from the origin to the points corresponding to the positive numbers is called the positive direction on the line. A directed line is a straight line upon which an origin, a unit of length, and a positive direction have been assumed. An arrowhead is usually placed upon a directed line to indicate the positive direction. If A and B are any two points of a directed line such that ' 0A = a, OB = b, then the length of the segment AB is always given hj h — a; that is, the length oi AB is the difference of the numbers correspond- ing to B and A. This statement is evidently equivalent to the following definition : 23^ 24 ANALYTIC GEOMETRY For all positions of two points A and B on a directed line, the length AB is given by (1) AB = OB - OA, where is the origin. (1) (II) (111) (\Y) +3 -f-6 -4 0+3 -3 -h5 -6 -2 6 A B B A A B B A Illustrations. In Fig. I. ^5 =05 -0^ = 6-3 = 4-3;^^ ==0^-0^=^3-6 = - 3; II. AB:=OB-OA^-^-?> = -l; BA= 0^ - Oi? = 3- (- 4) =+7; III. ^i5= OjB-0^ = + 5-(-3)= + 8; j5.4=0^ -05 =- 3-5 = -8; IV. AB= OJ5-0^=-6-(-2) = -4; 5^=0^-05 = -2-(-6)= +4. The following properties of lengths on a directed line are obvious : ^^ (2) AB = -BA. (3) AB is positive if the direction from A to B agrees with the positive direction on the line, and negative if in the contrary direction. The phrase " distance between two points " should not be used if these points lie upon a directed line. Instead, we speak of the length AB, remembering that the lengths AB and BA are not equal, but that AB = — BA. 17. Cartesian* coordinates. Let X'X and Y'Y be two directed Yf lines intersecting at 0, and P let P be any point in their plane. Draw lines through P parallel to X'X and TY respectively. Then, if OM = a, 0N = b, the numbers a, b are called the Cartesian coordinates of P, a the abscissa and b the ordinate. The directed lines X'X and Y'Y are called the * So called after Ren^ Descartes, 1596-1650, who first introduqed the idea of coordinates into the study of Geometry. CARTESIAN COORDINATES 25 axes of coordinates, X^X the axis of abscissas, Y^Y the axis of ordinates, and their intersection O the origin. The coordinates a, b of P are written (a, h), and the symbol P (a, b) is to be read : " The point P, whose coordinates are a and b:' Any point P in the plane determines two numbers, the coordi- nates of P. Conversely, given two real numbers a' and b\ then a point P' in the plane may always be constructed whose coordi- nates are (a', b^). For lay off OM' = a', ON' = b', and draw lines parallel to the axes through AI' and N'. These lines intersect at P'(a',b'). Hence Every point determines a pair of real numbers, and conversely, a pair of real numbers determ,ines a point. The imaginary numbers of Algebra have no place in this repre- sentation, and for this readon elementary Analytic Geometry is concerned only with the real numbers of Algebra. 18. Rectangular coordinates. A rectangular system of coordi- nates is determined when the axes X'X and Y'Y are perpendicular (oyTJ H G) X' (10, (-5 \-4} 4) to each other. This is the usual case, and will be assumed unless otherwise stated. 26 ANALYTIC GEOMETRY The work of plotting points in a rectangular system is much simplified by the use of coordinate or plotting paper, constructed by ruling oft" the plane into equal squares, the sides being parallel to the axes. In the figure, p. 25, several points are plotted, the unit of length being assumed equal to one division on each axis. The method is simply this : Count off from along Z'Z a number of divisions equal to the given abscissa, and then from the point so determined a number of divisions equal to the given ordinate, observing the Rule for signs : Abscissas are positive or negative according as they are laid off to the right or left of the origin, Ordinates are positive or negative according as they are laid Second First off abovc or bclow the axis of X. ^~' ^ ' ^ Rectangular axes divide the plane into four X' Third -^ portions called quadrants ; these are numbered (^,_) as in the figure, in which the proper signs of the coordinates are also indicated. PROBLEMS 1. Plot accurately the points (3, 2), (3, - 2), (- 4, 3), (6, 0), (- 5, 0), (0, 4). 2. Plot accurately the points (1, 6), (3, - 2), (- 2, 0), (4, - 3), (- 7, - 4), (- 2, 4), (0, - 1), ( V3, V2), (- V5, 0). 3. What are the coordinates of the origin? Ans. (0, 0). 4. In what quadrants do the following points lie if a and h are positive numbers: (-a, 6)? (-a, -6)? (6, -a)? (a, &)? 5. To what quadrants is a point limited if its abscissa is positive ? nega- tive ? its ordinate is positive ? negative ? 6. Plot the triangle whose vertices are (2, — 1), (— 2, 5), (- 8, — 4). 7. Plot the triangle whose vertices are (- 2, 0), (5 V3 - 2, 5), (- 2, 10). 8. Plot the quadrilateral whose vertices are (0, — 2), (4, 2), (0, 6), (-4,2). p^ CARTESIAN COORDINATES 27 9. If a point moves parallel to the axis of x, which of its coordinates remains constant ? if parallel to the axis of y ? 10. Can a point move if its abscissa is zero ? Where ? Can it move if its ordinate is zero? Where? Can it move if both abscissa and ordinate are zero? Where will it be? 11. Where may a point be found if its abscissa is 2? if its ordinate 12. Where do all those points lie whose abscissas and ordinates are equal ? 13. Two sides of a rectangle of lengths a and h coincide with the axes of X and y respectively. What are the coordinates of the vertices of the rec- tangle if it lies in the first quadrant ? in the second quadrant ? in the third quadrant ? in the fourth quadrant ? 14. Construct the quadrilateral whose vertices are (— 3, 6), (— 3, 0), (3, 0), (3, 6). What kind of a quadrilateral is it ? 15. Join (3, 5) and (-3, - 5); also (3, - 5) and (- 3, 5). What are the coordinates of the point of intersection of the two lines ? 16. Show that (x, y) and (x, — y) are symmetrical with respect to X'X] (x, y) and (— x, ?/) with respect toY'Y; and (x, y) and (— x, —y) with respect to the origin. 17. A line joining two points is bisected at the origin. If the coordinates of one end are (a, — 6), what will be the coordinates of the other end ? 18. Consider the bisectors of the angles between the coordinate axes. What is the relation between the abscissa and ordinate of any point of the bisector in the first and third quadrants ? second and fourth quadrants ? 19. A square whose side is 2 a has its center at the origin. What will be the coordinates of its vertices if the sides are parallel to the axes ? if the diago- nals coincide with the axes ? Ans. (a, a), (a, — a), (— a, — a), (— a, a); (a V2, 0), (- a V2, 0), (0, a V2), (0, - a V2). 20. An equilateral triangle whose side is a has its base on the axis of x and the opposite vertex above X'X. What are the vertices of the triangle if the center of the base is at the origin ? if the lower left-hand vertex is at the ^„..(|.0),(-|,0),(0,^); (0,0),(a,0),(?,2^). 28 ANALYTIC GEOMETRY 19. Angles. The angle between two intersecting directed lines is defined to be the angle made by their positive directions. In the figuies the angle between the directed lines is the angle marked 6. If the directed lines are parallel, then the angle between them is zero or it according as the positive directions agree or do not agree. Evidently the angle between two directed lines may have any value from to tt inclusive. Eeversing the direction of either directed line changes to the supplement it — 6. If both directions are reversed, the angle is unchanged. ^ = e. When it is desired to assign a positive direction to a line intersecting X^X, we shall always assume the upward direction as positive (see figures). X X' \B X CS) Theorem 1. If a and f3 are the angles between a line directed upward and the rectangular axes OX and OY, then (I) cos ^ = sin a. Froof. The figures are typical of all possible cases. In Fig. 1, and hence ^-? cos ^ 2 cos ex I = sin a. (by 6, p. 20) CARTESIAN COORDINATES 29 In Fig. 2, and hence 111 Fig. 3, cosyS / 7r\ . = cos [ a — — I = sm a. ^ ^ (by 4 and 6, p. 19) cos y8 = 1 = sin or. Q.E.D. The positive direction of a line parallel to X'X will be assumed to agree with the positive direction of X'X, that is, to the right. TT Hence for such a line a = 0, p = —, and the relation (I) still holds, since TT cos y8 = cos — = = sin = sm a. PROBLEMS 1 . Show that for lines directed downward cos )3 = — sin a. 2. What are the values of a and jS for a line directed N.E. ? N. W. ? S.E. ? S.W. ? (The axes are assumed to indicate the four cardinal points of the compass.) 3. Find the relation between the a's and /3's of two perpendicular lines directed upward. Ans. a' 20. Orthogonal projection. The orthogonal projection of a point upon a line is the foot of the perpendicular N let fall from the point upon the line. Thus in the figure M is the orthogonal projection of P on Z'X; jj/ N is the orthogonal projection of P on FT; X' \ M X P' is the orthogonal projection of P'on X'X. J 4iV If A and B are two points of a directed line, and M and N their projections upon a second directed line CD, then 'MN is callSd the projection of AB upon 02>. r' 30 ANALYTIC GEOMETRY Theorem II. First theorem of projection. If A and B are points upon a directed line making an angle y loith a second directed line CD, then the (II) projection of the length AB upon CJD = AB cos y. • Froof. In the figures let a = the numerical length of AB, I = the numerical length of ^^S" or BT; then a and I sue positive numbers giving the lengths of the respec- tive lines, as in Plane Geometry. Now apply the definition of the cosine to the right triangles ABS and ABT (p. 18). r. A ATT ^ ^ ^ ^ ^ ^ B B T II N M N (1) (2) (3) (4) (5) (6) In Fig. 1, 1 = a cos BA S = a cos y, MN=l, AB = a. .', MN = AB COS y. In Fig. 2, I = a cos ABT = a cos (tt — y) = — a cos y, (by 5, p. 20) MN=^ I, AB=—a.. .'. MN = AB cos y. In Fig. 3, I = a cos ^ i^T" = a cos (tt — y) = — a cos y, MN = -l, AB = a. .'. MN = AB cos y. ^ In Fig. 4, Z = a cos yl^r = «^ cos y, MN = -l, AB=-a. .-. MN = AB cos y. In Fig. 5, y = 0, ikfiV = 1, AB = a. Hence AfiV = ^jB = yl^ cos (since cos = 1). .'. MN = AB cos y. In Fig. 6, y = TT, iV/i\r = -l, AB=a. Hence MN^= — AB = AB cos tt (since cos tt = — 1). .'. ikfiV = ^5c0Sy. Q.E.D. CARTESIAlST COORDINATES 31 Consider any two given points Pi{^i, 2/1) » -^2(^2, 2/2)- Then in the figure 7I/1M2 = projection of PxP^ on X'X, NiNi = projection of PiPg on Y'Y. Pi(xi,y,) ■^ fa: 2, 2/2) y, Butby (1), p. 24, M1M2 = OM^ - OMi = 0^2 - «i, N^N^ = ON2 - ON^ = ^2 - 2/1- Hence Theorem III. Given amj tivo points P^ix^, y^, ^2(2^2, 2/2); ^^^^ a?2 — a?! = projection of P^P^ on X'X; projection of PxP^ on F' F. (Ill) 21. Lengths. We may now easily prove the important Theorem IV. The length I of the line Y joining two points P^ (a?i, y^, P^ {x^, 2/2) is given by the formula jy- (IV) I = V(xi-X2)^+(2/i-2/2)'- Proof Draw lines through Pi and -^ P2 parallel to the axes to form the right triangle PiSP^. Y Then SP^ = M^M^ = x^ - x^, P2S = N^N^ = 2/1 - 2 /2> PiP2=V^^'4-^'; arid hence ?AC«2,2/2) Z = V(a;i-(r2)'+(2/i-2/2)'. (by III) (by III) Q.E.D. 32 ANALYTIC GEOMETEY The method used in deriving (IV) for any positions of Pi and P2 is the following : Construct a right triangle by drawing lines parallel to the axes through Pi and P^- 'J-'he sides of this triangle are equal to the projections of the length P1P2 upon the axes. But these projec- tions are always given by (HI)? or by (III) with one or both signs changed. The required length is then the square root of the sum of the squares of these projections, so that the change in sign mentioned may be neglected. A number of different figures should be drawn to make the method clear. Ex. 1. Find the length of the hne joinmg the points (1, 3) and (—5, 5). Solution. Call (1, 3) Pi, and (- 5, 5)P2. Then ^1 = 1» 2/1 = 3, and Xa = — 5, 2/2 = 5; and substituting in (IV), we have Vio Y (-5 I^ <, --> ^ ""^ >. "-., (1, 3) Y 1 = V{1 + 5)^ + (3 - 5)^ 2 VlO. X It should be noticed that we are simply- finding the hypotenuse of a right triangle whose sides are 6 and 2. Remark. The fact that formulas (III) and (lY) are true for all positions of the points Pi and P2 is of fundamental importance. The application of these formulas to any given problem is there- fore simply a matter of direct substitution, as the example worked out above illustrates. In deriving such general formulas, since it is immaterial in what quadrants the assumed points lie, it is most convenient to draw the figure so that the points lie in the first quadrant, or, in general, so that all the quantities assumed as known shall he positive. PROBLEMS 1. Find the projections on the axes and the length of the lines joining the following points : (a) (- 4, — 4)_and (1, 3). _ Ans. Projections 5, 7; length = Vri. (b) (- V2, V3) and (V3, V2). Ans. Projections V3 + V2, V2 - V3,- length = VlO. CARTESIAN COORDINATES 33 (c) (0, 0) and ( - , -— — j . Ans. Projections - , - ^3 ; length = a. (d) {a -\- h^ c ■\- a) and (c + a, 6 + c). " Ans. Projections c — h^h — a\ length = ■>/(& — c)2 + (a — &)2, 2. Find the projections of the sides of the following triangles upon the axes: (a) (0, 6), (1, 2), (3, - 5). (b) (1, 0). (-1, -6), (-1, -8). (c) (a, 6), (6, c), (c, d). 3. Find the lengths of the sides of the triangles in problem 2. 4. Work out formulas (III) and (IV), (a) if Xi = Xg; (b) if y^ — yi. 5. Find the lengths of the side;S'of the triangle whose vertices are (4, 3), (2, -2), (-3, 5). / • 6. Show that the points (1, 4), (4, 1), (5, 5) are the vertices of an isosceles triangle. 7. Show that the points (2, 2^, (-2, - 2), (2 Vs, - 2 V3) are the vertices of an equilateral triangle. / 8. Show that (3, 0), (6, 4), (— 1, 3) are the vertices of a right triangle. What is its area ? 9. Prove that (- 4, - 2), (2, 0), (8, 6), (2, 4) are the vertices of a paral- IMogram. Also find the lengths of the diagonals. 10. Show that (11, 2), (6, - 10), (-6, - 5), (- 1, 7) are the vertices of a square. Find its area. 11. Show that the points (1, 3), (2, Vo), (2, — V6) are equidistant from the origin, that is, show that they lie on a circle with its center at the origin and its radius VlO. 12. Show that the diagonals of any rectangle are equal. 13. Find the perimeter of the triangle whose vertices are (a, 6), (— a, 6), (-a, -6). 14. Find the perimeter of the polygon formed by joining the following points two by two in order : (6, 4), (4, - 3), (0, - 1), (- 5, - 4), (- 2, 1). 16. One end of a line whose length is 13 is the point (-4, 8); the ordi- nate of the other end is 3. What is its abscissa ? Ans. 8 or — 16. 16. What equation must the coordinates of the point (x, y) satisfy if its distance from the point (7, — 2) is equal to 11 ? • . 34 ANALYTIC GEOMETRY 17. What equation expresses algebraically the fact that the point (x, y) is equidistant from the points (2, 3) and (4, 5)? 18. If the angle XOY (Fig., p. 24) equals w, show that the length of the line joining Pi(xi, ?/i) and P^ix^, y-i) is given by I = V(xi - Xa)^ + (?/i - 2/2)^ + 2 (Xi - X2) (z/i - 2/2) cos w. 19. If w = — , find distance between the points (—3, 3) and (4, — 2). Ain^. V39. 20. If w = — , find the perimeter of the triangle whose vertices are (1, 3), (2, 7), (- 4, - 4). An8, V2I + V223 + Vl09. 21. If w = -, find the perimeter of triangle (1, 2), (- 2, - 4), (3, - 5). Aus. 3 V5 + 2 V3 + V26 - 5 V3 + V53 - 14 V3. 22. Prove that (6, 6), (7, - 1), (0, -2), (-2, 2) lie on a circle whose center is at (3, 2). 23. If w = ^ — , find the distance between ( Vs, V2), (- Vi, V3). Ans. VlO + V2. 24. Show that the sum of the projections of the sides of a polygon upon either axis is zero if each side is given a direction established by passing continuously around the perimeter. 22. Inclination and slope. The inclination of a line is the angle between the axis of x and the line when the latter is given the upward direction (p. 28). The slope of a line is the tangent of its inclination. The inclination of a line will be J denoted by a^ ai, org? «'? etc. ; its slope -^ — :> by m, m^ mg, m', etc., so that m = tan a, mi = tan ai, etc. The inclination may be any angle from to TT inclusive (p. 28). The T/N r' slope may be any real number, since the tangent of an angle in the first two quadrants may be any number positive or negative. The slope of a line parallel to X'X is of course zero, since the. inclination is or tt. For a line parallel to Y' Y the slope is infinite. CARTESIAN COORDINATES 35 Theorem V. The slope tn of the line passing through two points Pii^ij yi)} Pii^i, y^ is given by Uy-t Uyo Proof. (1) Similarly, (2) But M1M2 = 0*2 — iCj = P1P2 COS a. .'. P1P2 cos a = X2 — Xi. ^1^2 = 2/2-2/1 = P1P2 cos /3. .'. P^P^ cos p = 1/2- yi- cos (3 = sin a. (by (III), p. 31) (by (II), p. 30) (by (III), p. 31) (by (II), p. 30) (by (I), p. 28) Hence, from (2), (3) Dividing (3) by (1), tan a = m = PyP^. sin a = yz — y^. ^2 - 2/1 _ x^ — x^ Remark. Formula (V) may be verified by constructing a right triangle whose hypot- enuse is P1P2) s-s on p. 31, whence tan a (= tan Z SP1P2) is found directly as the ratio of the opposite side, SP^ = 2/2 — 2/i) ^^ t^® adjacent side, PiS = X2 — Xi* 2/1-2/2 Q.E.D. Y' P2/ y---^s Am. / ^ * To construct a line passing through a given point Pi whose slope is a positive frac- Pj a units above S, and must lie to the left of P^ tiori - , we mark a point S h units to the right of P^ and a point draw PiPg- If the slope is a negative fraction, — ^ , then either S or Pj niust lie below S. 86 ANALYTIC GEOMETRY Theorem VI. If two lines are parallel, their slopes are equal; if perpendicular, the slope of one is the negative reciprocal of the slope of the other, and conversely. Proof Let a-^ and a^ be the inclinations and m^ and m^^ the slopes of the lines. If the lines are parallel, a^ = a^. .'. m^ = mg. If the lines are perpendicular, as in the figure, IT TT «'2 = ^1 + 2' °^ a^ = oc^- ^^ tan ( cx^ — -^ j (by 4 and 6, p. 19) (by 1, p. 19) 1 Q.E.D. 2 m The converse is proved by retracing the steps with the assump tion, in the second part, that ^2 is greater than aj. PROBLEMS 1. Find the slope of the line joining (1, 3) and (2, 7). Ans. 4. 2. Find the slope of the line joining (2, 7) and (—4, — 4). Ans. y-. 3. Find the slope of the line joining (Vs, V2) and (— V2, Vs). Ans. 2V6-5. 4. Find the slope of the line joining {a + h, c -\- a), (c -{- a, b + c). b — a Ans. c-b 5. Find the slopes of the sides of the triangle whose vertices are (1, 1), (- 1, - 1), ( V3, - V3). ^^^ ^^ 1+V3 ^ I-V3 ' i-Vs' 1+V3 6. Prove by means of slopes that (- 4, - 2), (2, 0), (8, 6), (2, 4) are the vertices of a parallelogram. 7. Prove by means of slopes that (3, 0), (6, 4), (— 1, 3) are the vertices of a right triangle. 8. Prove by means of slopes that (0, -2), (4, 2), (0, G), (-4, 2) are the vertices of a rectangle, and hence, by (IV), of a square. UNIVERSITY OF ^_ v^T>4_,«^j,gj^^ COORDINATES 37 9. Prove by means of their slopes that the diagonals of the square in problem 8 are perpendicular. 10. Prove by means of slopes that (10, 0), (5, 6), (5, - 5), (- 5, 5) are the vertices of a trapezoid. 11. Show that the line joining (a, b) and (c, - d) is parallel to the line joining (-a, — h) and (- c, d). 12. Show that the line joining the origin to (a, b) is perpendicular to the line joining the origin to (—6, a). 13. What is the inclination of a line parallel to Y'Y? perpendicular to Y'Y? 14. What is the- slope of a line parallel to Y'Y? perpendicular to Y'Y? 15. What is the inclination of the line joining (2, 2) and (—2, — 2)? Ans. —' 4 16. What is the inclination of the line joining (— 2, 0) and (— 5, 3)? Ans. 17. What is the inclination of the line joining (3, 0) and (4, V3) ? Am 18. What is the inclination of the line joining (3, 0) and (2, Vs) ? Ans. —-' 4 A ^ Ans. - 27t Ans.—- 19. What is the inclination of the line joining (0, — 4) and (— V3, — 5) ? Ans. ^- 20. What is the inclination of the line joining (0, 0) and (— V3, 1) 57r Ans. ^, 21. Prove by means of slopes that (2, 3), (1, — 3), (3, 9) lie on the same straight line. 22. Prove that the points (a, 6 + c), (6, c -\- a), and (c, a + 6) lie on the same straight line. 23. Prove that (1, 5) is on the line joining the points (0, 2) and (2, 8) and is equidistant from them. 24. Prove that the line joining (3, — 2) and (5, 1) is perpendicular to the line joining (10, 0) and (13, - 2). 38 ANALYTIC GEOMETRY 23. Point of division. Let Pi and P^ be two fixed points on a directed line. Any third point on the line, as P or P\ is said Pi P P2 P' "to divide the line into two segments," and is called a point of division. The division is called internal or external according as the point falls within or without PiP^- The position of the point of division depends upon the ratio of its distances from Pj and Pa- Since, however, the line is directed, some convention must be made as to the manner of reading these distances. We therefore adopt the rule : If P is a point of division on a directed line passing through Pi and P2, then P is said to divide P1P2 into the segments PiP P P and PP2. The ratio of division is the value of the ratio* -^— • We shall denote this ratio by X, that is, ^ If the division is internal, PiP and PP2 agree in direction and therefore in sign, and X is therefore positive. In external divi- sion X is negative. The sign of A. therefore indicates whether the point of division P is within or without the segment P1P2 ; and the numerical value determines whether P lies nearer Pi or Pg. The distribution of X is indicated in the figure. -1<\<0 XrO X>0 X=00 - 1 + A y = Vi + A2/2 l + A Proof. Given X = PP. Let a be the inclination of the line PiP upon the axis of x. Then, by the first theorem of projection [(II), p. 30], MiM = P^P cos or, MM2 = PP2 cos a. M^M _ P,P _ MM2~ PP.~ ' M^M = X — x^j Mi M M.X Project Pi, P, P2 Dividing, But Substituting, MM. X. (by hypothesis) (by (III), p. 31) A. Clearing of fractions and solving for x, x^ -f- Aa^g Similarly, y Q.E.D. 1 + A yi -h A..?/2 1 + A * Corollary. Middle point. The coordinates (x, y) of the middle point of the line joining Pi(xi, y^), P^ix^, 2/2) are found hy taking the averages of the given abscissas ^and ordinates ; that is, - |(a?i + ajg), y = | (y^ + y^). 3C P,P For if P is the middle point of P^P^, then A = — ^ = 1. 40 ANALYTIC GEOMETRY Ex. 1. Find the point P dividing Pi (-1, -6), P^ (3, 0) in the ratio \=-\. Solution. Applying (VII), xi = - 1, yi=— 6, = 3, y2 = 0. X = y = -1- i 3 1- -6- ■0 f -6_ T" Ans. Hence Pis (- 2i, -8). Ex. 2. Find the coordinates of the point of intersection of the medians of a triangle whose vertices are (xi, yi), (X2, ^2), (a^s, 2/3). Solution. By Plane Geometry we have to find the point P on the median AD such that AP = | AD, that is, AP : PD : : 2 : 1, or X = 2. By the Corollary, D is [|(X2 + X3), i (2/2 + Vs)]- To find P, apply (VII), remembering that A corre- sponds to (xi, 2/1) and D to (X2, 2/2)- AC^i^Vi) This gives x 2/ = Xi + 2 • i (X2 + X3) 2-i(2/2 + ?/3) 2/1 . r3J3,2/3>) 1 + 2 .-. X = 1 (xi + X2 + X3), 2/ = I (2/1 + 2/2 + 2/3)- ^ns. Hence the abscissa of the intersection of the medians of a triangle is the average of the abscissas of the vertices, and similarly for the ordinate. The symmetry of these answers is evidence that the particular median chosen is immaterial, and the formulas therefore prove the fact of the intersec- tion of the medians. PROBLEMS 1. Find the coordinates of the middle point of the line joining (4, — 6) and (-2, -4). Ans. (1, - 5). 2. Find the coordinates of the middle point of the line joining {a-\-b,c + d) and (a — 6, d — c). Ans. (a, d). 3. Find the middle points of the sides of the triangle whose vertices are (2, 3), (4, — 5), and (—3, — 6) ; also find the lengths of the medians. 4. Find the coordinates of the point which divides the line joining (—1,4) and (—5, — 8) in the ratio 1 : 3. Ans. (— 2, 1). 6. Find the coordinates of the point which divides the line joining (-3, -5) and (6, 9) in the ratio 2:5. Ans. (- f, -1). CARTESIAN COORDINATES 41*^^ 6. Find the coordinates of the point which divides the line joining (2, 6) and (— 4, 8) into segments whose ratio is — |. Ans. {— 22, 14). 7. Find the coordinates of the point which divides the line joining (—3, — 4) and (5, 2) into segments whose ratio is — f. Ans. (—19, — 16). 8. Find the coordinates of the points which trisect the line joining the points (- 2, - 1) and (3, 2). Ans. (- ^, 0), (4, 1). 9. Prove that the middle point of the hypotenuse of a right triangle is equidistant from the three vertices. 10. Show that the diagonals of the parallelogram whose vertices are (1,2), (- 5, - 3), (7, - 6), (1, - 11) bisect each other. 11. Prove that the diagonals of any parallelogram mutually bisect each other. 12. Show that the lines joining the middle points of the opposite sides of the quadrilateral whose vertices are (6, 8), (— 4, 0), (— 2, — 6), (4, — 4) bisect each other. 13. In the quadrilateral of problem 12 show by means of slopes that the ■* lines joining the middle points of the adjacent sides form a parallelogram. 14. Show that in the trapezoid whose vertices are (— 8, 0), (—4, — 4), (—4, 4), and (4, — 4) the length of the line joining the middle points of the non-parallel sides is equal to one half the sum of the lengths of the parallel sides. Also prove that it is parallel to the parallel sides. 15. In what ratio does the point (—2, 3) divide the line joining the points (-3, 5) and (4, -9)? Ans. |. 16. In what ratio does the point (16, 3) divide the line joining the points (-5, 0)and (2, 1)? Ans. - |. 17. Given the triangle whose vertices are (— 5, 3), (1, — 3), (7, 6); show that a line joining the middle points of any two sides is parallel to the third side and equal to one half of it. 18. If (2, 1), (3, 3), (6, 2) are the middle points of the sides of a triangle, what are the coordinates of the vertices of the triangle ? Ans. (-1,2), (5,0), (7,4). 19. Three vertices of a parallelogram are (1, 2), (-5, —3), (7,-6). What are the coordinates of the fourth vertex ? Ans. (1, - 11), (- 11, 5), or (13, - 1). 20. The middle point of a line is (6, 4), and one end of the line is (5, 7). What are the coordinates of tlie other end ? Ans. (7, 1). 21. The vertices of a triangle are (2, 3), (4, - 5), (- 3, - 6). Find the coordinates of the point where the medians intersect (center of gravity). 42 ANALYTIC GEOMETRY 22. Find the area of the isosceles triangle whose vertices are (1, 5), (5, 1), (—9, — 9) by finding the lengths of the base and altitude, 23. A line AB is produced to C so that BC = | AB. If the points A and B have the coordinates (5, 6) and (7, 2) respectively, what are the coordinates of C ? Ans. (8, 0). 24. Show that formula (VII) holds for oblique coordinates, that is, Z XOY may have any value. 25. How far is the point bisecting the line joining the points (5, 5) and (8, 7) from the origin ? What is the slope of this last line ? Ans. 2 Vl3, |. 24. Areas. In this section the problem of determining the area of any polygon the coordinates of whose vertices are given will be solved. We begin with Theorem VIII. The area of a triangle whose vertices are the origin, P^ (a^i, y-^, and Pc, (x^, y^ is give7i by the formula (VIII) Area of triangle OP^P^ = i(oo^y2 — ^22/i)- f.fe2/J ■^''•''•^* In the figure let a = = AXOP„ P- -- Z XOP^, e^ -- Z P,OP^. ^ = = y8-a. 3fa My JL (1) By 18, p. 20, (2) Area A OP^P^ = ^0P^- OP^ sin = \0P^- OP, sin (yS - a) [by (1)] (3) = \ OPi • OP2 (sin ft cos a — cos ft sin or). But m the figure sm ft = -^-^ = -^^^j cos B = = — ^> ^ OP, OP, '^ OP, OP, M^P^ y^ OM. X. sm a = — ^— ^ = -^^-^, cos a = = — - • OPi OPi OPi OP^ Substituting in (3) and reducing, we obtain Area A OP^P, = ^ {x^y, - x,y^. q.e.d. CARTESIAN COORDINATES 43 Ex. 1. Find the area of the triangle whose vertices are the origin, (—2, 4), and (- 5, - 1). Solution. Denote ( - 2, 4) by Pi, ( - 5, - 1) by Pg. Then cci = - 2, 2/1 = 4, X2 = - 5, 2/2 = - 1- (-\i) YJk / \ f \ \/ \ a,i) y \ I ^. ^ X r-s -t) Substituting in (VIII), Area = i [- 2 • - 1 - (- 5) • 4] = 11. Then Area = 11 unit squares. If, however, the formula (VIII) is applied by denoting (—2, 4) by P2, and (_ 5^ - 1) by Pi, the result will be - 11. The two figures are as follows : (1) (2) The cases of positive and negative area are distinguished by Theorem IX. Passing around the perimeter in the order of the vertices 0, P^, P^, if the area is on the left, as in Fig. 1, then (VIII) gives a posi- tive restdt; if the area is on the right, as in Fig. 2, then (YIII) gives a negative result. Proof When the area is on the left as in Fig. 1, then in (1), p. 42, we have fi > a, and hence is positive. Therefore sin $ is positive and the product in (2), „ p p. 42, which gives the area of OP^P^, is also positive. But when the area is on the right, as in Fig. 2, we have 13 < a, and hence 6 is negative. Then sin is negative, and hence also the product in (2), p. 42, which gives the area of OP^P^. q.e.d. Formula (VIII) is easily applied to any polygon by regarding its area as made up of triangles with the origin as a common vertex. Consider any triangle. .^1 .e+ (1) (2) 44 ANALYTIC GEOMETRY Theorem X. The area of a triangle whose vertices are P^ (x-^, yi), A(^2j 2/2); ^3(^3, 2/3) «« ff^^en by (X) Area A 1*1 jP2jP3=i (a:5i2/2 — a?22/i+^2?^3—»^32/2 +^32/1 — ^12/3)- This formula gives a positive or negative result according as the area lies to the left or right in passing around the perimeter in the order P^P^Pz- Proof Two cases must be distin- guished according as the origin is within or without the triangle. Fig. 1, origin within the triangle. By inspection, (5) Area A P^P^P^ = A OP^P^ + A OP^P.^ + A OP^P^, since these areas all have the same sign. 1 p yig. 2, origin without the triangle. By inspection, (6) Area A P^P^P^ = A OP.P^ + A OP,P, + A OP,P„ since OP^P^, OP^Pi have the same sign, but OP^P-s the opposite sign, the algebraic sum giving the desired area. By (VIII), A OP,P, = ^(x,7/, - X0,), A OP^P^ = ^ (x^y^ - x^y^), A OP^P^ = \ (x^y^ - x^y^). Substituting in (5) and (6), we have (X). Also in (5) the area is positive, in (6) negative. An easy way to apply (X) is given by the following Rule for finding the area of a triangle. First step. Write down the vertices in two columns, abscissas in one, ordinates in the other, repeating the coordinates of the first vertex. Second step. Multiply each abscissa by the ordinate of the next row, and add results. This gives x^y^ 4- ^^yz + ^zVi- Third step. Multiply each ordinate by the abscissa of the next row, and add residts. This gives y^x^ + 2/2^3 + Vz^x- Fourth step. Subtract the result of the third step from that of the second step, and divide by 2. This gives the required area, namely, formula (X). Q.E.D. Xi •2/1 X2 Vi a^3 Vz Xi 7 y\ CARTESIAN COORDINATES 45 -2 % It is easy to show in the same manner that the rule applies to any polygon, if the following caution be observed in the first step : Write down the coordinates of the vertices in an order agreeing with that established by passing continuously around the perimeter , and repeat the coordinates of the first vertex. Ex. 2. Find the area of the quadrilateral whose vertices are (1, 6), (-3, -4), (2, -2), (-1,3). Solution. Plotting, we have the figure from which we choose the order of the vertices as indi- .. ^ cated by the arrows. Following the rule : _ i 3 First step. Write down the vertices in — 3 — 4 order. 2 — 2 Second step. Multiply each abscissa by the ordinate of the next row, and add. This gives lx3 + (-lx-4) + (-3x-2) + 2x6=3 25. Third step. Multiply each ordinate by the abscissa of the next row and add. This gives 6x-l + 3x-3 + (-4x2) + (-2xl)=-25. Fourth step. Subtract the result of the third step from the result of the second step, and divide by 2. . 25 4-25 ^„ .^ . .-. Area = = 25 unit squares. Ans. The result has the positive sign, since the area is on the left. PROBLEMS 1. Find the area of the triangle whose vertices are (2, 3), (1, 5), (— 1, — 2). Ans. y. 2. Find the area of the triangle whose vertices are (2, 3), (4, —5), ( — 3, —6). Ans. 29. 3. Find the area of the triangle whose vertices are (8, 3), (— 2, 3), (4, — 5). Ans. 40. 4. Find the area of the triangle whose vertices are (a, 0), (— a, 0), (0, b). Ans. ah. 5. Find the area of the triangle whose vertices are (0, 0), (xi, yi), (X2, 2/2)- Ans. ^jyiH^^. 0^ 46 ANALYTIC GEOMETRY 6. Find the area of the triangler^whose vertices are (a, 1), (0, 6), (c, 1). 2 7. Eind the area of the triangle whose vertices are (a, 6), (&, a), (c, - c). 8. Find the area of the triangle whose vertices are (3, 0) , (0, 3 V3), (6, 3 V3). An8. 9V3. 9. Prove that the area of the triangle whose vertices are the points (2, 3), (5, 4), (—4, 1) is zero, and hence that these points all lie on the same straight line. 10. Prove that the area of the triangle whose vertices are the points (a, 6 + c), (&, c ■\- a)^ (c, a + 6) is zero, and hence that these points all lie on the same straight line. 11. Prove that the area of the triangle whose vertices are the points (a, c + a), (— c, 0), (—a, c — a) is zero, and hence that these points all lie on the same straight line. 12. Find the area of the quadrilateral whose vertices are (-2, 3), (-3, -4), (5, -1), (2, 2). Ans. 31. 13. Find the area of the pentagon whose vertices are (1, 2), (3, — 1), (6, - 2), (2, 5), (4, 4). Ans. 18. 14. Find the area of the parallelogram whose vertices are (10, 5), (— 2, 5), (- 5, - 3), (7, - 3). Ans. 96. 15. Find the area of the quadrilateral whose vertices are (0, 0), (5, 0), (9, 11), (0, 3). Ans. 41. 16. Find the area of the quadrilateral whose vertices are (7, 0),- (11, 9), (0, 5), (0, 0). Ans. 59. 17. Show that the area of the triangle whose vertices are (4, 6), (2, — 4), (—4, 2) is four times the area of the triangle formed by joining the middle points of the sides. 18. Show that the lines drawn from the vertices (3, — 8), (— 4, 6), (7, 0) to the medial point of the triangle divide it into three triangles of equal area. 19. Given the quadrilateral whose vertices are (0, 0), (6, 8), (10, — 2), (4, — 4) ; show that the area of the quadrilateral formed by joining the middle points of its adjacent sides is equal to one half the area of the given quadrilateral. CARTESIAN COORDINATES 47 25. Second theorem of projectiqj|j Lemma I. If M^^ M^, M^ are any imree points on a directed line, then in all cases T Jfi M-i M, Mo M^ Ml Proof. Let be the origin. By (1), p. 24, M^M^ = OM^ - OM^, M^Ms = OMs - OM^. Adding, M^M^ + M^M^ = OM^ - OM^. But by (1), p. 24, M^M^ = OM^ - OM^. Q.E.D. This result is easily extended to prove Lemma II. If Mi, M^, M^, •••, ikf„_i, M^ are any n points on a directed line, then in all cases MiM^ = M1M2 -h M^M^ + M^M^ + • • • + M„_,M„, the lengths in the right-hand member being so written that the second point of each length is the first point of the next. The line joining the first and last points of a broken line is called the closing line. Cii/, M>, M^D Thus in Fig. 1 the closing line is PyP^ ; in Fig. 2 the closing line is P1P5. 48 ANALYTIC GEOMETRY Theorem XI. Second theorem of projection. If each segment of a broken line be given the direction determined in passing continuously from one extremity to the other, then the algebraic sum of the pro- jections of the segments upon any directed line equals the projection of the closing line. Proof in Fig. 1 The proof results immediately from the Lemmas. For M1M2 = projection of P1P2 j M^M^ = projection of P2P3 5 MiM^ = projection of closing line PiPs- But by Lemma I Ml Mo + M^M^= M^Ms, and the theorem follows. ^ Similarly in Fig. 2. q.e.d. Corollary. If the sides of a closed polygon be given the direction established by passing continuously around the perimeter, the sum of the projections of the sides upon any directed line is zero. For tlje closing line is now zero. Ex. 1. Find the projection of the line joining the origin and (5, 3) upon a line passing through ( — 5, 0) whose inclination is — • 4 Solution. In the figure, applying the second theorem of projection, proj. of OP on AB = proj. of OM 4- proj. of MP It It = OM cos — + MP cos — 4 4 (by first theorem of projection, p, 30) = 5V2-i-fV2 = 4V2. Ans. The essential point in the solution of problems like Ex. 1 is the replacing of the given line, by means of Theorem XI, by a broken line with two seg- ments which are parallel to the axes. I CARTESIAN COORDINATES 49 Ex. 2. Find the perpendicular distance from the Ime passing through (4, 0), whose inclination is — , to the point (10, 2). ^ Solution. In the figure draw OC perpendicular to the given line AB. 27t ZXAS or 120°. .-. Z XOS = 30°, Z SOY = 60°. Required the perpendicular dis- tance RP. Project the broken line OMP upon OC. Then, by the second theorem of projection, (1) But in the figure (2) proj. of OP = proj. of OM + proj. of MP = OM cos Z XOS + MP cos ZSOY = 10.iV3 + 2.i = 1 + 5 Vs. proj. of OP = 05 + -87 = OA cos^XOS + RP = 4 . i Vs + RP. From (1) and (2), EP + 2V3 = 1 + 5V3. RP = l-\-SVs. Ans. PROBLEMS 1. Four points lie on the axis of abscissas at distances of 1, 3, 6, and 10 respectively from the origin. Find P1P4 by Lemma II. 2. A broken line joins continuously the points (—1, 4), (3, 6), (6, — 2), (8, 1), (1, — 1). Show that the second theorem of projection holds when the segments are projected on the X-axis. 3. Show by means of a figure that the projection of the broken line join- [ ing the points (1, 2), (5, 4), (- 1, - 4), (3, - 1), and (1, 2) upon any line is zero. 4. Find the projection of the line joining the points (2, 1) and (5, 3) upon 7t a line passing through the point (—1, 1) whose inclination is Ans. 3 V3 + 2 50 ANALYTIC GEOMETRY 5. What is the projection of the line joining these same points upon any line whose inclination is ^ ? Why ? 6. Find the projection of the line joining the points (-1, 3) and (2, 4) upon any line whose inclination is f it. Ans. — V 2. 7. Find the projection of the broken line joining the points (- 1, 4), It (3, 6), and (5, 0) upon a line whose inclination is -• Verify your result by finding the projection of the closing line. Au8. V 2, 8. Find the projection of the broken line-joining (0, 0), (4, 2), and (6, - 3) upon a line whose inclination is -r- • Ans. — — — o 2 9. Show that the projection of the sides of the triangle (2, 1), (- 1, 5), (—3, 1) upon a line whose inclination is — is zero. 10. Find the perpendicular distance from the point (6, 3) to a line passing It 7 through the point (—4, 0) with an inclination of —• Ans. — p- 11. Find the perpendicular distance from the point (—5, — 1) to a line passing through the point (6, 0) and having an inclination of f it. Anjs. 6V2. 12. A line of inclination — passes through the point (5, 0). Find the per- pendicular distance to the parallel line passing through the point (0, 2). 5 + 2 V3 Ans. CHAPTER III THE CURVE AND THE EQUATION 26. Locus of a point satisfying a given condition. The curve* (or group of curves) passing through 'all points which satisfy a given condition, and through no other points, is called the locus of the point satisfying that condition. For example, in Plane Geometry, the following results are proved : The perpendicular bisector of the line joining two fixed points is the locus of all points equidistant from these points. The bisectors of the adjacent angles formed by two lines is the locus of all points equidistant from these lines. To solve any locus problem involves two things : 1. To draw the locus by constructing a sufficient number of points satisfying the given condition and therefore lying on the locus. 2. To discuss the nature of the locus, that is, to determine properties of the curve, t Analytic Geometry is peculiarly adapted to the solution of both parts of a locus problem. 27. Equation of the locus of a point satisfying a given condition. Let us take up the locus problem, making use of coor- dinates. If any point P satisfying the given condition and there- fore lying on the locus be given the coordinates (x, y), then the given condition will lead to an equation involving the variables X and y. The following example illustrates this fact, which is of fundamental importance. * The word " curve" wiU hereafter signify any contimious line, straight or curved. t As the only loci considered in Elementary Geometry are straight lines and circles, the complete loci may be constructed by ruler and compasses, and the second part is relatively unimportant. 51 '-^ 52 ANALYTIC GEOMETRY Ex. 1. Find the equation in x and y if the point whose locus is required shall be equidistant from A{— 2, 0) and B{— 3, 8). Solution. Let P (x, y) be any point on the locus. Then by the given condition (1) PA = PB. But, by formula IV, p. 31, PA = V{x + 2)^ + {y- -0)2, P5=V(a: + 3)2 + (?/- Substituting in (1), -8)2. V(x + 2)'^ + (2/ - 0)2 (2) = V(x + 3)2 + {y- 8)2. Squaring and reducing, (3) 2x - 162/ -h 69 = 0. In the equation (3), x and y are variables representing the coordinates of any point on the locus, that is, of any point on the perpendicular bisector of the line AB. This equation has two important and characteristic properties : 1. The coordinates of any point on the locus may be substituted for x and y in the equation (3), and the result will be true. For let Pi (xi, yi) be any point on the locus. Then PiA = PiB, by defi- nition. Hence, by formula IV, p. 31, (4) V(xi + 2)2 + 2/i2 = V(xi + 3)2 + (^1 - 8)2, or, squaring and reducing, (5) 2 xi - 16 yi + 69 = 0. Therefore Xi and yi satisfy (3). 2. Conversely, every point whose coordinates satisfy (3) will lie upon the locus. For if Pi(Xi, yi) is a point whose coordinates satisfy (3), then (5) is true, and hence also (4) holds. q.e.d. In particular, the coordinates of the middle point C ot A and B, namely, x=-2i,y=:4: (Corollary, p. 39), satisfy (3), since 2 ( - 2i) - 16 x 4 + 69 = 0. This example illustrates the following correspo?idence between Pure and Analytic Geometry as regards the locus problem : Locus problem Pure Geometry Analytic Geometry The geometrical condition (satis- An equation in the variables x fied by every point on the and?/ representing coordinates locus). (satisfied by the coordinates of every point on the locus). THE CURVE AND THE EQUATION 53 This discussion leads to the fundamental definition : The equation of the locus of a point satisfying a given condition is an equation in the variables x and y representing coordinates such that (1) the coordinates of every point on the locus will satisfy the equation; and (2) conversely, every point whose coordinates satisfy the equation will lie upon the locus. • This definition shows that the equation of the locus must be tested in two ways after derivation, as illustrated in the example of this section and in those following. From the above definition follows at once the Corollary. A point lies upon a cun^e when and only when its coordinates satisfy the equation of the curve. 28. First fundamental problem. To find the equation of a curve which is defined as the locus of a point satisfying a giuen condition. , The following rule will suffice for the solution of this problem in many cases : Rule. First step. Assume that P (x, y) is any point satisfying the given condition and is therefore on the curve. Second step. Write down the given condition. Third step. Express the . given condition in coordinates and simplify the result. The final equation, containing x, y, and the' given constants of the problem, will be the required equation. Ex. 1. Find the equation of the straight line passing through Pi (4, — 1) 3 7t and having an inclination of Solution. First step. Assume P{x, y) any point on the line. Second step. The given condition, since the incli- • Stt nation a is — , may be written 4 (1) Slope of PiP = tan a = - 1. Third step. From (V), p. 35, .N \ /_ 1 1 \ -.6>0? ifO>6>a? 3. Find the equation of a line parallel to OX and (a) at a distance of 3 units above OX. (b) at a distance of 6 units below OX. (c) at a distance of 7 units above (— 2, — 3). (d) at a distance of 5 units below (4, — 2). 4. What is the equation of XX? of YY'? 6. Find the equation of a line parallel to the line x = 4 and 3 units to the right of it. Eight units to the left of it. 6. Find the equation of a line parallel to the line y = -2 and 4 units below it. Five units above it. 7. How does the line y = a-61ieifa>6>0? if6>a>0? 8. What is the equation of the axis of x ? of the axis oly? (c) Pi is (- 2, 3) and 7n = ? Ans. V2x-2y-\-6+2V2 = 0. 3^ 5Q " ' ANALYTIC GEOMETKY 9. What is the equation of the locus of a point which moves always at a distance of 2 units from the axis of x ? from the axis of y? from the line X =— 6? from the line y = 4? 10. What is the equation of the locus of a point which moves so as to be equidistant from the lines x = b and x = 9? equidistant from y = S and y=-7? 11. What are the equations of the sides of the rectangle whose vertices are (5, 2), (5, 5), (- 2, 2), (- 2, 5)'? In problems 12 and 13, Pi is a given point on the required line, m is the slope of the line, and a its inclination. 12. What is the equation of a line if (a) Pi is (0, 3) and ?n = - 3 ? Ans. 3x + ?/-3 = 0. (b) Pi is (- 4, - 2) and m = i? Ans. x- ?jy -2 = 0. 2 V3 (d) Pi is (0, 5) and m =: ^? Ans. V3x-2?/ + 10 = 0. (e) Pi is (0, 0) and m 1:= - I ? Ans. 2x + Sy = 0. (f) Pi is (a, b) and m = ? Ans. y = b. (g) Pi is (— a, b) and ni = 'X)? Ans. x =— a. 13. What is the equation of a line if (a) Pi is (2, 3) and a = 45° ? Ans. x - y -\- 1 = 0. (b) Pi is (- 1, 2) and a = 45° ? Ans. x - y -]- S = 0. (c) Pi is ( - a, - b) and a: = 45° ? Ans. x - y = b - a. (d) Pi is (5, 2) and a = 60°? Ans. Vsx - y -^ 2 - ^Vs = 0. (e) Pi is (0, - 7) and or = 60° ? Ans. -VSx-y -7 = 0. (f) Pi is (- 4, 5) and a = 0°? Ans. y = 6. (g) Pi is (2, - 3) and a = 90°? Ans. x = 2. (h) Pi is (3, - 3 V3) and a: = 120° ? Ans. ^Sx + y = 0. (i) Pi 'is (0, 3) and a = 150°? Ans. Vsx -\-Sy -9 = 0. (j) Pi is (a, 6) and a = 135° ? Ans. x -{- y = a -\- b. 14. Are the points (3, 9), (4, 6), (5, 5) on the line 3x + 2y = 25? 15. Find the equation of the circle with (a) center at (3, 2) and radius = 4. Ans. x^ + y^ - 6 x — 4y - S = 0. (b) center at (12, - 5) and r = 13. Ans. x^ + y^ - 24x -\-10y = 0. (c) center at (0, 0) and radius = r. Ans. x^ -\- 7/'^ = r^. (d) center at (0, 0) and r = 5. Ans. x'^ + 7/^ = 25. (e) center at (3 a, 4 a) and r = 5 a. Ans. x^ + y^ _ 2 a (3 cc + 4 y) = 0. (f) center at (6 + c, 6 — c) and r = c. Ans. x2 + ?/2 - 2 (& + c) X - 2 (6 - c) ?/ + 2 62 4. c2 = 0. THE CURVE AND THE EQUATION 67 16. Find the equation of a circle whose center is (5, — 4) and whose circumference passes through the point (—2, 3). 17. Find the equation of a circle having the line joining (3, — 5) and (— 2, 2) as a diameter. ' 18. Find the equation of a circle touching each axis at a distance 6 units from the origin. 19. Find the equation of a circle whose center is the middle point of the line joining (—6, 8) to the origin and whose circumference passes through the point (2, 3). 20. A point moves so that its distances from the two fixed points (2, — 3) and (—1, 4) are equal. Find the equation of the locus and plot. Ans. Sx-7ij + 2 = 0. 21. Find the equation of the perpendicular bisector of the line joining (a) (2, 1), (- 3, - 3). Ans. lOx + 8y -\- IS = (b) (3, 1), (2, 4). Ans. x - 3 y + 5 = 0. (c) (-1, -1), (3, 7). Ans. x-\-2y-7 = 0. (d) (0, 4), (3, 0). Ans. 6x-Sy-\-7 =0. (e) (xi, vi), {X2, 2/2). Ans. 2 (xi - X2) X + 2 (yi - 2/2) 2/ + X2^ - x^ + y^^ - y^ = 0. 22. Show that in problem 21 the coordinates of the middle point of the line joining the given points satisfy the equation of the perpendicular bisector. 23. Find the equations of the perpendicular bisectors of the sides of the triangle (4, 8), (10, 0), (6, 2). Show that they meet in the point (11, 7). 24^ Express by an equation that the point (h, k) is equidistant from (- 1, 1) and (1, 2) ; also from (1, 2) and (1, - 2). Then show that the point (f, 0) is equidistant from (- 1, 1), (1, 2), (1, - 2). 29. General equations of the straight line and circle. The methods illustrated in the preceding section enable us to state the following results : 1. A straight line parallel to the axis of y has an equation of the form x = constant. 2. A straight line parallel to the axis of x has an equation of the form y = constant. 58 ANALYTIC GEOMETRY Theorem I. The equation of the straight line passing through a point B (0, h) on the axis of y and having its slope equal to m is (I) y = fiidc + h. Proof First step. Assume that P (x, y) is any point on the line. Second step. The given condition may be written Slope of PB==m. Third step. Since by Theorem V, p. 35, Slopeof Pi5 = ^^^, [Substituting (x, y) for (xi, yi) and (0, 6) for {X2, 2/2)] ^1 y — h then = m, or 2/ = mx + b. q.e.d. Theorem II. The equation of the circle whose center is a given point (a, (3) and whose radius equals r is (II) 00^ + y^ -2aiic-2py + a^ -\- P^ -r^ = 0. Proof First step. Assume that P (x, y) is any point on the locus. Second step. If the center (a, /?) be denoted by C, the given condition is PC = r. Third step. By (lY), p. 31, PC = V(cc - ay +(y- pf. .-. ^{x - af + {y- ^y = r. Squaring and transposing, we have (II). q.e.d. Corollary. The equation of the circle whose center is the origin (0, 0) and whose radius is r is a?2 + i/2 = J.2, The following facts should be observed : Any straight line is defined by an equation of the first degree in the variables x and y. Any circle is defined by an equation of the second degree in the variables x and y, in which the terms of the second degree consist of the sum of the squares of x and y. THE CURVE AND THE EQUATION 69 30. Locus of an equation. The preceding sections have illus- trated the fact that a locus problem in Analytic Geometry leads p,t once to an equation in the variables x and y. This equation having been found or being given, the complete solution of the locus problem requires two things, as already noted in the first section (p. 51) of this chapter, namely, 1. To draw the locus by plotting a sufficient number of points whose coordinates satisfy the given equation, and through which the locus therefore passes. 2. To discuss the nature of the locus, that is, to determine properties of the curve. These two problems are respectively called : 1. Plotting the locus of an equation (second fundamental problem). 2. Discussing an equation (third fundamental problem). For the present, then, we concentrate our attention upon some given equation in the variables x and y (one or both) and start out with the definition : The locus of an equation in two variables repTCsenting coordinates is the curve or group of curves passing through all points whose coordinates satisfy that equation,"* and through such points only. From this definition the truth of the following theorem is at once apparent : Theorem III. If the form of the given equation he changed in any way (^for example, by transposition, by multiplicatidn by a constant, etc.), the locus is entirely unaffected. * An equation in the variables x and y is not necessarily satisfied by the coordinates of any points. For coordinates are real numbers, and the form of the equation may be such that it is satisfied by no real values of x and y. For example, the equation is of this sort, since, when x and ij are real numbers, x^ and y^ are necessarily positive (or zero), and consequently x^ + y^ + lis always a positive number greater than or equal to 1, and therefore not equal to zero. Such an equation therefore has no locus. The expression "the locus of the equation is imaginary" is also used. An equation may be satisfied by the coordinates of a finite number of points only. For example, x^ + y'^=0 is satisfied by x=0, y = 0, but by no other real values. In this case the group of points, one or more, whose coordinates satisfy the equation, is called the locus of the equation. 36^ 60 ANALYTIC GEOMETRY We now take up in order the solution of the second and third fundamental problems. 31. Second fundamental problem. Rule to plot the locus of a given equation. First step. Solve the given equation for one of the variables in terms of the other. ^ Second step. By this formula compute the values of the vari- able for which the equation has been solved by assuming real values for the other variable. Third step. Plot the points corresponding to the values so determined.^ Fourth step. If the points are numerous enough to suggest the general shape of the locus, draw a smooth curve through the poi^its. Since there is no limit to the number of points which may be computed in this way, it is evident that the locus may be drawn as accurately as may be desired by simply plotting a sufficiently large number of points. Several examples will now be worked out and the arrangement of the work should be carefully noted. Ex. 1. Draw the locus of the equation 2x-Sy + 6 = 0. Solution. First step. Solving for y, y = |x + 2. Second step. Assume values for x and compute y, arranging results in the form : YJi ji ^ Y y K y r [y (0, 2; y ^ > y (--3 0) X 2\ Thus, if 1, y = 2 . 1 + 2 2, y .r 1 . 2 + 2 etc. 31, Third step. Plot the points found. Fourth step. Draw a smooth curve through these points. X y X y 2 2 1 2| - 1 n 2 3^ -2 f 3 4 -3 4 42- -4 _ 2 .3 etc. etc. etc. etc. * The form of the given equation will often be such that solving for one variable is simpler than solving for the other. Always choose the simpler solution. t Remember that real values only may be used as coordinates. THE CURVE AND THE EQUATION 61 Ex. 2. Plot the locus of the equation 2/ zz x2 — 2 x — o. Solution. First step. The equation as given is solved for y. Second step. Computing y by assuming values of x, we find the table of v^alues below : X y X y -3 -3 .1 -4 - 1 2 — o -2 5 3 -3 12 4 5 -4 21 5 12 etc. etc. 6 21 etc. etc. i^ I / / / f / V / \ / \ / f y L i V \ / / s^ / ' I Third step. Plot the points. Fourth step. Draw a smooth curve through these points. This gives the curve of the figure. Ex. 3. Plot the locus of the equation x2 + ^2 4_ e X - IG = 0. First step. Solving for y, y = ± VlG - 6 X - x2. Second step. Compute y by assuming values of x. X y X y ±4 ±4 1 ±3 , - 1 ±4.6 2 — 2 ±4.9 3 imag. -3 ±5 4 " -4 ±4.9 5 u -5 ±4.6 G u -6 ±4 7 u -7 ±3 -8 -9 imag. I Fa I I X' "y" o J~^ 62 ANALYTIC GEOMETRY For example, if x = 1, y = ± Vl6 -6-1 = ± 3 ; if X = 3, y = ± Vl6 - 18 - 9 = ± V- 11, an imaginary number ; if X = - 1, y = ± Vl6 + 6-1 = ± 4.6, etc. Third step. Plot the corresponding points. Fourth step. Draw a smooth curve through these points. PROBLEMS 1. Plot the locus of each of the following equations, (a) X + 2 ?/ = 0. (b)x + 22/ = 3. (c) 3 X - ?/ + 5 = 0. (d) 2/ = 4 x2. (e) x2 + 4 y = 0. (f) 2/ = x2 - 3. (g) x2 + 4 y - 5 = 0. (h) 2/ = x2 + X + 1. (i) a; = 2/2 + 22/-3. (j) 4x = y3. (k) 4 X = 2/3-1. (1) y = x^- 1. (m) y = x^ — X. (n) ?/ = x^ — x2 — 5. (P) X2 + y2 := 9. (q) x2 + 2/2 = 25. (r) x2 + 2/2 + 9a.. = 0. (s) x2 + 2/2 + 4 2/ = = 0. (t) x2 + 2/2 - 6 X - -16: = 0. (u) x2 + 2/2 - 6 2/ - -16: = 0. (v) 4 2/ = x4 - 8. (w) 4 X = 2/* + 8. /^\ ., _ ^ ^''^^-1+X2 "i"^:- (z)x = -?-. r (o) x2 + 2/2 = 4. 1 + 2/ 2. Show that the following equations have no locus (footnote, p. 59). (a) x2 + 2/2 + 1 = 0. (f ) x2 + 2/2 + 2 X + 2 2/ + 3 = 0. (b) 2x2 + 32/2^-8. (g) 4x2 4-2/2 + 8x + 5 = 0. (c) x2 + 4 = 0. (h) ?/* + 2 x2 + 4 = 0. (d) x* + 2/2 + 8 = 0. (i) 9x2+4y24-i8x+82/ + 15=0. (e) (X + 1)2 + 2/2 + 4 = 0. (i) x2 + X2/ + 2/2 + 3 = 0. Hint. Write each equation in the form of a sum of squares, or solve for one variable and apply Theorem III, p. 11, to the quadratic under the radical. 32. Principle of comparison. In Ex. 1, p. 60, and Ex. 3, p. 61, we can determine the nature of the locus, that is, discuss the equation, by making use of the formulas (I) and (II), p. 58. The method is important and is known as the principle of comparison. THE CURVE AND THE EQUATION 63 The nature of the locus of a given equation may he determined by comparison with a general known equation, if the latter becomes ■ identical with the given equation by assigning particular values to its coefficients. The method of making the comparison is explained in the following Rule. First step. Change the form* of the given equation (if necessary) so that one or more of its terms shall be identical with one or more terms of the general equation. Second step. Equate coefficients of corresponding terms in the two equations, supplying any terms missing in the given equation with zero coefficients. Third step. Solve the equations found in the second step for the values f of the coefficients of the general equation. Ex. 1. Show that 2x — 3y + 6 = is the equation of a straight line (Fig., p. 60). " Solution. First step. Compare with the general equation (I), p. 68, (1) y = mx + b. Put the given equation in the form of (1) by solving for y, (2) y=fx + 2. Second step. The right-hand members are now identical. Equating cCefl&cients of x, (3) m = f . Equating constant terms, (4) 6 = 2. Third step. Equations (3) and (4) give the values of the coefficients m and 6, and these are possible values, since, p. 34, the slope of a line may- have any real value whatever, and of course the ordinate b of the point (0, b) in which a line crosses the F-axis may also be any real number. There- fore the equation 2x-3y + 6 = represents a straight line passing through (0, 2) and having a slope equal to §. q.e.d. *This transformation is called "putting the given equation in the form" of the general equation. tThe values thus found may be impossible (for example, imaginary) values. This may indicate one of two things, — that the given equation has no locus, or that it cannot be put in the form required. 64 ANALYTIC GEOMETRY Ex. 2. Show that the locus of (5) x2 + 2/2 + Gx- 16 = is a circle (Fig., p. 61). Solution. First step. Compare with the general equation (II), p. 68, (6) x2 + 2/2 - 2 ax - 2 jSy + a2 + /32 - r2 = 0. The right-hand members of (5) and (6) agree, and also the first two terms, x2 + ?/2. Second step. Equating coefficients of x, (7) - 2 a = 6. Equating coefficients of y, (8) -2/3 = 0. Equating constant terms, (9) a2 + /32 - r2 == - 16. ^ Third step. From (7) and (8), a = - 3, /3 = 0. Substituting these values in (9) and solving for r, we find r2 = 25, or r = 5. Since a, /3, r may be any real numbers whatever, the locus of (5) is a circle whose center is (— 3, 0) and whose radius equals 5. PROBLEMS 1. Plot the locus of each of the following equations. Prove that the locus is a straight line in each case, and find the slope m and the point of inter- section with the axis of y, (0, 6). (a) 2x4-2/-6 = 0. (b) X-Sy + S = 0. (c) x + 2y = 0. (d) 5 X - 6 ?/ - 5 = 0. (e) i^-fy-i = 0. (f) ?-f-l=0 5 6 is) 7 X - 8 ?y = 0. (h) f^-fy-l = 0. Ans. m =— 2, b = 6. Ans. m = 1, 6 = 2|. Ans. m =— I, b = 0. Ans. m. = I, 6 = — |. Ans. m = l,b=-j\. Ans. m = |, 6 = -6. Ans. m= I, 6 = 0. Ans. rn=l,b = -l^. THE CURVE AND THE EQUATION 65 2. Plot the locus of each of the equations following, and prove that the locus is a circle, finding the center (or, /3) and the radius r in each case.^ (a) x2 + 2/2 _ 16 = 0. Ans. {a, p) = (0, 0); r = 4. (b) x2 + y2 _ 49 = 0. Ans. (or, ^) = (0, 0); r = 7. (c) aj2 + ?/2 - 25 = 0. Ans. {a, p) = {0,0); r = 5. (d) x2 + ?/2 + 4x = 0. Ans. (or, /3) = (- 2, 0); r = 2. (e) x2 + 2/'^ - 8 7/ = 0. Ans. {a, p) = (0, 4); r = 4. _ (f) x2 + 2/2 + 4x-8?/ = 0. Ans. {a, p) = {- 2,4); r = ^20. (g) x2 + 2/2 _ 6x 4- 4 y - 12 = 0. Ans. (a, ^3) = (3, - 2); r = 5. (h) x2 + 2/2 - 4x + 92/ - I = 0. Ans. (a, ^3) = (2, - |); r = 5. (i) 3x2 + 32/2-6x- 82/ = 0. ^ns. (a, ^) = (1,|); r = f. The following problems illustrate cases in which the locus problem is completely solved by analytic methods, since the loci may be easily drawn and their nature determined. 3. Find the equation of the locus of a point whose distances from the axes XX' and YY' are in a constant ratio equal to f . A ns. The straight line 2 x — 3 y = 0. 4. Find the equation of the locus of a point the sum of whose distances from the axes of coordinates is always equal to 10. Ans. The straight line x + y — 10 = 0. 5. A point moves so that the difference of the squares of its distances from (3, 0) and (0, — 2) is always equal to 8. Find the equation of the locus and plot. Ans. The parallel straight lines 6x + 42/ + 3 = 0, 6x + 42/-13 = 0. 6. A point moves so as to he always equidistant from the axes of coor- dinates. Find the equation of the locus and plot. Ans. The perpendicular straight lines x-\-y = 0,x — y = 0. 7. A point moves so as to be always equidistant from the straight lines jc - 4 = and 2/ + 5 = 0. Find the equation of the locus and plot. Ans. . The perpendicular straight lines x-y-9 = 0, x + y-\-l=0. 8. Find the equation of the locus of a point the sum of the squares of whose distances from (3, 0) and (-3, 0) always equals 68. Plot the locus. Ans. The circle x2 + 2/2 = 25. 9. Find the equation of the locus of a point which moves so that its dis- tances from (8, 0) and (2, 0) are always in a constant ratio equal to 2. Plot the locus. Ans. The circle x2 + 2/^ = 16. 10. A point moves so that the ratio of its distances from (2, 1) and (- 4, 2) is always equal to |. Find the equation of the locus and plot. Ans. The circle 3 x2 + 3 2/2 - 24 x - 4 2/ = 0. i.:> 66 ANALYTIC GEOMETRY In the proofs of the following theorems the choice of the axes of CQordinates is left to the student, since no mention is made of either coordinates or equations in the problem. In such cases always choose the axes in the most convenient manner possible. 11. A point moves so that the sum of its distances from two perpendicular lines is constant. Show that the locus is a straight line. Hint. Choosing the axes of coordinates to coincide with the given lines, the equation ia x + y= constant. 12. A point moves so that the difference of the squares of its distances from two fixed points is constant. Show that the locus is a straight line. Hint. Draw XX^ through the fixed points, and Y Y'' through their middle point. Then the fixed points may be written (a, 0), (- a, 0), and if the "constant difference " be denoted by k, we find for the locus 4 ax = ^' or 4 ax = - fc. 13. A point moves so that the sum of the squares of its distances from two fixed points is constant. Prove that the locus is a circle. Hint. Choose axes as in problem 12. 14. A point moves so that the ratio of its distances from two fixed points is constant. Determine the nature of the locus. Ans. A circle if the constant ratio is not equal to unity and a straight line if it is. The following problems illustrate the Theorem. If an equation can be put in the form of a product of variable factors equal to zero, the locus is found by setting each fac- tor equal to zero and plotting the locus of each equation separately. 16. Draw the locus of 4 x2 - 9 2/2 := 0. Solution. Factoring, (1) {2x-Sy){2x + 3y) = 0. Then, by the theorem, the locus consists of the straight lines (2) 2x-3y = 0, (3) 2x-\-Sy = 0. Proof. 1. The coordinates of any point (xi, ?/i) which satisfy (1) will satisfy either (2) or (3). For if (xi, yi) satisfies (1), (4) (2xi-32/i)(2xi + 32/i) = 0. THE CURVE AND THE EQUATION 67 This product can vanish only when one of the factors is zero. Hence either 2a;i- 32/1 = 0, and therefore (xi, yi) satisfies (2) ; or 2 xi + 3 yi = 0, and therefore (xi, yi) satisfies (3). 2. A point (xi, yi) on either of the lines defined by (2) and (3) will also lie on the locus of (1). For if (xi, yi) is on the line 2 x - 3 y = 0, then (Corollary, p. 53) (5) 2xi-3?/i = 0. Hence the product (2 xi — 3 yi) (2 xi + 3 yi) also vanishes, since by (5) the first factor is zero, and therefore (xi, yi) satisfies (1). Therefore every point on the locus of (1) is also on the locus of (2) and (3), and conversely. This proves the theorem for this example. q.e.d. 16. Show that the locus of each of the following equations is a pair of straight lines, and plot the lines. (a) z^-y^=: 0. (j) 3x2 + xy - 22/2 + 6x - 4y = 0. (b) 9X2 -2/2 = 0. (k) x2 - 2/2 + X + ?/ = 0. (c) x2 = 92/2. (1) x2 - X2/ + 5x - 5 2/ = 0. (d) x2 - 4 X - 5 = 0. (m) x2 - 2 X2/ + 2/2 + 6 X - 6 2/ = 0. (e) 2/2-6?/ = 7. (n) x2 -42/2+ 5x+ 102/ = 0. (f) 2/2 - 5xy + 62/ = 0. (o) x2 + 4x2/ + 4 2/2 + 5x + IO2/ + 6 = 0. (g) X2/ - 2x2 - 3x = 0. .(p) x2 + 3x2/ + 2 2/2 + X + 2/ = 0. (h) X2/ - 2 X = 0. (q) x2 - 4 X2/ - 5 2/2 + 2 X - 10 2/ = 0. (i) X2/ = 0. (r) 3 x2 - 2 X2/ - 2/2 + 5 X - 5 y = 0. 17. Show that the locus of Ax"^ + Bx + C = is a pair of parallel lines, a single line, or that there is no locus according as A = B^ — 4: AC is positive, zero, or negative. 18. Show that the locus of ^x2 + Bxy + Cy^ = is a pair of intersecting lines, a single line, or a point according as A = B^ — 4: AC is positive, zero, or negative. ' ' ' ' ^ '• • -a^w^ , ^/ \ 33. Third fundamental problem. Discussion of an equation. The method explained of solving the second fundamental prob- lem gives no knowledge of the required curve except that it passes through all the points whose coordinates are determined as satisfying the given equation. Joining these points gives a curve more or less like the exact locus. Serious errors may be 68 ANALYTIC GEOMETRY made in this way, however, since the nature of the curve between any two successive points plotted is not determined. This obj ection is somewhat obviated by determining before plotting certain prop- erties of the locus by a discussion of the given equation now to be explained. The nature and properties of a locus depend upon the form of its equation, and hence the steps of any discussion must depend upon the particular problem. In every case, however, the fol- lowing questions should be answered. 1. Is the curve a closed curve or does it extend out infinitely far? 2. Is the curve symmetrical with respect to either axis or the origin ? The method of deciding these questions is illustrated in the following examples. Ex. 1. Plot the locus of IP (1) x2 + 4 2/2 = 16. Discuss the equation. Solution. First step. Solving for x, (2) x = ±2 V4 - ?/2. Second step. Assume values of y and compute x. This gives the table. Third step. Plot the points of the table. Fourth step. Draw a smooth curve through these points. X y X y ±4 ±4 ±3.4 1 ±3.4 -1 ±2.7 H ±2.7 -H 2 -2 imag. 3 imag. -3 Discussion. 1. Equation (1) shows that neither x nor y can be indefi- nitely great, since x^ and 4 ?/2 are positive for all real values and their sum must equal 16. Therefore neither x^ nor 4?/2 can exceed 16. Hence the curve is a closed curve. A second way of proving this is the following : From (2), the ordinate y cannot exceed 2 nor be less than — 2, since the expression 4 — ^/2 beneath the radical must not be negative. (2) also shows that X has values only from — 4 to 4 inclusive. yu ;^ THE CURVE AND THE EQUATION 69 2. To determine the symmetry with respect to the axes we proceed as follows : The equation (1) contains no odd powers of x or y ; hence it may be writ- ten in any one of the forms (3) (x)2 4- 4 (- 2/)2 = 16, replacing (x, y) by (x, - y) ; (4) (- x)2 + 4 (?/)2 = 16, replacing (x, y) by (- x, y) ; (5) (- x)2 + 4 (- y)2 = 16, replacing (x, ?/) by (- x, - y). The transformation of (1) into (3) corresponds in the figure to replacing each point P(x, y) on the curve by the point Q(x, — y). But the points P and Q are symmetrical with respect to XX\ and (1) and (3) have the same locus (Theorem HI, p. 59). Hence the locus of (1) is unchanged if each point is changed to a second point symmetrical to the first with respect to XX\ Therefore the locus is symmetrical with respect to the axis of x. Similarly from (4), the locus is symmetrical with respect to the axis ofy, and from (5), the locus is symmetrical with respect to the origin. The locus is called an eUipse. Ex. 2. Plot the locus of (6) y2_4a;_f.i5 = o. Discuss the equation. Solution. First step. Solve the equation for x, since a square root would have to be extracted if we solved for y. This gives (7) X = i(y2 + 15). X y 33- 4 ±1 ^ ±2 6 ±3 U ±4 10 ±5 12| ±6 etc. etc. Second step. Assume values for y and compute x. 70 ANALYTIC GEOMETRY Since y"^ only appears in the equation, positive and negative values of y give the same value of x. The calculation gives the table on p. 69. For example, if 2/ = ± 3, then a; = i (9 + 15) = 6, etc. Third step. Plot the points of the table. Fourth step. Draw a smooth curve through these points. Discussion. 1. From (7) it is evident that x increases as y increases. Hence the curve extends out indefinitely far from both axes. 2. Since (6) contains no odd powers of y, the equation may be written in the form - \o ^ / \ , nr r. (-?/)2-4(x) + 15 = by replacing (x, y) by (x, — y). Hence the locus is symmetrical with respect to the axis of x. The curve is called a parabola. Ex. 3. Plot the locus of the equation (8) xy-2y-4 = 0. '{? Solution. First step. Solving for y, 4 (9) x-2 Second step. Compute ?/, assuming values for x. When X y X y -2 -2 1 -4 -1 -f n -8 -2 -1 If -16 -4 -f 2 cp -5 -4 2i 16 : 2| 8 -10 s 3 4 etc. . etc. 4 2 5 4 6 1 12 0.4 etc. etc. 2, 2/ = 1 = 00. In such cases we assume values differing slightly from 2, both less and greater, as in the table. Third step. Plot the points. Fourth step. Draw the curve as in the figure in this case, the curve having two branches. 1. From (9) it appears that y diminishes and approaches zero as x increases indefi- nitely. The curve therefore extends indefi- nitely far to the right and left, approaching constantly the axis of x. If we solve (8) for X and write the result in the form 4 X = 2 y it is evident that x approaches 2 as ?/ increases indefinitely. Hence the locus extends both upward and downward indefinitely far, approaching in each case the line x =2. THE CURVE AND THE EQUATION 71 v/ 2. The equation cannot be transformed by any one of the three substitutions (X, y) into (x, - y), (X, y) into (-x, y), (X, y) into (- x, - ?/), without altering it in such a way that the new equation will not have the same locus. The locus is therefore not symmetrical with respect to either axis, nor with respect to the origin. — — — — — — oo — — — — — — — — ^ \ v^ ^^ ^ '^ p oo ^ =— u ,0) 5: *> N 1 -00 _ _ _ _ _ _ _ This curve is called an hyperbola. Ex. 4. Draw the locus of the equation (10) 4y = x^ X y X y 1 i -1 -\ H -II -H -'n 2 2 -2 ' -2 n ^% -2i -3|| 3 H -3 -6| 3i lOfl -3f -lOff Solution. First step. Solving for y, y = 1x3. Second step. Assume values for x and compute y. Values of x must be taken between the integers in order to give points not too far apart. For example, if x = 2|, 2/ = |.i|5=J^2_5=3i|, etc. 72 ANALYTIC GEOMETRY Kk i 1 1 X ? ; / / V ' ,/^ /O V // // r\ f/ ' f-x,- J - - --I — - Third step. Plot the points thus found. Fourtli step. The points determine the curve of tlie figure. •; Discussion. 1. From the given equation (10), x and y increase simultaneously, and therefore the curve extends out indefinitely from both axes. 2. In (10) there are no even powers nor constant term, so that by changing signs the equation may be written in the form 4(-2/) = (-x)3, replacing (x, ?/) by {-x, - y). Hence the locus is symmetrical with respect to the origin. The locus is called a cubical parabola. 34. Symmetry. In the above examples we have assumed the definition : If the points of a curve can be arranged in pairs which are symmetrical with respect to an axis or a point, then the curve itself is said to be symmetrical with respect to that axis or point. The method used for testing an equation for symmetry of the locus was as follows : if (x, y) can be replaced by {x, — y) through- out the equation without affecting the locus, then if (a, b) is on the locus, (a, — h) is also on the locus, and the points of the latter occur in pairs symmetrical with respect to XX\ etc. Hence Theorem IV. If the locus of an equation is unaffected by replacing y hy — y throughout its equation, the locus is symmetrical with respect to the axis of x. If the locus is unaffected by changing x to — x throughout its equation, the locus is symmetrical with respect to the axis of y. If the locus is unaffected by changing both x and y to — x and — y throughout its equation, the locus is symmetrical with respect to the origin. These theorems may be made to assume a somewhat different form if the equation is algebraic in x and y (p. 17). The locus of an algebraic equation in the variables x and y is called an algebraic curve. Then from Theorem lY follows THE CURVE AND THE EQUATION 73 Theorem V. Symmetry of an algebraic curve. If no odd powers of y occur in an equation^ the locus is symmetrical with respect to XX'; if no odd powers of x occur, the locus is symmetrical with respect to YY'. If every term is of even* degree, or every term of odd degree, the locus is symmetrical with respect to the origin. 35. Further discussion. In this section we treat of three, more questions which enter into the discussion of an equation. 3. Is the origin on the curve ? This question is settled by Theorem VI. The locus of an algebraic equation passes through the origin when there is no constant term in the equation. Proof The coordinates (0, 0) satisfy the equation when there is no constant term. Hence the origin lies on the curve (Corol- lary, p. 53). Q.E.D. 4. What values of x and y are to be excluded ? Since coordinates are real numbers we have the Rule to determine all values of x and y which must be excluded. First step. Solve the equation for x in terms of y, and from this result determine all values of y for which the computed value of x will be imaginary. These values of y must be excluded. Second step. Solve the equation for y in terms of x, and from this result determine all values of x for which the computed value of y will be imaginary. These values of x must be excluded. The intercepts of a curve on the axis of x are the abscissas of the points of intersection of the curve and XX\ The intercepts of a curve on the axis of y are the ordinates of the points of intersection of the curve and YT. Rule to find the intercepts. Substitute y — and solve for real values of x. This gives the intercepts on the axis of x. Substitute x = and solve for real values of y. This gives the intercepts on the axis of y. * The constant term must be regarded as of even (zero) degree. 74 ANALYTIC GEOMETRY The proof of the rule follows at once from the definitions. The rule just given explains how to answer the question: 5. What are the intercepts of the locus ? 36. Directions for discussing an equation. Given an equation, the following questions should be answered in order before plot- ting the locus. 1. Is the origin on the locus? (Theorem VI). 2. Is the locus sijmmetrical ivith respect to the axes or the origin? {^Theorems IV and V). 3. What are the intercepts? (Mule, p. 73). 4. What values of x and y must he excluded ? (^Rule, p. 73). 5. Is the curve closed or does it pass off indefinitely far? (§ 33, p. 68). Answering these questions constitutes what is called a general discussion of the given equation. Ex. 1. Give a general discussion of the equation (1) x2 - 4 2/2 + 16 2/ = 0. Draw the locus. X Sr n k ^ * ^ ^^ X ■Vs ^ ^ ^ ^ Vw _^ ^ ^ — ' ^ ^ !l= i fo, y= 2 X ^ ^" 'o ^ X ^ \^ *^ k^ ^ L^ ^ ^ . * ■^ * r' X 1. Since the equation contains no constant terra, the origin is on the curve. 2. The equation contains no odd powers of x\ hence the locus is symmet- rical with respect to YY'. 3. Putting ?/ =: 0, we find cc = 0, the intercept on the axis of x. Putting" ic = 0, we find ?/ = and 4, the intercepts on the axis of y. 4. Solving for x, (2) x.= ±2V2/2_4y. THE CURVE AND THE EQUATION 75 Hence all values of y between and 4 must be excluded, since for such a value y2 _ 4y is negative (Theorem IH, p. 11). Solving for y, (3) 2/ = 2 ± i Vx--2 + 16. Hence no value of x is excluded, since x^ + 16 is always positive. 5. From (3), y increases as x increases, and the curve extends out indefinitely far from both axes. Plotting the locus, using (2), the curve is found to be as in the figure. The curve is an hyperbola. PROBLEMS 1. Give a general discussion of each of the following equations and draw the locus. (n) 9?/2_x3 = 0. \o) 9?/2 + x3 = 0. (p) 2xy + 3x-4 =:0. (q) x2 - X2/ + 8 = 0. (r) x2 + xy -4 = 0. (s) x2 + 2x?/-3y = 0.' ^>(t) 2x?/-?/3 + 4x = 0. '^ (u) 3x2-?/ + x = 0. (v) 4 2/2 - 2 X - y = 0. (w) x2 - 2/2 + 6 X = 0. (x) x2 + 4 2/2 + 8 ?/ = 0. ^y) 9x2 + 2/2 + I8x - Qy = 0. >(z) 9x2-2/2 + 18x + 6?/ = 0. 2. Determine the general nature of the locus in each of the following equations by assuming particular values for the arbitrary constants, but not special values, that is, values which give the equation an added peculiarity.* (a) 2/2 = 2 mx. (f ) x^ - y^ = a^. (b) x2 - 2 my = m2. (g) x^ + y^ = r^. x2 2/2 _ (h) x2 + 2/2 = 2rx. (°) ^+52-^- (i) x2 + 2/2 = 2r2/. (d) 2x2/ = «2. (J) ^' + y^ = 2ax + 2by. 3,2 ^2 (k) a2/2 = x^ (^) a^-b2 = ^- (1) a% = a:3. * For example, in (a) and (b) m= is a special value. In fact, in all these examples zero is a special value for any constant. (a) x2 - -42/ = 0. (b) 2/2 - -4x + 3 = 0. (c) x2 + 42/2 -16 = 0. (d) 9x2 + 2/'' - 18 = 0. (e) x2 - - 4 2/2 - 16 = 0. (f ) x2 - -42/2 + 16 = 0. (g) x^ - -2/2 + 4 = 0. (h) X2 - - 2/ + X = 0. (i) xy - -4 = 0. (j) 9 2/ + ic3 = 0. (k) 4x - 2/3 = 0. . (1) 6x -y'=^o. (m) 5x - 2/ + 2/=^ = 0. 76 ANALYTIC GEOMETRY 3. Draw the locus of the equation y'^ = [x — a) (x — 6) (x - c), (a) when a < 6 < c. (c) when a 1. The conic is now called an hyperbola (see p. 71). 6. Plot each of the following. (i) x = (J) ^ = (a) x2?/ - 5 = 0. (e) y (b) x2y - ?/ + 2 X = 0. (f) y (c) x?/2_4x4-6 = 0. (g) y (d) T'y -y + ^ = 0. (h) y x2-3x 4x2 x2-4 x-3 x + 1 x2-4 2/-1 y-2 (k)4x = -/ 8y X2+ X (1) X 37. Points of intersection. If two curves whose equations are given intersect, the coordinates of each point of intersection must satisfy both equations when substituted in them for the variables (Corollary, p. 53). In Algebra it is shown that all values satisfying two equations in two unknowns may be found by regarding these equations as simultaneous in the unknowns' and solving. Hence the Rule to find the points of intersection of two curves ivhose equa- tions are given. THE CURVE AND THE EQUATION 77 First step. Consider the equations as simultaneous in the coordi- nates, and solve as in Algebra. Second step. Arrange the real solutions in corresponding pairs. These will he the coordinates of all the points of intersection. Notice that only real solutions correspond to common points of the two curves, since coordinates are always real numbers. Ex. 1. Find the points of intersection of (1) a: -7?/ + 25 = 0, (2) x2 + 2/2 = 25. Solution. First step. Solving (1) for X, (3) x.= 1y-2b. Substituting in (2), {ly- 25)2 + 2/2 = 25. Reducing, 2/^ — 7 ?/ + 12 = 0. .-. 2/ = 3 and 4. Substituting in (3) [not in (2)], X = — 4 and + 3. Second step. Arranging, the points of intersection are (—4, 3) and (3, 4). Ans. In tlie figure the straight line (1) is the locus of equation (1), and the circle the locus of (2). Ex. 2. Find the points of intersection of the loci of (4) (5) 2x2 + 32/2 = 35, 3x2-42/ = 0. Solution. First step. Solving (5) for x2, (6) x2 = f2/. Substituting in (4) and reducing, 92/2 + 82/ -105 = 0. .-. y = S and — %^ Substituting in (6) and solving, X = ± 2 and ± | V- 105. Second step. Arranging the real values, we find the points of intersection are (+ 2, 3), (- 2, 3). Ans. In the figure the ellipse (4) is the locus of (4), and the parabola (5) the locus of (5). / I it (■■JsX^ — -J^ ^'^y_ 7% t \ z ^ -I \ ^ \-* x^ ^ it^ s; 7 ^)s f r' "^ 78 ANALYTIC GEOMETRY PROBLEMS Find the points of intersection of the following loci. , 7x-ll2/ + l = 01 ""■ x-y = ^}' Ans.i6,l). ^- l^Vy^tl}' ^'''- (0'2), (-1, -I). ^ 2/2 ^ 16 X ^ 4- y-.x = Of' ^^'- (^'^)' (l^'l^)- x2 + 2/2 _ 4 X + 6 y - 12 = 0^ ^' 2y = Sx + 3 }• ^^MTV,m(-3, -3). ^ x2-?/2 = 16^ ^' x2 = 8y }• Ans. i±4V2,i). ^ x^ + y^ = 41'] ^- xy = 20 r ^^^- (±5' ±4), (±4, ±6). x2 + 2/2_6a;_2y_i5 = o 1 , , ^ .. , ^- 9x2 + 92/2 + 6x-62/-27 = 0r ^^5- (" 2, 1), (- H, - ff). 10. ^ . For what values of 6 are the curves tangent ? /-«. ( ^ , ), 6 = ±7VTa y2 -- 2 rix ^ ^^- x^ = 2pyf' - ^'^'- (0, 0), (2j), 2p). ,- 4x2 + ?/2^5>l 12- y2 33 8x r ^'''- (-2), (1,-2). x2 z= 4 ay 1 - ^^- y^ ^^^ r ^ns. (2a, a), (-2a, a). x2 + 4a2j x2 + ?/2^100^ !*• ^2-^ h ^ws. (8,6), (8, -6). 2 J ,_ x^ + y^ = 5a^-] • x2 = 4ay r ^''^- (2a, «), (-2a, a). ,„ 6%2+ (j2y2 = ^262>| ^®' X2 + 2/2 = (^2 |- ^^s- («, 0), (- a, 0). THE CURVE AND THE EQUATION 79 17. The two loci — -^ = 1 and ?- + ^ = 4 intersect in four points. 4 9 4 9 Find the lengths of the sides and of the diagonals of the quadrilateral formed by these points. Ans. Points, (± VlO, ± | V6). Sides, 2 VlO, 3 Vc. Diagonals, V94. Find the area of the triangles and polygons whose sides are the loci of the following equations. 18. 3x + y + 4 = 0, 3x-5?/ + 34 = 0, 3a;-2?/ + l = 0. Ans. 36. 19. x-\-2y = 6,2x + y = 1,y = x-\-l. ' Ans. f. 20. x + y = a,x-2y = ia, y-x-\-1a = 0. Ans. 12 a^. 21. x = 0,y = 0,x = 4,y =-6. Ans. 24. 22. x-y = 0, x-\-y = 0, x — y = a, x + y = b. Ans. — . 2 23. 2/ = 3x-9, ?/ = 3x + 5, 2y =x-6, 2y = a; + 14. Ans. 56. 24. Find the distance between the points of intersection of the curves 3x - 2 ?/ + 6 = 0, x2 + 2/2 = 9. Ans. if ^^• 25. Does the locus of y^ = 4x intersect the locus of 2x + 3y + 2 = 0? Ans. Yes. 26. For what value of a will the three lines 3x + ?/ — 2 = 0, ax + 2?/ — 3 = 0, 2x — 2/ — 3 = meet in a point ? Ans. a = 5. 27. Find the length of the common chord of x^ -\- y^ = lii and ?/2 = 3 x + 3. Ans. 6. 28. If the equations of the sides of a triangle are x + Ty + 11 = 0, 3x + 2/— 7 = 0, X — 3?/ + l = 0, find the length of each of the media ns. Ans. 2 V5, I V2, 1 VlTO. Show that the following loci intersect in two coincident points, that is, are tangent to each other. 29. ?/2-10x-6?/-31 =0, 2?/-10x = 47. 30. 9x2-4?/2 + 54x-162/ + 29 = 0, 15x-8?/ + ll = 0. 38. Transcendental curves. The equations thus far consid- ered have been algebraic in x and ?/, since powers alone of the variables have appeared. We shall now see how to plot certain so-called transcendental curves, in which the variables appear otherwise than in powers. The Rule, p. 60, will be followed. 80 ANALYTIC GEOMETKY Ex. 1. Draw the locus of (1) y = \ogiox. Solution. Assuming values for x, y may be computed by a table of loga- rithms, or, remembering the definition of a logarithm, from (1) will follow (2) X = lO^'. Hence values may also be assumed for y, and x computed by (2). This is done in the table. In plotting, unit length on XJT' is 2 divisions, unit length on YY' is 4 divisions. General discussion. 1. The curve does not pass through the origin, since (0, 0) does not satisfy the equation. 2. The curve is not symmetrical with re- spect to either axis or the origin. 3. In (1), putting x = 0, 2/ = log = — 00 = intercept on YY\ In (2), putting y = 0, X = 100 = I = intercept on XX'. Ya X y X y 1 .1 - 1 3.1 i .01 -2 10 1 .001 -3 100 2 .0001 -4 etc. etc. etc. etc. 4. From (2), since logarithms of negative numbers do not exist, all nega- tive values of x ofre excluded. From (2) no value of y is excluded. 5. From (2), as y increases x increases, and the locus extends out indefi- nitely from both axes. From (1), as X approaches zero, y approaches negative infinity ; so we see that the curve extends down indefinitely and approaches nearer and nearer to YY\ THE CURVE AND THE EQUATION 81 Ex. 2. Draw the locus of (3) y = 8mx if the abscissa x is the circular measure of an angle (Chapter I, p. 19). Solution. Assuming values for x and finding the corresponding number of degrees, we may compute y by the table of Natural Sines, p. 21. For example, if X = 1, since 1 radian = 57°. 29, y = sin 57°. 29 = .843. [by (3)] It will be more convenient for plotting to choose for x such values that the corresponding number of degrees is a whole number. Hence x is expressed in terms of ;r i«i the table. For example, if X y X y 7t 6 .50 It ~ 6 -.50 7t 3 .86 It ~ 3 -.86 It 2 1.00 It ~ 2 -1.00 27r "3" .80 27r 3 -.80 57r 6 .50 hit 6 -.50 Tt - It , y = sni - 2Tt — , ?/ = sm sin 60° 2Tt -sin ?^ (4, p. 19) = - sin 120°= -sin 60° (5, p. 20) = - .86. In plotting, three divisions being taken as the unit of length, lay off • ^0 = OB = 7r = 3.1416, and divide AO and OB up into six equal parts. The course of the curve beyond B is easily determined from the relation sin (2 ;r + x) = sin x. Hence y = smx = sin (2 ;r + x), that is, the curve is unchanged ifx + 27tbe substituted for x. This means, however, that every point is moved a distance 2 tt to the right. Hence the arc r. __ f^ ,j V y ■■1 # ^ V ^ *■ > S; s. , -J TT- f' f / 1 1 e. 't" '3, / ' 1 1 ^^ y X 1 ^ ^ 4\J '■ \ / TT TT IT Itt 5jr n* A^ \ " " SL 1 1 1 ^ -- £. i- .i>. -- a._ .T p-: — *s - - - — f/ ^ ^ ^ s •^ ^ ~ ~ ~ p ?/= -1 /? ~^ r V' - APO may be moved parallel to XX' until A falls on B, that is, into the position BEC, and it will also be a part of the curve in its new position. 82 ANALYTIC GEOMETRY Also, the arc OQB may be displaced parallel to XX' until falls upon C. In this way it is seen that the entire locus consists of an indefinite number of congruent arcs, alternately above and below XX'. General discussion. 1. The curve passes through the origin, since (0, 0) ^satisfies the equation. 2. Since sin (— x) = — sin x, changing signs in (3), or y =— sm X, y = sin(-x). Hence the locus is unchanged if (x, y) is replaced by (— x, —y), and the curve is symmetrical with respect to the origin (Theorem IV, p. 72) . 3. In (3), if X = 0, y = sin = = intercept on the axis of y. Solving (3) for x, (4) x = sin-'^y. In (4), if y = 0, = me, n being any integer. Hence the curve cuts the axis of x an indefinite number of times both on the right and left of 0, these points being at a distance of tt from one another. 4. In (3), X may have any value, since any number is the circular meas- ure of an angle. In (4), y may have values from - 1 to + 1 inclusive, since the sine of an angle has values only from — 1 to +1 inclusive. 5. The cuTve extends out indefinitely along XX' in both directions, but is contained entirely between the lines ?/=: + l, y = —i. The locus is called the wave curve, from its shape, or the sinusoid, from its equation (3). Ex. 3. Draw the locus of y = tan x. There is no difficulty in obtaining the curve of the' figure and in verifying the properties indicated by a dis- cussion similar to the pre- ceding examples. THE CURVE AND THE EQUATION 83 PROBLEMS Plot the loci of the following equations. 1. 2/ = cosx. 5. y = ta,n-^x. 9. y = sin2x. 2. y = cotx. 6. 2/ = 2^. 10. 2/ = tan-. 3 y-s&cx '^- y = 21ogioX. 3. y-secx. ^ 11 y = 2co8X. 4. y = sm-^x. 8. 2/ = (l+ajf. 12. y = sin x + cos x. 39. Graphical representation in general. Any equation con- taining two variables may be represented graphically by a curve called the graph of the equation by considering the variables as coordinates and plotting the locus in the usual way. This method of representing a given law is widely used in all branches of science. Ex. 1. Draw the graph of the Simple Interest Law, which shall represent the relation between amount and time for a given principal and rate per cent. The law is proven in Algebra to be (1) J. = P(l + m), where A = amount, P = principal, r = rate, n = number of years. Solution. For convenience, take P = one dollar.* Let One division on OX = 1 year. One division on OF = 1 dollar, abscissas = values of n, ordinates = values of A. (0,1) Then the required graph is the locus of O (1,0) (n,o)X (2) ?/ = rx + 1. The locus of (2) is a straight line passing through (0, 1) and having a slope equal to r (Theorem I, p. 58). This graph may be used to solve interest problems. For if the number of years n is given, we merely have to measure off the corresponding ordinate A of the straight line, and this will give the amount of one dollar at the given rate for n years. * Any other case is obtained by multiplying all the ordinates in the figure by P. 84 ANALYTIC GEOMETRY Ex. 2. In Physics it is shown that the vokime (v), pressure (p), and absolute temperature (t) of a given mass of a perfect gas are connected by the law (3) pv = kt, k being a constant dependent upon the particular gas. Draw the graph if the temperature is assumed constant. Solution. Assume one division on OX = unit of pressure, one division on OY = unit of volume, abscissas = pressures, ordinates = volumes. Then the required graph is the locus o^ (4) xy = constant. The curve is one branch* of an hyperbola extending to the right and upward indefinitely, approaching in each case the corresponding axis. Such curves are called isothermals (equal temperatures), and jtl^e figure is called the Pressure-Volume Diagram. PROBLEMS 1. Draw the graph of the Simple Interest Law if the variables are (a) n and P. (c) A and P. (e) P and r. (b) n and r. (d) A and r. 2. Draw the graph of the law of Ex. 2 if the variables are (a) p and t. (b) v and t. 3. The amount (A) of any principal (P) at compound interest (r%) for n years is given by the Compound Interest Law A=P{1 + r)». Draw the graph of this law if the variables are (a) A and P. (c) A and n. (e) P and n. (b) A and r. ■ (d) P and r. (f) r and n. Hint. Take tlie logarithm of both sides when convenient for computation. * Since negative volumes have no physical meaning, in many cases only a portion of the entire locus can be made use of in the representation. CHAPTER IV THE STRAIGHT LINE AND THE GENERAL EQUATION OF THE FIRST DEGREE 40. The idea of coordinates and the intimate relation connect- ing a curve and an equation, which results from the introduction of coordinates into the study of Geometry, have been considered in the preceding chapters. Analytic Geometry has to do largely with a more detailed study of particular curves and equations. In this chapter we shall consider in detail the straight line and the general equation of the first degree in the variables x and y representing coordinates. 41. The degree of the equation of a straight line. It was shown in Chapter III (Theorem I, p. 58) that (1) y = mx + b is the equation of the straight line whose slope is m and whose intercept on the F-axis is ^ ; m and b may have any values, positive, negative, or zero (p. 34). But if a line is parallel to the F-axis, its equation may not be put in the form (1) ; , for, in the first place, the line has no intercept on the F-axis, and, in the second place, its slope is infinite and hence cannot be substituted for m in (1). The equation of a line parallel to the F-axis is, however, of the form (2) X = constant. The equation of any line may be put either in the form (1) or (2). As these equations are both of the first degree in x and y we have Theorem I. The expiation of any straight line is of the first degree in the coordinates x and y. 85 86 ANALYTIC GEOMETRY 42. The general equation of the first degree, A3o+By-\-C=0, The equation (1) Ax-{-By-\-C = 0, where A, B, and C are arbitrary constants (p. 1), is called the general equation of the first degree in x and y because every equa- tion of the first degree may be reduced to that form. Equation (1) represents all straight lines. For the equation y = mz + b may be written mx — y -{■ b - 0, which is of the form (1) it A = m, B = — 1, C= b; and the equation x = constant may be written X — constant = 0, which is of the form (1) if J. = 1, ^ = 0, C= — constant. Theorem II. (Converse of Theorem I.) The locus of the general equation of the first degree Ax -{- By -{- C =^ is a straight line. X? Proof Solving (1) for y, we obtain AC (2) 2/ = -^--^- This equation has the same locus as (1) (Theorem 111, p. 59). By Theorem I, p. 58, the locus of (2) is the straight line whose A C slope is m = — — and whose intercept on the F-axis is ft = — — • B B If, however, £ = 0, it is impossible to write (1) in the form (2). But if ^ = 0, (1) becomes ^x + C = 0, C or x= 7 • A The locus of this equation is a straight line parallel to the F-axis (1, p. 57). Hence in all cases the locus of (1) is a straight line. Q.E.D. Corollary I. The slope of the line Ax -[- By -{- C = () is m =— —; that is, the coefficient of x with its sign changed B divided by the coefficient of y. 1^ THE STRAIGHT LINE 87 Corollary II. The lines Ax -{- Bi/ -\- C = and A^x-\- B'y + C = are parallel when and only when the coefficients of x and y are proportional; that is, A__B A'^B'' For two lines are parallel when and only when their slopes are equal (Theorem VI, p. 36) ; that is, when and only when _A__A^ B~ B'' Changing the signs and applying alternation, we obtain Corollary III. The lines Ax-^By -\-C = and A'x -\- B'y + C =^ are perpendicular when and only when AA' + BB' = 0. For two lines are perpendicular when and only when the slope of one is the negative reciprocal of the slope of the second (Theorem VI, p. 36) ; that is, A _B' ~B-A^' or AA' + BB' = 0. Ly - Corollary IV. The intercepts of the line Ax + By -^C = on the X- and Y-axes are respectively a = 7 and = — - • A B For the intercept on the X-axis is found (p. 73) by setting y = (i and solving for X, and the intercept on the F-axis has been found in the above proof. Corollaries I and IV are given chietly for purposes of reference. In a numerical example the intercepts are found most simply by applying the general rule already given (p. 73) ; and the slope is found by reducing the equation to the form 2/ = mx + &, when the coefficient of x will be the slope. 88 ANALYTIC GEOMETRY Theorems I and II may be stated together as follows : The locus of an equation is a straight line when and only when the equation is of the first degree in x and y. Theorem II asserts that the locus of every equation of the first degree is a straight line. Then, to plot the locus of an equation of the first degree it is merely necessary to plot two points on the locus and draw the straight line passing through them. The two simplest points to plot are those at which the line crosses the axes. But if those points are very near the origin it is better to use but one of them and some other point not near the origin whose coordinates are found by the Eule on p. 60. Theorem III. When two equations of the first degree, (3) Ax + By ^ C = ; and (4) A^x'+B'y + C^ = 0, have the same locus, then the corresjjonding coefficients are propor- tional; that is, A'~~ B'~ C' Proof The lines whose equations are (3) and (4) are by hypothesis identical and hence they have the same slope and the same intercept on the F-axis. Since they have the same slope, A A' B^B'' (Corollary I, p. 86) and since they have the same intercept on the F-axis, B~ B' by alternation we obtain = — . (Corollary lY, p. 87) A B ^ C B A' = J'''''^C' = B''' , , ABC and hence - = -=-. q.^..^. THE STRAIGHT LINE, 89 Ex. 1. Find the values of a and h for which the equations 2ax + 2?/-5 = and 4x-3?/ + 76 = will represent the same straight line. Solution. These two equations will represent the same straight line if (Theorem III) 2a _ ^ _ -5 . 4 ~ -3~ 76 ' and hence the required values are obtained by solving 2a_^ _2___ -b^ 4 ~ 33 ^" 33 ~ "75" for a and h. This gives a = -t, 6 = ||. 43. Geometric interpretation of the solution of two equations of the first degree. If we solve the equations (1) Ax + By -\-C = and (2) A^x -{- B^y -\- C^ = 0, we obtain the coordinates of the points of intersection of the lines whose equations are (1) and (2) (Eule, p. 76). But if these lines are parallel they do not intersect, and if they are identical they intersect in all of their points. The relation between the position of the lines whose equations are (1) and (2) and the number of solutions of the simultaneous equations (1) and (2) may be indicated as follows : _, . . ^ _ Number of solutions Fosition of Lines of equations Intersecting lines. One solution. Parallel lines. No solution. Coincident lines. An infinite number. It is sometimes as convenient to be able to determine the number of solutions of two equations of the first degree without solving them as it is to be able to determine the nature of the roots of a quadratic equation without solving it. The following theorem enables us to do this. 90 ANALYTIC GEOMETRY Theorem IV. Two equations of the first degree^ Ax-\- By + C = and A'x -{- B'y -\- C^ = 0, have, in general, one solution for x and y ; hut if A__B_ A'~ B'' there is no solution unless A__B__C_ A'~B'~C'' when there is an infinite number of solutions. The proof follows at once from Corollary II, p. 87, and Theorem III. PROBLEMS V 1 . Find the intercepts of the following lines and plot the lines. (a) 2 x + 3 2/ = 6. Ans. 3, 2. (b) ^ + ^ = 1. ' Ans. 2, 4. 2 4 (c) - - ^ = 1. Ans. 3, - 5. (d) - + -^ = 1. Ans. 4, -2. 4 — 2 2. Plot the following lines. (a) 2x-32/+5 = 0. (c) ^ + | = 1. (b) 2/-5-4ic = 0. (d) --y- = \. 3. Find the equations, and reduce them to the general form, of the lines for which (a) m = 2, 6 = - 3. ^ns. 2 x - y - 3 = 0. (b) m = - i, 6 = f . Ans. x + 2 ?/ - 3 = 0. (c) m = f , 6 = - f . Ans. 4 x - 10 y - 25 = 0. (d) a = - , 6 = — 2. Ans. x-y -2 = 0. 4 q _, (e) or = — - , 6 = 3. Ans. x + y - 3 = 0. Hint. Substitute \ny = mx + b. THE STRAIGHT LINE 91 A. Find the number of solutions of the following pairs of equations and plot the loci of the equations. (^) I4 X + 6 2/ + 9 = 0. ^'''- ^° '^^^*^°°- (b) { / Ans. One. x + y = 1. '2-3x (d) -^ . ^ ^ Ans. No solution. I^lz X — lo ' ('2 3x^1/ (c) -^ - ,0 ■ ^ns. An infinite number. ^ ' 1^6 X + 2 ?/ = 4. 20 = 0. 2/4-6 = 0. 5. Plot the lines 2x — 32/ + 6 = and x — y = 0. Also plot the locus of (2 X - 3 2/ + 6) + A; (X - y) = for A: = 0, ±1, ±.2. , 6. Select pairs of parallel and perpendicular lines from the following. fLi:y = 2x-S. ,,JX2:?/=-3X+2. . TUT T , T ^^nU:y = 2x + 7. ^^'- Li\\Ls;L,±L,. L2>4 : y = 1 X + 4. rXi:x + 3y = 0. (b) ^ X2 : 8x + y + 1 = 0. Ans. Li ± L3. [Ls-.dx-Sy + 2 = 0. fXi : 2 X - 5 y = 8. (c) -^ X2 : 5 y + 2 X = 8. Ans. X2 -L L3. U3:35x-14y = 8. 7. Show that the quadrilateral whose sides are 2x — 3y + 4 = 0, 3x — y — 2 = 0, 4x — 6?/ — 9 = 0, and 6x — 22/ + 4 = 0isa parallelogram. 8. Find the equation of the line whose slope is — 2 which passes through the point of intersection of y = 3 x + 4 and y = — x + 4. Ans. 2x + 2/ — 4 = 0. 9. What is the locus of y = wx + 6 if 6 is constant and m arbitrary ? if m is constant and b arbitrary ? 10. Write an equation which will represent all lines parallel to the line (a) y = 2 X + 7. (c) y - 3 X - 4 = 0. (b)?/ = -x + 9. (d) 22/-4x + 3 = 0. 11. Write an equation which will represent all lines having the same intercept on the F-axis as (a), (b), (c), and (d) in problem 10. 12. Find the equation of the line parallel to2x — 3?/ = whose intercept on the F-axis is — 2. Ans. 2x — 3y — 6 = 0. 13. What is the locus of Ax + By + C = it B and C are constant and A arbitrary ? if -4 and iJ are constant and C arbitrary ? 92 ANALYTIC GEOMETRY 44. Straight lines determined by two conditions. In Ele- mentary Geometry we have many illustrations of the determina- tion of a straight line by two conditions. Thus two points determine a line, and through a given point one line, and only one, can be drawn parallel to a given line. Sometimes, however, there will be two or more lines satisfying the two conditions ; thus through a given point outside of a circle we can draw two lines tangent to the circle, and four lines may be drawn tangent to two circles if they do not intersect. Analytically such facts present themselves as follows. The equation of any straight line is of the form (Theorem II, p. 86) (1) Ax-\-mj-{-C = 0, and the line is completely determined if the values of two of the coefficients A, B, and C are known in terms of the third. For example, it A = 2B and C — —3B, equation (1) becomes 2Bz + Btj-3B==0, or 2x-\-y — 3 = 0. Any geometrical condition which the line must satisfy gives rise to an equation between one or more of the coefficients A, B, and C. Thus if the line is to pass through the origin, we must have (7=0 (Theorem VI, p. 73) ; or if the slope is to he 3, then =3 (Corollary I, p. 8G). B Two conditions which the line must satisfy will then give rise to two equations in A, B, and C from which the values of two of the coefficients may be determined in terms of the third, and the line is then determined. If these equations are of the first degree, there will be only one line fulfilling the given conditions, for two equations of the first degree have, in general, only one solution (Theorem IV, p. 90). If one equation is a quadratic and the other of the first degree, then there will be two lines fulfilling the conditions, provided that the solutions of the equations are real. And, in general, the number of lines fulfilling the two given conditions will depend on the degrees of the equations in the A, B, and C to which they give rise. ^ THE STRAIGHT LINE 93 I Rule to determine the equation of a straight line which satisfies two conditions. First step. Assume that the equation of the line is Ax -{- By ^ C =^ ^. Second step. Find two equations between A, B, and C each of which expresses algebraically the fact that the line satisfies one of the given conditions. Third step. Solve these equations for two of the coefficients A, B, and C in tei^ms of the third. Fourth step. Substitute the results of the third step in the equct/- tion in the first step and divide out the remaining coefficient. The result is the required equation. Ex. 1. Find the equation of the line through the two points Pi (5, — 1) and P2(2, -2). Solution. First step. Let the required equation be (1) Ax + B2j-hC = 0. Second step. Since Pi lies on the locus of (1) (Corollary, p. 53), (2) 6A-B+C = 0', and since Pg lies on the line, (3) 2A-2B-i-C = 0. Third step. Solving (2) and (3) for A and B in terms of C, we obtain A=-IC, B = IC. Fourth step. Substituting in (1), -lCx+lC2j + C = 0. Dividing by C and simplifying, the required equation is x-3y-S = 0. Ex. 2. Find the equation of the line, passing through Pi (3, — 2) whose slope is — i. Solution. First step. Let the re- quired equation be (4) Ax + By+C^O. Second step. Since Pi lies on (4), (5) 3A-2B+C = 0; and since the slope is — i^ (6) -^=-1. 94 ANALYTIC GEOMETRY Third step. Solving (5) and (6) for A and C in terms of B, we obtain A = IB, C = IB. Fourth step. Substituting in (4), lBx + By-^lB = 0, or a; + 42/ + 5 = 0. PROBLEMS 1. Find the equation of the line satisfying the following conditions and plot the lines. (a) Passing through (0, 0) and (8, 2). (b) Passing through (-1, 1) and (- 3, 1). (c) Passing through (—3, 1) and slope = 2. (d) Having the intercepts a = o and & = — 2. (e) Slope = — 3, intercept on X-axis = 4. (f ) Intercepts a = — 3 and 6 = — 4. (g) Passing through (2, 3) and (- 2, - 3). (h) Passing through (3, 4) and (- 4, - 3). (i) Passing through (2, 3) and slope = — 2. (j) Having the intercepts 2 and — 5. 2. Find the equation of the line passing through the origin parallel to the line 2 cc — 3 2/ = 4. Ans. 2 x — 3 ?/ = 0. 3. Find the equation of the line passing through the origin perpendicular to the line 5x4-2/ — 2 = 0. Ans. x — 6y = 0. 4. Find the equation of the line passing through the point (3, 2) parallel to the line 4x — y — 3 = 0. Ans. 4x — y — 10 = 0. 5. Find the equation of the line passing through the point (3, 0) perpen- dicular to the line 2x + y — 5 = 0. Ans. x — 2y — 3 = 0. 6. Find the equation of the line whose intercept on the Y-axis is 5 which passes through the point (6, 3). Ans. x -f 3y — 15 = 0. 7. Find the equation of the line whose intercept on the JT-axis is 3 which is parallel to the line x — 4?/-|-2 = 0. Ans. x — 4?/ — 3 = 0. 8. Find the equation of the line passing through the origin and through the intersection of the lines x — 2y -\- S = and x + 2y — 9 = 0. Ans. X — y = 0. 9. Find the equation of the straight line whose slope is m which passes through the point Pi (xi, yi). Ans. y — yi = m{x — Xi). Ans. x-4y = 0. Ans. y-i = o. An^. 2 X - ?y + 7 = 0. Ans. 2x-3y -6 = 0. Ans. 3x-l-y-12 =0. Ans. 4x + 3?/-M2 = 0. Ans. Sx-2y = 0. Ans. x-y + 1 = 0. Ans. 2x + y-l = 0. Ans. ^--y- = i. 2 5 THE STRAIGHT LINE 95 10. Find the equation of the straight line whose intercepts are a and b. Ans. - + l = l. a 11. Find the equation of the straight line passing through the points Pi{xu y\) and Pa (0^2, 2/2). , Ans. (2/2 -Vi)^- {X2 -Xi)y + x^yi - xiy^ = 0. 12. Show that the result of the last problem may be put in the form x-xi ^ y-yi X2-X1 y^-yi Hint. Add and subtract x^y^, factor, transpose, and express as a proportion. 45. The equation of the straight line in terms of its slope and the coordinates of any point on the line. In this section and in those immediately following, the Rule in the preceding section is applied to the determination of general forms of the equations of straight lines satisfying pairs of conditions which occur frequently. These general forms will then enable us to write the equations of certain straight lines with the same ease that the equation y — mx + h enables us to write the equation of the straight line whose slope and intercept on the F-axis are given. Theorem V. Point-slope form. The equation of the straight line which passes through the point Pi (xi, y^ and has the slope m is (V) - y -.y^z=m{pc — cCi). Proof. First step. Let the equation of the given line be (1) Ax-{-By -\-C =^0. Second step. Then, by hypothesis, (2) Axi + 5?/i + C = and (3) -f = ^- Third step. Solving (2) and (3) for A and C in terms of By we obtain A =— mB and C = B (inxi — y^). 96 ANALYTIC GEOMETRY Fourth step. Substituting in (1), we have — mBx -\- By -\- B (^mx-i — y^) = 0. Dividing by B and transposing, y — yi = m{x — x^). ^ q.e.b. If Pi lies on the T-axis, cci = and y^ = b, so that this equa- tion becomes y = mx + b. 46. The equation of the straight line in terms of its intercepts. We pass now to the consideration of a line determined by two points, and we consider first the case in which the two points lie on the axes. This section does not, therefore, apply to lines par- allel to one of the axes or to lines passing through the origin, as in the latter case the two points coincide and hence do not deter- mine a line. Theorem VI. Intercept form. If a and b are the intercepts of a line on the X- and Y-axes 7'espectively, then the equation of the line is (VI) - + 1^ = 1. ^ ^ ah Proof First step. Let the equation of the given line be (1) Ax + By + C = 0. Second step. By definition of the intercepts (p. 73), the points (a, 0) and (0, b) lie on the line; hence (2) ^a + <^ = 0, (3) Bb-\-C = 0. Third step. Solving (2) and (3) for A ar«d B in terms of C. we obtain A = C and B = --C. a Fourth step. Substituting in (1), we have - - Cx - - Cy + C = 0. a b Dividing by C and transposing, ^ , ?/ _ 1 r T — -L- Q.E.D. THE STRAIGHT LINE 97 Ex. 1. Write the equation of the locus of 2x — 62/ + 3 = 0in terms of its intercepts and plot the line. Solution. Transposing the constant term, we have Dividing by — 3, 2x -3 X + 2y = l, _. + ? = >■ 2 2 This equation is of the form (VI). Hence a = — I and b = |. Plotting the points (— |, 0) and (0, I) and joining them by a straight line, we have the required line. 47. The equation of the straight line passing through two given points. Theorem VII. Two-point form. The equation of the straight line passing through Pi (xi, y{) and P^, (x^, ])%) is /yil) ^ - ^1 ^ y - Z/1 ^ ^ a?2 — a?i 2/2 — 2/i Proof. Let the equation of the line be (1) Ax-{-By^-C = 0. Then, by hypothesis, (2) Axi + By, + C = and (3) Ax^ + %2 + C = 0. To follow the Rule, p. 93, we must solve (2) and (3) for A and B in terms of C, substitute in (1), and divide by C ; that pro- cedure amounts to eliminating A, B, and C from (1), (2), and (3), and that elimination may be more conveniently performed as follows : Subtract (2) from (1) ; this gives A{x-x,)+B{y-y,)=0, or (4) A{x-a:,) = -B{y-yi). 98 ANALYTIC GEOMETKY Similarly, subtracting (2) from (3), we obtain (5) A (x, -x,) = -B (2/2 - 2/i). Dividing (4) by (5), we find = Q.E.D. *^2 "^i 2^2 yi Corollary. The condition that three points, Pi(xi, i/i), ^2(^2? 2/2)? and Pg (ccg, 2/3) should lie on a line is that ^z — ^i ^ yz — yi ^2 — ^1 2/2 yi For this is the condition that P3 should lie on the line (VII) passing through i^jj^ndL>P2 (Corollaiy, p. 53). The TYiethod of proving the corollary should be remembered rather than the corollary itself, as then the condition may be immediately written down 'from (VII). PROBLEMS 1 . Find, by substitution in the proper formulas, the equations of the lines satisfying the conditions in problem 1, p. 94. 2. Find the equations of the lines fulfilling the following conditions and plot the lines. (a) Passing through the origin, slope = 3. Ans. Sx — y = 0. (b) Passing through (3, - 2) and (0, - 1). Ans. x + 3?/ + 3 = 0. (c) Having the intercepts 4 and — 3. ' Ans. Sx — 4:y — 12 = 0. (d) F-intercept = 5 and slope = 3. Ans. 3x — y + 5 = 0. (e) Passing tiirough (1, — 2) and (3, — 4). Ans. x + y + 1 = 0. (f ) Having the intercepts — 1 and — 3. Ans. Sx-\-y + 3 = 0. (g) Passing through (- |, |) and slope = - f . Ans. ix + 6y—7 = 0. (h) Passing through (0, 0) and slope = m. Ans. y = mx. /T' 3. Find the equations of the sides of the triangle whose vertices are (-3,2), (3, - 2), and (0, - 1). Ans. 2x + 32/ = 0, a; + 3?/ + 3 = 0, and x + ?/ + 1 = 0. 4. Find the equations of the medians of the triangle in problem 3 and show that they meet in a point. Ans. x = 0,7x-\-9y -\- S = 0, and 5x + 9y -{■ S = 0. Hint. To show that three lines meet in a point, find the point of intersection of two of them and prove that it lies on the third. THE STRAIGHT LINE 99 5. Show that the medians of any triangle meet in a point. Hint. Taking one vertex for origin and one side for the X-axis, the vertices may then be called (0, 0), (a, 0), and (6, c). 6. Determine whether or not the following sets of points lie on a straight line. (a) (0, 0), (1, 1), (7, 7). Ans. Yes. (b) (2, 3,), (- 4, - 6), (8, 12). Ans. Yes. (c) (3, 4), (1, 2), (5, 1). Ans. No. (d) (3, - 1), (- 6, 2), (- I, 1). Ans. No. (e) (5, 6), (1,1), (-1,-1). ^ns. Yes. (f) (7, 6), (2, 1), (6, - 2). Ans. No. 7. Reduce the following equations to the form (VI) and plot their loci. (a) 2x-f 32/-6 = 0. (d) 3x + 4y + 1 = 0. (b) x-32/ + 6 = 0. (e) 2x-4y-7=0. (c) 3x-42/ + 9 = 0. (f) 7x-6y-Z = 0. 8. Find the equations of the lines joining the middle points of the sides of the triangle in problem 3 and show that they are parallel to the sides. • Ans. 4x + 6?/ + 3 = 0, x + 3?/=:0, and x-\-y = 0. 9. Find the equation of the line passing through the origin and through the intersection of the lines x + 2 y = 1 and 2x-4?/-3 = 0. Ans. X + 10 y = 0. 10. Show that the diagonals of a square are perpendicular. Hint. Take two sides for the axes and let the length of a side be a. 11. Show that the line joining the middle points of two sides of a triangle is parallel to the third. Hint. Choose the axes so that the vertices are (0, 0), (a, 0), and (6, c). 12. Find the equation of the line passing through the point (3, - 4) which has the same slope as the line 2 x - y = 3. Ans. 2x-y -10 = 0. 13. Find the equation of the line passing through the point (-1, 4) which is parallel to the line 3 x + y + 1 = 0. Ans. 3x + y -1 = 0. 14. Two sides of a parallelogram are2x + 3y-7 = and x-3y + 4 = 0. Find the other two sides if one vertex is the point (3, 2). Ans. 2x-f3y- 12 = and x-3y + 3 = 0. 15. Find the equation of the line passing through the point (- 2, 3) which is perpendicular to the line x-{-2y = 1. Ans. 2x -y + 1 = 0. 100 ANALYTIC GEOMETRY 16. Show that the three lines x — 2y = 0, x-{-2y — 8 = 0, and x + 2y — S + k{x — 2 y) = meet in a pomt no matter what value k has, ' 17. Derive (V) and (VII) by the Rule on p. 53, using Theorem V,' p. 35. 18. Derive (VI) and (VII) by the Rule on p. 53, using the theorem that the corresponding sides of similar triangles are proportional. 19. Derive y = mx + h and (V) by the Rule on p. 53, using the definition of the tangent of an acute angle in a right triangle. ' 20. Derive the equation of the straight line in terms of the perpendicular distance p from the origin to the line and the angle w which that perpendicular makes with the positive direction of the X-axis. Hint. Find the intercepts in terms of p and w by solving the right triangles in the figure and substitute in (VI). Ans. X cos w + 2/ sin w — p = 0. V^ 21. What is the locus of (V) if Xi and 2/1 are constant and m arbitrary ? 22. What is the locus of (VI) if a is constant and 6 arbitrary ? if 6 is con- stant and a arbitrary ? ,.,j^- ' 23. Write an equation which represents all lines passing through (2, — 1). 24. Write an equation representing all lines whose intercept on the X-axis is 3. 25. Write in two different forms the equation of all lines whose intercept on the Y-axis is — 2. 26. Write an equation representing all lines whose slope is — \. 27. If the axes are oblique and make an angle of w, then the equation of a straight line in terms of its inclination a and intercept on the Y-axis h is sin a sin (w — a) « + &. 28. If the angle between the axes is w, the equation of the line passing through Pi(iCi, 2/1) whose inclination is a is sin a , , y — y^ — - — ^^ {x — Xx). -^ ^^ sm (w - a) ^ '^ 29. Show that equations (VI) and (VII) hold for obhque coordina-tes. ril THE STRAIGHT LINE 101 48. The normal form of the equation of the straight line. In the preceding sections the lines considered were determined by two points or by a point and a direction. Both of these methods of determining a line are frequently used in Elementary Geometry, but we have now to consider a line as determined by two conditions which belong essentially to Analytic Geometry. Y AV A r\. > ' X <^ 7- B •^ Ox ^^ ^A A Vx I't ,A & ^-v / > M Let ABhe any line, and let ON be drawn from the origin perpen- dicular to J. ^ at C. Let the positive direction on ON he from toward N, — that is, from the origin toward the line, — and denote the positive directed length OC by ^ and the positive angle XON, measured, as in Trigonometry (p. 18), from OX as initial line to ON as terminal line, by (o* Then it is evident from the figures that the position of any line is determined hy a pair of values of p and oo, both p and w being positive and w < 2 tt. On the, other hand, every line determines a single positive value of p and a single positive value of a> which is less than i\^\ Y> ' \ ^ ^A ^ A \ X n \ -K-rf 2 7r, unless ^ = 0. When p = 0, however, AB passes through the origin, and the rule given above for the positive direction on ON becomes meaningless. From the figures we see that we can choose for , (3) proj. of DP on ON = DP cos ( — — w J = ?/ sin w. For the angle between the directed lines DP and ON equals that between OrandOiV=|-a;. Substituting from (2) and (3) in (1), we obtain £c cos 0) + y sin o) — ^ = 0. q.e.d. To reduce a given equation (4) Ax+By + C^O to the normal form, we must determine w and p so that the locus of (4) is identical with the locus of (5) X cos 0) + 2/ sin CO — ^ = 0. * The designation of this equation is made clear by the definition of the normal in Chapter IX. ;3 THE STRAIGHT LINE 103 Then we must have corresponding coefficients proportional (Theorem III, p. 88). cos 0) _ sin (n _— p •'• A ~ B ~ ~C" Denote the common value of these ratios by r ; then (6) cos o> = rA, (7) sin CD = rB, and (8) -p = rC. To find r, square (6) and (7) and add ; this gives sin2 io -f cos^ to = r2(^2 _^ B""). But sin^ w 4- cos^ w = 1 ; and hence r^(^^ + B"^) = 1, or (9) r = ■ ^ Equation (8) shows which sign of the radical to use ; for since p is positive, r and C must have opposite signs, unless C = 0. If C = 0, then, from (8), ^ = 0, and hence w < tt (p. 101) ; then sin a> is positive, and from (7) r and B must have the same signs. Substituting the value of r from (9) in (6), (7), and (8) gives A . B C cos (0 = ■ ? sm 0) = , ? p = ■ Hence (5) becomes (10) / • x + y + , . = 0, ^ ^ ±VlMr^ ±-y/A^-\-B^ iVI^:^ which is the normal form of (4). The result of the discussion may be stated in the following Rule to reduce Ax -\- By -\- C = to the normal form. First step. Find the numerical value of V^ ^ -f B"^. Second step. Give the result of the first step the sign opposite to that of C, or, if C = 0, the same sign as that of B. Third step. Divide the given equationlby the result of the second step. The result is the required equation. I 104 ANALYTIC GEOMETRY The advantages of the normal form of the equation of the straight line over the other forms are twofold. In the first place, every line may have its equation in the normal form; whether it is parallel to one of the axes or passes through the origin is immaterial. In the second place, as will be seen in the following section, it enables us to find immediately the distance from a line to a point. PROBLEMS \ 1. In what quadrant will ON (Fig., p. 101) lie if sin w and cos w are both positive? both negative? if sinw is positive and cosw negative? if sinw is negative and cos w positive ? ^ 2. Find the equations and plot the lines for which (a) w = 0, p = 5. Arts, x =± ^. Atis. ?/ + 3 = 0. Ans. V2x+ V2?/- 6 = 0. Ans. ic— V32/ + 4 = 0. 7 Tf i\^ (e) w = -— , j9 = 4. Ans. V2 x - V2y -8 = 0. 3. Reduce the following equations to the normal form and find p and w. (a) 3a; + 42/-2 = 0. Ans. p = |, w = cos-i f = sin- 1 f . (b) 3x - 4y -^ 2 = 0. Ans. p = f, a> = cos-i | = sin-i (- f). (c) 12 a; - 5 ?/ = 0, Ans. p = 0, w = cos-i (- ff) = sin-^i j-^. (d) 2x + 5?/ + 7 = 0. ^ Ans. p = — , a; = cos-i(' — ^"j ^z sin- Y ~ Y + V2y \_V29/ V-V29^ (e) 4aj - 3?/ + 1 = 0. Ans. p = 1, w = cos-i(- ij = sin-if. (f) 4x-5 2/ + 6 = 0. D(b) c-. 2 , p = 3. .(c) 0,. 7t p = 3. ;..,(d) 0,. _27t ' p = 2. Ans. p = \ _ V41 / \ a- VIT/ Vii V_V4i/ WV41 4. Find the perpendicular distance from the origin to each of the follow- ing lines. (a) 12x+5?/-26 = 0. Ans. 2. (b) x-\-y + l=0. Ans. 1V2. (c) 3x-2?/-l = 0. Ans. j\^Ts. ri' THE STRAIGHT LINE 105 5. Derive (VIII) when (a) - + 2/ sin w — p = 0, then the equation of the other leg is X cos (tt — co) + y sin (tt — w) — _p = 0, or — X cos (o + y sin (a — p = 0. Let (a, 0) be any point in the base. Then the distances from the legs to (a, 0) are respectively a cos w — _p and — a cos w — p, so that the sum of these distances is — 2 _p, that is, a constant. PROBLEMS 1 . Find the distance from the line (a) xcos45° + 2/sin45°- V2 = to (5, - (b) |x-fy-l = to (2,1). (c) 3x + 4y + 15 = to (-2, 3). 7). (d) 2x-7y + 8 = to (3, (e) x-Sy = to (0,4). 5). Ans. -2V2. Ans. Ans. Ans. 49 Ans. + V53 12 + V10 2. Do the origin and the point (3, - 2) lie on the same side of the line x-2/ + l = 0? Ans. Yes. ' 108 ANALYTIC GEOMETKY 3. Does the line 2x + 3?/ + 2 = pass between the origin and the point (-2, 3)? Ans. No. 4. Pind the lengtlis of the altitudes of the triangle formed by the lines 2x + 3y = 0, x + 3^ + 3 = 0, and x + y + 1 - Q. Ans. — ^, — =:, and V2. V13 Vio 5. Find the distance from the line Ax -{- By -{- G = to the point Pi(xi, yi). ■ ^^ Ax i + Byi+C 6. Prove Theorem IX when / ^ n tt,, ,7r^ . ^ S 7t , ,. S 7t - (a)p = 0, a;<--; (b)-{. + VlO Ly = mx-3. +VTT^ 12. Derive the normal equation of the line by means of Theorem IX. 13. Prove that the altitudes on the legs of an isosceles triangle are equal. 14. Prove that the three altitudes of an equilateral triangle are equal. 15. Prove that the sum of the distances from the sides of an equilateral triangle to any point is constant. Hint. Take the center of the triangle for origin, with the JC-axis parallel to one side. THE STRAIGHT LINE 109 fO 16. Find the areas of the triangles formed by the following lines. (a) 2 X - 3 ?/ + 30 = 0, X = 0, X + 2/ = 0. Ans. 30. (b) x + 2/ = 2, 3x + 4y- 12 = 0, x-?/+6 = 0. Ans. f (c) 3x - 4y + 12 = 0, X - 3?/ + 6 = 0, 2x - ?/ = 0. Am. 33. (d) x + 3y-3 = 0, 5x-2/-15 = 0, x-^+l = 0. Ans. 8.' 17. Plot the following lines and find the area of the quadrilaterals of which they are the sides. (a) X = ?/, y == 6, X + y = 0, 3x + 2 ?/ - 6 = 0. Ans. 16i. (b)x + 22/-5 = 0, 2/ = 0, x + 42/ + 5 = 0, 2x + ?/-4 = 0. Ans. 18. (c)2x-42/-f8 = 0, x + y = 0, 2x- 2/^4 = 0, 2x4-y-3 = 0. Ans. 4tVo. 50. The angle which a line makes with a second line. The angle between two directed lines has been defined (p. 28) as the angle between their positive directions. When a line is given by means of its equation, no positive direction along the line is fixed. In order to distinguish between the two pairs of equal angles which two intersecting lines make with each other we define the angle which a line makes with a second line to be the positive angle (p. 18) from the second line to the Jii^st line. Thus the angle which Li makes with L^ is the angle 0. We speak always of the '■' angle which one line makes with a second line," and the use of the phrase " the angle between two lines " should be avoided if those lines are not directed lines. We have thus added a third method of designating angles to those given on p. 18 and p. 28. ; Theorem X. The angle 6 which the line L, : A,x ^B,7j + C, = makes with the line is given hy (X) tan^ = AxA2 + B1JB2 110 ANALYTIC GEOMt:TRY Proof. Let a-^ and a^ be the inclinations of L^ and L^ respec- tively. Then, since the exterior angle of a triangle equals the sum of the two opposite interior angles, we have In Fig. 1, ai = $ -\- a^, or 6 = a^ — a^, In Fig. 2, a2 = 'Tr — 6 + ai, or = tt + (ai — a^). And since (5, p. 20) tan (tt -\- ) = tan <^, we have, in either case, tan = tan (a^ — a^) tan ai — tan org 1 + tan ai tan a^ (by 13, p. 20) But tan ai is the slope of Zj and tan ^2 is the slope of La ; hence (Corollary I, p. 86) tan Reducing, we get tan B, B, -( :-^)( -I-) A,B, - A,B, A^A^ + B^B^ Q.E.D. Corollary. If mi and m^ are the slopes of two lines, then the angle $ which the first line makes with the second is given by tan^ = 1 + minii THE STRAIGHT LINE 111 Ex. 1. Find the angles of the triangle formed by the lines whose equations are L:2x-Sy-6 = 0, M:6x-y-6 = 0, N:6x + iy -25 = 0. Solution. To see which angles formed by the given lines are the angles of the triangle, we plot the lines, obtaining the triangle ABC. A is the angle which M makes with X, so that M takes the place of Li in Theorem X and L of Lz- Hence Ai = 6, Bi=-1; ^2 = 2, B2=-3. Then tan 1 - ^'^^ ~ ^^^' - ArA2 + B1B2 -2 + 18 12 + 3 16 "15 and hence ^ = tan-i(H). B is the angle which L makes with N, and by Corollary III, p. 87, 5 = — • C is the angle which N makes with Jf, so that if tan C A2B1 - AiB2 we must set Hence and A1A2 + B1B2 Ai = 6, Bi = i; A2 = Q, B2=-l. 24 + 6 _ 30 _ 15 ~32~ tanC = -4 C = tan-i(if). 16 We may verify these results. For if J5 = -, then A = --C; and hence (6, p. 20, and 1, p. 19) tan ^ = cot C true for the values found. tan C , which is Ex. 2. Find the equation of the line through (3, 5) which makes an angle of — with the line X - y + 6 = 0. ^ • Solution. Let mi be the slope of the required line. Then its equation is (Theorem V, p. 95) (1) y-5 = mi(x-3). YJ, k 4 / 3 / ~~. ^\ ^^'1 5) \ ~~ \ \ \ 1 \ \ \ 112 ANALYTIC GEOMETRY The slope of the given line is m^ = 1, and since the angle which (1) makes with the given line is — , we have (by the Corollary), ^ \ 3 ' ^ ^ ' tan^ = :^^^ 1, d'^Xl'^- 3 1 + nil /" 1 + mi ^ whence mi = ^ = — (2 + V3). 1 - V3 Substituting in (1), we obtain y-6 = -{2+V3){x-S),_ or (2 + Vi) X 4- y - (11 + 3 V3) = 0. In Plane Geometry there would be two solutions of this problem, — the line just obtained and the dotted line of the figure. Why must the latter be excluded here? PROBLEMS 1. Find the angle which the line 3x—y-\-2 = makes with 2x + 7/ — 2=0; also the angle which the second line makes with the first, and show that these angles are supplementary. . 3 tt tt 4 ' 4 2, Find the angle which the line * (a) 2 X — 5 ?/ + 1 = makes with the line x — 2y + S = 0. • (b) X + y + 1 = makes with the line x — y + 1 — 0. (c) 3x — 4?/ + 2=0 makes with the line x -\- 3y — 7 = 0. (d) 6x — Sy + S = makes with the line x = 6, (e) x — 'ly + l = makes with the line x + 22/ — 4 = 0. In each case plot the lines and mark the angle found by a small arc. Ans. (a)tan-i(-J,); (b)|; (c) tan-i(V-) ; (d) tan-i(- i) ; (e) tan-^/,). ^Z. Find the angles of the triangle whose sides are x + 3y — 4 = 0, 3ic - 2?/ + 1 = 0, and a: - y + 3 = 0. Ans. tan-i(- V), tan-i(i), tan-i(2). Hint. Plot the triangle to see wliicli angles formed by the given lines are the angles of the triangle. -^4. Find the exterior angles of the triangle formed by the lines 5x — y + S=0, y = 2,x-4y + 3 = 0. Ans. tan-i(5), tan-i(- i), tan-i(- V)- 5. Find one exterior angle and the two opposite interior angles of the triangle formed by the lines 2x-3?/-6=0, 3x+42/-12 = 0, x-Sy+6=0. Verify the results by formula 12, p. 20. THE STRAIGHT LINE 113 6. Find the angles of the triangle formed by3x+2?/-4=0, x-3y+6=0, and 4x — 3y— 10=0. Verify the results by the formula tan J. + tan B + tan C = tan J. tan 5 tan C, if A -\- B -\- C = 180°. 7. Find the line passing through the given point and making the given angle with the given line. . (a) (2, 1), -, 2x-3?/ + 2 = 0. Ans. 5x - ij - 9 = 0. (b) (1, - 3), — , X + 22/ + 4 = 0. Ans. 3x + y = 0. (c) (2, - 5), -, X + 3 ?/ - 8 = 0. Ans. x-2ij -12 = 0. /■IK , V 7. A ?n + tan0 , (d) (xi, 2/i), (f>, y = mx + b. Ans. y - yi = - (x - Xi). 1 — m tan

, Ax -\- By -^ C = 0. Ans. y -y^ = -—^ (x - Xi). A. tan -^ B 8. Show from a figure that it is impossible to draw a line through the inter- section of two lines and "making equal angles with those lines" in the sense in which we have defined " the angle which one line makes with a second line." Prove the same thing by formula (X). How are the bisectors of the angles of two lines to be defined ? 9. Given two lines Xi:3x — 4y — 3 = and X2:4x — 3?/ + 12 = 0; find the equation of the line passing through their point of intersection such that the angle it makes with Li is equal to the angle L^ makes with it. Ans. 7x-7?/ + 9 = 0. 51. Systems of straight lines. An equation of the first degree in X and ij which contains a single arbitrary constant will repre- sent an infinite number of lines, for the locus of the equation will be a straight line for any value of the constant, and the locus will be different for different values of the constant. The lines represented by an equation of the first degree which contains an arbitrary constant are said to form a system. An equation which represents all of the lines satisfying a single con- dition must contain an arbitrary constant, for there is an infinite number of lines satisfying a single condition ; hence a single geo- metrical condition defines a system of lines. Thus the equation y=2x-\-b, where b is an arbitrary constant, represents the system of lines having the slope 2; and the equation y — 5 — m (x — 3), where m is an arbitrary constant, represents the system of lines passing through (3, 5). 114 ANALYTIC GEOMETRY Second rule to find the equation of a straight line satisfying two conditions. First step. Write the equation of the system of lines satisfying one condition. Second step. Determine the arbitrary constant in the equation found in the first step so that the other condition is satisfied. Third step. Substitute the result of the second step in the result of the first step. This gives the required equation. This rule is, in general, easier of application than the rule on p. 93. It has already been applied in solving Ex. 2, p. Ill, and will find constant application in the following sections. The number of lines satisfying the conditions imposed will be the number of real values of the arbitrary constant obtained in the second step. Ex. 1. Find the equations of the straight lines having the slope f and intersecting the circle «2 + ^/^ = 4 in but one point. Solution. First step. The equation represents the system of lines whose slopes are | (Theorem I, p. 58). Second step. The coordinates of the inter- section of the line and circle are found by solv- ing their equations simultaneously (Rule, p. 76). \yi ^111 Substituting the value of y in the line in the V^ \. equation of the circle, we have x2-f (|« + &)2 = 4, or 25x2 4. 24 6x + (16 &2 _ 64) = 0. ^: / YA The roots of this equation, by hypothesis, must be equal; hence the discriminant must vanish (Theorem II, p. 3) ; that is, 576 &2_ 100 (16 62- 64) = 0, whence Third step. Substitute these values of b in the equation of the first step. We thus obtain the two solutions and 2/ = f X - f . THE STRAIGHT LINE 115 PROBLEMS • 1. Write the equations of the systems of lines defined by the following conditions. (a) Passing through (—2, 3). (b) Having the slope — f . (c) Distance from the origin is 3. (d) Having the intercept on the F-axis = — 3. (e) Passing through (6, — 1). (f ) Having the intercept on the JT-axis = 6. (g) Having the slope i. (h) Having the intercept on the F-axis = 5. (i) Distance from the origin = 4. 2. What geometric conditions define the systems of lines represented by the following equations ? (a) 2»-3y + 4A: = 0. (b) kx-Sy -7 = 0. (c) X + y - k = 0. (d) x + k = 0. (e) x + 2ky -S = 0. (f ) 2kx-3y + 2 = 0. (g) X cos a + 2/ sin a + 5 = 0. Hint. Reduce the given equation to one of the well-known forms of the equation of the first degree. 3. Determine k so that (a) the line 2x — Sy + k = passes through (—2, 1). Ans. k = 7. (b) the line 2kx — 5y + S = has the slope 3. Ans. k = J/. (c) the line x-\- y — k = passes through (3, 4). Ans. k = 7. (d) the line Sx — 4y -\- k = has intercept on X-axis = 2. Ans. k =— 6. (e) the line x — Sky + 4: = Q has intercept on F-axis = — 3. Ans. k = — ^. (f) the line 4x — 3y-f6A: = 0is distant three units from the origin. Ans. k = ± ^. 4. Find the equations of the straight lines with the slope — j\ which cut the circle x^ -^ y"^ = 1 in but one point. Ans. 5 x -f- 12 y = ± 13. 5. Find the equations of the lines passing through the point (1, 2) which cut the circle x^ -{-y^ = 4:in but one point. Ans. y = 2 and 4 x -|- 3 ?/ = 10. 6. Find the equation of the straight line passing through (—2, 5) which makes an angle of 45° with the F-axis. Ans. x + y — 3 = 0. 116 ANALYTIC GEOMETRY 7. Find the equation of the straight line which passes through the point (2, — 1) and which is at a distance of two units from the origin. Ans. X = 2 and Sx — 4y = 10. 8. Find the equation of the straight line whose slope is f such that the distance from the line to the point (2, 4) is 2. Ans. 3 x — 4 y = 0. 52. The system of lines parallel to a given line. Theorem XI. The system of lines parallel to a given line Ax + By + C = is represented by (XI) Aic-\- By -\-k = Oj where k is an arbitrary constant. Proof. All of the lines of the system represented by (XI) are parallel to the given line (Corollary II, p. 87). It remains to be shown that all lines parallel to the given line are represented by (XI). Any line parallel to the given line is determined by some point Pi (xi, ?/i) through which it passes. If Pi lies on (XI), then AX]_ + By^^ -f A: = 0; and hence k — — Ax^ — By^. That is, the value of k may be chosen so that the locus of (XI) passes through any point Pi. Then (XI) represents all lines parallel to the given line. q.e.d. It should be noticed that the coefficients of x and y in (XI) are the same as those of the given equation. Ex. 1. Find the equation of the line through the point Pi (3, — 2) paral- lel to the line Li:2x -^y - A = 0. Solution. Apply the Rule, p. 114. First step. The system of lines parallel to the given line is 2x-Sy -\-k = 0. Second step. The required line passes through Pi; hence • 2-3-3(-2) + fc = 0, and therefore k =—12. Third step. Substituting this value of k, the required equation is 2x-3y -12 = THE STRAIGHT LINE 117 53. The system of lines perpendicular to a given line. Theorem XII. The system of lines perpendicular to the given is represented by (XII) BQe-Ay + k = 0, where h is an arbitrary constant. Proof. All of the lines of the system represented by (XII) are perpendicular to the given line, for (Corollary III, p. 87) AB — BA = 0. It remains to be shown that all lines perpen- dicular to the giv.en line are represented by (XII). Any line perpendicular to the given line is determined by some point ^i(^ij l/i) through which it passes. If Pi lies on (XII), then whence Bxi — Ayi -{- k = 0, k = Ayi — Bxi. That is, the value of k may be chosen so that the locus of (XII) passes through any point Pi. Then (XII) represents all lines perpendicular to the given line. q.e.d. Notice that the coefficients of x and y in (XII) are respectively the coefficients of y and x in the given equation with the sign of one of them changed. Ex. 1. Find the equation of the line through the point Pi (— 1, 3) perpen- dicular to the line Li: 6x — 2y ■}■ S = 0. Solution. Apply the Rule, p. 114. First step. The equation of the system of lines perpendicular to the given line is 2x-\- 5y + k = 0. Second step. The required line passes - —^ — I — I — r**kJ I — [— through Pi ; hence "/^l M I I Ml^ 2(-l) + 6-3 + A: = 0, or A; = - 13. Third step. Substitute this value of k. The required equation is then 2x + 62/-13 = 0. 118 ANALYTIC GEOMETRY PROBLEMS 1. Find the equation of the straight line which passes through the point (a) (0, 0) and is parallel to (c — 3 y + 4 = 0. Ans. x — Sy = 0. (b) (3, — 2) and is parallel to x -\- y -\- 2 = 0. Ans. x + y — 1 = 0. (c) (— 5, 6) and is parallel to 2x + 'ky - S = 0. Ans. x + 2y -7 = 0. (d) (—1, 2) and is perpendicular toSx — iy + l = 0. Ans. Ax + Sy -2 = 0. (e) (— 7, 2) and is perpendicular tox-3?/ + 4 = 0. Ans. Sx + y + 19 = 0. 2. Find the equations of the lines drawn through the vertices of the triangle whose vertices are (- 3, 2), (3, - 2), and (0, — 1), which are parallel to the opposite sides. Ans. The sides of the triangle are 2x + 3y = 0,x-{-Sy + S = 0,x-j-y-}-l = 0. The required equations are 2x + Sy + S = 0,x + Sy-S = 0,x + y-l = 0. 3. Find the equations of the lines drawn through the vertices of the triangle in problem 2 which are perpendicular to the opposite sides, and show that they meet in a point. Ans. Sx-2y -2 = 0, 3x-y + 11 = 0, x-y- 6 = 0. 4. Find the equations of the perpendicular bisectors of the sides of the triangle in problem 2, and show that they meet in a point. Ans. Sx-2y = 0,3x-y-6 = 0,x-y + 2 = 0. 6 . The equations of two sides of a parallelogram are 3a; — 4y+6 = and X + 5 2/ — 10 = 0. Find the equations of the other two sides if one vertex is the point (4, 9). Ans. Sx — 4y + 24: = and x + 5 ?/ — 49 = 0. 6. The vertices of a triangle are (2, 1), (— 2, 3), and (4, — 1). Find the equations of (a) the sides of the triangle, (b) the perpendicular bisectors of the sides, and (c) the lines drawn through the vertices perpendicular to the opposite sides. Check the results by showing that the lines in (b) and (c) meet in a point. 7. Show that the perpendicular bisectors of the sides of any triangle meet in a point. 8. Show that the lines drawn through the vertices of a triangle perpen- dicular to the opposite sides meet in a point. 9. Find the value of C in terms of A and B if Ax + By + C = passes through a given point Pi (xi, yi) ; show that the equation of the system of lines through Pi may be written A{x — Xi) + B{y — yi) = 0. THE STRAIGHT LINE 119 54. The system of lines passing through the intersection of two given lines. Theorem XIII. The system of lines passing through the intersec- tion of two given lines and ig : ^2^ + ^2^/ + C'g = is represented by the equation (XIII) Aix + Biy + Ci + ^ (^2£c + Bty + C2) = O, where k is an arbitrary constant. Proof All of the lines represented by (XIII) pass through the intersection of L^ and L^. For let P^ (x^, yi) be the intersection of Li and Xg. Then (Corollary, p. 53) A^x,-\-B,y^^C^ = and A 2X1 + B^yi + C2 = 0. Multiply the second equation by k and add to the first. This gives ^1^1 + B,y, + Ci + A^ (^23^1 + B0, + C2) = 0. But this is the condition that Pj lies on (XIII). That all lines through the intersection of Xi and L^ are repre- sented by (XIII) follows as in the proofs of Theorems XI and XII. Q.E.D. Corollary. If L^ and L^ are parallel, then (XIII) represents the system of lines parallel to Li and L^. For if Li and L2 are parallel, then Ax_Bx A2 B2 and hence Ai _Bi kA^ kBi By composition, Ai + kA2_Bi + kB2 Ai Bx Hence L and (XIII) are parallel (Corollary II, p. 87) Notice that (XIII) is formed by multiplying the equation of L^ by k and adding it to the equation of L^. 120 ANALYTIC GEOMETRY Ex. 1. Find the equation of the line passing through Pi (2, 1) and the intersection of ii : 3 ic — 5 ?/ — 10 = and L2 : x ■{- y + 1 = 0. Solution. Apply the Rule, p. 114. The system of lines passing through the intersection of the given lines is represented by Sx - 5 y - 10 -^ k {x + y -\- 1) = 0. If Pi lies on this line, then 6_5-10 + fc(2 + l + l) = 0; whence A; = |. Substituting this value of k and simplifying, we have the required equation 21x-ll?/-31 = 0. Ex. 2. Find the equation of the line passing through the intersec- tion of Li:2x + y + 1 =0 and L-zix - 2y + 1 = and parallel to Ls-Ax-Sy-1 = 0. Solution. Apply the Rule, p. 114. The equation of every line through the intersection of the first two given lines has the form 2x + ?/ + l + A;(x-2?/ + l) = 0, or (2 + A:)x + (1 - 2 /c) ?/ + (1 + A:) = 0. If this line is parallel to the third line (Corollary II, P- 87), 2+fc _ l-2fc 4 ~ -3 ' whence k = 2. Substituting and simplifying, we obtain 4x-3y + 3 = 0. The geometrical significance of the value of k in Theorem XIII is given most simply when L^ and L2 are in normal form. Theorem XIV. The ratio of the distances from Li : X cos 0)1 + 2/ sin toi — ^1 = and L2 : x cos (1)2 -{- y sin (02 — 2^2 = ^ to any 'point of the line L : X cos (01 + 2/ sin wi — pi -\- k (x cos wo + ?/ sin wg — ^2) = is constant and equal to — k. THE STRAIGHT LINE 121 Proof. Let Pi {x^ ?/i) be any point on L. Then Xi cos 0)1 + yx sin ini— pi-\- k (x^ cos 2 + yi sin wg — p^ = 0, and hence - h = "'i cos.». + y. sin.», -y, Xi cos 0)2 + 2/i sm 0)2 — ^2 The numerator of this fraction is the distance from Zi to Pi, and the denominator is the distance from L^ to Pi (Theorem IX, p. 106). Hence — A; is the ratio of the distances from L^ and L^ to any point on L. q.e.d. Corollary. If k = ± 1, then L is the bisector of one of the angles formed by L^ and L^. That is, the equations of the bisectors of the angles between tivo lines are found by reducing their equations to the normal form and adding and subtracting them. For when Jfc = ± 1 the numerical values of the distances from L\ and L2 to any point of L are equal. The angle formed by L^ and L^ in which the origin lies, or its vertical angle, is called an internal angle of L^ and igj ^^^ either of the other angles formed by Xi and L^ is called an external angle of those lines. From the rule giving the sign of the distance from a line to a point (p. 105) it follows that L lies in the internal angles of L^ and L^ when k is nega- tive, and in the external angles when k is posi- tive. If the origin lies on L^ or L^, the lines must in each case be plotted and the angles in which k is posi- tive found from the figure. PROBLEMS 1. Find the equation of the line passing through the intersection of 2x-3?/ + 2 = and 3x-4?/— 2 = 0, ^ithout finding the point of intersec- tion,) which (a) passes through the origin. (b) is parallel to5x — 2y + 3 = 0. (c) is perpendicular to3x — 2?/ + 4 = 0. Ans. {B.)bz-ly = 0; (b) 5x - 2^/ -50 = ; (c) 2x + By - 58 = 0. 122 ANALYTIC GEOMETRY 2. Find the equations of the lines which pass through the vertices of the triangle formed by the lines 2x — Sy+l = 0, x — y = 0, and 3x + 4?/ — 2 = which are (a) parallel to the opposite sides. (b) perpendicular to the opposite sides. Ans. (a) 3x + 4 2/- 7 = 0, 14x-21?/+ 2 = 0, 17x - ITy + 5 = 0; (b) 4:X-Sy-1=0, 21x + Uy -10 = 0, 17x + 17?/ - 9 = 0. 3. Find the bisectors of the angles formed by the lines Ax — 3y — 1 = and 3x — 4^ + 2 = 0, and show that they are perpendicular. Ans. 7 X -7y -\- 1 = and x + ?/ — 3 = 0. 4. Find the equations of the bisectors of the angles formed by the lines 6x-12 2/ + 10 = and 12x - 5?/ + 15 = 0. Verify the results by Theorem X. 5. Find the locus of a point the ratio of whose distances from the lines 4x-3i/ + 4 = 0and5x+12y-8=:0isl3to5. Ans. 9x + 9y-i = 0. 6. Find the bisectors of the interior angles of the triangle formed by the lines 4x-3y = 12, 5x-12i/-4 = 0, and 12x-5y-13 = 0. Show that they meet in a point. Ans. 7x-9y-16 = 0, 7x + 7y-9 = 0, 112x-642/-221=0. 7. Find the bisectors of the interior angles of the triangle formed by the lines 5x-12y = 0, 5 x + 12 y + 60 = 0, and 12 x - 5 ^ - 60 = 0, and show that they meet in a point. Ans. 2y + 6 = 0, 17 x + 7 ?/ = 0, 17 x - 17 y - 60 = 0. 8. The sides of a triangle are 3x + 4?/ — 12 = 0, Sx — iy = 0, and 4x + 3?/ + 24 = 0. Show that the bisector of the interior angle at the vertex formed by the first two lines and the bisectors of the exterior angles at the other vertices meet in a point. 9. Find the equation of the line passing through the intersection of x + y — 2 = and x — y + Q = and through the intersection of2x — y + 3 = andx -3?/ + 2 = 0. Ans. 19x -{- Sy + 26 = 0. Hint. The systems of lines passing through the points of intersection of the two pairs of lines are x + y — 2+k{x-y + G) = and 2x — y + 3 + k'(x — 3y + 2)=0. These lines will coincide if (Theorem III, p. 88) l + k _ 1-k _ -2 + 6Jc 2+k'~-l-3k' 3+2k' ' Letting p be the common value of these ratios, we obtain l + k=2p + pk% l-k = -p-3pk', and -2 + 6fc=3p + 2p/y. From these equations we can eliminate the terms in pk' and p, and thus find the value of k which gives that line of the first system which also belongs to the second system. THE STRAIGHT LINE f23 10. Find the equation of the line passing through the intersection of 2X + 52/ — 3 = and Sx - 2y — 1 = and through the intersection of x-y = and x + 3y-6 = 0. Ans. 43 x - 35 y - 12 = 0. A figure composed of four lines intersecting in six points is called a complete quadrilateral. The six vertices determine three diagonals of which two are the diagonals of the ordinary quadri- lateral formed by the four lines. 1 1 . Find the equations of the three diagonals of the complete quadrilateral formed by the lines x -{■ 2y = 0, 3x-4?/ + 2 = 0, x - y -\- S = 0, and 3x-2?/ + 4 = 0. Ans. 2x-?/ + l = 0, a; + 2 = 0, 5x-6y + 8 = 0. 12. Show that the bisectors of the angles of any two lines are perpen- dicular. 13. Find a geometricarinterpretation of k in (XI) and (XII). 14. Find the geometrical interpretation of ^ in (XIII) when Li and L^ are not in normal form. 15. Show that the bisectors of the interior angles of any triangle meet in a point. 16. Show that the bisectors of two exterior angles of a triangle and of the third interior angle meet in a point. 55. The parametric equations of the straight line. The angles a and ^ between a line directed upward "* and the coordi- nate axes (p. 28) are called the direction angles of the line. Their cosines, cos a and cos fi, are called the direction cosines of the line and satisfy the relation (1) cos^a-f cos2^ = 1. For (Theorem I, p. 28) cos ^ = sin or and sin^ a + cos^ a = 1. Given a line with direction angles a and /3 passing through Pi(xi, 7/i). Let P(x, y) be any point on this line and denote the variable directed length P^P by p. The projections of PyP on the axes are respectively (Theorem III, p. 31) ic — iCi and y — ?/i, or (Theorem II, p. 30) p cos a and p cos ^. * If the line is horizontal we suppose that it is directed to. the right, so a= and i3 = -• 124 ANALYTIC GEOMETRY Hence x — Xi = p cos a and y — y^ = p cos ^; whence x = x-^-{- p cos a. y = yi-\-p (^os /3. Hence we have Theorem XV. Parametric form. The coordinates of any pomt P (x, y) on the line through a given point Pi (xi, y^ whose direction angles are a and /3 are given by ^ ^ \y = yi + pcosfi, where p denotes the variable directed length PiP. Equations (XV) are called the parametric equations of the straight line because they express the variable coordinates of any point (x, y) on the line in terms of a single variable parameter p. As p varies from — oo to + oo the point P {x, y) describes the line in the positive direction. These equations are important in deal- ing with problems which involve the distances from a point 7\ on a line to the intersections of that line with a given curve. Theorem XVI. Symmetric form. The equation of a straight line in terms of the coordinates of a point Pj (xi, y^ on the line and its direction cosines is ^ ^ cos a "" cos p Hint. Solve (XV) for p and equate the two values obtained. Theorem XVII. The direction cosines of the line Ax -^ By -[-C = -^ o are cos o = . cos p when the sign of the radical is the same as that of A . Proof. Let Pi {x^. y^ be a point on the given line. Then (Corollary, p. 53) ^^^ ^ ^^^ + (7 = 0. THE STRAIGHT LINE 125 SubtractiDg from the given equation, we obtain A(x-x{)-hB(t/-y,)=0. Transpose the second term and divide by —AB] this gives a; — a^i ^ y — yi -B A Dividing this equation by (XVI), we have cos a _ cos /3 -B ~ A ' Let r denote the common value of these ratios. Then cos a = — Br and cos (3 = Ar. Squaring and adding, cos^ a + cos2 13 = (.1^ 4- B^) r\ Then from (1), p. 123, r = y and hence — B A (2) cos a — --==^= and cos (3 The sign of the radical must be the same as that of yl. q.e.d. [For since the line is directed upward, i3< — , and hence cos/3 is positive.] Corollary. If cos cc and cos f3 are proportional to two numbers a and h, then a o ^ cos a = ? cos p = ± Va^ + ft2 + Va^ + ft2 The sign of the radical must be the same as that of b. To reduce the equation of a given straight line to the symmet- rical or parametric form it is necessary to know the coordinates of some point on the line (which may be found by the Rule, p. 60) and its direction cosines (which are given by Theorem XVII). Then we can write the required equations by Theorem XV or XVI. 126 ANALYTIC GEOMETRY P X ' y 2. 1 5-1 6 Ex. 1. Plot the line whose parametric equations are x = 2 — | /o and Solution. Comparing with (XV) we see that Pi (2, 1) is a point on the line. A second point will enable us to plot the line. We have at once the table Hence the line joining the points Pi (2, 1) and P2(— 1, 5) is the required line. (x,2/), or(2-fp, l + 4p), are the coordinates of that variable point P on the line whose distance from Pi is the variable p. Ex. 2. Given the circle C : x^ -{■ y^ = 2,6 and the line whose parametric equations are x = 5 — f p and y = — 3 + | /? ; find the product of the dis- tances from Pi (5, — 3) to the points of intersection of the .line with C, and the middle point of the chord formed by the line. Solution. By Theorem XV the coordinates of any point on the line are (5-fp, -3 + fp), where p denotes the distance from Pi to that point. If that point lies on C (Corollary, p. 53), (5-3^)2 + (_3 + |p)2 = 25, or, simplifying, (3) /) + 9 = 0. The roots of this quadratic are the directed lengths pi = P1P2 and pi = P1P3, where P^ and P3 are the points of intersection of the line and circle. For if P2 (5 - | pi, - 3 + f pi) is on the circle, (5-|Pi)^+(-3 + |pi)2 = 25, or Pi' Upi + 9 = 0. Hence pi, and similarly p2, is a root of (3). The product of these distances is therefore 9 (Theorem I, p. 3). Half the sum of these roots is PiP, or 2/- (Theorem I, p. 3). For p = ^^ we have x = f | and y = ||, so the middle point of the chord is the point THE STRAIGHT LINE 127 PROBLEMS 1. Plot the following lines : ('^ {:=2;S:. . <^» 1 V5 2 V5 2. Prove that if cos a and cos /3 are the direction cosines of a line directed upward, then — cos a and — cos /3 are the direction cosines of the same line directed downward. 3. Find the coordinates of the points on the line H ~ -. ^ , for which p = S, — 2, and 4. Verify the geometric significance of p for each of these points by Theorem IV, p. 31. 4. Find the product of the distances from Pi (2, 1) to the intersections of the line x = 2 — ^p and y = 1 + | p with the circle x^ -\- y^ = 25, and explain the sign of the result. Ans. — 20. 6. Given the ellipse x^ + 4:y^ = 16 and the line x = Xi - ^ p and y = Vi + ^ P ; find the equation whose roots are the distances from Pi{^i, Vi) to the points of intersection of the line and ellipse. Ans. -p^ - ^"^^ ~ ^^^' p + xi2 + 4yi2 - 16 :^ 0. 25 5 6. Find the condition that Pi in problem 5 should be the middle point of the chord on which it lies. Hint. The two values of p must be numerically equal, with opposite signs. 7. Given the parabola y^ = 4x and the line a; = 2 + p cosar, y = -4: -I- p cos /3 ; find the condition which cos a and cos j8 must satisfy if the line meets the parabola in but one point. Ans. cos2 a + 4 cos a cos /3 + 2 cos^/S = 0. 8. If a and 6 are two numbers such that a^ + 62 = 1, prove that a and b are the direction cosines of some line. 9. Derive equation (XVI) from Theorem V (p. 95) and Theorem I (p. 28). 10. Prove that the common value of the ratios in (XVI) is the length PiP. Hint. Square (XVI), apply the Theorem on the sum of the antecedents and of the consequents, and then take the square root. 11. Derive equations (XV) from (XVI) by means of problem 10. ^6 128 ANALYTIC GEOMETRY MISCELLANEOUS PROBLEMS 1. Find the point on the line 3 x — 5 ?/ + 6 = which is equidistant from the points (3, — 4) and (2, 1). 2. Find the equation of the line through the intersection of the lines 7x + 2/-3 = and 3x + 6y — 11 = which is perpendicular to the line joinhig their intersection to the origin. 3. Find the equation of the line through the point (2, 5) such that the portion of the line included between the axes is bisected at that point. 4. Find the equation of the line through the point (2, — 3) such that the portion of the line included between the lines 3x + 2/ — 2 = and x + 5?/ + 10 = is bisected at that point. 6. Prove that the diagonals of a rhombus are perpendicular. 6. If the F-axis makes an angle of w with the X-axis, find the equation of the straight line in terms of its intercept h on the Y-axis and its inclination a. 7. If the Y-axis makes an angle of w with the JT-axis, find the equation of the straight line whose inclination is a which passes through Pi(xi, 2/i)- 8. If the Y-axis makes an angle of w' with the X-axis, find the normal form of the equation of the straight line. 9. Find the tangent of the angle which one line makes with another if the axes are oblique. 10. Show that all of the lines for which m = b pass through the same point, and find the coordinates of that point. 1 1 . Show that all of the lines for which - + - = constant pass through the a b same point, and find the coordinates of that point. 12. Prove that all of the lines Ax -\- By + C = for which A -\- B + C = pass through the same point, and find the coordinates of that point. 13. Find the points in which the lines 2x — Sy = 0, x-t-4y — 2 = 0, 2x-3y + X(x+4?/-2) = 0, 2x-3?y-X(x- 4 ?/ - 2) = cut the X-axis. Show that the last two points divide the line joinhig the first two points inter- nally and externally in the same numerical ratio. 14. Prove that Ax -\- By + C = represents a straight line by showing that if Pi and P2 lie on the locus of the equation, the point which divides PiP2 in the ratio X lies on the locus of the equation. 15. Find the bisectors of the exterior angles of the triangle formed by 2 X - 3 y -f- 120 = 0, X + ?/ = 0, and 3x-f-42/-6 = 0. Show that these lines meet the opposite sides in three points on the same straight line. THE STRAIGHT LINE 129 16. Find the equation of tlie line passing through the intersection of Az + By -^ C = and A'x + B'y + C = which (a) passes through the origin, (b) is parallel to the X-axis, (c) is parallel to the F-axis. 17. Show that the lines {A + X^^^ + (^ + '^B')y + {C + XC) = pass through a point if X is a variable parameter and the other letters are constant. 18. Let Aix + Biy + Ci = 0, A^x + B^y + Cs = 0, and A^x + ^sy + C3 = be three given lines forming a triangle. Show that the equation of any line Ax -\- By -\- C = may be written in the form a (Aix + Biy + d) + )8 {A^x + ^22/ + C2) + 7 (^3^ + Bsy + C3) = 0, where a, /3, and 7 are definite constants. Hint. Use Theorem III, p. 88. 19. Find the ratio in which the line 2x — 6y + 8 = divides the line join- ing the points Pi(l, 3) and P2(7, 2). Hint. The coordinates of the point dividing PjPa 'i^to segments whose ratio is A are — , ) ; determine A so that this point lies on the given line. 1 + A. l + A/' 20. Find the ratio in which the line x + Sy — 6 = divides the line joining (- 3, 2) and (6, 1). 21. Determine m so that the line y = mx — 7 divides the line joining (3, 2) and (1, 4) in the ratio 3:2. 22. Find the equation of the line passing through the point (2, — 3) which divides the line joining (6, 3) and (2, — 1) in the ratio 2 : 5. 23. Show that the ratio of the distances from the line Ax -}- By -^ C = 0to the points Pi{Xi, yi) and P2(X2, 2/2) is -^ -^ -• Ax2 + By2 + C 24. Show that the line Ax + By + C = divides the line joining Pi(xi, yi) and P2 (X2, 2/2) into segments whose ratio is — - — ~ . 25. Show by the preceding example that any line cuts the sides of a tri- angle P1P2, P2-P3, and P3P1 in the points L, M, iV^such that PiL P2M PsN ^ ^ LP2 ^ MPs ^ NPi ~ 26. Plot the line 2x — Sy + 6 = and indicate all of the points for which 2x-3y + 5>0. 27. Find the area of the triangle formed by AiX + Biy + Ci = 0, A2X + B2y 4 C2 = 0, and A^x + B^y -f O3 = 0. CHAPTER V THE CIRCLE AND THE EQUATION x^ + y^ + Dx + Ey + F = 56. The general equation of the circle. If (a, ^) is the center of a circle whose radius is r, then the equation of the circle is (Theorem II, p. 58) (1) x^-{-y^-2ax-2py-\-a^-\- ft^-r^ = Oj or (2) (oc-ay + (y-py = r\ ' In particular, if the center is the origin, a = 0, ^ = 0, and (2) reduces to (3) ijc^ + y^ = r\ Equation (1) is of the form (4) x''-{-2f-^Dx-{-Ey-\-F= 0, where (5) D = -2a, E=- 2 13, Siud F=:a^ + /3^- 7^. Can we infer, conversely, that the locus of every equation of the form (4) is a circle ? By comparing (4) with (1) we obtain (5). Whence (6) a=--, ^ = --, and r^ = These values of a and f3 are real, and if D^ + E^ — 4 F is posi- tive, the value of r is real and the locus of (4) is a circle. To plot the locus of (4) by points (Rule, p. 60), we solve for y. This gives E I /E^ —AF (7) y = -'^±^~x^-Dx + {-^— The discriminant of the quadratic under the radical in (7) is © = D2 - 4(- 1) f 1 = 2)2 + ^' - 4F, which is the numerator of r^ in (6). 130 THE CIRCLE 131 If © is positive, the quadratic under the radical is positive for values of x between the roots (Theorem III, p. 11) and the equa- tion has a locus, as we have seen. If is zero, the roots of the quadratic are real and equal (Theo- rem II, p. 3). But for all other values of x the quadratic is negative (Theorem III, p. 11). The locus therefore consists of the single point D e\ For the quadratic in (7) equals zero when k = — — (p. 2), and hence, from (7), E the corresponding value of t/ is —-^- This also follows from (6) if we suppose r approaches zero, for then the circle consists only of its center ( — ^ » ~ "^ ) ' If is negative, the quadratic in (7) is negative for all values of X (Theorem III, p. 11) except the roots, which are imaginary (Theorem II, p. 3). Hence there is no locus. The expression ® = D^ -\- E^ — 4:F is called the discriminant of (4). When = the locus of (4) is often called a point-circle or a circle whose radius is zero. We have thus proved Theorem I. The locus of the equation (I) oe^ + y^ + Utoo + Ey + F = 0, whose discriminant is ® = D^ -{- E^ — 4: F, is determined as follows: (^a) When is positive the locus is the circle whose center is ( — — > — 77 ) and whose radius is r = -i- Vz)^ -^ ^^ — 4i^= ^ V©. ^ / ( D e\ (b) When © is zero the locus is the point-circle I — "^' ~~ "o 7' (c) When is negative there is no locus. ^ ' Corollary. When E = the center of (I) is on the X-axis, and when D = the center is on the Y-axis. Whenever in what follows it is said that (I) is the equation of a circle it is assumed that is positive. 132 ANALYTIC GEOMETRY Ex. 1. Find the locus of the equation x^ + y^ — 4:X + Sy — 5 = 0. Solution. The given equation is of the form (I), where D = - 4, ^ = 8, F = -6, and hence e = 16 + 64 + 20 = 100>0. The locus is therefore a circle whose center is the point (2, — 4) and whose radius is -J- VlOO = 5, The equation Ax^ -\- Bxy + Cy"^ -{-DX + Ey + F =0 is called the general equation of the second degree in X and y because it contains all possible terms in x and y of the second and lower degrees. Theorem II. The locus of the general equation of the second YA t^ ^ ^ N X ' / \ i-' / \ \ (-2, i) \ / \ / \ / ^ ^ ^ y r* degree^ Ax^ + Bxy ^Cif^Bx^Ey^F=^, 1)2 + ^2 is a circle when and only when A is positive. C,B = 0, and 4.AF Proof The equation of every circle must have the form (I) ; hence the coefficients of x^ and y^ must be equal and the xy term must be lacking ; that is, the locus of (II) can be a circle only when A =C and B = 0. If these conditions be satisfied, (II) may be written in the form x"" + y^ -}- jx -h jy -\- -j = 0, whose locus is a circle when and only when its discriminant D'^-^ E^-4.AF IS positive. Q.E.D. 57. Circles determined by three conditions. The equation of any circle may be written in either one of the forms (x-ay + (y-py = r' or x'^-\-y^ + Dx + Ey-{- F=0. Cf*, THE CIRCLE 133 Each of these equations contains three arbitrary constants. To determine these constants three equations are necessary, and as any equation between the constants means that the circle sat- isfies some geometrical condition, it follows that a circle may be determined to satisfy three conditions. Rule to determine the equation of a circle satisfying three conditions. First step. Let the required equation he (1) (^_a)2-f(y-/3)2 = r2 or (2) x^-^y^ + Dx-\-Ey^F=0, as 77? ay be more convenient. Second step. Find three equations between the constants a, fi, and r [or D, E, and F] which expy^ess that the circle (1) \_or (2)] satisfies the three given co7iditions. Third step. Solve the equations found in the second step for a, /?, and r [or D, E, and F~\. Fourth step. Substitute the results of the third step in (1) [or (2)]. The result is the required equation. Ex. 1. Find the equation of the circle passing through the three points Pi(0, 1), P2(0, a), and P3(3, 0). Solution. First step. Let the required equa- tion be (3) x'^ + y"^ + Dx + Ey + F = 0. Second step. Since Pi, P2, and Pg lie on (3), their coordinates must satisfy (3). Hence we have 1 + ^ + P = 0, 36 + 6 ^ -I- P = 0, (4) (5) and (6) 9 + 3D + P = 0. Third step. Solving (4) , (5) , and (6) , we obtain E = -7, F=6, D = -B. Fourth step. Substituting in (3), the required equation is x2 + y' 7y + G = 0. 134 ANALYTIC GEOMETRY since it By Theorem I we find that the radius is | V2 * and the center is the point (I, I). Ex. 2. Find the equation of the circle passing through the points Pi (0, — 3) and P^ (4, 0) which has its center on the line ic + 2 ?/ = 0. Solution. First step. Let the required equation be (7) x2 + ?/2 + Dx + Ey + F = 0. Second step. Since Pi and P2 lie on the locus of (7), we have (8) 9-3^ + P = and (9) 16 + 4 D + P = 0. The center of (7) is ( , — 7- ) > and lies on the given line, -f-(-f)-. or (10) D-\-2E = 0. Third step. Solving (8), (9), and (10), we obtain D = - -V-, E = h and P = - -2/-. Fourth step. Substituting in (7), Xve obtain the required equation, x2 + 2/2 - -V- « + 1 y - ¥ = 0. or 5 x2 + 5 2/2 _ 14 x + 7 ?/ - 24 = 0. The center is the point (|, — ^^), and the radius is ^ V29. > PROBLEMS 1. Find the equation of the circle whose center is (a) (0, 1) and whose radius is 3. (b) (—2, 0) and whose radius is 2. (c) (—3, 4) and whose radius is 5. (e) (or, 0) and whose radius is a. (f ) (0, p) and whose radius is /3. is) (Oj — jS) and whose radius is /3. Arts. x2 + ?/2 — 2 y - 8 = 0. Ans. «2 + 2/2 ^ 4 a; _ 0. Ans. x^ + y'2 -{-6x - Sy = 0. Ans. a;2 + 2/2 — 2 ax =: 0. Ans. x^ + y^ - 2 Py = 0. Ans. a;2 + 2/2 + 2 ]82/ = 0. * The radius is easily obtained, since Vi is the length of the diagonal of a square whose side is one unit. We may construct a line whose length is y/n by describing a semicircle on a line Avhose length is n + 1 and erecting a perpendicular to the diameter one unit from the end. The length of that perpendicular will be -y/n. THE CIRCLE ^ 135 2. Find the locus of the following equations. L (a) a;2 + 2/2 _ 6x - 16 = 0. (f) x2 + y2 _ e^; + 4y - 5 = 0. P — X^(b) 3x2 + 3y2_i0x-24 2/ = 0. (g) (x + 1)2 + (y _ 2)2 = 0. ' (c) x2 + 2/2 ^ 0. (h) 7 x2 + 7 i/2 - 4 X - y = 3. ^ (d) x2 + 2/2 _ 8x - ?/ + 25 = 0. (i) x2 + 2/^ + 2 ox + 2 6?/ + a2 + 62 = q. (e) x2 + 2/2 - 2x + 2 2/ + 5 = 0. (j) x2 + 2/2 + 16x + 100 = 0. 3. Find the equation of the circle which (a) has the center (2, 3) and passes through (3, — 2). Ans. X2 + 2/2 -4X-62/-13 = 0. '^ (b) passes through the points (0, 0), (8, 0), (0, - 6). Ans. x2 + ^2 _ 3 a; _^ (5 y _ 0^ (c) passes through the points (4, 0), (— 2, 5), (0, — 3). Ans. 19x2 +-19^2^ 2x- 47 2/ - 812 = 0. (d) passes through the points (3, 5) and (—3, 7) and has its center on the JT-axis. Ans. x2 + y2 _|_ 4 j. _ 46 = Q. (e) passes through the points (4, 2) and (—6, — 2) and has its center on "* the F-axis. Ans. x2 + 2/2 + 5 2/ - 30 = 0. (f) passes through the points (5, — 3) and (0, 6) and has its center on the line2x -3?/ - 6 = 0. Ans. 3x2 + 82/2 - 114x - 642/ + 276 = 0. (g) has the center ( — 1, — 5) and is tangent to the JT-axis. Ans. x2 + ^2 _^ 2 X + 10 2/ + 1 = 0. (h) passes through (1, 0) and (5, 0) and is tangent to the F-axis. ^715. x2 + 2/^-6x±2V5y + 5 = 0. (i) passes through (0, 1), (5, 1), (2, - 3). Ans. 2x2 + 2^2_iox + 2/-3 = 0. (j) has the line joining (3, 2) and (— 7, 4) as a diameter. Ans. x2 + 2/^ + 4 X - 6 2/ - 13 = 0. (k) has the line joining (3, — 4) and (2, — 5) as a diameter. Ans. x2 + 2/2 - 5 X + 9 y + 26 = 0. (1) which circumscribes the triangle formed by x — 6 = 0, x4-22/ = 0, and X - 2 2/ = 8. Ans. "2 x2 + 2 y2 _ 21 x + 8 ?/ + 60 = 0. (m) passes through the points (1, — 2), (— 2, 4), (3, — 6). Interpret the result by the Corollary, p. 98. (n) is inscribed in the triangle formed by4x + 32/ — 12 = 0, 2/ — 2 = 0, tx - 10 = 0. Ans. 36 x2 + 36 2/2 - 516 x + 60 2/ + 1585 = 0. 4. Plot the locus of x2 + 2/2 - 2 x + 4 7/ + A: = for A; = 0, 2, 4, 5 - 2, - 4, 8. What values of k must be excluded ? Ans. lc>h. 136 ANALYTIC GEOMETRY 5. What is the locus of x"^ + y^ + Dx + Ey -{- F = ii DandE are fixed and F varies ? 6. For what values of k does the equation A"^ + y^ — 4x + 2 ky + 10 = have a locus ? Ans. k> -{■ V6 and A; < — V6. 7. For what values of k does the equation x^ + y^ -\- kx + F = have a locus when (a) F is positive ; (b) F is zero ; (c) F is negative ? Ans. (a) ^ > 2 Vf and k<-2 ^F ; (b) and (c) all values of k. 8. Find the number of point-circles represented by the equation in problem 7. Ans. (a) two ; (b) one ; (c) none. 9. Find the equation of the circle in oblique coordinates if w is the angle between the axes of coordinates. Ans. {x - a:)2 + (?/ - /3)2 + 2 (x — a) {y - ^) cos w = r'^. 10. Write an equation representing all circles with the radius 6 whose centers lie on the X-axis ; on the Y-axis. 11. Find the number of values of k for which the locus ot^ (a) x^ + y"^ -}- 4:kx - 2 y + 6 k = 0, (b) x^ + y^ + 4:kx - 2y - k = 0, (c) x2 -f 2/2 + 4 fcx - 2 2/ + 4 A: = is a point-circle. Ans. (a) two ; (b) none ; (c) one. 12. Plot the circles a;^ + ?/2 + 4 x - 9 = 0, x'^ + y^ - ix - 9 = 0, and x2 -f- 2/2 _|_ 4x _ 9 + A:(x2 + 2/2 _ 4x - 9) = for A: = ± 1, ±3, ± i, - 5, — ^. Must any values of k be excluded ? 13. Plot the circles x2 + 2/2 + 4x = 0, x2 + 2/2 _ 4^; = 0, and x2 + y2 + 4x -f A; (x2 -|- 2/2 — 4 x) = for the values of k in problem 12. Must any values of A: be excluded ? 14. Plot the circles x2 -1- ?/2 + 4x + 9 = 0, x2 + ?/2 - 4x + 9 = 0, and x2 + 2/2 + 4x4-9 + A:(x2 + 2/2-4x4- 9) = for A; = - 3, - |, -5, - |, — I, — f , — 1. What values of k must be excluded ? ' 58. Systems of circles. An equation of the form x^-\-y'^ + Dx + Ey + F=0 will define a system of circles if one or more of the coefficients contain an arbitrary constant. Thus the equation x^ -\- 7/^ — r^ = represents the system of concentric circles whose centers are at the origin. Very interesting systems of circles, and the only systems we shall consider, are represented by equations analogous to (XIII), p. 119. THE CIRCLE 137 Theorem III. Given two circles, ' Ci : ^2 + 7/2 + D^x + E^y + i^'i = and C^:x^ -{-if -{- D^x + E^y + F^ = 0; then the locus of the equation (III) oc^ + y^ + n^oc + Eiu + F^ + l^iic'' + 2/^ + i>2^ + E^y + JPa) = O t5 (Z circle except when k = — X. In this case the locus is a straight line. Proof. Clearing the parenthesis in (III) and collecting like terms in x and y, we obtain {l + l<)x^-\-{l-\-k)y^ + {Dy^kR,)x + {E, + kE^)y + {F, + kF,) = 0. Dividing by 1 + ^ we have "^ +y ^ ij^k ^^ 1 + /^ ^ i^rk ^• The locus of this equation is a circle (Theorem I, p. 131). If, however, k = — 1, we cannot divide by 1 + A^. But in this case equation (III) becomes (A - D,)x + (E, -E,)y+ (F, - F,) = 0, which is of the first degree in x and y. Its locus is then a straight line called the radical axis of Ci and C^. q.e.d. Corollary I. The center of the circle (III) lies upon the line joiiiing the centers of Ci and Cg and divides that line into seg- ments whose ratio is equal to k. For by Theorem I (p. 131) the center of Ci is Pi( - — i. - -^ ) and of C-i is f D2 Ei\ \ 2 2 / P2 ( » j • The point dividins? P1P2 into segments whose ratio equals k is (Theorem VII, p. 39) the point — » -—— > or, L— 1-f-A^ l-rfl! —J . ,., . ( Dx + kDi Ei-{-kE2\ ^ . . . ^^ ^ . .tttn sjmphfymg, (^ - ^ > - - ^ j ' which is the center of (III). 138 ANALYTIC GEOMETRY Corollary II. The equation of the radical axis of C-^ and C^ is (A -D,)x+ (E, - E,)y + (Fi - F,) = 0. Corollary III. The radical axis of two circles is perpendicular to the line joining their centers. Hint. Find the line joining the centers of C^ and C2 (Theorem VII, p. 97) and show that it is perpendicular to the radical axis by Corollary III, p, 87. . The system (III) may have three distinct forms, as illustrated in the following examples. These three forms correspond to the relative positions of C^ and Cg, which may intersect in two points, be tangent to each other, or not meet at all. Ex. 1. Plot the system of circles represented by x2 + 2/2 4- 8x - 9 + ^-(x2 + ?/2 - 4a; - 9) = p^, • Solution. The figure shows the circles a:2 + 7/2 + 8 X - 9 = and x'^ -h y^ - 4:X - 9 = plotted in heavy lines and the circles corresponding to A; = 2, 5, 1, 4,-4, - f , and - I ; these circles all pass through the intersection of the first two. The radical axis of the two circles plotted in heavy lines, which corre- sponds to k = — 1, is the F-axis. THE CIRCLE 139 Ex. 2. Plot the system of circles represented by x2 + ?/2 -f 8x + k{x^ + 2/2 - ix) = 0. - ^ -= ^ Y, ^ -- -- N -- / / y \ /^rVJ ^ s \ / / / /' ^\ r N \, ^ 1 ( / / ^\ 'P^ Ni \ \ \l ( r A ^3 \ X \ ^■^ fc3 J X \ \ \ \ ^!^ ^Ki /' 1 /I \ \ \^ ^f \^ ^-^ y / / \ \ \ ^ \- ■^ -^ y / - ^ -^ / r \. ^> ^ ^ -- Solution. The figure shows the circles x^ + y^ + Sx = and x^ + y'^-4:X = plotted in heavy lines and the circles corresponding to A: = 2, 3, I, 5, 1, 1, - 7, i, - 4, - 3, and - f These circles are all tangent to the given circles at their point of tangency. The locus for Z: = 2 is the origin. Ex. 3. Plot the system of circles represented by ' x2 + 2/2_iOx + 9 + A;(x2 + 2/2 + 8cc + 9) = 0. Solution. The figure shows the circles x2 + y2 _ 10 X + 9 = and x2 + ?/2 _^. gx + 9 plotted in heavy lines and the circles corresponding to Jc = h 17, h - 10, - tV. and - V- 140 ANALYTIC GEOMETRY These circles all cut the dotted circle at right angles, as will be shown later. For k = f the locus is the point-circle (3, 0), and for A: = 8 it is the point-circle (—3, 0). In all three examples the radical axis, for which A; = — 1, is the F-axis. Theorem IV. When the circles Ci : ^2 _^ 2/' + A^ -h E,ij + Fi = and C^ : x^ -\- y^ -\- D^x -f Ec^y + F^ = intersect in two points P^ (xi, y^) and P^ (x^, y^, then the system of circles represented by a;2 + y2 _,_ ^^^ _j_ ^^y _j_ 2^^ _p ;^(^2 _^ y2 _^ ^^^ ^ ^^^ + ^2) = consists of all circles passing through P-^ and P^. Proof. First, every circle of the system passes through Pj and Pa- I'or, since Pj lies on C^ and Cg, we have ^1 + 2/1' + A^i + A2/1 + i^i = and x^ -^ 2/i2 ^ ^^^^ ^ ^^^^ + P2 = 0. Multiply the second equation by h and add to the first; this gives which is the condition that P^ lies on any circle of the system. In the same manner we can show that every circle of the system passes through P^. In the second place, every circle which passes through P^ and P2 is in the system. For any such circle is determined by Pi, Pg and a point P3 (ccg, y^) not on the line P^P^- Then if P^ lies on a circle, of the system, we have ^3' + 2/3' + D^x^ + E^y^ + i^i + /v (rrg^ + y,^ + D^x^ + i^J^T/g + P2) = 0, and hence ^ ^ _ ^a^ + ^a^ + A^3 + A.Vs + ^1. a^s' + 2/3' + Aa^s + -E^22/3 -f P2 That is, a value of k can be determined so that the corresponding circle passes through P3. Since P3 is any point not on P^P^, that circle is any circle which passes through Pi and P^; and hence every circle which passes through Pj and Pg belongs to the system. q.e.d. THE CIRCLE 141 Corollary. The radical axis of two iiitersecting circles is their common chord. In like manner we may prove Theorem V. When the circles Ci : ^2 + 2/2 + D^x + J5:iy + Fi = and Ca : cc^ + 2/^ + D^^ + E^y -^ F^ = are tangent at the j^oint Py (xi, t/j), then the system of circles repre- sented hy ^' 4- 2/' + Aa^ + E,y J^F^-\-k{x^ + t/ + D^x + E^y + ^2)= consists of all circles tangent to C^ and C^ at Pj. These theorems show how to construct the circles of the system in case Ci and Cg intersect or are tangent, but there is no analo- gous theorem if Ci and Cg do not intersect. In what follows we shall consider a method which applies to all three cases. Theorem VI. The equation of the sy stein (III), (p. 137), may he tvritten iii the form (VI) ic2 + y2 _,_ j^f^ _,_ 2?^ ^ Q^ where k' is an arbitrary constant, if the axes of x and y be respec- tively chosen as the line of centers and the radical axis of Ci and Cg. Proof No matter how the axes be chosen, the equations of Ci and C2 have the forms C^:x^ + 7/ + D,x + Fiy + Fi = and C2 : £c2 + ?/2 4- D^x + E^y + F^ = 0. If the centers of Ci and C\ lie on the Z-axis, then Fi = and F2 = (Corollary, p. 131). The equation of the radical axis (Corollary II, p. 138) then becomes {D,-D,)x^{F,-F,)=0. If this line is the F-axis, whose equation is x = Q, we must have Fi — F2 = 0, and "hence F^ = F^. Substituting F for Fj and Fa and setting E^ = and F2 = in (III), we obtain x"^ -{- y"" + D^x -{- F -{- k (.t2 + t/^ + D^x + F) = 0. 142 ANALYTIC GEOMETRY Collecting like powers of x and y and dividing by 1 + yfc, we obtain x^ ^-tf ^ ic + F = 0. X -]- ki The coefficient of a; changes with k and may be denoted by a single letter; if we set A + hP, 1-^k -"' we obtain equation (VI). q.e.d. Corollary. The centers of the circles of the system (YI) lie on the X-axis. The study of the system of circles (III), p. 137, may then be effected by the study of the system (VI), whose equation is in a simpler form than that of (III). Theorem VII. If r' is the radius of that circle of the system X^ + 2/2 + k'x + F = whose center is (a\ 0), then k'^ — 4: F Proof For by Theorem I (p. 131) we have r'^ = and a'2 = -— • Hence r'^ = a'"^ — F. 4 Corollary I. When F is negative, r' is the hypotenuse of a right triangle ivhose legs are a! and V— F .^ Corollary II. When F is zero, then r' = a'. Corollary III. When F is positive, a' is the hypotenuse of a right triangle ivhose legs are r' and '\F . We may readily construct circles of the system (VI) by the use of these corollaries. With the preliminary remark that the centers of all of the circles of the system lie on the Z-axis (by the Corollary), we shall consider the three cases separately. * When F is negative, —F is positive, and hence y/-F is a real number. THE CIRCLE 143 Case I. F < 0. In this case r'^ = a'^ — F is positive for all real values of a', and hence every point on the A'-axis may be used as the center of a circle belonging to the system. On OF lay off OA = V— i^. With any point P' on the Z-axis as center and with. P'A as a radius, describe a circle ; this circle will belong to the system. For let OP' = a'; then P'A = r' by Corollary I. The system is then composed of all circles whose centers lie on the Z-axis which pass through A (0, + V— F). It is evi- dent that the circles will also pass through B (0, — V— F). Case II. F =0. In this case r''^ = a'^, and hence all points on the Z-axis may be used as centers. Further, the circles of the system will all pass through the origin (Theorem YI, p. 73). Hence the circle whose center is any point P' on the Z-axis and whose radius is P'O will belong to the system. It is evident that all of the cir- cles of the system are tangent to the F-axis at the origin and also to each other. Case III. F> 0, In this case r'^ = a'^ — F is positive only when a' is numerically greater than \F, and hence points on the Z-axis for which a' is numerically less than V^ cannot be used as centers. With as a center and with Vf as a radius, describe a circle, the dotted circle in the figure. Let P' be any point on the Z-axis outside of this circle. Draw P'A tangent to the dotted circle. With P' as center and P'A as radius, describe a circle; this circle will belong to the system. For let P'0 = a'', then, since OA = Vf, and since ^ is a right angle, P'A =r' by Corollary III. Two intersecting jircles whose tangents at a point of intersection are perpendicu- lar are said to be orthogonal ; hence the system is composed of all circles whose centers are on the Z-axis which cut the dotted circle 144 ANALYTIC GEOMETRY orthogonally. If P' falls at C or D, the radms will be zero ; that is, the point-circles C and D belong to the system and are called its limiting points. Hence Theorem VIII. The circles of the system represented by a;2 4. 2/2 + h'x -\-F=0 have their centers on the X-axis, and (a) pass through (0, + V— F) and (0, — V— F) if F is negative ; (b) are tangent to each other at the origin ifF= 0; (c) are orthogonal to the circle x^ + y'^ = F if F is positive. The constructions given in the proof were used in drawing the figures on pages 138 and 139. It is evident from the figures, and can be proved analytically, that there are no point-circles if F is negative, that there is one point-circle if F is zero, and that there are two if F is positive. 59. The length of the tangent. Theorem IX. Given a point Pi (xi, y-^ and the circle C -.x"^ + !/-{- Dx-\-Ey -^F^O, then the product of any secant through P^ and its external s.eg- Tnent is Proof Let the equations of any line through Px be (Theorem XV, p. 124) x = Xx + p cos a, 2/ = 2/1 + P cos /?. Then if the point (x, y) or (x^ -f />j p cos a, yi + p cos (3) lies on C, we have (Corollary, p. 53) (xi + p cos a)2 + (?/i -f p cos (3y + D(xi-\- p cos a) 4- -^ (2/1 + f> cos l3)-\-F=0. THE CHICLE 146 Simplifying, arranging according to powers of p, and using (1), p. 123, we have p2 + p [(2 Xi + D) cos « + (2 2/1 + £) cos y8] + ^i' + 2/i' + Dx^ + Eij, + Fi = 0. The roots of this quadratic are the lengths of the secant PA and its external segment P1P2' Hence the product of I\Ps and P1P2 is (Theorem I, p. 3) As this expression does not contain cos a or cos /3 it is imma- terial in what direction the secant be drawn. q.e.d. Corollary. The square of the length of the tangent from P^ to C is given by (IX). For when the secant swings around on Pi until it becomes tangent to C, P1P3 and P1P2 both become equal to P1P4. Theorem X. The ratio of the squares of the lengths of the tan- gents di'awn from any point of the circle Cj,:x'-\-y''^- D,x + E,y + ^i + k(x^ + 2/' + D2X + E,y + P2) = to the circles C,:x' + y''-\- D,x -^E,y-]-F, = and C^-.x^ + 2/ -\- D^x + E^y 4- Pg = is constant and is equal to — k. Proof Let Pi(a:i, y^ be any point on Cj^. Then ^i^ + Vi + A^i + ^i2/i + ^1 + ^ (^1' + Vi + D^^i + E^y, + P2) = 0. Dividing by the parenthesis and transposing, we obtain ^i" + yi' + A^i + -gi 3/1 + Pi ^ j^ ^l' + 2/l' + A^l + P2 2/l + ^2 By the Corollary the numerator of this fraction is the square of the length of the tangent from Pj to C^, and the denominator is the square of the length of the tangent from Pj to Cg. Hence the ratio of the squares of the lengths of those tangents is con- stant and equal to — A;. q.e.d. 146 Ai^ALYTIC GEOMETRY Corollary I. The locus of a point from which the ratio of the squares of the lengths of the tangents to the circles C^ and C^ is constant and equal to — k is the circle C^. Theorem X proves only one part of the Corollary. It remains to be proved that all points such that the ratio of the squares of the lengths of the tangents from these points to C\ and (J% equals — Ic lie on C'^.. Corollary II. The locus of points from which tangents to two circles are equal is the radical axis of those circles. PROBLEMS 1. By means of Theorem VIII plot the following systems of circles. (a) ic2 + 2/2 + 4a; - 1 + A;(x2 + 2/2 - 2x - 1) = 0. (b) x2 + 2/2 -f 4x + 1 + A;(x2 + 2/2 - 2x + 1) = 0. (c) x2 + 2/2 + 4x + A;(x2 + y^-2x) = 0. (d) x2 + 2/2 + 2x - 4 + fc (x2 + 2/2 + Cx - 4) = 0. IP (e) x2 + 2/2 + 2x + 9 + fc (x2 + 2/2 - 4x + 9) = 0. (f ) x2 + 2/2 - 6 X + A; (x2 + 2/2 + 8 x) = 0. 2. Find the lengths of the tangents from the point (a) (5, 2) to the circle x2 + ^2 _ 4 = q. (b) (- 1, 2) to the circle x^ + y^-6x-2y = 0. (c) (2, 5) to the circle 2x2 + 2 2/2 + 2x + 42/-l=0. (d) (1, 2) to the circle x^-\-y^ = 25. What does the imaginary answer in (d) mean ? Ans. Point is within the circle. 3. Determine the nature of the following systems. (a) x2 + 2/2 + 2 X - 4 2/ + A; (x2 + y2 _ 2 X + 4 y) = 0. (b) x2 + 2/2 + 4 X - y + fc (x2 + 2/2 - 4 X + 2/ - 4) = 0. (c) x2 + 2/^+ 2 X - 4 2/ + 1 + A; (x2 + 2/2 - 2 X + 4 y + 1) = 0. 4. Find the equation of the circle passing through the intersections of the circles x^ + y^ — 1 = and x^ -{- y^ + 2x = which passes through the point (3,2). ^ns. 7x2 + 7 2/2- 24x- 19 = 0. 6. Find the equation of the circle passing through the intersections of x2 -I- 2/2 — 6 X = and x2 + 2/2 — 4 = which passes through (2, — 5). Ans. x2 + 2/^-3x-2 = 0. 6. Find the equation of that circle of the system x^ -\- y^ — Ax — 3 + A; (x2 + 2/'"^ — 4 2/ — 3) = whose center lies on the line x — y — 4:ih=X). Ans. X2 + 2/2 -6x + 2|/-3='0- THE CIRCLE 147 7. Find the equation of the circle passing through the intersections of x^ -{■ y^ — i X + 2y = and x^ + ^z^ — 2y — 4 = whose center lies on the line 2x + 4y-l = 0. Ans. x'^ + y^ - 3x + y - 1 = 0. 8. Find the equations of the circles passing through the intersections of aj2 _|_ y2 _ 4 — and x^ + y^ + 2x — S = whose radii equal 4. Ans. x^-\-y^ — 6x-7 = and x^ -{- y^ -\- 8x = 0. 9 . Find the radical axes of the circles x^ + y2 — 4 x = 0, x^ + ?/2 + 6 x — 8 ?/ = 0, and x2 + 2/2 + 6x — 8 = taken by pairs, and show that they meet in a point. 10. Find the radical axes of the circles x^-]-y^-9=0, 3x2+3 2/2-6 x+82/ — 1=0, and x2+2/2_)_8 y=0 taken by pairs, and show that they meet in a point. 11. Show that the radical axes of any three circles taken by pairs meet in a point. 12. By means of problem 11 show that a circle may be drawn cutting any three circles at right angles. 13. By means of problem 11 prove that if several circles pass through two fixed points their chords of intersection with a fixed circle will pass through a fixed point. 14. The square of the tangent from any point Pi of one circle to another is proportional to the distance from the radical axis of the two circles to Pi. 15. If Ci and C^ (Theorem III) are concentric, then all the circles of the systeni (III) are concentric. 16. Show that when Ci and C2 (Theorem III) are concentric the equation of the system (III) cannot be written in the form given in Theorem VI. 17. Show that the radical axis of any pair of circles in the system (III) is the same as the radical axis of Ci and C2. 18. How may problem 11 be stated if the three circles are point-circles ? MISCELLANEOUS PROBLEMS 1 . Find the equation of the circle which circumscribes the triangle formed byx + 2?/ = 0, 3x-2y = 6, and x - y = 6. 2. Find the equation of the circle inscribed in the triangle in problem 1. 3. Find the angle between the radii of the circles x^ -\-y^ = 26 and x^ -\-y^ — \6x-hS9 = which are drawn to a point of intersection. Hint. Find the radii, the length of the line of centers, and apply 17, p. 20. 4. Find the angle between the radii of the circles x^ + y^ + DiX + Eiy + Pi = and x2 + 2/2 ^ j)^x + P22/ + P2 = which are drawn to a point of intersection. 148 ANALYTIC GEOMETRY 5. Find the condition that the angle in problem 4 should be a right angle. 6. Show that an angle inscribed in a semicircle is a right angle. 7. Prove that the perpendicular dropped from a point on a circle to a diameter is a mean proportional between the segments of the diameter. 8. If w is the angle between the oblique axes OX and OY, then the locus of x2 + 2 cos (oxy + ?/2 + Dx + jE"?/ + F = is a circle. 9. Given a circle C:x2+2/2+Zte+^2/+ 2^=0 and a line i:J.x+%+C=0; show that the system of curves x"^ -\- y^ -\- Bx + Ey + F + k {Ax + By -j-C)=0 consists of all circles whose centers lie on the line through the cent.er of C perpendicular to L. 10. Find the radical axis of any two circles of the system in problem 9. 1 1 . Find a geometric interpretation of k in the equation in problem 9. 12. What does the equation of the system in problem 9 become if (a) the Y-axis is the line L and the X-axis passes through the center of C ? (b) the, origin is the center of C and the Y-axis is chosen parallel to L? 13. Show how to construct the circles of the system x^+y^—r^-{-k{x—a)=0 when (a) r < a ; {h) r = a; and (c) r>a. -i 14. Show that the discriminant of (III) is r2^k^ - {(P' - r^ - r2^) ^ + r^ (1 + A:)2 ' where ri is the radius of Ci, r^ of C2, and d is the length of the line joining the centers of C\ and C2. 15. From problem 14 show that if there are no point-circles in (III), then Ci and C2 intersect ; if there is one point-circle in (III), then C\ and C2 are tangent ; if there are two point-circles in (III), C\ and Ci do not intersect. CHAPTER VI POLAR COORDINATES 60. Polar coordinates. In this chapter we shall consider a second method of determining points of the plane by pairs of real numbers. We suppose given a fixed point 0, called the pole, and a fixed line OA, passing through 0, called the polar axis. Then any point P determines a length OP = p and an angle A OP = 0. The numbers p and are called the polar coordinates of P. p is called the radius vector and 6 the vectorial \ angle. The vectorial angle is positive \ or negative as in Trigonometry (p. 18). "^ The radius vector is positive if P lies on the terminal line of 6, and negative if P lies on that line produced through the pole 0. Thus in the figure the radius vector of P is positive, and that of P' is negative. It is evident that every pair of real numbers (p, 0) determines a single point, which may be plotted by the Rule for plotting a point whose polar coordinates (p, 0) are given. First step. Construct the terminal line of the vecto- rial angle 0, as in Trigo- nometry. Second step. If the radius vector is positive, lay off a length OP = p on the terminal line of 6; if negative, produce the 149 150 ANALYTIC GEOMETRY terminal line through the pole and lay off OP equal to the numer- ical value of p. Then P is the required point. In the figure on p. 149 are plotted the points whose polar coordinates are (''-f>(^-T>(-'X><«-).-(^.-f> Every point P determines an infinite number of pairs of numbers (p, 6). The values of d will differ by some mul- tiple of 7t, so that if is one value of 6 the others will be of the form + kTt, where k is a positive or negative integer. The values of p will be the same numerically, but will be positive or negative, if P lies on OB, according as the value of 6 is chosen so that OB or OC is the terminal line. Thus, if OB = p the coordinates of B may be written in any one of the forms (p, 0), (— p, tt + 0), (/), 2 TT + 0), {— p, (p—7t), etc. Unless the contrary is stated, we shall always suppose that 6 is positive, or zero, and less than 2 ir ; that is, < ^ < 2 tt. PROBLEMS 1. Plot ..ep„lnu(4.|),(e -),(-., VO.(^,f).(-/-^). (5, 7t). 2. Plot the points (6, ±j), (-2, ± |), (3, 7t), (-4, tt), (6, 0), (-6, 0). \ 4/ \ 2/ 3. Show that the points (/>, 6) and (p, — d) are symmetrical with respect to the polar axis. 4. Show that the points (p, 6), (— p, 0) are symmetrical with respect to the pole. 5. Show that the points {— p, 7t — 6) and (p, 6) are symmetrical with respect to the polar axis. 61. Locus of an equation. If we are given an equation in the variables p and 6, then the locus of the equation (p. 59) is a curve such that : 1. Every point whose coordinates (p, 6) satisfy the equation lies on the curve. 2. The coordinates of every point on the curve satisfy the equation. POLAR COORDINATES 161 The curve may be plotted by solving the equation for p and finding the values of p for particular values of 6 until the coor- dinates of enough points are obtained to determine the form of the curve. The plotting is facilitated by the use of polar coordinate paper, which enables us to plot values of 6 by lines drawn through the pole and values of p by circles having the pole as center. The tables on p. 21 are to be used in constructing tables of values of p and 0. In discussing the locus of an equation the following points should be noticed. 1. The intercepts on the polar axis are obtained by setting ^ = and 6 = TT and solving for p. But other values of 9 may make p = and hence give a point on the polar axis, namely, the pole. 2. The curve is symmetrical with respect to the pole if, when — p is substituted for p, only the form of the equation is changed. 3. The curve is symmetrical with respect to the polar axis if, when — ^ is substituted for 0, only the form of the equation is changed. 4. The directions from the pole in which the curve recedes to infinity, if any, are found by obtaining those values of for which p becomes infinite. 5. The method of finding the values of $ which must be ex- cluded, if any, depends on the given equation. Ex. 1. Discuss»and plot the locus of the equation p = 10 cos 0. Solution. The discussion enables us to simplify the plotting and is there- fore put first. 1. For ^ = p = 10, and foY0 = 7t p = — 10. Hence the curve crosses the polar axis 10 units to the right of the pole. 2. The curve is symmetrical with respect to the polar axis, for cos(- 0) = COS0 (4, p. 19). p d 9 10 Tt 2 7t l2 9.7 nit 12 - 2.6 7f 6 8.7 2;r 3 - 5 7t 37r - 7 4 7 4 7t 5 hit 6 - 8.7 3 llTT - 9.7 57t 12 12 2.6 It -10 152 ANALYTIC GEOMETRY 3. As cos d is never infinite, the curve does not recede to infinity. Hence the curve is a closed curve. 4. No values of 6 make p imaginary. Computing a table of values we obtain the table on p. 151. As the curve is symmetrical with respect to the polar axis, the rest of the curve may be easily constructed without com- puting the table farther; but as the curve we have already constructed is symmetrical with respect to the polar axis, no new points are obtained. The locus is a circle. Ex. 2. Discuss and plot the locus of the equation p"^ = aP- cos 2 0. Solution. The discussion gives us the following properties. 1. For ^ = or 7t p = ±a. Hence the curve crosses the polar axis a units to the right and left of the pole. 2. The curve is symmet- rical with respect to the pole. 3. It is also symmetrical with respect to the polar axis, for cos (— 2 ^) = cos 2 6 (4, p. 19). 4. p does not become infinite. 5. p is imaginary when cos 2 ^ is negative, cos 2 d is negative when 2 ^ is in the second or third quadrant; that is, when i^>2»>* or I^>2»>^. 2 2 2 2 Hence we must exclude values of such that ^^^ n^ ^ J 77r ^ 57r —->e>- and — >d> 4 4 4 4 The accompanying table of values is all that e P e p 12 ±a ±.93 a It 6 It 4 ±.7a need be computed when we take account of 2, 3, and 5. POLAR COORDINATES 153 The complete curve is obtained by plotting these points and the points symmetrical to them with respect to the polar axis. The curve is called a lemniscate. In the figure a is taken equal to 9.5. Ex. 3. Discuss and plot the locus of the equation 2 ^ "" 1 + cos ^ ' Solution. 1. For ^ = p = 1, and iov d = it p the polar axis one unit to the right of the pole. 2. The curve is not symmetrical with respect to the pole, be inferred from 1 ? 3. The curve is symmetrical with respect to the polar axis, since cos(- d) = cos^ (4, p. 19). 4. p becomes infinite when 1 + cos ^ = or cos 6 =— \ and hence 6 = 7t. The curve recedes to infinity in but one direction. 5. p is never imaginary. On account of 3 the table of values is computed only to ^ = tt, and the rest of the curve is obtained from the symmetry with respect to the polar axis. The locus is a parabola. CO ; so the curve crosses How may this e p Q ? 1 lit 2.7 IT 1.02 12 12 7t 1.07 27r 3 4 6 7t 4 1.2 3 7r 4 6.7 TC 1.3 hit 14 3 G 57r 12 1.6 ll;r 50 * 12 Tt 2 2 It CO PROBLEMS Discuss and plot the loci of the following equations. 1. p = 10. e = tan-il. 5. psin^ = 4. 2. /t) = 5. Q=- Z. p = 16.C0S d. 4. p COS ^ = 6. 6 6. p 7. p = COS^ COS 6 154 ANALYTIC GEOMETRY 1 - 2 cos ^ 9. p = a sin 6. 10. p = a{l — cos^). 11. /)2 sin 2 ^ n: 16. 12. p2 = 16sin2^. 13. p^cos^2d = a^. 14. p = asm2 6. p = acos2d. 8 15. p = 1 — e cos for e = 1, 2, i. 16. pcose = asin^d. 17. /9C0S ^ = a cos 2d. 18. /) = a(4 + 6cos^) for 6 = 3, 4, 6. 19. p = 1 + tan ^ 20. p = asece ±h for a > 6, a = 6, a• sin ^. The locus of this equation is called a cardioid. p{P,d) PROBLEMS 1. Chords passing through a fixed point on a circle are extended their own lengths. Find the locus of their extremities. Ans. A circle whose radius is a diameter of the given circle. 2. Chords of the circle p = 10 cos 6 which pass through the pole are extended 10 units. Find the locus of the extremities of these lines. Ans. p = 10 (1 + cos d). 3. Chords of the circle p = 2a cos 6 which pass through the pole are extended a distance 2 6. Find the locus of their extremities. Ans. p = 2 (6 + a cos 6). , POLAR COORDINATES 169 4. Find the locus of the middle points of the lines drawn from a fixed point to a given circle. Hint. Take the fixed point for the pole and let the polar axis pass through the center of the circle. Ans. A circle whose radius is half that of the given circle and whose center is midway between the pole and the center of the given circle. 6. A line is drawn from a fixed point meeting a fixed line in Pi. Find the locus of a point P on this line such that OPi • OP = a^. Ans. A circle. 6. A line is drawn through a fixed point meeting a fixed circle in Pi and Pa. Find the locus of a point P on this line such that OP = 2 ^^^ ' ^^^ . Ans. A straight line. OPi + OP2 CHAPTER VII TRANSFORMATION OF COORDINATES 65. When we are at liberty to choose the axes as we please we generally choose them so that our results shall have the sim- plest possible form. When the axes are given it is important that we be able to find the equation of a given curve referred to some other axes. The operation of changing from one pair of axes to a second pair is known as a transformation of coordinates. We regard the axes as moved from their given position to a new position and we seek formulas which express the old coordinates in terms of the new coordinates. 66. Translation of the axes. If the axes be moved from" a first position OX and OF to a second position O'Z' and O'F' such that O'X' and O'F' are respectively parallel to OX and OY, then the axes are said to be translated from the first to the second position. Let the new origin be 0\h, k) and let the coordinates of any point P before and after the translation be respectively (x, y) and (x\ ?/'). Projecting OP and OO'P on OX, we obtain (Theorem XI, p. 48) X = x^ -\- h. Similarly, y = y' -{- k. Hence, Theorem I. If the axes be translated to a netv origin (h, k), and if (x, y) and (x', y') are respectively the coordinates of any point P before and after the translation, then Y' Y' ' ■ N 1 d ^^^ B (h. ky X Y X /" / » A M X (I) •£c = a?' + h, y = y^ ■\-k. 160 TRANSFORMATION OF COORDINATES 161 Equations (I) are called the equations for translating the axes. To find the equation of a curve referred to the new axes when its equation referred to the old axes is given, we substitute the values of x and y given by (I) in the given equation. For the given equation expresses the fact that P (a?, y) lies on the given curve, and since equations (I) are true for all values of (a:, y), the new equation gives a relation between x' and y^ which expresses that P{x\ 2/') lies on the curve and is therefore (p. 53) the eqiiar tion of the curve in the new coordinates. Ex. 1. Transform the equation x2 + y2_6x + 4y-12 = when the axes are translated to the new origin (3, — 2). Solution. Here ^ = 3 and k=—2, so equations (1) become X = x' -\- S, y = y' - 2. Substituting in the given equation, we obtain (x' + 3)2 + (z/'-2)2-6(x' + 3) + 4(^-2) -12 = 0, or, reducing, x'^ + y'^ = 25. This result could easily be foreseen. For the locus of the given equation is (Theorem I, p. 131) a circle whose center is (3, -2) and whose radius is 5. When the origin is translated to the center the equa- tion of the circle must necessarily have the form obtained (Corollary, p. 58). PROBLEMS 1. Find the new coordinates of the points (3, - 5) and (- 4, 2) when the axes are translated to the new origin (3, 6). 2. Transform the following equations when the axes are translated to the new origin indicated and plot both pairs of axes and the curve. Ans. Sx' -4y' = 0. Y £ " ^ -^ ^ ■^ N ' / \ \ \ X 0' (3, 2) X' / / \ / V ^ y Ans. iC'2 + 2/'2 ^ Ans. y"^ = 6 x'. (a) 3x-4y = 6, (2,0). (b) x2 + 2/2 - 4iC - 2 ?/ = 0, (2, 1). (c) y2-6x + 9 = 0, (1,0). (d) x2 + 2/2_i = o, (-3, -2). (e) ?/2_2fcx + fc2 = 0, (-, oV (f) x2-4?/2+8x+24?/-20=0, (-4, 3). Ans. x'2-4i/'2 = Ans. Ans. x'2+?/'2_6a;'_4y'+12=0. y'2 - 2 kx'. 162 ANALYTIC GEOMETRY 3. Derive equations (I) if 0' is in (a) the second quadrant ; (b) the third quadrant ; (c) the fourth quadrant. 67. Rotation of the axes. Let the axes OX and F be rotated about through an angle 6 to the positions OX' and 0Y\ The equations giving the coordinates of any point referred to OX and OF in terms of its coordinates referred to OX' and OY' are called the equations for rotating the axes. Theorem II. The equations for rotating the axes through an angle 6 are (II) 05 = a?' cos 5 — y^ sin 9y 2/ = 0?' sin ^ + y^ cos 9, Proof. Let P be any point whose old and new coordi- ^^^ ^ nates are respectively (x, y) ^ and (x', y'). Draw OP and draw PM' perpendicular to OX'. Project OP and OM'P on OX. The proj. of OP on OX = x. The proj. of OM' on OX = x' cos 6. (Theorem III, p. 31) (Theorem II, p. 30) The proj. of M'P on OX = y'Gos['^ + e\ (Theorem II, p. 30) = — y sm &. Hence (Theorem XI, p. 48) X == x' cos — y' sin 6. (by 6, p. 20) In like manner, projecting OP and OM'P on OF, we obtain y =z x' COS 0\-\-y'cosO x' sin -\- y' cos 0. Q.E.D. If the equation of a curve in x and y is given, we substitute from (II) in order to find the equation of the same curve referred to OX' and OY'. TRANSFORMATION OF COORDINATES 163 Ex. 1. Transform the equation x^ — y^ = 16 when the axes are rotated It through Solution. Since . 7t 1 /- 1 sm - = - V 2 = -— 4 2 V2 and cos — = — -i 4 V2 equations (II) become x' - y' x' + y' Vi V^ Substituting in tlie given equation, we obtain Wttf \ s yirt- : !s s^ Z 1 " \ s Z / - -^ s Z t ~- - 4^ 5 41 ^ X ^ Js -r : 7 z s^ t " - -7 z s _s - 7- z s^^ I -/-A - ( ^ ' -y' V _ / ^' -^y' Y or, simplifying, x^ + y'V V2 ^ ^ V2 x'?/' + 8 = 0. 16, PROBLEMS 1. Find the coordinates of the points (3, 1), (—2, 6), and (4, — 1) when It the axes are rotated through — • 2. Transform the following equations when the axes are rotated through the indicated angle. Plot both pairs of axes and the curve. Tt (a) X - y = 0, (b) x2 + 2xy + 2/2 = 8, ^. 4 (c) 2/2 = 4x, -|. (d) X2 + 4X2/ + 2/2 3,16,^. 4 (e) x2 + 2/2 = r2, e. (f) x2 + 2 x?/ + 2/2 + 4 X - 4 2/ = 0, - -. 4 Ans. y' = 0. J.ns. x'2 = 4. -4 ns. x'2 = 4 2/'. ^ns. 3x'2-2/'2 = 16. Ans. x'2 + 2/'2 = r2. 3. Derive equations (II) if 6 is obtuse. 68. General transformation of coordinates. If the axes are moved in any manner, they may be brought from the old position to the new position by translating them to the new origin and then rotating them through the proper angle. 164 ANALYTIC GEOMETRY ^j-j-j. (i€ = o(^' COS O-y^ ^ ^ 12/ = a?' sin 9 + y' Y ^'\ ^' X" /^ v^ ^' Theorem III. If the axes he translated to a new origin (Ji, k) and then rotated through an angle 0, the equations of the transforma- tion of coordinates are oe' cos $ — y' &in 6 -\- 7i, cos 6 + 7c. Proof To translate the axes to O'X" and O'Y" we have, by (I), x = x" -\- h, where (x", t/") are the coordi- nates of any point P referred to O'X" and O'Y". To rotate the axes we set, by (II), , x" = x' cos 6 — y' sin 6, y" = x' sin -\- y' cos 0. Substituting these values of x" and y", we obtain (III). q.e.d. 69. Classification of loci. The loci of algebraic equations (p. 17) are classified according to the degree of the .equations. This classification is justified by the following theorem, which shows that the degree of the equation of a locus is the same no matter how the axes are chosen. Theorem IV. The degree of the equation of a locus is unchanged by a transformation of coordinates. Froof Since equations (III) are of the first degree in x' and y', the degree of an equation cannot be raised when the values of X and y given by (III) are substituted. Neither can the degree be lowered; for then the degree must be raised if we transform back ta the old axes, and we have seen that it cannot be raised by changing the axes.^ As the degree can neither be raised nor lowered by a trans- formation of coordinates, it must remain unchanged. q.e.d. * This also follows from the fact that when equations (III) are solved for x' and ; results are of the first degree in x and y. the TRANSFORMATION OF COORDINATES 165 70. Simplification of equations by transformation of coordi- nates. The principal use made of transformation of coordinates is to discuss the various forms in which the equation of a curve may be put. In particular, they enable us to deduce simjde forms to which an equation may be reduced. Rule to simplify the form of an equation. First step. Substitute the values of x and y given by (I) [or (II)] and collect like powers of x' and y'. Second step. Set equal to zero the coefficients of two ter'ms obtained in the first step which contain h and k (or one coeffi- "cient containing 6). Third step. Solve the equations obtained in the seco7id step for h and k* (or 9). Fourth step. Substitute these values for h and k (or &) in the result of the first step. The result will be the required equation. In many examples it is necessary to apply the rule twice in order to rotate the axes, and then translate them, or vice versa. It is usually simpler to do this than to employ equations (III) in the Rule and do both together. Just what coefficients are set equal to zero in the second step will depend on the object in view. It is often convenient to drop the primes in the new equation and remember that the equation is referred to the new axes. Ex. 1. SimpUfy the equation y2 _ gx + 6 y + 17 = by translating the axes. Solution. First step. Set x = x'' -\- h and y = y' -\- k. This gives {y' + A;)2 - 8 (x' + 7i) + 6 (?/' + A:) + 17 = 0, or (1) y'2_8x' + 2A: ?/'+ A:2 f = 0. + 6 y'+ k^ -8h + 6A: + 17 * It may not be possible to solve these equations (Theorem IV, p. 90). t These vortical bars play the part of parentheses. Thus 2 i- + 6 is the coefficient of y' and k^-8h + Gk+ 17 is the constant term. Their use enables us to collect like powers of x^ and y' at -the same time that we remove the parentheses in the preceding equation. 166 ANALYTIC GEOMETRY Second step. Setting the coefficient of y' and the constant term, the only , coefficients containing h and A;, equal to zero, we obtain (2) 2 fc + 6 =: 0, (3) A;2- 8/1 + 6 A; + 17 = 0. Third step. Solving (2) and (3) for h and fc, we find A; =-3, /i=:l. Fourth step. Substituting in (1), remember- ing that h and Ic satisfy (2) and (3), we have y'l _ 8 X' = 0. The locus is the parabola plotted in the figure which shows the new and old axes. Ex. 2. Simplify x2 + 4?/2-2x-16?/ + l=0 by translating the axes. Solution. First step. Set x =x' + h, y=y' +k. This gives Y' ^r- ^ / / / X s (i -3) X' \ \ s s S (4) x"^+^y"^-\-2h X' + ^k y'+ h^ -.16 + 4A:2 ~2h -16k + 1 = 0. Second step. Set the coefficients of x' and y' equal to zero. This gives 2h-2 = 0, Sk-16 = 0. Third step. Solving, we obtain ^1 = 1, A: = 2. Fourth step . Substituting in (4) , we obtain x'2 + 4 y'2 ^ 16. Plotting on the new axes, we obtain the figure. Ex. 3. Remove the xy-term from x'^ -}- ixy + y^ = 4hj rotating the axes. Solution. First step. Set x = x' cos ^ — y' sin 6 and y = x'sme + y' cos ^, whence + 4 sin cos e + sin2 e ?/2 = 4. x'2 - 2 sin ^ cos d + 4 (cos2 d - sin2 d) -{- -2 &m d cos, 6 or, by 3, p. 19, and 14, p. 20, (5) (1 + 2sin2^)x'2 + 4cos2^.xV+(l - 2 sin2^)y'2 ^ 4. x'y'-\- sin2 — 4sin^cos^ 4- cos2 6 TRANSFORMATION OF COORDINATES 167 Second step. Setting the coefficient of xfy' equal to zero, we have cos 2 ^ = 0. Third step. Hence 2^ = ->[ 1 1 1 \\ itrtttP - S A z . -7- - S ^ z S ^ ^-Z - "^^ S ^Z3I ^^v^ j\ ^^2 - ^ ^ "" ? 7\ ~Z s: s : 7 4 s^ z ^i_ s ' z V ^ - '7 A INNlK Fourth step. Substituting in (5), we obtain, since sin — = 1 (p. 21), The locus of this equation is the hyperbola plotted on the new axes in the figure. ' From cos 2 ^ = we get, in general, 26 = — \- nit, where n is any positive 7t Tt or negative integer, or zero, and hence 6 = — + n — - Then the xy-term may 4 2 be removed by giving d any one of these values. For most purposes we choose the smallest positive value of 6 as in this example. Ex. 4. Simplify x3 4-6x2 + 12x — 4y + 4 = by translating the axes. Solution. First step. Set x = x' -{- h, y = y' + k. We obtain (6) x'^-i-Sh x'^+ 3/i2 x'-4?/+ h^ = 0. + 6 -\-l2h + 6/i2 + 12 + 12 ;i - ik + 4 Second step. Set equal to zero the coefficient of x'2 and the constant term. This gives 3 /i + 6 = 0, 7^3 4. 6 ^2 + 12 /i - 4 A; -f 4 = 0. Third step. Solving, h = -2, k = -l. Fourth step. Substituting in (6), we obtain x'3 _ 4 ?/' = 0, YA r| 1 ' i ^ yc .Y / ^' (-2 rl) x' f 1 1 1 1 ^ — — — - whose locus is the cubical parabola in the figure. 168 ANALYTIC GEOMETRY PROBLEMS 1. Simplify the following equations by translating the axes. Plot both pairs of axes and the curve. (a) x2 + 6 X + 8 rr 0. (b) x^-4y + S = 0. (c) x2 + 2/2 + 4 X - 6 ?/ - 3 = 0. (d) 2/2 - 6 X - 10 ?/ + 19 = 0. (e) x2 - 2/2 + 8x -Uy - 33 = 0. (f) x2 + 42/2 - 16xV 24y + 84 = 0. (g) 2/^ + 8x-40 = 0. (h) x3 - 2/2 + 14 2/ - 49 = 0. (i) 4x2 -4x2/+ 2/2 -40x^2«^2/-|; 99 = 0. 2. Remove the xy-term from the following equations by rotating the axes. Plot both pairs of axes and the curve. (a) x2 - 2 xy + y^ = 12. (b) x2 - 2 X2/ + 2/2 + 8 X + 8 2/ = 0. (c) xy = 18. (d) 25 x2 + 14 X2/ + 25 y^ = 288. (e) 3x2 -10x2/ + 3 2/2 = 0. (f ) 6 x2 + 20 V3 xy + 26 y2 = 324. Ans. x'2 = 1. Ans. x'2 1= 4 2/'. Ans. x'2 + 2/'2 = 16. Ans. 2/'2 = 6x'. Ans. x'2 _ ^'2 = 0. Ans. x'2 + 4 2/'2 = 16. Ans. 8 x' + 2/'"^ = 0. Ans. 2/'2 = X'3. Ans. (2x'- 2/0^-1=0 ^ns. 2/'2 = 6. ^ns. y/2y'^ + 8x' = 0. ^)IS. x'2 _ 2/'2 = 36. ^ws. 16 x'2 + 92/^2=144 Ans. x'2 _ 4 2/'2 = 0. Ans. 9x'2-2/'2 = 81. 71. Application to equations of the first and second degrees. In this section we shall apply the Rule of the preceding section to the proof of some general theorems. Theorem V. Bi/ rtioving the axes the general equation of the first degree, Ax + By +C = Oj may he transformed into x' = 0. Proof Apply the Rule on p. 165, using equations (III). Set X — x^ cos — y' sin 6 -\- h, ^y = x' sin -{- y' cos +- k. This gives (1) A cos 6 -\-BsmO x' — A sin y' + Ah -^ Boose + Bk + C = 0. TRANSFORMATION OF COORDINATES 169 Setting the coefficient of ?/' and the constant term equal to zero gives (2) - ^ sin (9 + j5 cos ^ = 0, (3) Ah + Bk-^C = 0. From (2), tan ^ = "T ' or tan" From (3) we can determine many pairs of values of h and /c. One pair is A \ Substituting in (1) the last two terms drop out, and dividing by the coefficient of x' we have left £c' = 0. q.e.d. We have moved the origin to a point (h, k) on the given line L, since (3) is the condition that {h, k) lies on the line, and then rotated the axes until the new axis of y coincides with L. The particular point chosen for (h, k) was the point 0' where L cuts the Z-axis. This theorem is evident geometric- ally. For ic' = is the equation of the new F-axis, and evidently any line may be chosen as the F-axis. But the theorem may be used to prove that the locus of every equation of the first degree is a straight line, if we prove it as above, for it is evident that the locus of ic' = is a straight line. Theorem VI. The term in xy may always he removed from an equation of the second degree, Ax^ + Bxy -\-Cy^-^Dx+Ey-\-F=Oj by rotating the axes through an angle 6 such that (VI) tan2^ = A- C 170 ANALYTIC GEOMETRY Proof. Set and This gives (4) y1cos2<9 -\-B sin ^cos 6 + C sin2 e X = x' COS — y' sin y = x' sin -{- y' cos ^. — B sin ^ cos ^ + C cos^ ^ a;' — Z) sin $ -{- Ecosd x'^-2AsmeG0se + B(Gos^O-sm^e) + 2Csiiie^ose + Dcose ■i- EsinO Setting the coefficient of x'y' equal to zero, we have (C -A)2 sin ecosO + B (cos^ - sin^ 0) = 0, or (14, p. 20), (C - ^) sin 2 ^ + 5 cos 2 ^ = 0. B y'-\-F=0. Hence tan2<9 A -C If satisfies this relation, on substituting in (4) we obtain an equation without the term in xy. q.e.d. Corollary. In transforming an equation of the second degree hy rotating the axes the constant term is unchanged unless the new equation is multiplied or divided hy some constant. For the constant terra in (4) is the same as that of the given equation. Theorem VII. The terms of the first degree m,ay he removed from, an equation of the second degree, Ax^ H- Bxy + C//2 + Dx + Ey -\- F = 0, hy translating the axes, provided that the discriminant of the terms of the second degree, A = ^^ — 4 yl C, is not zero. Proof Set X = x' -\- h, y = y' -\- k. This gives (5) Ax''' + Bx'y' 4- Cy'^ ^-2Ah + Bk x' ~h Bh y' + Ah' + 2Ck -\-Bhk + E + Ck^ -\-Dh ■\-Ek + F 0. I TRANSFORMATION OF COORDINATES 171 Setting equal to zero the coefficients of x' and y\ we obtain (6) 2Ah-{- Bk-{- D = (), (7) Bh + 2Ck^E = 0. These equations can be solved for h and k unless (Theorem IV, p. 90) 2A_B B ~2C or B'^-^AC = 0. If the values obtained be substituted in (5), the resulting equa- tion will not contain the terms of the first degree. q.e.d. Corollary I. If an equation of the second degree he transformed by translating the axes, the coefficients of the terms of the second degree are unchanged unless the new equation be multiplied or divided by some constant. For these coefficients in (5) are the same as in the given equation. Corollary II. When A is not zero the locus of an equation of the second degree has a center of symmetry. For if the terms of the first degree be removed the locus will be symmetrical with respect to the new origin (Theorem V, p. 73) . If A= ^2_4 j[(7— 0, equations (6) and (7) may still be solved for h and k 2A B D if (Theorem IV, p. 90) -:— = — ; = — > when the new origin {h, k) may be any point on the line 2Ax-\- By -\- D = 0. In this case every point on that line will be a center of symmetry. For example, consider x2 + 4a;^/ + 4?/2-|-4a; + 8?/ + 3 = 0. For this equation equations (6) and (7) become In these equations the coefficients are all proportional and there is an infinite number of solutions. One solution is A = — 2, A: = 0. For these values the given equation reduces to cc2 + 4 xy + 4 ?/2 — 1 = 0, or (x + 2 y + 1) (x + 2 2/ — 1) = 0. The locus consists of two parallel lines and evidently is symmetrical with respect to any point on the line midway between those lines. 172 ANALYTIC GEOMETRY MISCELLANEOUS PROBLEMS '^ 1. Simplify and plot. (a) y^-6y + 6 = 0. (e) x^ + 4xy + y"^ = S. (b) x2 + 2xy + 2/2 - 6x - 6i/ + 5 = 0. (f) x^ - 9 y^ - 2 x - S6y + 4 = 0. X{c) ?/2 + 6x- 102/4-2 = 0. (g) 25 2/2 -16x2 + 502/ -119 = 0. (d) x2 + 42/2 _ 8x - \Qy = 0. (h) x2 + 2x2/ + ?/2 _ Sx = 0. 2. Find the point to which the origin must be moved to remove the terms of the first degree from an equation of the second degree (Theorem VII). 3. To what point (h, k) must we translate the axes to transform (1 - e2) x2 + 2/2 - 2px + p2 = into (1 - e2) x2 + 2/2 - 2 e'^px - e'^p^ = ? 4. Simplify the second equation in problem 3. 6. Derive from a figure the equations for rotating the axes through + — ■jt 2 and 1 and verify by substitution in (II), p. 162. 2 6. Prove that every equation of the first degree may be transformed into 2/' z= by moving the axes. In how many ways is this possible ? 7. The equation for rotating the polar axis through an angle is 6 = 6' -V . 8. The equations of transformation from rectangular to polar coordi- nates, when the pole is the point (^, k) and the polar axis makes an angle of with the X-axis, are X = h + p cos {6 + 0), y = k + psm{6 -\- (p). 9. The equations of transformation from rectangular coordinates to oblique coordinates are X = x' + 2/^ cos w, y = y' sin w, if the X-axes coincide and the angle between OX' and OY' is w. 10. The equations of transformation from one set of oblique axes to any other set with the same origin are , sin (w — 0) , sin (w — ^) X = x' — ^^ ^ + y ^^ ^» sm w sin w , sin . , sin \Li sm oj sm (a where w is the angle between OX and OF, is the angle from OX to OX', and \p is the angle from OX to OY'. CHAPTER yill CONIC SECTIONS AND EQUATIONS OF THE SECOND DEGREE 72. Equation in polar coordinates. The locus of a point P is called a conic section* if the ratio of its distances from a fixed point F and a fixed line DD is constant. F is called the focus, DD the directrix, and the constant ratio the eccentricity. The line through the focus perpendicular to the directrix is called the principal axis. Theorem I. If the pole is the focus and the polar axis the princi- pal axis of a conic section^ then the polar equation of the conic is •* (I) P = ep 1 — e cos ^ where e is the eccentricity and p is the distance from the directrix to the focus. P(P'0) Froof. Let P be any point on the conic. Then, by definition, FP EP = '- From the figure, FP = p and EP = HM = p -{- p cos 0. Substituting these values of FP and ^^ EP, we have or, solving for p, p + p cos ep e; p — cos Q.E.D. * Because these curves may be regarded as the intersections of a cone of revolution with a plane. 173 174 ANALYTIC GEOMETRY From (I) we see that 1. A conic is symmetrical with respect to the principal axis. For substituting cos (— 6) = cos d. 6 for 6 changes only the form of the equation, since 2. In plotting, no values of 6 need be excluded. The other properties to be discussed (p. 151) show that three cases must be considered according as e = 1. The parabola e = 1. When e = 1, (I) becomes P p = J ^ 1 - cos ^ and the locus is called a parabola. 1. For ^ = p = 00, and for TT p = V The parabola therefore crosses the principal axis but once at the point 0, called the vertex, which is ^ to the left of the focus F, or mid- way between F and BD. 2. p becomes infinite when the denominator, 1 — cos 0, vanishes. If 1 — cos ^ = 0, then cos = 1', and hence. ^ = is the only value less than 2 tc for which p is infinite. 77" /^.^^ 3. When increases from to — > Z then cos 6 decreases from 1 to 0, 1 — cos 6 increases from to 1, p decreases from oo to j), and the point P (p, 6) describes the parabola from infinity to B. TT When 6 increases from — to tt, then cos decreases from to — 1, 1 — cos Q increases from 1 to 2, IP p decreases from /> to ^) and the point P {p, 6) describes the parabola from B to the vertex 0, CONIC SECTIONS 175 On account of the symmetry with respect to the axis, when 6 increases from tt to —^i P(p,_0) describes the parabola from to B'-y and when 6 increases from —^ to 2 tt, from B' to infinity. When e < 1 the conic is called an ellipse, and when e > 1, an hyperbola. The points of similarity and difference in these curves are brought out by considering them simultaneously. TJie ellipse, e < 1 . ep e The hyperbola, e > 1. 1. For ^ = ^ 1. For 6^ = p = ep P- 1 — el — e 1 — el — e As e < 1, the denominator, and hence As e > 1, the denominator, and hence p, is positive, so that we obtain a point p, is negative, so that we obtain a A on the ellipse to the right of F. point A on the hyperbola to the left e •> of F. As = 1 when e ^ (numerically) when e > 1^ less than FH. ^^^^ P >P > ^^ ^ ^^^ ^^ *^® ^®^* ^^ ^' T^ n ep e For ^ = ;r p=:;-f— = :^— p. pis ^ ^ ep e YoT d = 7t p = —^— = p. pis 1+e 1+e 1+e 1+e positive, and hence we obtain a point positive, and hence we obtain a sec- A' to the left of F. ond point A' to the left of F. As <1, then p1; and hence e there are no values of 6 for which p becomes infinite. 3. When then .6 increases from to cos 6 decreases from 1 to 0, 1 — e cos 6 increases from 1 — e to 1 ; hence p decreases from to ep, 1 — e and P {p, 6) describes the ellipse from ^toC. The hyperbola^ e > 1. 2. p becomes infinite if 1 — e cos ^ = 0, or cos 6 = -• As e>l, then -<1; and hence e there are two values of 6 for which p becomes infinite. 3. When 6 increases from to cos-i ( - )» then cos 6 decreases from 1 to - , e 1 — e cos 6 increases from 1 — e to ; exi hence p decreases from to — oo, 1 — e and P (p, 6) describes the lower half of the left-hand branch from A to infinity. When e increases from cos-i (i)-f- When 6 increases from — to 7t, 2 then cos 6 decreases from to — 1, 1 — e cos d increases from 1 to 1 + e ; hence p decreases from ep to — — , 1 + e and P (p, d) describes the ellipse from C to A\ The rest of the ellipse, A'C'A, may be obtained from the symmetry with respect to the principal axis. The ellipse is a closed curve. then cos d decreases from - to 0, e 1 — e cos ^ increases from to 1 ; hence p decreases from oo to ep, and P (p, 6) describes the upper part of the right-hand branch from infinity toC. When 6 increases from — to vt, 2 then cos 6 decreases from to — 1, 1 — e cos ^ increases from 1 to 1 + e ; hence p decreases from ep to ep l + ~e and P (p, 6) describes the hyperbola from C to A\ The rest of the hyperbola, A'C to infinity and infinity to A, may be obtained from the symmetry with respect to the principal axis. The hyperbola has two infinite branches. CONIC SECTIONS 177 PROBLEMS 1. Plot and discuss tlie following conies. Find e and p, and draw the focus and directrix of each. ^ 3 (e) P (a) p = T- - cos 5 (b) p = 2 1- - i cos (? (C) p = 8 1- - 2 cos ^ IA\ ^ — 5 (g) p = (h) p = 3 - cos 6 2 - 3cos^' 2 2 - cos^" 12 2 - 2 cos ^ 3 - 4 cos ^ 2. Transform the equations in problem 1 into rectangular coordinates, simplify by the Rule on p. 165, and discuss the resulting equations. Find the coordinates of the focus and the equation of the directrix in the new variables. Plot the locus of each equation, its focus, and directrix on the new axes. Ans. (a) ?/2=:4x, (1,0), x = - 1. ." x2 y2 , / 4 ^\ 16 ^ ' 6_4 IJL ' V 3 / 3 / 3 ^ ' 6_4 6_4 ' V 3 J 3 (d) 2/2 = 5x, (1,0), x = -%. x2 ?/2 ^ / 3 -\ 27 64 8 /2 . /18 ^\ 8 5' w5-f-. (-?■»)• —I x2 ?/2 , /48 ^\ 2"; ^ ' 1296 144 ' V 7 / 7 3. Transform (I) into rectangular coordinates, simplify, and find the coor- dinates of the focus and the equation of the directrix in the new rectangular coordinates if (a) e = 1, (b) e^l. Am. (a) 2/2 = 2px, (|, o), a:=-| (b) e2p (1 - e2)2 1 _ e2 + _^.l, (-^, 0),x = ^ e2p2 'V 1 _ e2 / 178 ANALYTIC GEOMETRY 4. Derive the equation of a conic section when (a) the focus lies to the left of the directrix ; (b) the polar axis is parallel to the directrix. Ans. (a) p = ^^ ; (b) p - ^^ 1 + e cos 6 1 — e sin 5. Plot and discuss the following conies. Find e and p, and draw the directrix of each. 7 (^) P = ^;-, ^- (^) P 1 + cos e 3 + 10 cos ^ 5 {h)p = - — .. {d)p = 1 - sin ^ 3 - sin e 73. Transformation to rectangular coordinates. Theorem II. If the origin is the focus and the X-axis the princi- pal axis of a conic section, then its equation is (II) (1 - e^) 0^2 _^ ^2 _ 2 e'^px - eY = 0, where e is the eccentricity and xz=z—p is the equation of the directrix. Proof Clearing fractions in (I), p. 173, we obtain P — ep cos 6 = ep. Set p = ± Vx^ + y^ and p cos ^ = a; (p. 155). This gives ± V a;^ -{- y^ — ex = ep, or ± Vx^ + y^ = ex -\- ep. Squaring and collecting like pov^ers of x and y, we have the required equation. Since the directrix DD (Fig., p. 173) lies p units to the left of F its equation is ic ——p. q.e.d. 74. Simplification and discussion of the equation in rectangu- lar coordinates. The parabola, e = 1. When 6 = 1, (II) becomes y^ — 2px —p^ = 0. Applying the Eule on p. 165, we substitute (1) x = x' -\- h, y = y' + k, obtaining (2) y'^ - 2px' + 2ky'-{-k:'- 2ph -p^ = 0. CONIC SECTIONS 179 Set the coefficient of y^ and the constant term equal to zero and solve for h and h. This gives (3) 7,=-|, 7, = 0. Substituting these values in (2) and dropping primes, the equa- tion of the parabola becomes y^ = 2jjx. From (3) we see that the origin has been removed from F to O, the vertex of the ^ parabola. It is easily seen that the new coordinates of the focus are [ ^ , 1 , and x the new equation of the directrix is X = — ^' Hence Li Theorem III. Jf the origin is the vertex and the X-axis the axis of a parabola, then its equation is (III) y^ = 2poe. The focus is the point ( — , ), and the equation of the directrix V \ I IS X = — —• A general discussion of (III) gives us the following properties of the parabola in addition to those already obtained (p. 174). 1. It passes through the origin but does not cut the axes elsewhere. 2. Values of x having the sign opposite to that X of p are to be excluded (Rule, p. 73). Hence the curve lies to the right of YY' when p is positive and to the left when p is negative. 3. No values of y are to be excluded ; hence the curve extends indefinitely up and down. Theorem IV. If the origin is the vertex and the Y-axis the axis of a parabola, then its equation is (IV) a?2 = 2py. 180 ANALYTIC GEOMETRY V 7 \ / \ ^ V , X' V-^ X D Y' D The focus is the point ( 0, "^ I j and the equation of the directrix . V Proof Transform (HI) by rotating the 77" axes through — — • Equations (II), p. 162, give us for B =~ — x = y\ y=-x'. Substituting in (III) and dropping primes, we obtain x^ = 2py. Q.E.D. After rotating the axes the whole figure is turned through — in the positive direction. The parabola lies above or below the X-axis according as p is positive or negative. Equations (III) and (IV) are called the typical forms of the equation of the parabola. Equations of the forms Ax^ + Ey = and Cy^ + Dx = 0, where A, E, C, and D are different from zero, may, by transpo- sition and division, be written in one ^ of the typical forms (III) or (IV), so that in each case the locus is a ^ parabola. Ex. 1. Plot the locus of x^ + 4 ?/ = and find the focus and directrix. Solution. The given equation may be written x2 = - 4 y. Y' 'y-f D ^ P X' y^o ^\ X ya ^ D y^i X' /^ "^ N / ^(C ,-i; \ / \ 1 \ i i 1 \ Y' Comparing with (IV), the locus is seen to be a parabola for which p = —2. Its focus is therefore the point (0, — 1) and its directrix the line y = 1. Ex. 2. Find the equation of the parabola whose vertex is the point 0' (3, — 2) and whose directrix is parallel to the F-axis, if p = 3. CONIC SECTIONS 181 Solution. Referred to O'X' and O'Y' as axes, the equation of the parabola is (Theorem III) (4) y'^ = ex\ The equation for translating the axes from to (7 are (Theorem I, p. IGO) X = x' + 3, y = y' -2, whence (5) x' = x-^,y' = y + 2. Substituting in (4), we obtain as the re- quired equation (2/ + 2)2 = 6(^-3), or 2/2_6a.^4y ^22 = 0. Referred to O'X' and 0'Y\ the coordinates of T are (Theorem III) (|, 0) and the equa- tion of DB is x' = — |. By (5) we see that, referred to OX and OF, the coordinates of F are (|, - 2) and the equation of DD is X = ir nf rV > ^ y / / 5: F .O' b,- 2) .Y . s. V s ^ Z.I PROBLEMS 1 . Plot the locus of the following equations. Draw the focus and direc- trix in each case. (a) y^ = 4:X. (b) 2/2 + 4x {6) y^ -6x = 0. 0. (e) x2 + 10 y = 0. (c) x2-82/ = 0. (f) 2/2 + x = 0. 2. If the directrix is parallel to the F-axis, find the equation of the parabola for which (a) p = 6, if the vertex is (3, 4). (b) p = — 4, if the vertex is (2, — 3). (c) p = 8, if the vertex is (—5, 7). (d) p '= 4, if the vertex is (/i, k). 3. The chord through the focus perpendicular to the axis is called the latus rectum. Find the length of the latus rectum of y^ = 2px. Ans. 2 p. 4. What is the equation of the parabola whose axis is parallel to the axis of y and whose vertex is the point (or, )3) ? Ans. (x — a)^ = 2p{y — ^). 5 . Transform to polar coordinates and discuss the resulting equations (a) 2/2 = 2px, (b) x2 = 2py. 6. Prove that the abscissas of two points-on the parabola (III) are propor- tional to the squares of the ordinates of those points. Ans. (?/ - 4)2 = 12 (X - 3). Ans. (2/ + 3)2=-8(x-2). Ans. {y - 7)2 = 16 (x + 5). Ans. {y -kf = 8{x- h). 182 ANALYTIC GEOMETRY 75. Simplification and discussion of the equation in rectan- gular coordinates. Central conies, e%l. When e^l, equation (II), p. 178, is (1 - e2)a;2 + 2/2-2 e''2)x - eY = 0. To simplify (Rule, p. 165), set (1) x = x' -{-h, y = y' + k, which gives (2) (1 - e^)x''' + 2/'' -f 2 h (1 - e^) = 0. ic' + 2 7^2/' + (1-6^)^2 - 2 e'^ph — e^ Setting the coefficients of cc' and y^ equal to zero gives 2 hi\ -e^)-2e^p = 0, 2k = 0, '^ - whence (3) _ * = r^' * = o- Substituting in (2) and dropping primes, we obtain or (4) — r^ + -fr = i- (1 - 6^)2 1-^2 This is obtained by transposing the constant term, dividing by it, and then dividing numerator and denominator of the first fraction by 1 — e^. The ellipse, e < 1. The hyperbola, e > 1. From (3) it is seen that h is posi- From (3) it is seen that h is nega- tive when e < 1. Hence the new ori- tive when e > 1. Hence the new ori- gin lies to the right of the focus F. gin lies to the left of the focus F. Further, > 1 numerically, so ' 1 - e2 h>p numerically ; and hence the new origin lies to the left of the directrix DD. CONIC SECTIONS 183 The locus of (4) is symmetrical with respect to YY' (Theorem V, p. 73). Hence is the middle point oi AA'. Construct in D Y d' E -A B^ ^t e' ^?Sv 1 \ \ \ X' A F F' y X D V. Y B^ y D' either figure F' and D'D' symmetrical respectively to F and DB with respect to YY\ Then F' and X)'i)' are a new focus and ^ directrix. For let P and P' be two points on the curve, symmetrical with respect to YY'. Then from the symmetry PF — P'F' and PE — P'E'. But since, by definition, PF P'F' — — = e, then = e. Hence the same conic is traced by P', using F' as focus PMi P Mi and D'D' |is directrix, as is traced by P, using F as focus and DD as directrix. Since the locus of (4) is symmetrical with respect to the origin (Theorem V, p. 73), it is called a central conic, and the center of symmetry is called the center. Hence a central conic has two foci and two directrices. The coordinates of the focus F in either figure are (-A») For the old coordinates of F were (0, 0). Substituting in (1), the new coordi- nates are x = — h, y' = — k, or, from (3), / — - — ^. V i The coordinates of F' are therefore The new equation of the directrix DD is x - _^ 184 ANALYTIC GEOMETRY For from (1) and (3), x = x' + e^ y = y l-e2 (Theorem II) and dropping primes, we obtain x = P Substituting in x = —p P Hence the equation oi D'D' is x = j We thus have the Lemma. The equation of a central conic whose center is the origin and whose principal axis is the X-axis is (4) _^!_ + ^L = l. e^p^ Its foci are the points I ± e^p e^p^ 0^ 1-e' and its directrices are the lines x = :t IP The ellipse, e < 1. For convenience set (5) a l-e2 1 e^p a2 and 6^ are the denominators in (4) and c is the abscissa of one focus. Since -.e 1. Eor convenience set (6) a = ep l-e2' 62: e^p a^ and - b^ are the denominators in (4) and c is the abscissa of one focus. Since e > 1, 1- e2 is negative; and hence a, b^, and c are positive. We have at once a2 + 62 e2p2 (1 - e2)2 e4p2 (1 - e2)2 e2p2 . e2p 2ri2 a^p l-e2 c2 and a2__ c ~ (1 - e2)2 l-e2 Hence the directrices (Lemma) are the lines x = ± —• c By substitution from (6) in (4) we obtain X2_y2^_^ a2 62 CONIC SECTIONS 185 The ellipse^ e < 1. The intercepts are x = ±a and y = ±h. AA' = 2 a is called the major axis and BB' = 26 the minor axis. Since a^ — h^ = c^ is positive, then a>b, and the major axis is greater than the minor axis. ^1 ^ D / /" h B D 1 V <—a-\ — * X' A F ^.-C—^F'JA X D \ Y f____«^____. d' Hence we may restate the Lemma as follows. Theorem V. The equation of an ellipse whose center is the origin and whose foci are on the X-axis is (V) a' where 2 a is the major axis and 2 b the minor axis. If c'^ = a^ — b^, then the foci are (± c, 0) and the directrices are x = ± —- c Equations (5) also enable us to express e and p, the constants of (I), p. 173, in terms of a, 6, and c, the constants of (V). For c _ e^p ev _ a~l-e2"l-e2~ (7) and (9) 62 c l-e2 e^p P The hyperbola, e > 1. The intercepts are x = ± a, but the hyperbola does not cut the F-axis. AA' = 2 a is called the transverse axis and BB' = 26 the conjugate axis. Hence we may restate the Lemma as follows. Theorem VI. The equation of an hyperbola whose center is the origiru and whose foci are on the X-axis is x^ r a^ o2 where 2 a is the transverse axis and 2 b the conjugate axis. If c^ = a^-\- b^, then the foci are (± c, 0) and the directrices are x = i — • c Equations (6) also enable us to express e and p, the constants of (I), p. 173, in terms of a, 6, and c, the constants of (VI). For (8) 1^ '"P . ^ -- and l-e2 e2p2 (10) — = ^ ^ c l-e2 l-e2 e^p l-e2 = 1>. 186 ANALYTIC GEOMETRY The ellipse, e < 1. In the figure OB = b, OF" = c, and since c^ = a'^ — 6^, then BF' = a. Hence to draw the foci, with JB as a center and radius OA, describe arcs cutting XX' at F and F'. Then F and F' are the foci. It a = b, then (V) becomes x2 + 2/2 = a2, whose locus is a circle. Transform (V) by rotating the axes through an angle of (Theo- 2 rem II, p. 162). We obtain Theorem VII. The equation of an ellipse whose center is the origin and whose foci are on the Y-axis is «2_^^2 r2 The hyperbola, e>l. In the figure OB = 6, OA' = a ; and since c2 = a2 + 62^ then BA' = c. Hence to draw the foci, with as a center and radius BA', describe arcs cutting XX' at F and F'. Then F and F' are the foci. If a = 6, then (VI) becomes x2-y2 = a2, whose locus is called an equilateral hyperbola. Transform (VI) by rotating the axes through an angle of (Theo- rem II, p. 162). We obtain Theorem Vni. The equation of an hyperbola whose center is the origin and whose foci are on the Y-axis is (VII) (VIII) ^ — JL-IL = 1, Y \ B' A B'\ i f / / a/ y / F' c A2 X' n 10 F \B' £ D D Y where 2 a is the major axis and 2 bis the minor axis. If d^ = a^ — 62, the foci are (0, ± c) and the directrices a2 are the lines y =± —• where 2a is the transverse axis and 2h is the conjugate axis. If c'^ = a^ + b^, the foci are (0, ± c) and the directrices are the lines y = ±^. CONIC SECTIONS 187 The ellipse^ e < 1. The essential difference between (V) and (VII) is that in (V) the de- nominator of x2 is larger than that of 2/^, while in (VII) the denominator of y^ is the larger. (V) and (VII) are called the typical forms of the equation of an ellipse. The hyperbola^ e > 1. The essential difference between (VI) and (VIII) is that the coeffi- cient of ?/2 is negative in (VI), while in (VIII) the coefficient of x^ is nega- tive. (VI) and (VIII) are called the typical forms of the equation of an hyperbola. An equation of the form where A, C, and F are all different from zero, may always be written in the form (11) a /3 By transposing the constant term and then dividing by it, and dividing numerator and denominator of the resulting fractions by A and C respectively. The locus of this equation will be 1. An ellipse if a and (3 are both positive, a^ will be equal to the larger denominator and b^ to the smaller. 2. An hyperbola if a and ^ have opposite signs, a^ will be equal to the positive denominator and b^ to the negative denomi- nator. 3. If a and p are both negative, (11) will have no locus. Ex. 1. Find the axes, foci, directrices, and eccentricity of the ellipse 4 x^ + y2 = 16. Solution. Dividing by 16, we obtain - + ^ = 1 4 16 The second denominator is the larger. By comparison with (VII), 62 = 4, a2 = 16, c2 = 16 - 4 = 12. Hence 6 = 2, a = 4, c = Vl2. The positive sign only is used when we extract the square root, becaus.e a, 6,'and c are essentially positive. r> , JJ D __/■ ^ > V / \ / \ 7? B' X \ / \ / \ "'■i / ■a r\ T\ T' /s^ ^ y 188 ANALYTIC GEOMETRY Hence the major axis^^^' = 8, the minor axis BB' = 4, the foci F and F' are the points (0, ± Vl2), and the equations of the directrices DD and D'ly a2 16 ^4 are y = ± — = ± —= = ± - Vl2. c Vl2 3 Vl2 4 1 From (7) and (9), e = — — and p = -— =: = - V12. 4 V12 3 PROBLEMS 1. Plot the loci, directrices, and foci of the following equations and find e and p. (a) x2 + 9?/2 = 81. (e) dy^ - 4x2 = 36. (b) 9 x2 _ 16 y2 = 144. (f ) x2 - 2/2 = 25. (c) 16 x2 + 2/2 = 25. (g) 4 x2 + 7 2/2 = 13. (d) 4x2 + 92/2 = 36. (h) 5x2 _ 32/2 = 14. 2. Find the equation of the ellipse whose center is the origin and whose foci are on the JT-axis if (a) a = 5, 6 = 3. Ans. 9 x2 + 25 2/2 = 225. (b)a'=6, e=:i. Ans. 32x2 + 362/2 = 1152. (c) 6 = 4, c = 3. Ans. 16 x2 + 25 2/2 = 400. (d) c = 8, e = f. Ans. 5x2 + 9 2/2 = 720, 3. Find the equation of the hyperbola whose center is the origin and whose foci are on the X-axis if (a) a = 3, h = 5. Ans. 2bx^-9y^ = 225. (b) a = 4, c = 5. Ans. 9x^-16y^ = 144. (c) e = f, a = 5. Ans. 5x2-42/2 = 125. (d) c = 8, e = 4. Ans. 15x2-2/2 = 60. 4. Show that the latus rectum (chord through the focus perpendicular to 2 62 the principal axis) of the ellipse and hyperbola is a 5. What is the eccentricity of an equilateral hyperbola ? A71S. V2. 6. Transform (V) and (VI) to polar coordinates and discuss the resulting equations. 7. Where are the foci and directrices of the circle ? 8. What are the equations of the ellipse and hyperbola whose centers are the point (a, /3) and whose principal axes are parallel to the X-axis ? Ans ('^-^)^ , (y-^)^ ^i. i^-o^)^ (y-/3)2 ^ a2 62 ' a2 62 * CONIC SECTIONS 189 76. Conjugate hyperbolas and asymptotes. Two hyperbolas are called conjugate hyperbolas if the transverse and conjugate axes of one are respectively the conjugate and transverse axes of the other. They will have the same center and their principal axes (p. 173) will be perpendicular. If the equation of an hyperbola is given in typical form, then the equatlofi of the conjugate hyperbola is found by changing the signs of the coefficients of x^ and y'^ in the given equation. For if one equation be written in the form (VI) and the other in the form (VIII), then the positive denominator of either is numerically the same as the negative denominator of the other. Hence the transverse axis of either is the conjugate axis of the other. Thus the loci of the equations (1) 16x2-2/2^16 and -16x2 + 2/2=16 are conjugate hyperbolas. They may be written «2 2/2 a;2 2/2 ___^la„d-- + - = l. The foci of the first are on the X-axis, those of the second on the T-axis. The transverse axis of the first and the conjugate axis of the second are equal to 2, while the conjugate axis of the first and the transverse axis of the second are equal to 8. The foci of two conjugate hyperbolas are equally distant from the origin. For c2 (Theorems VI and VIII) equals the sum of the squares of the semi- transverse and semi-conjugate axes, and that sum is the same for two conjugate hyperbolas. Thus in the first of the hyperbolas above c^ = 1 -f 16, while in the second c2=16 + l. If in one of the typical forms of the equation of an hyperbola we replace the constant term by zero, then the locus of the new equation is a pair of lines (Theorem, p. Q>^) which are called the asymptotes of the hyperbola. Thus the asymptotes of the hyperbola (2) b'^x'- - aSf = a%'' are the lines (3) ^2^2 _ ^2^2 ^ 0, or (4) bx -\- ay = and bx — ay = 0. 190 ANALYTIC GEOMETRY Both of these lines pass through the origin, and their slopes are respectively ,5) -^nd^. An important property of the asymptotes is given by Theorem IX. The branches of the hyperbola approach its asymp- totes as they recede to infinity. Proof. Let P^ (x-^, y\) be a point on either branch of (2) near the first of the asymptotes (4). The distance from this line to Pi (Fig., p. 191) is (Eule, p. 106) (6) d= ^-^±^y^ . Since Pi lies on (2), bW - «Vi' = «^'^'- Factoring, bx^ + a?/i = ■_ — "^ oxi — ayi a%^ Substituting in (6), d = . - 4- VZ>2 + a^ (bxi — ayi) As Pi recedes to infinity, a^i and ?/i become infinite and d approaches zero. For bxi and ayi cannot cancel, since Xi and pi have opposite signs in the second and fourth quadrants. Hence the curve approaches closer and closer to its asymptotes. Q.E.D. Two conjugate hyperbolas have the same asymptotes. For if we replace the constant term in both equations by zero, the resulting equations differ only in form and hence have the same loci. Thus the asymptotes of the conjugate hyperbolas (1) are respectively the loci of 16x2 - 2/2 = and -lGx^ + y'^ = 0, which are the same. An hyperbola may be drawn with fair accuracy by the fol- lowing Construction. Lay o& OA = OA' = a on the axis on which the foci lie, and OB = OB' = b on the other axis. Draw lines through A, A', B, B' parallel to the axes, forming a rectangle.^* Draw the * An ellipse may be drawn with fair accuracy by inscribing it in such a rectangle. CONIC SECTIONS 191 diagonals of the rectangle and the circumscribed circle. Draw the branches of the hyperbola tangent to the sides of the rec- tangle at A and A' and approaching nearer and nearer to the di- agonals. The conju- gate hyperbola may be drawn tangent to the sides of the rec- tangle at B and B' and approaching the diagonals. The foci of both are the points in which the circle cuts the axes. The diagonals will be the asymptotes, because two of the vertices of the rec- tangle ( ± a, ± 6) will lie on each asymptote (4) . Half the diagonal will equal c, ^ the distance from the origin to the foci, because c^= a^ -\- b^. 77. The equilateral hyperbola referred to its asymptotes. The equation of the equilateral hyperbola (p. 186) is (1) x2 - 2/2 = ot2. - Its asymptotes are the lines X — y — and ic + y = 0. These lines are perpendicular (Corollary III, p. 87), and hence they may be used as coordinate axes. Theorem X. The equation <^f an equilateral hyperbola referred to its asymp- totes is (X) 2ocy = a^. It Proof. The axes must be rotated through to coincide with the asymptotes. Hence we substitute (Theorem II, p. 162) x' -\-y' -x' + y' V2 V2 in (1). This gives Or, reducing and dropping primes, 1xy = a2. Q.E.IX 192 ANALYTIC GEOMETRY 78. Focal property of central conies. A line joining a point on a conic to a focus is called a focal radius. Two focal radii, one to each focus, may evidently be drawn from any point on a central conic. - Theorem XI, The sum of the focal radii from any point on an ellipse is equal to the major axis 2 a. Theorem Xn. The difference of the focal radii from any point on an hyperbola is equal to the transverse axis 2 a. T> Y' d' E II ^ r"^^ e' < f^ ^ \ X' A F jr' 1^ X D V b' r' r| D J E ly h' H 1 \r X' F' JA / ^ y' \i X D \ Proof. Let P be any point on the ellipse. By definition (p. 173), r = e- PE, r' = e . PE\ Hence r + r' = e {PE + PE") = e • HH\ From (7), p. 185, e= -» Proof. Let P be any point on the hyperbola. By definition (p. 173), r = e- PE, r' = e- PE'. Hence r' - r = e {PE' - PE) = e • HH\ From (8), p. 185, e= -, and from the equations of the direc- and from the equations of the direc- trices (Theorem V), trices (Theorem VI), «2 c a2 HH' = 2 — . c Hence r+r' = -.2- = 2a. « ' Q.E.D. Hence»-'-r = --2- = 2a. « " Q.K.n. 79. Mechanical construction of conies. Theorems XI and XII afford simple methods of drawing ellipses and hyperbolas. Place two tacks in the drawing board at the foci F and F' and wind a string about them as indicated. If the string be held fast at A, and a pencil be placed in the loop FPF' and be moved so as to keep the string taut, then PF + PF' is constant and P describes an ellipse. If the major axis is to be 2 a, then the length of the loop FPF' must be 2 a. CONIC SECTIONS 193 If the pencil be tied to the string at P, and both strings be pulled in or let out at A at the same time, then PF" — PF will be constant and P will describe an hyperbola. If the transverse axis is to be 2 a, the strings must be adjusted at the start so that the difference between PF' and PF equals 2 a. To describe a parabola, place a right triangle with one leg EB on the directrix DD. Fasten one end of a string whose length is AE at the focus F, and the other end to the triangle at A. With a pencil at P keep the string taut. Then PF = PE ; and as the triangle is moved along Di> the point P will describe a parabola. PROBLEMS 1. Find the equations of the asymptotes and hyperbolas conjugate to the following hyperbolas, and plot. (a) 4ic2 - 2/2 = 36. (c) 16x2 _ y2 4. 64 = 0. (b) 9x2 - 25y2 = loO. (d) 8x2 - 16y2 + 25 = 0. 2. Prove Theorem IX for the asymptote which passes through the first and third quadrants. 3. If e and e' are the eccentricities of two conjugate hyperbolas, then 1 1 . 4. The distance from an asymptote of an hyperbola to its foci is numer- ically equal to 6. 5. The distance from a line through a focus of an hyperbola, perpen- dicular to an asymptote, to the center is numerically equal to a. 6. The product of the distances from the asymptotes to any point on the hyperbola is constant. 7. The focal radius of a point Pi(xi, y{) on the parabola 2/2 = 2px is 194 ANALYTIC GEOMETRY 8. The focal radii of a point Pi (xi, yi) on the ellipse b^x^ + a^y^ = a^b^ are r = a — exi and r' = a -\- exi. 9. The focal radii of a point on the hyperbola b^x^ — a^y^ = a'^b^ are r = exi — a and r' = exi + a when Pi is on the right-hand branch, or r = — exi — a and r' = — exi + a when Pi is on the left-hand branch. 10. The distance from a point on an equilateral hyperbola to the center is a mean proportional between the focal radii of the point. 11. The eccentricity of an hyperbola equals the secant of the inclination of one asymptote. 80. Types of loci of equations of the second degree. All of the equations of the conic sections that we have considered are of the second degree. If the axes be moved in any manner, the equation will still be of the second degree (Theorem IV, p. 164), although its form may be altered considerably. We have now to consider the different possible forms of loci of equations of the second degree. By Theorem VI, p. 169, the term in xy may be removed by rotating the axes. Hence we only need to consider an equation of the form (1) Ax"" -\-Cif-\-Dx + Ey + F = 0. It is necessary to distinguish two cases. Case I. Neither A nor C is zero. Case II. Either ^ or C is zero. A and C cannot both be zero, as then (1) would not be of the second degree. Case I When neither A nor C is zero, then ^ = B^ — 4, AC is not zero, and hence (Theorem VII, p. 170) we can remove the terms in X and y by translating the axes. Then (1) becomes (Corollary I, p. 171) (2) Ax"" -{- Cy'^ -{- F' = 0. We distinguish two types of loci according as A and C have the same or different signs. CONIC SECTIONS 195 Elliptic type, A and C have the same sign. 1. F' 7!^0* Then (2) may be iC2 y2 written 1 = 1, a /3 F' F' where a = » /3 = A C Hence, if the sign of F' is different from that of A and C, the locus is an ellipse; but if the sign of F' is the same as that of A and C, there is no locus. 2. F' = 0. The locus is a point. It may be regarded as an ellipse whose axes are zero and it is called a degenerate ellipse. Hyperbolic type, A and C have dif- ferent signs. 1. F' ^0.* Then (2) may be written h — = 1, a p F' F' where a = , 8 = A^ C Hence the locus is an hyperbola whose foci are on the F-axis if the signs of F' and A are the same, or on the X-axis if the signs of F' and C are the same. 2. F' = 0. The locus is a pair of intersecting lines. It may be regarded as an hyperbola whose axes are zero and it is called a degenerate hyperbola. Case II When either A or C is zero the locus is said to belong to the parabolic type. We can always suppose ^1 = and C 9^ 0, so that (1) becomes ^ - (3) Ci/ + Dx-\- Ey + F=0. For if ^ 5^ and C = 0, (1) becomes Ax"^ + Dx + Ey -\- F=(i. Rotate the axes It (Theorem II, p. 162) through — by setting x = —y',y~ x'. This equation becomes Ay'^ + Ex' — Dtf -\- F=Q), which is of the form (3). By translating the axes (3) may be reduced to one of the forms (4) Cy^-\-Dx = ox (5) . Cif + F' =0. For substitute in (3), x = x' + h, y = y' -\- k. This gives (6) Cy'^^-Dx' -^2Ck\y' ^-Ck'^ =0. -\- E I -\-Dh + Ek + F If we determine h and k from 2Ck + E=0, Ck^ + Dh -\- Ek + F = 0, [then (6) reduces to (4). But if Z> =: 0, we cannot solve the last equation for h, so [that we cannot always remove the constant terra. In this case (6) reduces to (5). * Read " F' not equal to zero " or "F^ diiferent from zero." 196 ANALYTIC GEOMETRY Comparing (4) with (III), p. 179, the locus is seen to be ^.parab- I ¥' ola. The locus of (5) is the pair of parallel lines y = ±^ — — when F' and C have different signs, or the single line y = when F' = 0. li F' and C have the same sign, there is no locus. When the locus of an equation of the second degree is a pair of parallel lines or a single line it is called a degenerate parabola. We have thus proved Theorem XIII. The locus of an equation of the second degree is a conic, a point, or a pair of straight lines, ivhich may he coincident. By moving the axes its equation m^ay he reduced to one of the three forms .^ Ax" + Cif + F* = 0, Cy'^ -^ Dx = 0, Cy^ + F* = 0, '* where A , C, and D are different from zero. Corollary. TJie locus of an equation in which the term in xy is lacking, ^^^2 _^ ^f + Dx + Ey + F = 0, will helong to the parabolic type if A = or C = 0, the elliptic type if A and C have the same sign, the hyperbolic type if A and C have different signs. PROBLEMS 1 . To what point is the origin moved to transform (1) into (2) ? - ( - — -— ^ V 2A' 20/ Ans. . V 2A 2C. 2. To what point is the origin moved to transform (3) into (4) ? into (5) ? 3. Simplify Ax^ + Dx-\-Ey + F=Ohy translating the axes (a) if ^ 7^ 0, (b) it E = 0, and find the point to which the origin is moved. (b) Jx2 + F' = 0, (-^'O) 2A * In describing the final form of the equation it is unnecessary to indicate by primes what terms are different from those in (1). CONIC SECTIONS 197 4. To what types do the loci of the following equations belong? (a) 4:X^ + y^ - ISx -{- 7 y - 1 = 0. (e) x^ + 1 y^ - 8x + 1 = 0. (b) 2/2 + 3» - 4y + 9 = 0. (f) x^ + y2-6x + 8y = 0. (c) 121a;2-44y2 + 68x-4 = 0. (g) 3x2-42/2 -62/ + 9 = 0. (d) x2 + 4?/-3 = 0. (h) x2-8x + 9?/ -11 = 0. (i) The equations in problem 1, p. 172, which do not contain the xy-term. 81. Construction of the locus of an equation of the second degree. To remove the xi/-teim. from (1) Ax^ + Bxi/ -\- Cy-^Dx-\- Ey-\- F=0 it is necessary to rotate the axes through an angle 6 such that (Theorem VI, p. 169) (2) tan2^ = j4^, while in the formulas for rotating the axes [(II), p. 162] we need sin 6 and cos $. By 1 and 3, p. 19, we have (3) cos 2 ^ = ± Vl -f- tan^ 2 From (2) we can choose 2 ^ in the first or second quadrant so the siffn in (3) must be the same as in (2). will then be acute; and from 15, p. 20, we have /I -cos 2^ ^ , ^/l + cos2^ (4) sm ^ = + \j , cos 6> = + V In simplifying a numerical equation of the form (1) the com- putation is simplified, if A = B^ — 4:AC=^0, by first removing the terms in x and ?/ (Theorem VII, p. 170) and then the a^y-term. Hence we have the Rule to construct the locus of a numerical equation of the second degree. First step. Compute A = B^ — 4^C. Second step. Simplify the equation by (a) translating and then rotating the axes if ^^ 0', (b) rotating and then translating the axes if A = 0. 198 ANALYTIC GEOMETRY Third step. Determine the nature of the locus hy inspection of the equation (§ 80, p. 194). Fourth step. Plot all of the axes used and the locu^. In the second step the equations for rotating the axes are •found from equations (2), (3), (4), and (11)^ p. 162. But if the xy-tQim. is lacking, it is not necessary to rotate the axes. The equations for translating the axes are found by the Eule on p. 165. Ex. 1. Construct and discuss the locus of x2 + 4 x?/ + 4 y2 4. 12 X - 6 2/ = 0. Solution. First step. Here A = 42-4-l-4 = 0. Second step. Hence we rotate the axes through an angle d such that, '''''^' ■ 4 4,, tan2^ = — = \? 1-4 3 Then by (3), cos2^ = -|, 2 1 and by (4), sin^ = — — and cos^ — V5 Vs (1) The equations for rotating the axes [{II), p. 162] become x' — 2y' Ix' -\-y' ^ = rr-, y = V5 V5 Substituting in the given equation,* we obtain a;'2 -2/' = 0. V5 It is not necessary to translate the axes. Third step. This equation may be written x"^ = -^y'. 3 Hence the locus is a parabola for which p = — z:, and whose focus is on the F'-axis. "^^ * When A = the terms of the second degree form a perfect square. The work of substitution is simplified if the given equation is first written in the form (x + 2y)^ + 12x-(iy = 0. It will be shown in Chapter XII that when A = the locus is always of the parabolic type. CONIC SECTIONS 199 Fourth step. The figure shows both sets of axes,* the parabola, its focus and directrix. In the new coordinates the focus is the point ( 0, — ^ ) and the direc- ^ 2V5^ Q trix is the line y' = (Theorem 2V5 The old coordinates of :^ f^w^ V _ N ^. . Jt !v. V 2t XJt ' ""^^^ 7~t - ^^U "tt^^ -' t ^^. IV, p. 179). the focus may be found by substi- tuting the new coordinates for x' and y' in (1), and the equation of the directrix in the old coordinates may be found by solving (1) for y' and substituting in the equation given above. Ex. 2. Construct the locus of 5 a;2 + 6 xy + 5 2/2 + 22 (c - 6 y + 21 = 0. Solution. First step. A = 6^-4'6-S7^0. Second step. Hence we translate the axes first. It is found that the equa- tions for translating the axes are X = x' - i, y = y' -\- S, and that the transformed equation is 5x'^ + 6 xY + 5 /2 ^ 32. From (2) it is seen that the axes must be rotated through — . Hence we set Y N 1 Y Y V^ \l. / j \ ^ s, / 1 \ >< V ^ \, \ ^ / > >< \ z \ ^ \ ' — y z — \ •' y = X" + y" V2 V2 and the final equation is 4.x'"^ + y'"^ = \Q. Third step. The simplified equa- tion may be written 16 1. Hence the locus is an ellipse whose major axis is 8, whose minor axis is 4, and whose foci are on the F''-axis. Fourth step. The figure shows the three sets of axes and the ellipse. *The inclination of OX' is 0, and hence its slope, tan 6, may be obtained from (4). 2 . 1 method given in the footnote, p. 35. ^1 ■ • , , „ sin this example tan 6 = In = 2, and the X'-axis may be constructed by the 200 ANALYTIC GEOMETRY PROBLEMS 1. Simplify the following equations and construct their loci, foci, and directrices. (a) 3x2 -4x2/ + 8x-l=0. Ans. x'"^ - 4 y'"^ + 1 = 0. (b) 4x2 + 4 x?/ + 2/2 + 8 X - 16 ?/ = 0. Anz. 5 x'2 - 8 V5 y' = 0. (c) 41 x2 - 24 X2/ + 34 ?/ + 25 =: 0. Ans. x'2 + 2 ?/'2 _^ 1 ^ 0. (d) 17x2 -12x?/ + 8?/2-68x + 24?/ -12 = 0. An&. x''2 + 4 2/"2 _ 16 = 0. (e) 2/2 + 6 X - 6 y + 21 = 0. Ans. y"^ + 6 x' = 0. (f ) x2 - 6 X2/ + 9 2/2 + 4 X - 12 ?/ + 4 = 0. Ans. y'"^ = 0. (g) 12xy - 52/2 + 48?/ - 36 = 0. Ans. 4x''2 _ g^/'^a = 3(3. (h) 4x2 _l2xy + 92/2 + 2x- 3 2/ -12 = 0. Ans. 52 2/''2 - 49 = 0. (i) 14x2 - 4x2/ + 112/2 -88x + 34?/ + 149 = 0. Ans. 2x''^ + Sy''^ = 0. (j) 12x2 + 8x2/ + 18 2/2 + 48 x + 16 2/ + 43 = 0. ^ns. 4x2 + 22/2 = 1. (k) 9x2 + 24x2/ + 162/2 -36x- 482/ + 61 = 0. ^ns. x"2 + 1 = 0. (1) 7 x2 + 50 X2/ + 7 2/2 = 50. Ans. 16 x'2 - 9 2/^2 = 25. (m) x2 + 3X2/ - 32/2 + 6x + 92/ + 9 = 0. ^ns. 3x''2 - 7 2/^2 ^ q. (n) 16x2 - 24x2/ + O?/^ - 60x - 80 2/ + 400 = 0. Ans. y'"^ - 4 x'' = 0. (o) 95x2 + 56a;y _ IO2/2 - 56x + 2O2/ + 194 = 0. ^MS. Qx"'^-y"-'^^Vl = Q. (p) 5 x2 - 5 xy - 7 2/2 - 165 x + 1320 = 0. Ans. 15 x''2 _ \\ y"% _ 330 = 0. 82. Systems of conies. The purpose of this section is to illustrate by examples and problems the relations between conies and degenerate conies and between conies of different types. A system of conies of the same type shows how the degenerate conies appear as limiting forms, while a system of conies of dif- ferent types shows that the parabolic type is intermediate between the elliptic and hyperbolic types. Ex. 1. Discuss the system of conies represented by x2 + 4 2/2 = k. Solution. Since the coefficients of x2 and y^ have the same sign, the locus belongs to the elliptic type (Corollary, p. 196). When k is positive the locus is an ellipse ; when fc = the locus is the origin, — a degenerate ellipse / and when k is negative there is no locus. CONIC SECTIONS 201 In the figure the locus is plotted for k = 100, 64, 36, 16, 4, 1, 0. It is seen that as k approaches zero the ellipses become smaller and finally degenerate into a point. As soon as k becomes negative there is no locus. Hence the point is a limiting case between the cases when the locus is an ellipse and when there is no locus. Ex. 2. Discuss the system of conies represented by 4 x^ — 16 y^ = k. Solution. Since the coefficients of x^ and y^ have opposite signs, the locus -3 L^ k ^ ■^ — - Y_ — — _ — — — — ^ ^ ^ V ^ N ^ s ^ *«», ^ •^ ^ ^ ^ ^ :> *^; s ^ 1 ^ ^ -*- k- -21 6^ $ ^ ^ ^* \ S ^ ^ frs ^ / »"^ N N ^ ^ >s ..^ k= -6 ^^ yfi ^ / A \ \ N v^ ^ >s 1:/-^ ^ ^^\/ ^^ / V \ \ \ v "S ^ \^^ j/\ (^ ^ / f A" J / / x* ^ ^ k V I / / / X ^ >' :^ «> S \ \ / y / < ^ ^ •^ —J <^ :^ s N ^^ / / > -ii ^ ^ ^ ^ =^^ \ ^ .^ ^ ^ ^ ^ r •^ ^ ^ % ;5: s, y' i t- ^ z^ ^ -^ v; ^ ^ s, .^ ^ --' **^ V ^ ^ H r' 1 S*J belongs to the hyperbolic type. The hyperbolas will all have the same asymptotes (p. 189), namely, the lines cc ± 2 y = 0. The given equation may be written k k 4 16 The locus is an hyperbola whose foci are on the X-axis when k is positive and 202 ANALYTIC GEOMETRY on the F-axis when k is negative. For A: = the given equation shovs^s that the locus is the pair of asymptotes. In the figure the locus is plotted for k = 256, 144, 64, 16, 0, - 64, - 256. It is seen that as k approaches zero, whether it is positive or negative, the hyperbolas become more pointed and lie closer to the asymptotes and finally degenerate into the asymptotes. Hence a pair of intersecting lines is a lim- iting case between the cases when the hyperbolas have their foci on the X-axis and on the Y-axis. Ex. 3. Discuss the system of conies represented hj y'^ = 2kx ■\- 16. Solution. As only one term of the second degree is present, the locus belongs to the parabolic type (Corollary, p. 196). The given equation may be simplified (Rule, p. 165) by translating the axes to the new origin ( , Y We thus obtain \ k y Q The locus is therefore a parabola whose vertex is ( , 0) and for which p = k. It will be turned to the right when k is positive, and to the left when k is negative. But if A: = 0, the locus is the degenerate parabola y = ± 4. ^ Y ^ -^ '■**s, V, ^ ^ ^ '-' ^ " ■■ -— - "^ N s. y y ^ ^ , - - ^— ^ "*^ «^ \ /" -^ ^ — " ■» . — - k^l> ■^ a* ^ k = ^ — " "^ ^ :^ =? ^ ^ V -^ ^ / 7 \ "^ ■^ ^- ■#■ ^ \v y / / Y ■t \ N s< X K r y •^ f M Vl to \ > V; - JT ^ i i j X V. '^ \ ^ \j y / y ^ ^ ^ ^ :^ \ \ / ^ ^ ^^ ^ kH 1 ^ ^ ^ '" , ^ ff» ^ ^ ^ c=: ^ - - — P ^ r-^ / y \ ^^ *^ ^ ~~ — . . "^ ^ y s s "** ^ .^ — -^ ^ e '-' ^ X" V ■^ -^ ^ r M ^ ■ In the figure the locus is plotted for A; = ± 4, ± 2, ± 1, db |, 0. It is seen that as k approaches zero, whether it is positive or negative, the vertex recedes from the origin and the parabola lies closer to the lines ?/ = ± 4 and finally degenerates into these lines. The degenerate parabola consisting of two parallel lines appears as a limiting case between the cases when the parab- olas are turned to the right and to the left. CONIC SECTIONS 203 Ex. 4. Discuss the system represented by Solution. x2 25 - A; 9 - A: When fc < 9 the locus is an ellipse whose foci are {± c, 0) where c2 = (25 -k) - {d -k) = 16 (theorem V, p. 185). When 9 < fc < 25 the locus is an hyperbola whose foci are (± c, 0), where c^ = (25 - fc) - (9 - fc) = 10 (Theorem VI, p. 185). When A; > 25 there is no locus. Since the ellipses and hyperbolas have the same foci, (± 4, 0), they are called confocal. Clearing of fractions, we obtain (9 - A:)x2 + (25 -k)y^ = {9- k) (25 - k). . Hence when A; = 9 or 25 the locus is a degenerate parabola y^-OoTX^ = 0. In the figure the locus is plotted for k = - 56, - 24, 0, 7, 9, 11, 16, 21, 24, 25. As k increases and approaches 9 the ellipses flatten out and fmally 1 1 1 1 1 1\[ 1 l\J l: \i\f\ 1/ 1 — N )<^\'T~ "^E^2 ^^ ^\ •^'^ I *=■ 2^ / ' ^ y X S^P~ ~H-2 5 / v/"! \ i 2^/ s^ ^^»J Z S, tjL'^ = oil A \^^ / "^^^C s.^'x 7 ^t$^^ \-"% t t'^^%^--- -~ty\.^% 4 -t- ^t- Cu ^'-^ ^ ^ IT X^ " _F^y ^ , L'^NZl-- -qrt"^7 i 5,"^^ ?> \^\ \ t>ti -"'' V 5 L.J X^ ^Z -t~'^^^ ^^^^5 t X S^\=z^ ^^^ ^■^^- -^"^ ^^i ^ ^S -^ JL s^ ^^ ■'^ /^U^ -fc-' N'^ ^ - == r--i ,. E ^ 1 -degenerate into the JT-axis, and as k decreases and approaches 9 the hyper- bolas flatten out and degenerate into the X-axis. Hence the locus of the parabolic type, y'^ = 0, appears as a limiting case between the ellipses and hyperbolas. As k increases and approaches 25. the two branches of the hyperbolas lie closer to the F-axis, and in the limit they coincide with the F-axis. Ex. 5. Plot and discuss the locus of A:x2 + 2 2/2 - 8 x = 0. Solution. If A: = 0, the locus is a parabola. If k is not zero, the locus is an ellipse or hyperbola according as k is positive or negative, passes through the origin for all values of k. The locus 204 ANALYTIC GEOMETRY Simplifying by translating the axes (Rule, p. 165), it is found that if the origin is ( - , j the equation becomes From this the axes may be determined and the locus sketched. In the figure the locus is plotted for fc = 1, f, ^, 0, — 1, — i. If A; is posi- tive and approaches zero, the ellipses become longer and lie closer to the \ Y\ , y^ S ^ ft y ■b- ^ •v ^ s. 4- A f. P y "^ V N ^^ y y^ \^ =s ( -^ ' V s, ^• \ ^ y ^ S ^>' \ \ A ^ ^ N y J ^ > 7: <" \ r ^t \ f»» \ J ' ) \ } ) ) X / ^ > ^ ^ ^ ^ ^ y ,/ y ^ s> y /^' y % N "^ ^ X / / N ^ ■"v ^ - >• y / 's V V - y X ^ y ^ y' parabola. If k is negative and approaches zero, the right-hand branches of the hyperbolas lie closer to the parabola and the left-hand branches recede from the origin. This shows that tUe parabola is a limiting form between the ellipse and hyperbola. How does the locus behave if k approaches + co or — co ? >- PROBLEMS 1. Plot on separate sheets the foci and directrices of the conies plotted in examples 1, 2, and 3. Where are the foci and directrices Of the degenerate conic in each system ? Verify the results analytically. 2. Plot the following systems of conies and show that the conies of each system belong to tlie same type. Draw enough conies so that the degenerate conies of the system appear as limiting cases. <^>S+ k. (c) 16 = k. (b) 2/2 = 2 kx. (d) x2 3. Problem 1 for the systems in problem 2. r 9 2ky -6. CONIC SECTIONS 205 4. Plot the system — + — = 1 for positive values of k. What is the locus k 16 if k = lQ? Show how the foci and direatrices behave as k increases or decreases and approaches 16. Where are the foci and directrices of a circle ? 5 . Plot the system in problem 4 for positive and negative values of k. Show hov^the conies change as k approaches zero when it is positive and negative. 6. Plot the following systems of conies and show that all of the conies of each system are confocal. Discuss degenerate cases and show that two conies of each system pass through every point in the plane. ^ ' 16-A; SQ-k ^ ' 64 - k 16 - k (b) y^ = 2kx-}- A;2. (d) x^ = 2ky + k^. 7. Plot and discuss the systems (a) 16 {X _ A;)2 + 9 2/2 = 144. (c) {y - k)"^ = 4 x. (b) xy = k. (d) 4 (X - k)'^ -9{y- ky = 36. 8. Plot the following systems and discuss the locus as k approaches zero and infinity. Show how the foci and directrices behave in each case. ^ (^-^2 .^ {x-kl_y^^ ^ ' k^ 36 ^ ^ A;2 36 9. Show that all of the conies of the following systems pass through the points of intersection of the conies obtained by setting the parentheses equal to zero. Plot the systems and discuss the loci for the values of k indicated. (a) (?/2-4x) + fc(z/2 + 4rc) = 0, fc3z:+l, -1. (b) {x^ + ?/2 _ 16) + fc(x2 - 2/2 _ 4) 3z: 0, A; = + 1, - 1, - 4. (c) (x2 + 2/2 - 16) + fc(a;2 - 2/2 - 16) = 0, fc = + 1, - 1. (d) (x2 + 16 2,2 - 64) + A: (x2 - 4 2/2 _ 36) = 0,k = -l,4, - -y. (e) x2 + 4 2/ + A: (x2 - 4 y + 16) = 0, A; = + 1, - 1. MISCELLANEOUS PROBLEMS 1. Construct the loci of the following equations, their foci and directrices. (a) 9x2 + 24x2/+ 162/2- 50x4-802/ -275 = 0. (b) 56x2-64x2/ + W)9y2._i76x + 282 y- 896 = 0. (c) 5x2 _ I2xy + 6x - S6y - 63 = 0. 2. Find the value of p if ^2 = 2px passes through the point (3, - 1). X2 2/2 3. Find the values of a and h if — + ^ = 1 passes through the points (3, - 6) and (4, 8). "^ ^^ 4. Find the equation of the locus of a point P if the sum of its distances from the points (c, 0) and (— c, 0) is 2 a, 206 ANALYTIC GEOMETRY 6. Find the equation of the locus of a point P if the difference of its distances from the points (c, 0) and ( — c, 0) is 2 a. 6. Find the equation of the locus of a point if its distances from the line x = — — and the point ( — , j are equal. 7. Show that a conic or degenerate conic may be found which satisfies five conditions, and formulate a rule by which to find its equation. Hint. Compare p. 93 and p. 133. 8. Find the equation of the conies which satisfy the following conditions. (a) Passing through (0, 0), (1, 2), (1, - 2), (4, 4), (4, - 4). (b) Passing through (0, 0), (0, 1), (2, 4), (0, 4), (- 1, - 2). (c) Passing through (3, 7), (4, 6), (5, 3) if ^ = -B and C = 0. (d) Passing through (1, 2), (3, 4), {4, 2), (2, - 1), (4, 2). (e) Passing through (0, 0), (0, 1), (1, 0), (6, 6), (5, 6). (f) Passing through (0, 0), (2, 0), (- 3, 2), (5, 2) with its axes parallel to the coordinate axes. 9. What is the nature of a conic which passes through five points, of which three or four are on a straight line ? The circle whose radius is a and whose center is the center of a central conic is called the auxiliary circle. 10. The ordinates of points on an ellipse and the auxiliary circle which have the same abscissas are in the ratio of 6 : a, 11. The area of an ellipse is Ttab. Hint. Divide the major axis into equal parts. With these as hases inscribe rectan- gles in the ellipse and auxiliary circle. Apply problem 10 and increase the number of rectangles indefinitely. 12. The auxiliary circle of an hyperbola passes through the intersections of the directrices and asymptotes. 13. Show that the locus of xy -^ Dx ■}- Ey -\- F = is either an equilateral hyperbola whose asymptotes are parallel to the coordinate axes or a pair of perpendicular lines. 14. Discuss the form of the locus oi x^ - y^ -\- Dx -{- Ey + F = 0. CHAPTER IX TANGENTS AND NORMALS 83. The slope of the tangent. Let Pi be a fixed point on a curve C and let P^ be a second point on C near P^. Let P^ approach P^ by moving along C. Then the limiting position PiT of the secant through Pi and P^ is called the tangent to C at Pi. It is evident that the slope of PiT is the limit of the slope of P1P2. The coordinates of Pg ^^y be written (xi + A, 2/1 + k), where h and k will be positive or negative numbers according to the relative positions of Pi and Pg. The slope of the secant through Pi and Pg is therefore (Theorem V, p. 35) 3/1 -fa -^k^ k (1) Xi — Xi h) h As P2 approaches Pi both h and k approach zero, and hence k approaches -? which is indeterminate. The actual value of the k limit of - may be found in any case from the conditions that' Pi and P2 lie on C (Corollary, p. 53), as in the example following. 207 208 ANALYTIC GEOMETRY Ex. 1. Find the slope of the tangent to the curve C : 8 y'= x^ at any point Pi {xu yi) on G. Solution. Let Pi (xi, yi) and P^ (xi + ^, 2/i + k) be two points on C. Then (Corollary, p. 53) (2) 8 2/i = Xi3 and 8 (^i + k) = (xi + h)\ or (3) 8 ?/i + 8 A; = Xi3 + 3 XiS^i + 3 x^h^ + /i^. Subtracting (2) from (3), we obtain ^k = ^Xx^h + Zxrh'^ + h^. Factoring, 8 A; = /i (3 Xi2 + 3 XiA + TjF) ; fc _ 3xi2_+_3xiM-^ h~ 8 y\ 1 H I f '/ ^^ P^. -J h X /^ "o / X / / f / 1 1 * / / / Y and hence Then, as P2 approaches Pi, h and k approach zero and the limit of - — limit of = ' h 8 8 ■ 3xi2 Hence the slope m of the tangent at Pi is m = — -• 8 C is symmetrical with respect to 0, and the tangents at symmetrical points are parallel since only even powers of Xi and yi occur in the value of m. The tangent at the origin is remarkabL In that it crosses the curve. The method employed in this example is general and may be formulated in the following Rule to determine the slope of the tangent to a curve C at a point Pi on C. First step. Let Pi (xi, 1/1) and P^ {xi + h, yi + k) he two points on C. Substitute their coordinates in the equation of C and subtract. Second step. Solve the result of the first step for — ?* the slope of the secant through Pi and P^. Third step. Find the limit of the result of the second step when h and k approach zero. This limit is the required slope. * The sohxtioii will contain h and k separately, so that the equation is not solved in the ordinary sense. TANGENTS AND NORMALS 209 Ex. 2. Find the slope of the tangent to the semicubical parabola 3 2/2 = x^ atPi(iCi, 2/i). Solution. First step. Let Pi(iCi, yi) and P^ixi + ^, 2/i + k) be two points on the curve. Then (Corollary, p. 53) (4) 3?/i2 = Xi3 and 3 yi2 + 6 kyi + 3 A;^ = xi^ + 3 x^^h + 3 Xih"^ + ^8. Subtracting, 6yik + 3A;2 = Bxi^h + 3xi;i2 _i_ ^3, Second step. Factoring, k{6yi + Sk) = h{Sxi^ + S Xih + ^^2). k _ Sxi^ + Sxih + h^ h~ Hence Qyi + Bk Third step. As h and k approach zero, limit of- = limit of- h 62/1 + 3A; 3xi2 62/1 Xj^ 22/1 Hence the slope of the tangent at Pi is m = At the origin m = - and is indeterminate. To find the value of m at the origin, we may either apply the rule a second time, setting cci = and 2/1 = 0, or eliminate 2/1 from the value of m by means of (4), thus obtaining a value which is determinate at the origin. PROBLEMS 1. Find the slopes of the tangents to the following curves at the points indicated. (a) 2/2 = 8 X, Pi (2, 4). (b) x2 + 2/2 = 25, Pi (3, -4). (c) 4x2 + 2/2 = 16, Pi(0, 4). (d) x2 - 92/2 = 81, Pi (15, - 4). Ans. 1. Ans. |. Ans. 0. 2. Find the slopes of the tangents to the following curves at the point Pi(xi, 2/1). (a) 2/2 = 6x. 3 Ans. — • (b) 16 2/ = x4 (c) x2 + 2/2 = 16. Ans. Ans. Xi« Atis. 2/1 Ans. 3xi2 + 2xi 2yi Ans. 8-4xi 2/1-1 Ans. Xi A nQ 2/1 ./I /to. Xi + 2yj_ Ans. 4-Xi 2-2/1 Ans. Xi + S 210 ANALYTIC GEOMETRY (d) x2 - 2/2 = 4. (e) 2/2 = x3 + x2. (f) 4x2 4-2/2-16x-22/ = 0. (g) xy = a\ (h) X2/ + 2/2 = 8. (i) x2 - y2 _ 8 X + 4 2/ = 0. (j) x2 + 2/2 + 6x-82/ = 0. 4-2/1 84. Equations of tangent and normal. We have at once the Rule to find the equation of the tangent to a curve C at a point Pi{^i, Vi) on C. First step. Find the slope m of the tangent to C at Pj {Rule, p. 208). Second step. Substitute Xi, yi, and m in the point-slope form of the equation of a straight line [(V), p. 95]. Third step. Simplify that equation hy means of the condition that Pi lies on C {Corollary, p. 53). Ex. 1. Find the equation of the tangent to C : 8 y = x^ at Pi (xi, 2/1). 3xi2 Solution. First step. From Ex. 1, p. 208, the slope is m = Second step. Hence the equation of the tangent is 2/ - 2/1 = -x- {X - Xi), or ^ (1) 3xi2x - 8 2/ - 3xi3 +'82/1 = 0. Third step. Since Pi lies on C, 8 2/1 = Xi^. Substituting in (1), we obtain (2) 3 Xi2x - 8 2/ - 2 Xi3 = 0. The normal to a curve C at a point Pi on C is the line through Pi perpendicular to the tangent to C at Pi. Its equation is found from that of the tangent by the Rule on p. 114, using Theorem XII, p. 117'. TANGENTS AND NORMALS 211 Ex. 2. Find the equation of the normal at Pi to the curve in Ex. 1. Solution. The equation of any line perpendicular to (2) has the form (Theorem XII, p. 117) (3) 8 x + 3 xth/ + k = 0. If Pi lies on this line, then (Corollary, p. 53) 8a;i + Sxih/i-\-k = 0, whence A: = — 8 Xi — 3 xi^yi. Substituting in (3), the equation of the normal is 8 X + 3 Xi2y - 8 xi - 3 iCi22/i = 0. PROBLEMS 1. Eind the equations of the tangents and normals at Pi(xi, ?/i) to the curves in (a) to (e), problem 2, p. 209. Ans. (a) yiy = 3 (x + Xi), 2/iX + 3 y = Xiyi + 3 ?/i. (b) Xi^x - 4 ?/ = 12 2/1, 4 X + Xi^y = 4 xi + Xi^^i/i. (c) xix + Viy = 16, yix - xiy = 0. (d) xix - yiy = 4, yix + x^y = 2 x^yx. (e) (3 Xi2 4- 2 xi) X - 2 2/1?/ - Xi3 = 0, 2 yiX + (3 Xi2 + 2 Xi) y = 3 Xi^y i + 4 Xi?/i. 2. Find the coordinates of a point on each of the curves in (/) to (j), problem 2, p. 209, and then find the equations of the tangent and normal at that point. 3. Find the equations of the tangents and normals to the following curves at the points indicated. (a) 2/2-8x + 4y=:0, (0, 0). Am. 2x - ?/ = 0, x + 2?/ = 0. (b) xy = 4, (2, 2). Ans. x + y = 4, x — y — ^. (c) x2-42/2 = 25, Pi(xi, ?/i). Ans. XiX-42/i?/=25, 42/iX+Xi2/=5xi?/i. (d) x2 + 2xy = 4, Pi(xi, ?/i). Ans. (xi + 2/i) X + Xxy - 4, XiX - (xi + ^i) y = x^ - Xi?/i - y^. (e) 2/2 = 2 px. Pi (xi, ?/i). ' Ans. 2/i2/=P(a^ + a:i), 2/iX+j32/ = Xi?/i+i32/i. (f) -0 + ^ = 1' Pi (^1,2/1). ^^^' -^ + -P" - ^' 62 a2 - a262 '''^" (g) 62x2 - a22/2 = a262, p^(xi, yi). Ans. 62a; jic _ a^y^y = a'^b^, a^yiX + b^Xiy = (a2 + 62) Xiyi. (h) x^-y^ -\-x^ = 0, (0, 0). Ans. y = ±x, x = Ty- 212 ANALYTIC GEOMETRY 85. Equations of tangents and normals to the conic sections. Theorem I. The equation of the tangent to the circle C : x^ -\- y^ = r"^ at the point P^ (x-^, y^) on C is (I) oc^oo + ynj = r p2 Proof. Let Pi (xi, ^i) and P2 (xi + A, S'l + k) be two points on the circle C. Then (Corollary, p. 63) (xi + W + (2/1 + W = r% (1) and or (2) Xi2 + 2xiA + /i2 4-yi2 + 22/iA;4-A;2 = r2. Subtracting (1) from (2), we have 2 Xi/i + ^2 + 2 vik + A:2 = 0. Transposing and factoring, this becomes A;(2yi + fc)=-/i(2xi + h), k _ 2xi 4- ^ whence 22/i + A; is the slope of the secant through Pi and P2. Letting Pj. approach Pi, h and k approach zero, so that m, the slope of the tangent at Pi, is ,. ., - 2xi + A xi m = limit of = = -. 2 2/1 + A; 2/1 The equation of the tangent at Pi is then (Theorem V, p. 95) Xi y -yi = or (X - Xi , 2/1 r2. Q.E.D. xix + yiy But by (1), xi2 4- yi2 so that the required equation is Xix + yiy Theorem II. The equation of the tangent to the locus of Ax^ + Bxg + Ci/ + Dx -{- Ey + F=0 at the point Pi (xi, 3/1) on the locus is (II) TANGENTS AND NORMALS 213 Proof. Let Pi (xi, 2/1) and P2 (xi + ^, ?/i + A:) be two points on the conic. Then (Corollary, p. 53) (3) • Axx^ + Bxiv^ + Cyi^ -\- Dxi + Eyi + F = and A{xi + h)^ -{■B{xi + h) {yi + A;) -f C (yi + k)^ + D(xi + /i) + -E;(2/i + A;) + P= 0. Clearing parentheses, we have (4) Axi^ + 2 ^XiA + Ah'^ + ^Xi2/i + -BxiA; + Byih + ^/iA; + Cyi^ + 2 C2/1A; -f Ck^ 4- Dxi + -D^ + ^2/1 + Ek + F = 0. Subtracting (3) from (4), we obtain (5) 2Axih + Ah^ + Bxik + 5?/i/i + Bhk + 2 CyiA; -f- Ck"^ -\- Dh -\- Ek = 0. Transposing all the terms containing h and factoring, (5) becomes k{Bxi + 2Cyi-\-Ck + E)=-h{2Axi -\- Ah + Byi + Bk-+ D), k 2Axi + Byi + D-{-Ah + Bk whence h Bxi + 2 Cyi + E -\- Ck This is the slope of the secant P1P2 [(1), p. 207]. Letting P2 approach Pi, h and k will approach zero and the slope of the ^^^S^^t i« ^ ^ _ 2Ax, + By,-i-D ^ ^ Bxi + 2 C2ji -{- e' The equation of the tangent line is then (Theorem V, p. 95) 2Axi-{-Byi + D. , ^' Bxi + 2Cyi + E^ '^ To reduce this equation to the required form we first cleaij of fractions and transpose. This gives (2 Axi + By I + D)x fiBxi + 2Cyi-\-E)y ^- (2 Axi^ + 2 Bxiyi + 2 Cyi^ + Dxi + ^2/i ) = 0. But from (3) the last parenthesis in this equation equals -{Dxi + Eyi + 2F). Substituting, the equation of the tangent line is {2Axi + Byi +-:©y:rH- {Bxi + 2Cyi + E)y + {Dxi + Eyi + 2F) = 0. Removing the parentheses, collecting the coefficients of A, B, O, D, E^ and P, and dividing by (2), we obtain (II). q.e.d. Theorem II enables us to write down the equation of the taiP" gent to the locus of any equation of the second degree. It is remembered most easily in the form of the following Rule. 214 ANALYTIC GEOMETRY Rule to write the equation of the tangent at Pi (xi, t/i) to the locus of an equation of the second degree. First step. Substitute x^x and y^y for x"^ and y"^, — — - — — for xy, ana — - — - and for x and y in the given equation. Second step. Substitute the numerical values of Xt and y^, if given, in the result of the first step. The result is the required equation. In like manner, or at once from this Kule, we have Theorem III. The equation of the tangejit at P^ (xi, y^) to the ellipse b^x^ + a^y^ = a%^ is b^oc^oc + a^y^y = a^b^ ; hyperbola b^x^ — a^y"^ — a%'^ is b'^oc^uc — a^yiy = a^b"^ ; parabola y"^ = 2px is y^y = _p (x + x-^. By the method on p. 210, we obtain Theorem IV. The equation of the normal at Pi(xi, y{) to the ellipse b^x^ + a^y^ = a%^ is ahj^oc — h^x^y = {a^ — b^) oc^yi ; hyperbola b'^x^ — a^y^ = a%^ is ahj^x + b'^x^y — {a^ + 6^) x^y^ ; parabola y"^ = 2px is y^x + j^y = x^y-i + vy^- PROBLEMS 1. Find the equations of the tangents and normals to the following conies at the points indicated. (a) 3x2-10?/2 = 17, (3, 1). (d) 2x2 -^2 '=14, (g^ _ 2). (b) y2 = 4aj, (9, _ 6). (e) x^^hy'^ = 14, (3, 1). (c) x2 + y2 ^ 25, (- 3, - 4). (f) x2 = 62/, (- 6, 6). (g) x2-x2/ + 2x-7 = 0, (3,2). (h) X2/-y2 + 6x + 82/-6 = 0, (-1,4). The directed lengths on the tangent and normal from the point of contact to the Z-axis are called the length of the tangent and the length of the normal respectively. Their projections on the Z-axis are known as the subtangent and subnormal. 2. Find the subtangents and subnormals in (a), (b), (d), and (e), prob- lem 1. Am. (a) - Lo, _«_ ; (b) - 18, 2 ; (d) - |, 6 ; (e) |, - f . 3. Find the lengths of the tangents and normals in (a), (b),Jd), and (e), probl&m 1. Ans. (a) \ VTsT, J^ VlsT ; (b) 6 VTo, 2 VlO ; ^ (d) I Vio, 2 VlO ; (e) \ V34, | Vsi". \ TANGENTS AND NORMALS 216 4. Find the subtangents and subnormals of (a) the ellipse, (b) the hyper- bola, (c) the parabola. Ans. (a) Xi- 62 a2 _ a; 2 xi; (b)^ , a^ Xi 2xi,p. xi a^ xi a^ 5. Show how to draw the tangent to a parabola by means of the sub- normal or subtangent. . 6. Prove that a point Pi on a parabola and the intersections of the tangent and normal to the parabola at Pi with the axis are equally distant from the focus. 7. Show how to draw a tangent to a parabola by means of problem 6. 8. The normal to a circle passes through the center. 9. If the normal to an ellipse passes through the center, the ellipse is a circle. 10. The distance from a tangent to a parabola to the focus is half the length of the normal drawn at the point of contact. 11. Find the equation of the tangent at a vertex to (a) the parabola; (b) the ellipse; (c) the hyperbola. 12. Find the subnormal of a point Pi on an equilateral hyperbola. Ans. Xi. 13. In an equilateral hyperbola the length of the normal at Pi is equal to the distance from the origin to Pi. 86. Tangents to a curve from a point not on the curve. Ex. 1. Find the equations of the tangents to the parabola y"^ = 4x which pass through P2 (— 3, — 2). Solution. Let the ;point of contact of a line drawn through P2 tangent to the parabola be Pi. Then by Theorem III the equation of that line is (1) ?/i?/ = 2 X + 2 Xi. Since P2 lies on this line (Corollary, p. 53), (2) -2yi=-6 + 2xi; and since Pi lies on the parabola. (3) ?/l2 = 4Xi. The coordinates of Pi, the point of contact, must satisfy (2) and (3). Solving them, we find that Pi may be either of the points (1, 2) or (9, - 6). 216 ANALYTIC GEOMETKY If (1, 2) be the point of contact, the tangent line is, from (1), 22/ = 2x + 2, or x-y + 1 = 0. If (9, — 6) be the point of contact, the tangent line is - 6 2/ = 2 X + 18, or x-Sy + 9 = 0. The method employed may be stated thus : Rule to determine the equations of the tangents to a curve C passing through P^ix^, y^ not on C. First step. Let P^ (x'l, ?/i) he the point of tangency of one of the tangents, and find the equation of the tangent to C at P^ {Rule, p. 210). Second step. Write the conditions that (x^, y^) satisfy the result of the first step and (cci, y-^ the equation of C, and solve these equa- tions for Xi and yi. Third step. Substitute each pair of values obtained in the second step in the result of the first step. The resulting equations are the required equations. PROBLEMS 1. Find the equations of the tangents to the following curves which pass through the point indicated and construct the figure. (a) x^ + y^ = 25, {7, -1). Ans. 3x - 4?/ = 25, 4x + 3?/ = 25. (b) 2/2 := 4x, (- 1, 0). Ans. y = x-\-\,y + x-\-l = 0. (c) 16x2 + 25y2 = 400, (3, - 4). Ans. y + 4 = 0, Sx - 2y = 11. (d) 8y = x3, (2, 0). Ans. y = 0,27 x-8y- 54 = 0. (e) x2 _}. i6?/2 _ 100 = 0, (1, 2). Ans. None. (f) 2xy + ?/2 = 8, (- 8, 8). Ans. 2x + 3?/ - 8 = 0, 4x + 3?/ + 8 = 0. (g) y^-\-4x-ey = 0, {-^, -1). Ans. 2x-Sy = 0, 2x-y-\-2 = 0. (h) x2 + 4 y = 0, (0, - 6). Ans. None. (i) x2 - 3 ?/2 + 2 X + 19 = 0, (- 1, 2). Ans. x + 3?/-5 = 0, x-3y + 7 = 0. (j) y2 = x^, (f, 0). Ans. y = 0, 3x - y - 4 = 0, 3x + y - 4 = 0. 2. Find the equations of the lines joining the points of contact of the tangents in (a), (b), (c), (f), (g), and (i), problem 1. Ans. (a)7x-y = 25; (b) x = 1 ; (c) 12x - 25y = 100; (f) x = l; (g) x-2y = 0; (i) y = 6. TANGENTS AND NORMALS 217 87. Properties of tangents and normals to conies. Theorem Y. If a point moves off to infinity on the parabola y^ tangent at that point approaches parallelism with the X-axis. Proof. The equation of the tangent at the point Pi(xi, yi) is (Theorem III, p. 214) 2/12/ =px+ pxi. Its slope is (Corollary I, p. 86) 2px, the P m = —• 2/1 As Pi recedes to infinity y^ becomes infinite, and hence m approaches zero, that is, the tangent approaches parallelism with the X-axis. q. e . d. Theorem VI. If a point moves off to infinity on the hyperbola b'^^-a'^y'^ = aW, the tangent at that point approaches coincidence with an asymptote. Proof. The equation of the tangent at the point Pi (Xi, yi) is (Theorem "* III, p. 214) (1) 62a;ix - a'^yiy = a'^bK &2xi Its slope is (Corollary I, p. 86) cC'yi As Pi recedes to infinity Xi and yi become infinite and m has the inde terminate form — . CO But since Pi lies on the hyperbola, 62xi2 - a2yi2 = a'^b'^ Dividing by a'^yi^, transposing, and extracting the square root, 6 Multiplying by bxi ayi 62Xi '1 2/1^ + 1. m = — — = ±-A — + 1. a^yi a \ 2/i2 218 ANALYTIC GEOMETRY From this form of m we see that as 2/1 becomes infinite m approaches ± -, the slopes of the asymptotes [(5), p. 190], as a limit. The intercepts of (1) are — and ■- As their limits are zero the limiting position of the tangent will pass through the origin. Hence the tangent at Pi approaches coincidence with an asymptote. q.e.d. These theorems show an essential distinction between the form of the parabola and that of the right-hand branch of the hyperbola. ^ Theorem vn. The tangent and normal to an ellipse bisect respectively the external and internal angles formed by the focal radii of the point of contact * Proof. The equation of the lines joining Pi (xi, yi) on the ellipse 62x2 + a22/2 = a262 to the focus F' (c, 0) (Theorem V, p. 185) is (Theorem VII, p. 97) yix + {c-xi)y-cyi = 0, and the equation of PiP is ^ yix - {c + Xi)y + cT/i = 0. The equation of the tangent AB is (Theorem III, p. 214) 62xix + a^yiy = a^b^. We shall show that the angle 6 which AB makes with PiF' equals the angle which PiP makes with AB. By Theorem X, p. 109, a2y^2 _ 52cxi -f b'^xi^ _ {a'^ yi^ + b^Xr^ - b^cxi b^xiyi + a^cyi - a^x^yi a^cyi - {a^ - 62) xi^i But since Pi lies on the ellipse. a2y^2 ^ 52a;^2 _ (j252^ and (Theorem V, p. 185) a262 - 62cxi Hence tan 6 = In like manner tan = a2-62 62(a2- :C2. CXi) cyi < a'^cyi - c2xi2/i cyi {a^ - cxi) - 62cxi - 62xi2 - a^i^ (62xi2 + a^yr^) + 62cxi h'^xiyi - a^cyi - a'^Xiyi a2c2/i + (a2-62)xi2/i aW + 62cxi _ 62 cyi a^cyi + c2xiyi * This theorem finds application in the so-called whispering galleries. TANGENTS AND NORMALS 219 Hence tan ^ = tan ; and since d and are both less than it, 6 = . • " h~ B-\- Ek-\ Letting P^ approach P^, the limit of —^ which is the slope of A the tangent, is seen to be Hence the equation of the tangent is (Theorem V, p. 95) A y = X, — ^ B ^ or . Ax + By = 0. q.e.d. If ^ = and B = 0, the terms of the lowest degree, if set equal to zero, will be the equation of the two or more lines which will then be tangent to C at the origin. For example, if the equation of C is x'^ — y'^ + x^ — 0, the two lines a;2 _ y2 — will be tangent to C at the origin (problem 3, (h), p. 211). TANGENTS AND NORMALS 223 89. Second method of finding the equation of a tangent. The tangent to a curve C at a point Pi may now be found as follows. Transform C by moving the origin to Pi (Theorem I, p. 160). The equation of the tangent at Pi in the new coordinates is then found immediately by Theorem XII. Transform it by translating the axes to their first position. The result is the equation of the tangent at Pi in the given coordinates. Ex. 1. Find the equation of the tangent to C : 4 x^ — 2 ?/2 _j. 3^3 = q at Pi(- 2, 2) which lies on C. Solution. Set (Theorem I, p. 160) x = x'-2, y = y' + 2. The equation of C becomes 4 (x' - 2)2 - 2 (y' + 2)2 + {x' - 2)3 = 0. Only the terms of the first degree are needed, and these may he picked out without clearing the parentheses. Tlie equation of the tangent is therefore 4 (- 4 x') - 2 . 4 ?/' + 12 x' = 0, or x' + 2 y' — 0. To transform to the old axes, set x' = X + 2, y' = y -2. "We thus obtain x + 2?/-2 = 0, which is the equation of the tangent to C at Pi. PROBLEMS 1. Find the equations of the tangents at the origin to (a) x2 + 2 xy + ?/2 - 6 X + 8 y = 0. (d) y = x^ - 2 x2 + x. (h) xy -y^ -\-x-Sy = 0. {e) x^ -\- y^ + x - y = 0. (c) x2 + 4x2/ - 3x + 4?/ = 0. (f) x3 + x2 - Sxy - 4y2 = q. 2. Find the equations of the tangents to the following curves at the points indicated by the method of section 89. (a) 9x2 - 2/2 + 2 X - 4 = 0, (2, 6). (b) x2 + 4xy + 62/-7=0, (-1, 3). (c) xy + 6x - 4?/ - 6 = 0, (2, 3). (d) y2a.4x + 2y + 8 = 0, (-4, 2). (e) y^ = x^ + 8, (2, 4). (f) 2/ = X* - 3x3 - 5x2 + 4x + 4, (0, 4). Ans. 19x-6y-2 = 0. Ans. 5x + 2/ + 2=0. Ans. 9x-2y-12 = 0. Ans. 2x + 3y + 2 = 0. Ans. Sx-2y + 2 = 0. An^. y = 4 X + 4. 224 ANALYTIC GEOMETRY 3 . Find the angle which the locus of xy-\-4y — 2x = makes at the origin with that of x^ -\- ixy + x -{- 8y = 0. j^^^^ 5 ■ 4* 4. Find the angle which the line 2x — Sy — 9 = makes with the locus of x2/ + 6x-4y-19 = 0at(3, -1). ^^^ ^ 4 MISCELLANEOUS PROBLEMS 1. Find the equations of the tangents and the normals to the following conies at the points indicated. (a) x^ + 4xy -ix-lOy + 7 = 0, (3, - 2). (b) x?/ -4x4-32/ -4 = 0, (-1,4). (c) x?/ + 2/2 + 2x + 2?/ = 0, (- 3, 3). (d) 2/2 + 4x + 62/ - 27 = 0, (5, - 7). (e) x2 + 3X2/ + 2/2 - 102/ - 1 = 0, (2, 3). (f) x2 - 8x + 32/ - 14 = 0, (1, 7). 2. Find the equation of one of the tangents to the ellipse x^ + 9?/^ — 4x 4- 9 2/ = which is parallel to the line 4x — 9y — 36 = 0. .3. For what point of the parabola y'^ = 2px is the length of the tangent equal to four times the abscissa of the point of contact ? 4. What is the length of the tangent to a parabola at an extremity of the latus rectum ? Restate the equation of the parabola in problem 14, p. 221, in terms of this length. 5. For what point on the parabola y^ = 2px is the normal equal to (a) twice the subtangent? (b) the difference between the subtangent and the subnormal? 6. Through a point of the ellipse h^"^ + a'^y^ = a'^b'^ and that point of the auxiliary circle with the same abscissa normals are drawn. What is the ratio of the subnormals ? 7. For what points of an hyperbola is the subtangent equal to the subnormal ? 8. The ordinate of a point on an equilateral hyperbola and the length of the tangent drawn from the foot of that ordinate to the auxiliary circle are equal. 9. A tangent to a parabola meets the directrix and latus rectum produced at points equally distant from the focus. 10. The semi-conjugate axis of a central conic is a mean proportional between the distance from the center to a tangent and the length of the normal drawn at the point of contact. TANGENTS AND NORMALS 225 11. Find the points of the ellipse for which the lengths of the tangent and normal are equal. 12. Any point on an equilateral hyperbola is the middle point of that part of the normal included between the axes of the hyperbola. 13. A circle is drawn through a point on the minor axis of an ellipse and through the foci. Show that the lines drawn through the given point and the points of intersection of the circle and ellipse are normal to the ellipse. 14. How many normals may be drawn through a given point to (a) an ellipse ? (b) an hyperbola ? (c) a parabola ? CHAPTER X RELATIONS BETWEEN A LINE AND A CONIC. APPLICA- TIONS OF THE THEORY OF QUADRATICS 90. Relative positions of a line and conic. If a line and conic are given, it is evident that (a) the line is a secant of the conic, (b) the line is tangent to the conic, or (c) the line does not meet the conic. The coordinates of the points of intersection of the line and conic are found by solving their equations (Rule, p. 76), which are of the first and second degrees respectively. To solve, we eliminate y* and arrange the resulting equation in the form (1) •' Ax^ + Bx + C = 0. Denote the roots by Xi and X2 and the discriminant B^—AAChj A. Analytically the three cases above present themselves as follows : (a) If A is positive, the line is a secant. For Xi and X2 are real and unequal (Theorem II, p. 3), and hence they are the abscissas of the points of intersection, which must be distinct. (h) If A is zero, the line is a tangent. For in this case a^i = x^, so that the points of intersection coincide. (c) If A is negative, the line does not meet the conic. For Xx and Xg are imaginary, and hence there are no points of intersection (p. 77). If ^ = 0, one root of (1) is infinite (Theorem IV, p. 15) and one point of inter- section is said to be " at infinity." li A = and B = 0, then both roots of (1) are infinite and the line is said to be ** tangent at infinity." If ^ = 0, B = 0, and C = 0, then (1) is satisfied by all values of x, and hence has an infinite number of roots. All of the points on the line lie on the conic; that is, the conic is degenerate and consists of straight lines of which the given line is one. *If one equation does not contain ?/, then x is found by solving that equation. But for our purposes it is unnecessary to complete the solution. 226 LINE AND CONIC 22T "*»^ ^ r/. / ^"v ^ 7 Ns7 %\ Z ^ X' J %% _J ^ y 7 y 2-^ ^ ^^ -7^ — y'— In solving the equations of the line and conic it might be easier to eliminate x than y. Then (1) would be a quadratic in y, but the result of the discussion would be the same. If one equation did not contain y, it would be necessary for our purposes to eliminate x instead of y, and vice versa. Ex. 1. Determine the relative positions of the line 3x — 2y + 6 = and the parabola y^ + 4x = 0. Solution. It is easier to eliminate x than y. Solving the equation of the line for x, we obtain 2y -Q x= — -. 3 Substituting in y2 + 4 x = 0, we get 3 2/2 + 8 y - 24 = 0. The discriminant of this quadratic is A = 82-4..8(-24) = 352. As A is positive, the line is a secant. Ex. 2. Determine the relative position of the line 4 x + y + 5 = and the ellipse 9 ic^ + ?/2 = 9. Solution. It is easier to eliminate y than x. From the first equation, y =- (ix+ 5). Substituting in the second and arranging, we get 25x2 + 40X + 16 = 0. The discriminant is A = 402 - 4 • 25 • 16 = 0. Hence the line is a tangent. Ex. 3. Determine the relative position of the loci ofx2 — 2/2^3a; — 3y = and X — y = 0. Solution. Eliminating y, we get x2 - x2 4- 3 X - 3 X = 0, or • x2 + . X + = 0. As this equation is true for all values of x, then all of the points on the line lie on the conic. The equation of the conic may evidently be written {x — y) (x + y + S) = 0. The locus of this equation is (Theorem, p. 66) the degenerate conic consisting of the pair of lines x-y = 0, x + y + S = Or of which one is the given line. ..[ U:4 \ ^ A' Mo X J y] [ Ans. Tangent. Ans. Do not meet. Ans. Secant. Ans. Tangent. Ans. Tangent. Ans. Secant. Ans. Do not meet. Ajis. Do not meet. Ans. Secant. Ans. Tangent. 228 ANALYTIC GEOMETRY PROBLEMS 1. Determine the relative positions of the loci of the following equations and plot their loci. (a) x + y + l = 0,x^ = 4y. (b) x-2y + 20 = 0^x^ + y^ = 16. (c) y2-4x = 0,2x + 3y-S = 0. (d) x2 + 2/2 - X - 2 ?/ = 0, X + 2 2/ = 5. (e) 2xy -Sx -y = 0,y + Sx-6 = 0. (f ) x^ + y^ - Gx - Sy = 0, X - 2y = 6. (g) 4 x2 + ^-2 _ 16 aj = 0, X + ?/ - 8 = 0. (h) x2 + ?/2 - 8x - 6 = 0, X + 8 = 0. (i) 8x2-6 2/2 + i6x-32 = 0, 2x-3y = 0. (j) x2 + xy + 2x + y = 0, 2x + y + 4 = 0. (k) x2 + 2 x?/ + ?/2 + 4 X - 4 ?/ = 0, X + ?/ = 1. Ans. Secant, with one point of intersection at infinity. (1) 4 x2 — ?/2 + 4 X + 1 = 0, 2 X — 2/ + 1 = 0, Ans. Line is part of conic, (m) x2 + 4x!/ + ?/2 + 4 X - 6 ?/ = 0, 2 X - 3 ?/ = 0. Ans. Tangent, (n) x2 - 4 ?/2 + 8 ?/ - 20 = 0, X - 2 ?/ + 2 = 0. Ans. Tangent at infinity, (o) x2-6x?/ + 0?/2 + x-3?/-2 = 0, x-32/=l. Ans. Line is part of conic, (p) 6 x2 — 5 x?/ — 6 ?/2 = 18, 2 X — 3 ?/ = 0. Ans. Tangent at infinity. 2. Find the middle points of the chords of the conies in (c), (f), and (i), problem 1, which are formed by the given line. Ans. (c) (V-, -3); (f) (-V-, - |) ; (i) (- f, -1). 3. Interpret Theorem II, p. 3, geometrically by determining the relative positions of the parabola y = Ax^ + Bx + C and the line y = 0. Construct the figure if (a) A = l, B=-l, C=0; (b) A = l, B=C=0; (c) A = l B=l, C=0. 91. Relative positions of lines of a system and a conic, and of a line and conies of a system. Given a system of lines (that is, an equation of the first degree containing a parameter k) and a conic, we can determine the values of k for which the lines of the system intersect, are tangent to, or do not meet the conic, as follows. Eliminate x or y, as may be more convenient, from the equa- tions of the system of lines and the conic, thus obtaining an equation either of the form (1) Aif -\- By -\- C = 01 Ax^ + Bx + = 0. LINE AND CONIC 229 The discriminant A will be in general a quadratic in k. Determine the values of k for which A is positive, zero, or negative (Theorem III, p. 11) and apply the results of the preceding section. The same process serves to separate the conies of a system (that is, the loci of an equation of the second degree containing a parameter k) into three classes according as they intersect, are tangent to, or do not meet a given line. Only here the values of A;, if any, for which the equation has no locus must be excluded. Ex. 1. Find the values of k for which the line y = 2x -\- k intersects, is tangent to, or does not meet the ellipse x2 + 4?/2 — 8x + 4?/ = 0. Solution. Eliminating y by substitution in the second equation, we obtain 17x2 + 16A;x + 4fc24.4A: = 0. The discriminant of this quadratic is A = (16 A:)2 _ 4 . 17 (4 A:2 + 4A;) =- 16(fc2 + 17 k). 17, By (a), (&), and (c), p. 226, (a) the line is a secant if - 16 {k^ + 17 fc) > ; (b) the line is a tangent if - 16 {k^ + 17 A:) = ; (c) the line does not meet the ellipse if — 16 (A;^ + 17 A;) < 0. Apply Theorem III, p. 11, to the quadratic — 16(fc2 + 17 A;). Since A = (— 16 • 17)^ is positive, ^ = — 16, and the roots are and (a) if - 17 < A; < 0, the quadratic - 16 (A:^ + 17 A;) > ; (b) if A: = or - 17, the quadratic - 16 {k^ + 17 A;) = ; (c) if A; < - 17 or A; > 0, the quadratic - 16 (A;2 + 17 A:) < 0. Hence (a) the line is a secant if — 17 < A: < 0. (b) the line is a tangent if A: = or — 17. (c) the line does not meet the ellipse if A: < — 17 or A; > 0. The lines of the system are all parallel. The figure shows the two tangent lines and indicates where the lines lie for different values of k. 230 ANALYTIC GEOMETRY PROBLEMS 1 . Determine the values of k for which the loci of the following equations (a) intersect, (b) are tangent, (c) do not meet. Construct the figure in each (a) 2/ = A:x — 1, x2 = 4 2/. Ans. (a) A;>1 or ^<-l; (b) A; = ± 1 ; (c) -l 5 or A; < - 5. (c) x'^ + y^ = k, 3 X - 4 2/ + 10 = 0. Ans. (a) A: > 4 ; (b) A: = 4 ; (c) < A; < 4. (d) y = A;x + 2, x2 — 8 y = 0. Ans. (a) For all values of k. (e) x2 + y2 _ 2 A;x = 0, y = x. Ans. (a) For all values except A; = ; (b) A; = 0. (f) 4 x? - 2/2 = 16, y = A:x. Ans. (a) - 2 < A; < 2 ; (b) A: = ± 2 ; (c) A; > 2 or A: < - 2. (g) y'^ = 2kx, X - 2 2/ + 2 = 0. Ans. (a) A; > 1 or A; < ; (b) A: = or 1; (c) < A: < 1. (h) x2 + 4 ^2 _ 8x = 0, y = kx + 2-^k. Ans. (a) All values except A: = ; (b) fc = 0. (i) xy = k, 2 X + y + 4 = 0. Ans. (a) A; < 2 ; (b) Ar=: 2 ; (c) A; > 2. (j) X2/. + 2/2 - 4 X + 8 2/ = 0, X - 2 2/ + A: = 0. Ans. (a) A; > 48 or A; < ; (b) A; = or 48 ; (c) 48 > A; > 0. (k) 4x2 + 2/2 - 6x + 6y = 0, 2/ = A;x + 1 - A;. Ans. (a) A; > 1 or A; < - If ; (b) A: = 1 or - if ; (c) - if < A; < 1. 2. Determine the values of k for which the loci of the following equations are tangent and construct the figure. (a) x2 - 4 2/ + 16 = 0, 2/ = A;. (b) 9x2 + 162/2 = 144^ y _ a- ^ ^^ (c) 4x2/ + y2 + 16 _ 0, X = A;. (d) x2 + 4x2/ + 2/2 = A:, 2/ = 2x + 1. (e) x2 + 2x2/ + 2/2 + 8x-62/ = 0, 4:X-^y = k. (f) x2 + 2x2/ -4x + 22/ = 0, 2x - 2/ + A; - 3 = 0. 92. Tangents to a conic. If in the preceding section the value of the discriminant of (1) is zero, then the line and conic are tan- gent. The equation obtained by setting that discriminant equal to zero is called the condition for tangency. Hence the condition for tangency of a line and conic is found by eliminating either X or y from their equations and setting the discriminant of the resulting quadratic equal to zero. Thus in Ex. 1, p. 229, the condition for tangency is A = - 16 (A;2 + 17 A;) = 0. Ans. k = 4. Ans. k=±5. Ans. k=±2. Ans. k=--i-,. Ans. k = 0. An^. A; = 3 or 13 LINE AND CONIC 231 X v Ex. 1. Find tlie condition for tangency of the line - + - = 1 and the parab- ola 2/2 = 2px. Solution. Eliminating z by solving the first equation for x and substituting in the second, we get 6^2 + 2 apy — 2 abp = 0. The discriminant of this quadratic is A = (2 ap)2 - 4 6 (- 2 abp) = 4 op (ap + 2 62). Hence the condition for tangency is 4 ap{ap + 2 62) = or ap {ap + 2 62) = 0. Rule to find the equation of a line tangent to a given conic and satisfying a second condition. First step. Write the equation of the system of lines satisfying the second condition. Second step. Find the condition for tangency of the equation found in the first step and the given conic. Third step. Solve the equation found in the second step for the value of the parameter of the system of lines and substitute the real values found in the equation of the system. The equations obtained are the required equations. Ex. 2. Find the equations of the lines with the slope \ which are tangent to the hyperbola x2 — 6 y2 + 12 y — 18 = and find the points of tangency. Solution. First step. The lines of the system (1) y = ix + fc . have the slope \ (Theorem I, p. 58). y\ y V ^ y^ ^ ^ ^ X s, ^ y ^ X" (^ b \ ^ ^ ^ t I ^ t^ X \ \ X ' X ^ ^ -f) s V fi "^ ^ T-e ■1 ^ X *• , i Second step. Solving (1) for x and substituting in the given equation, (2) y2 + (4 ^ _ 6) y + 9 - 2 A:2 = 0. Hence the condition for tangency is (4 fc - 6)2 - 4 (9 - 2 A:2) = 0. 232 ANALYTIC GEOMETEY Third step. Solving this equation, A; = or 2. Substituting in (1), we get the required equations, namely, (3) x-2?/ = 0, x-2?/ + 4 = 0. To find the points of tangency we substitute each value of k in (2), which then assumes the second form of (7), p. 4, namely, if fc = 0, (2) becomes (?/ - 3)2 = ; .-. y = Z; if A: = 2, (2) becomes (y + 1)2 = ;.-.?/ = - 1. Hence 3 and — 1 are the ordinates of the points of contact. Then, from (1), if fc = and y = 3, we have x = 6 ; if A: = 2 and ?/ = — 1, we have x = — Q. Hence, if A; = 0, the point of contact is (6, 3) ; if A; = 2, the point of contact is (— 6, — 1). The points of contact may also be found by solving each of equations (3) with the given equation. PROBLEMS 1. Determine the condition for tangency of the loci of the following equations. (a) 4x2+?/2-4x-8=0, y = 2x-\-k. Ans. k^ -\-2k -\1 =0. (b) xy + X - 6 = 0, x = ky -^-b. Aiis. fe2 + 14 A; + 25 = 0. (c) x2 — 2/2 = o?^ y — kx. Ans. A; = ± 1. (d) x2 + 2/2 = ^2^ 4 2/ - 3 X = 4 A;. Ans. 16 A:2 = 25 r'^. (e) x2 + ?/2 zz r2, 2/ = mx + 6. Ans. (2?w6)2-4(l + m2)(62_^2)^0. (f).^ + ^^ = l,-+^.l. ^n. ^ + ^ = 1. ^ -^ a2 . 62 ' a ^ a2 ^ ^2 / ,. x2 2/2 1 a: , 2/ , . a2 52 (g) —c.-7-. — ^i — + - = 1. Ans. =1. ^^' a2 62 ' a ^ a2 ^2 X2 W2 (h)* — + ^- = 1, 2/ = mx + /3. Ans. ahn^ + 6^ - /32 = q. a2 6^ 0)* ^ - r^ = 1' y = mx + p. Ans. a'^m^ - 62 - /32 = 0. a2 62 (j)* x2 4- 2/2 = r2, x cos w + 2/ sin w — p = 0. ^ns. p2 _ y2 _ 0. (k)* 2 x^ = a2 - + ^ = 1. Ans. aB = 2 a2. (1)* x2 + 2/2 ^ ^2^ ^x + % = 1. Ans. A^r'^ + ^2^2 ^^ 1. * In these problems it is assumed that the constants involved are not zero. LINE AND CONIC 233 2. Find the equations of thie tangents to the following conies which satisfy the condition indicated, and their points of contact. Verify the latter approx- imately by constructing the figure. (a) ?/2 — 4 a;^ slope = i. Ans. x — 2y + i = 0. (b) x2 + 2/2 ^ 16, slope = - f Ans. 4:X + Hy ±20 = 0. (c) 9 ic2 + 16 ?/2 = 144, slope = — |. Ans. x + 4 y ± 4 VlO = 0. (d) x2 — 4 y^ = 36, perpendicular to 6x — 4y + 9 = 0. Ans. 2x + Sy ±3Vl = 0. (e) x2+2?/2-x + ?/=0, slope=-l. Ans. x-\-y = l, 2x + 2y + l = 0. (f) x^=4:y, passing through (0,-1). Ans. y =±x — 1. (g) x^ = 8y, passing through (0,2). Ans. None, (h) 4 x2 - y2 = 16, slope = 2. Ans. y = 2x. (i) xy + y^ ~4:X-\-8y = 0, parallel to 2x -4y = 7. Ans. x = 2y, x-2y + 48 = 0. (j) 4 x2 + 2/2 - 6 X + 6 2/ = 0, passing through (1, 1). Ans. x-y=0, 19x + ll2/-30 = 0. (k) x2 + 2 X2/ + 2/2 + 8 X - 6 2/ = 0, slope = f . Ans. 4 X — 3 2/ = 0. (1), x2 + 2x2/ - 4x + 2 ?/ = 0, slope = 2. Ans. y = 2x, 2x -y -\-10 = 0. (m) y^ = 2px, slope = m. Ans. y = mx-\--^' 2 m (n) h^x'^ + a22/2 = a2&2, slope = m. Ans. y = mx ± Vo^iM^. (o) 2xy = a2, slope = m. Ans. y = mx ± a V— 2 m. 3. Find the highest and lowest points of the conic (a) x2 + 6 xy + 9 y2 _ 6 X =: 0. Ans. Highest (f , f). (b) x2 - 2x2/ - 4x - 4?/ - 8 = 0. Ans. (0, - 2), (- 4, - 6). (c) x2 - 2/2 - 4 X + 8 2/ - 16 = 0. Ans. None. JTmt. Find the points of contact of the horizontal tangents. 93. Tangent in terms of its slope. The method of the preced- ing section for finding a tangent with a given slope may be applied to general equations and yield formulas for the equation of a tangent in terms of its slope. Theorem I. The equation of a tangent to the parabola y'^ = 2px in terms of its slope m is (I) y = -,nx+£^. 234 ANALYTIC GEOMETRY Proof. Eliminating x from y = mx + k and y^ = 2px, we obtain m?/^ — 2py + 2pk = 0. Hence the condition for tangency is A z= (- 2py -4m (2pk) = 0, , whence k = 7^ — 2m Substituting in y = mx + k, we obtain (I). q.e.d. In like manner we prove Theorem II. The equation of a tangent in terms of its slope m to the circle x^ -\- y^ = r^ is y = moo Ht r Vl + m^ ; ellipse Px^ + a'^y'^ = a%'^ is y = thx^ i '\/a^ni^ + b^ ; hyperbola V^x^ — a^y^ = a^"^ is y = mx ±_ ^ahn^ — b^. PROBLEMS 1. Find the equations of the common tangents to the following pairs of conies. Construct the figure in each case. (a) 2/2 = 5x, 9x2 + 9?/2 ^ I6. Ans. 9x ± 12y + 20 = 0. (b) 9 »2 + 16 y2 ^ 144^ 7 x2 - 32 2/2 = 224. Ans. ±x-y±5 = 0. (c) x2 + 2/2 = 49, x2 + 2/2 - 20 2/ + 99 = 0. Ans. ± 4x - 32/ + 35 = 0, ± 3x - 4?/ + 35 = 0. Hint. Find the equations of a tangent to each conic in terms of its slope and then determine the slope so that the two lines coincide (Theorem III, p. 88). 2. Two tangents, one tangent, or no tangent can be drawn from a point Pi (a^i, Vl) to the locus of (a) y^ = 2px according as 2/1^ — 2pxi is positive, zero, or negative. (b) &2x2 + a^yi = a262 according as b'^Xi^ + a^yi^ — aW is positive, zero, or negative. (c) &2x2 _ diyi — CJ252 according as Hh:,^ — a'hfi^ — a262 is negative, zero, or positive. 3. Two perpendicular tangents to (a) a parabola intersect on the directrix. (b) an ellipse intersect on the circle x2 +2/2 = a2 4. jj\ (c) an hyperbola intersect on the circle x^ ^- y^ = a"^ — &2. LINE AND CONIC 235 94. The equation in p. In the following sections we shall suppose that the line is given in parametric form (Theorem XV, p. 124), (1) Cx = Xi + \y = yi + p cos a, P cos p. The geometric significance of these equations should be constantly borne in mind. A line is given which passes through P^ (x^, y^) and whose direction cosines (p. 123) are cos a and cos /3 (or whose slope is m ■■ sm a cos /3 by (I), p. The point {x, y) or cos a cos a (a^i + p cos o, 2/i + p cos ^) is that point on the line whose directed distance from P^ is the variable p. Suppose the conic is the parabola >■ (2) 2/2 - 7;9X = 0. If the point (xx + p cos a^y^-^p cos jS) on (1) lies on (2), then (Corollary, p. 53) {yx + p cos i3)2 - 2 p (xi + p cos a) = 0, or (3) cos2 /3 • /32 + (2 2/1 cos ^ -2p cos a)p + (yi^ - 2pxi) = 0. This equation is called the equation in p for the parabola. Its roots, p\ and /92, are the directed lengths P1P2 and P1P3 from Pi to the points of intersection of the line and parabola. For p is the distance from P^ to the point {x^ + p cos a, t/j 4- p cos )3); and when p satisfies equation (3) the point \X^ + p cos a,yi + p cos /3) lies on the parabola. Hence Theorem HI. The directed distances from Pi (xi, 2/1) to the points of inter- iction of the line X = Xi + /? cos a^ y = y\-\- p cos ^ md the parabola ij"^ = 2px are the roots of the equation in p, cos2 /3 • p2 4. (2 2/1 cos /3 - 2p cos a)p -\- {tji^ - 2pxi) = 0. 236 ANALYTIC GEOMETRY The equation in p for any conic is the equation whose roots are the distances from a point Pi to the points of intersection of tlie conic and the line through Pi whose direction angles are or and ^. The method used in proving Theorem III is general and justifies the Rule for forming the equation in p for any conic. Substitute Xi + p cos a for x and yi -i- p cos /3 for y in the equation of the conic and arrange the result according to powers of p. For convenience of reference we state the following theorems which are proved by this Rule. Theorem IV. The equation in pfor the central conic tP-x'^ ± a'^y'^ — a%'^ = is (IV) (62 cos2 a ± a2 cos2 ^)p2 + (2 hHx cos or ± 2 a'^yx cos /3)p + (62xi2 ± cc^y^^ - a^b'^) = 0. Theorem V. The equation in p for the locus of Ax^ + Bxy + (72/2 + Dx + ^y + F= is (V) {A cos2 or + E cos a cos jS + C cos2 ^) ^2 + [(2 Axi + Byi + D) cos a + {Bxi + 2 Cyi + E) cos /3] p + (^Xi2 + Bxm + Cyt' + Dxi + Eyi -h F) = 0* The relative position of the line (1) and a conic depends upon the discrimi- nant of the equation in p. For according as the roots of the equation in p are real and unequal, real and equal, or imaginary (Theorem II, p. 3), the line and conic will intersect, be tangent, or not meet at all. PROBLEMS 1. Find the equation in p for each of the following conies. {&) xy = 8. (e) 2x^ + xy + Sx-iy = 0. (b) x2 + 2/2 = 9. {i) x'^ + 2 xy -^ y'^ - 4x = 0. (c) 8x2 _ y2 ^16. (g) xy + 4x - 8y - 3 = 0. (d) x2 - 2/2 -f 4 X - 6 y = 0. (h) x2 + 4 xy + 2/2 - 3 x = 0. 2. What can be said of the coeflBcients and roots of the equation in p (a) if Pi (xi, 2/i) lies on the conic ? (b) if the line is tangent to the conic at Pi ? (c) if the line meets the conic at infinity ? (d) if Pi is the middle point of the chord formed by the line ? * Notice that the coeflacient of p2 is found by substituting cos a for x and cos j8 for y in the terms of the second degree in the given equation. The constant term is found by substituting x\ for x and y^ for y throughout the given equation. Compare the coeffi- cients of cos a and cos /3 within the brackets with equations (6) and (7), p. 171. LINE AND CONIC 237 3. Determine the relative position of the following lines and conies and construct the figures. (a) ?/2 _ 4 X + 4 = (b) 4 xy + 3 ?/2 - 4 X + 4 ?/ - 16 = f X = 3 + 5 p, < n . , Ans. Secant. r (c) 4x2 + 92/2 - 40a; -72y + 100 = (d) 3x2 + x?/ - 4 ?/2 - X + y = Vio I VIO Ans. Tangent. 1- V2 Ans. Do not meet. \y=-2-hlp. Ans. Line is part of conic. (e) 4x2-9?/2 = 36 = 2-|p, = 3 + 4p. Ans. - Secant with one point of intersection at infinity. 12/ 95. Tangents, We shall show how to find the equation of a tangent to a conic by means of the equation in p by considering the tangent to the parab- ola 2/2 — 2_px = at the point Pi (xi, 2/i)- Let (1) X = Xi + /) cos or, 2/ = :yi 4- /9 cos /3 be any secant through Pi intersecting the parabola at Pq. One root of the equation in p is pi = P1P2 and the other is p2 = 0. Hence (III), p. 235, becomes (Case I, p. 4) cos2/3 • p2 + (2 2/1 COS ^ - 2p cos a)p = 0. [Or tlie constant term is zero by the Corollary, p. 53.] When P2 approaches Pi the line becomes tangent (p. 207), and as pi becomes zero we must have (Case III, p. 5) (2) 2 2/iCos j8 — 2p cos a = 0. This is the condition that (1) is tangent to the parabola. Solving (1) for cos a and cos /3 and substi- tuting in (2), we obtain 2yiy-2px-2 y^ + 2pxi = 0. But since y-^ — 2 pxi this reduces to 2/12/ - p (x + xi) = 0, which is the form given in Theorem III, p. 214. 238 ANALYTIC GEOMETRY ± - V— 1, so the slopes of 96. Asymptotic directions and asymptotes. If the coeflacient of p2 in the equation in p is zero, then one root is infinite (Theorem IV, p. 15) ; and hence the line and conic have one point of intersection at an infinite distance from Pi. The direction of such a line is called an asymptotic direction. Theorem VI. The asymptotic directions of the hyperbola are parallel to the asymptotes, of the parabola are parallel to the axis, while the ellipse has no asymptotic directions. Proof. Set the coefficient of p^ in the equation in p for the hyperbola [(IV), p. 236] equal to zero. This gives 62 cos2 a -a^ cos2 /3 = 0. cos/3 b .-. m = =±-' cos a a Therefore the slopes of the asymptotic direc- tions are the same as those of the asymptotes [(5), p. 190]. ^^ Similarly for the parabola m = — = 0, so cos a that the asymptotic direction is parallel to the axis. -,, , „. . ,., cosiS b For the ellipse, m like manner, m = = ± - ^ ' cos a a the asymptotic directions are imaginary ; that is,'there are no asymptotic directions. q.e.d. Corollary. Every line having the asymptotic direction of a conic intersects the conic in but one point in the finite part of the plane. If both roots of the equation in p become infinite, the line is said to be "tangent to the conic at infinity" and is called an asymptote. Using this definition of the asymptotes, we have, in justification of the prelimi- nary definition on p. 189, the following theorem. Theorem VII. The equation of the asymptotes of the hyperbola &2x2 - a2?/2 = am is 62^2 - a2y2 ^ 0. Proof. Both roots of the equation in p for the hyperbola [(IV), p. 236] will be infinite if (Theorem IV, p. 15) 62 cos2 a — a2 cos2 /3 = and 2 62^1 cos a: — 2 a'^yi cos ^ = 0. From the first equation, cos /3 = ± - cos a. Substituting in the second, we get bxi T ayi = as the condition that Pi should lie on an asymptote. But this is the condition that Pi should lie on one of the lines 6x =F ay = or 62x2 — a'^y'^ = 0. Hence this equation is the equation of the asymptotes. q.e.d. LINE AND CONIC 239 The method of the proof justifies the Rule for finding the equation of the asymptotes of any hyperbola. First step. Derive the equation in p (Rule, p. 236). Second step. Set the coefficients of p'^ and p equal to zero. Third step. Eliminate co8 a and cos ^ from these equations and drop the subscripts on Xi and yi. PROBLEMS 1 . Find the equations of the tangents to the following conies drawn from the points indicated. (The method of section 95 can be applied whether Pi lies on the conic or not.) (a) xy = 16, (4, 4). Ans. x + y = S. (b) x2 + 2xy = 4, (2, 0). Ans. x-\-y = 2. (c) x2 = 4 y, (0, - 1). Ans. x2 - (y + 1)2 = 0, or x = ±{y + 1). (d) x2-32/2 + 2x+19 = 0, (-1, 2). Ans. (x + 3?/ - 6)(x - 3y + 7) = 0. 2. Determine the slopes of the asymptotic directions of the following conies. (a) x2 - xy - 6 2/2 - 8 X = 0. Ans. i, - |. (h) xy-y^ + 4x-6 = 0. Ans. 0,1. (c) x2 + 4 x?/ + 4 2/2 _ 2 X = 0. Ans. - ^. (d) 4 x2 + xy + 2/2 _ 3 =: 0. Ans. None. (e) 9 x2 - 6 xy + y2 _ 2 2/ + 5 = 0. Ans. 3. (f) x2 + 6xy + 4y2 = lo. Ans. - i, - 1. (g) xy + Dx + Ey + F=0. Ans. 0, oo. 3. Determine whether the loci of the equations in problem 2 belong to the elliptic, hyperbolic, or parabolic type. 4. Find the equations of the asymptotes of the following hyperbolas. (a) xy - y2 + 2 X = 0. (b) 2x2 -xy -4 = 0. (c) x2-6xy + 8y2 = 10. (d) xy - 4 X - 3 y = 0. (e) 2x2-7xy + 3y2 = 14. (f) x2 - 4 y2 + 2 X + 8 y = 0. Ans. y + 2 = 0, x-y + 2 = 0. Ans. x = 0, 2x — y = 0. Ans. x-4y=0, x-2y = 0. Ans. X = 3, y = 4. Ans. 2x-y=:0, x-3y = 0. Ans. x-2y + 3 = 0, x + 2y-l = 0. 5. Find the equations of the asymptotes of the hyperbolas (a), (b), (f), and (g) in problem 2. Ans. (a) 75x-25y + 296 = 0, 50x + 25y - 184 = 0; (b)y + 4 = 0, x-y + 4 = 0; (f)x + 4y = 0, x + y = 0; (g) X + jE; = 0, y + D = 0. 6. Prove that the parabola has no asymptotes. 240 ANALYTIC GEOMETRY 7. Show that the asymptotic directions of the locus of Ax^ + Bxy + Cy^ + Dx -i- Ey + F = sue determined by the locus of Ax^ + Bxy + Cy^ = 0. 8. By means of problem 7 show that the locus of the general equation of the second degree belongs to the hyperbolic, parabolic, or elliptic type according a,s A = B^ - 4: AC is positive, zero, or negative. 9. Show how to determine the direction of the axis of any parabola by means of problems 7 and 8. 97. Centers. The problem of this section is to determine the center of symmetry, if there is a center, of the locus of (1) Ax'^ + Bxy + Cy^ -\- Dx + Ey + F = 0. That is, we seek a point Pi{Xi,'yi) which is the middle point of every chord of (1) drawn through it. If Pi is the middle point of the chord P2P3 formed by the line x = Xi-\- pcosa, y = yi + p cos jS, "^ then the roots of the equation in p must be equal numer- ically with opposite signs. Hence the coefficient of p in (V), p. 236, must be zero (Case II, p. 4). (2) .-. (2 Axi + Byi + D) cos a + {Bxi + 2 C?/i + E) cos ^ = 0. If Pi is the middle point of every chord passing through it, (2) is satisfied by all values of cos a and cos /3. For cos /3 = and cos a = we get (3) 2 Axi + Byi + D = 0, Bxi + 2 Cyi -{- E = 0, and if equations (3) are satisfied, (2) is always satisfied. We can solve (3) for a single pair of values of Xi and 2/1 (Theorem IV, p. 90) unless — = —, OT A = B^-AAC = Q, B 2C and the locus of (1) will have a single center. But if A = there will be no center unless at the same time — = — , when every point on the line 2 C E '2. Ax -^ By -{- B =i will be a center. Hence we have Theorem VIII. The locus of Ax'^ + Bxy + Cy^ + Dx-}- Ey + F=0 will have a single center of symmetry if A = B^ — 4: AC is not zero. If A = there will he no center unless — = — , when all of the points on a line will , , 2C E oe centers. Corollary. The center will be the point of intersection of the lines 2Ax-\-By + n = 0, Bjo^'^Cy -\-E = 0. LINE AND CONIC 241 If the locus is of the elliptic or hyperbolic type (p. 195), there will be a single center. But if the locus belongs to the parabolic type, there is no center unless the locus degen- erates. If the locus is a pair of parallel lines, then every point on the line midway between them is a center. To find the center in a numerical example we proceed as in the above proof as far as equations (3) and then solve those equations. Ans. (0, 0). Ans. None. Ans. (-12, -4). PROBLEMS 1 . Find the centers of the following conies. (a) x2 + xy - 4 = 0. (b) x'^ -2xy + y^ -4:X = 0. (c) xy-2y^ + 4x-4.y = 0. (d) jc2 _ 8x2/ + 16 ?/2 + 2 X - 8 2/ - 3 = 0. Ans. Any point of the line x — 4y + l=0. (e) x2 + 4xy + y2 - 8x = 0. Ans. (- |, f). (f) 4x2 + I2xy + 92/2 _ 2x + 6 = o. Ans. None. (g) 4x2 + 12x2/ + 92/2 -4x- 62/ -8 = 0. Ans. Any point of the line 2x + 32/ — 1 = 0. 2. If all the coefficients of the general equation of the second degree except B are constant, and if B varies so that B^ — iAC approaches zero, how does the center of the locus behave ? 98. Diameters. The locus of the middle points of a system of parallel chords of a curve is called a diameter of the curve. Consider the ellipse 62x2 + a22/2 = a262 and the system of parallel lines whose direction angles are a and /3. The para- metric equations of that line through Pi{xi, 2/1) are (Theorem XV, p. 124) jc = xi + p cos a, 2/ = 2/1 + /o cos /S. If Pi is the middle point of the chord, then the roots pi = P1P2 and p2 = P1P3 of the equation in p [(IV), p. 236] must be equal numerically with opposite signs. Hence (Case II, 2 b'^Xi cos a + 2 a^yi cos j8 = 0. cos/3 p. 4) Dividing by 2 cos a and setting m (p. 235), we get cos a 62xi + a'^myi = as the condition that (Xi, ^i) is the middle point of a chord whose slope is m. This is the condition that Pi should lie on the line DD' : 62x + a^rny = 0. 242 ANALYTIC GEOMETRY Hence we have Theorem IX. The diameter of the ellipse 62x2 + a22/2 = a262 bisecting all chords with the slope m is (IX) b^a^ + a^my = O. This reasoning may be applied to any conic, and justifies the Rule for deriving the equation of a diameter of a conic bisecting all chords with the slope m. First step. Derive the equation in p {Rule, p. 236). Second step. Set the coefficient of p equal to zero. Third step. Replace Xi and yi by x and y respectively, and by m. The result is the required equation. By this means we prove Theorem X. The diameter bisecting all chords with the slope m of the hyperbola b'^x^ - a^y- = a-b^ is b^x — a'^my = O ; parabola 2px is my = p. Corollary. All the diameters of the parabola are parallel to its axis, and every line parallel to the axis is a diameter. Theorem XI. The diameter of the locus of Ax'2 + Bxy + Cy^ + Dx + Ey + F=0 bisecting all chords of slope m is (XI) 2Ax + By + n + m {Bx ■\- ^'Cy -\- E) = O. Corollary. The diameter passes through the center if the locus has a center, and every line through the center is a diameter. Hint. Apply the Corollary, p. 240, and Theorem XIII, p. 119. Cr LINE AND CONIC 243 PROBLEMS 1. Find the equation of the diameter of each of the following conies which bisects the chords with the given slope m. (a) x2 - 4 2/2 = 16, m = 2. Ans. x-Sy = 0. (b)2/2 — 4x, wi = — -J. Ans. y + 4 = 0. (c) xy = 6, m = 3. Ans. y -{-Sx = 0. (d) x2 - xy — 8 = 0, m = l. Ans. x - y = 0. (e) x2 - 4 y2 + 4 a; - 16 = 0, m=-l. Ans. x + 4?/ + 2 = 0. (f) xy + 2y^-4x-2y-\-e = 0,m = i. Ans. 2x + lly -16 = 0. 2. Find the equation of that diameter of (a) 4 x2 + 9 2/2 = 36 passing through (3, 2). Ans. 2x-3y = 0. (b) 2/^ = 4x passing through (2, 1). Ans. y = \. (c) xy = 8 passing through (— 2, 3). ^ns. 3 x + 2 y = 0. (d) x2 — 42/ + 6 = passing through (3, — 4). Ans. x = 3. (e) xy — 2/2 4. 2 X — 4 = passing through (5, 2). Ans. 4x — 9?/ — 2 = 0. 3 . Find the slope of (XI) if ^ - 4 ^ C = 0. How may the result be inter- preted by means of problems 8 and 9, p. 240 ? 4. What relation exists between m\ the slope of (XI), and m ? Ans. 2 Cmm' + 5 (m + m') + 2^ = 0. 5. What does the result of problem 4 become for 62 (a) the ellipse 62x2 4- ahj^ = a^b^ ? -Ans. mm' = • a2 62 (b) the hyperbola 62x2 - a2y2 = a^^ ? Ans. mm' = — . a2 (c) the parabola y^ = 2px? Ans. m' = 0. 6. By means of problem 5 discuss the relative directions of a set of parallel chords and the diameter bisecting them. 7. Find the equation of the chord of the locus of (a) x2 + 2/^ = 25 which is bisected at the point (2, 1). Ans. 2x + y — 5 = 0. (b) 4x2 — 2/2 = 9 which is bisected at the point (4, 2). Ans. 8 X - y - 30 = 0. (c) xy = 4 which is bisected at the point (5, 3). Ans. 3x + 5y- 30 = 0. (d) x2 — xy — 8 = which is bisected at the point (4, 0). Ans. 2x-y-8 = 0. 244 ANALYTIC GEOMETRY 99. Conjugate diameters of central conies. In every system of parallel chords of a central conic there is one which passes through the center and which is therefore a diameter (Corollary to Theorem XI). This diameter and the one bisecting the chords parallel to it are called conjugate diameters. The ellipse Let m be the slope of a diameter of the ellipse &2x2 + a2y2 = aW. From Theorem IX the slope of the conjugate diameter is (Corollary I, p. 86) 62 m' = 62 .-. mm = a2 Hence Theorem xn. If m and m' are the slopes of two conjugate diameters of the ellipse h^x^ + a^^ = a^b^, then &2 (XII) niTn' = — a^ Corollary. Conjugate diameters of the ellipse lie in different quadrants. For tn and m' have opposite signs since tlieir product is negative. The hyperbola Let m be the slope of a diameter of the hyperbola 62x2 - a22/2 = a262. From Theorem X the slope of the conjugate diameter is (Corollary I, p. 86) m' = .-. mm' 62 62 a2* Hence ^ Theorem Xm. If m and m' are the slopes of two conjugate diameters of the hyperbola b'^x'^ — a^y^ = a^b% then (XIII) rmnf = 62 "%. y' yf/ ^V^V ^,^ \v: PPf ^V\ l^^^J^ A- aJ^^ \v ^^ y^''^ "^^^nVV /^^ ^^K v/ y' vS Corollary. Conjugate diameters of the hyperbola lie in the same quad- rant, but on opposite sides of the asymptotes. For m and rw'-liave the same sign since their product is positive, and if one is numerically less than _ , the other must be a T numerically greater than _ which is the a slope of one asymptote [(5), p. 190]. LINE AND CONIC 245 The ellipse The length of a diameter of the ellipse, or of its conjugate diameter, is that part of the line included between the points of intersection of the line and the ellipse. Construction. To construct the diameter conjugate to a given diam- eter AB, draw a chord EF parallel to AB, and then draw the diameter CD bisecting EF, The hyperbola The length of that one of two con- jugate diameters of the hyperbola which does not meet the hyperbola is defined to be that part of the line included between the branches of the conjugate hyperbola (p. 189). Construction. To construct the diameter conjugate to a given diam- eter AB, draw a chord EF parallel to AB, and then draw the diameter CD bisecting EF. Theorem XIV. Given a point Pi (^1? Vi) on the ellipse b^x^ + a^y^ = a^b"^, the equation of the diameter conjugate to the diameter through Pi is (XIV) b^acix + aHjxy = O. Proof. The diameter through Pi passes through the origin (Corollary, p. 242), and hence its slope is (Theo- rem V, p. 35) m = — . Then, from X\ (XII), the slope of the conjugate y^x-i diameter is m' = ^ — The equa- a'^yx tion of the line through the origin with the slope m' is (Theorem V, p. 95) 62xi y = -^x, a^yi which may be written in the form (XIV). Q.E.D. Corollary. The points of intersec- tion of (XIV) with the ellipse are \ b a / \ b a / Theorem XV. Given a point Pii^h Vi) on the hyperbola 6^x2 _ a^y2 = a^b^, the equation of the diameter conjugate to the diameter through Pi is (XV) b^ociQc — ahjiy = O. Proof. The diameter through Pi passes through the origin (Corollary, p. 242), and hence its slope is (Theo- rem V, p. 35) m = — - Then, from Xi (XIII), the slope of the conjugate b^Xi diameter is m' = -s— • The equa- a^yi tion of the line through the origin with the slope m' is (Theorem V, p. 95) b'^xi a^yi which may be written in the form (XV). Q.E.D. Corollary. The points of intersec- tion of (XV) with the conjugate hyper- bola are \ b a / \ b a / These are found by the Rule, p. 76. These are found by the Rule, p. 76. 246 ANALYTIC GEOMETRY PROBLEMS 1. "What is the relation between the slopes of conjugate diameters of the equilateral hyperbola 2 xy = a^ ? Ans. m + m' = 0. 2. The tangents at the ends of a diameter of (a) an ellipse, (b) an hyper- bola, are parallel to the conjugate diameter. 3. The tangent at the end of a diameter of a parabola is parallel to the chords which the diameter bisects. 4. The sum of the squares of two conjugate semi-diameters of an ellipse equals a^ + 62. Hint. Let P^ (ajj, y-^ be any point on the ellipse. Find the squares of the distances from the center to P^ and to one of the points in the Corollary to Theorem XIV, add, and apply the Corollary, p, 53. 5. The difference of the squares of two conjugate semi-diameters of an hyperbola equals a^ — &2. Hint. See the hint to problem 4. 6. The angle between two conjugate diameters is sin-^^^, where a' and h' are the lengths of the conjugate semi-diameters. ^ 7. Conjugate diameters of an equilateral hyperbola are equal in length. 8. Conjugate diameters of an equilateral hyperbola are equally inclined to the asymptotes. 9. The lines joining the ends of conjugate diameters of an hyperbola are parallel to one asymptote and bisected by the other. 10. The product of the focal radii (problems 8 and 9, p. 194) drawn to any point on (a) an ellipse, (b) an hyperbola, equals the square of the semi- diameter conjugate to the diameter drawn through that point. 11. The asymptotes of an hyperbola are conjugate diameters of an ellipse which has the same axes as the hyperbola. 12. Show that the conjugate diameters of the ellipse in problem 11 are equal. MISCELLANEOUS PROBLEMS 1. Find the condition for tangency of (a) 2/2 = 2px and Ax ^- By + C = 0. (b) &2a;2 + a2y2 - a^p- and J.x + 2^?/ + O = 0. (c) 62^2 _ a27/2 ^ ^262 and Ax ^ By ^ C = 0. 2. Find the points on each of the conies where the tangents are equally inclined to the axes. When is the solution impossible ? LINE AND CONIC 247 3. Find the points on the ellipse where the tangents are parallel to the line joining the positive extremities of the axes. 4. The perpendicular from the focus of a parabola to a tangent intersects the diameter drawn through the point of contact on the directrix. 6. The perpendicular from a focus of a central conic to a diameter inter- sects the conjugate diameter on the directrix. 6. Tangents at the extremities of a chord of a parabola intersect on the diameter bisecting that chord. 7. Find the equation of (i) the ellipse, (b) the hyperbola, referred to conjugate diameters as axes of coordinates. (See problem 10, p. 172.) 8. Given the equation in p for an equation of the second degree; what may be said of the relative positions of the line, the conic, and the point Pi (a) if the constant term is zero ? (b) if the coefficient of p is zero ? (c) if the coefficient of p^ is zero ? (d) if the coefficient of p and the constant term are zero ? (e) if the coefficients of p^ and p are zero ? (f) if the coefficients of p^ and p and the constant term are zero ? (g) if the discriminant is positive, negative, or zero ? 9. Tangents to an hyperbola at the extremities of conjugate diameters intersect on the asymptotes. 10. The area of the parallelogram formed by tangents at the extremities of conjugate diameters of (a) an ellipse, (b) an hyperbola, is 4 ab, that is, it is equal to the area of the rectangle whose sides equal the axes. 11. The diagonals of the parallelogram circumscribing the ellipse in problem 10 are conjugate diameters. 12. Chords drawn from a point on (a) an ellipse, (b) an hyperbola, to the extremities of a diameter are parallel to a pair of conjugate diameters. 13. The directrix of a parabola is tangent to the circle described on any focal chord as a diameter. 14. The tangent at the vertex of a parabola is tangent to the circle described on any focal radius as a diameter. CHAPTER XI LOCI. PARAMETRIC EQUATIONS 100. The first fundamental problem (p. 53) of Analytic Geom- etry is to find the equation of a given locus. In this chapter we shall first give some additional problems which may be solved by the Rule on p. 53, using either rectangular or polar coordinates as may be more convenient. We shall then consider two classes of loci problems which are not readily solved by that Rule and which include nearly all of the important loci occurring in Ele- mentary Analytic Geometry. '^ PROBLEMS It is expected that tlie locus in each problem will be constructed and discussed after its equation is found. 1. The base of a triangle is fixed in length and position. Find the locus of the opposite vertex if (a) the sum of the other sides is constant. Arts. An ellipse. (b) the difference of the other sides is constant. * Ans. An hyperbola. (c) one base angle is double the other. Ans. An hyperbola. (d) the sum of the base angles is constant. Ans. A circle. (e) the difference of the base angles is constant. Ans. A conic. (f) the product of the tangents of the base angles is constant. Ans. A conic. (g) the product of the other sides is equal to the square of half the base. Ans. A lemniscate (Ex. 2, p. 152). (h) the median to one of the other sides is constant. Ans. A circle. 2. Find the locus of a point the sum of the squares of whose distances from (a) the sides of a square, (b) the vertices of a square, is constant. Ans. A circle in each case. 3. Find the locus of a point such that the ratio of the square of its dis- tance from a fixed point to its distance from a fixed line is constant. Ans. A circle. 248 LOCI 249 4. Find the locus of a point such that the ratio of its distance from a fixed point Pi (cci, yi) to its distance from a given line Ax + By + C = is equal to a constant k. Ans. {A^ + ^2 _ ^2^2) a;2 _ 2 k'^ABxy -\- {A^ -\- m - k^B^) y^ - 2 {A^xi + B^i-{-k^AC)x-2 {A'^yi + B^y^ + k^BC) y + (Xl2 + 2/i2) (^2 4. J52) _ ^2(72 = 0. 5. Find the locus of a point such that the ratio of the square of its dis- tance from a fixed line to its distance from a fixed point equals a constant k. Ans. x^ — k-{x — p)2 — k'^y'^ = if the F-axis is the fixed line and the X-axis passes through the fixed point, p being the distance from the line to the point. 6. Find the locus of a point such that (a) its radius vector is proportional to its vectorial angle. Ans. The spiral of Archimedes, p = a9. (b) its radius vector is inversely proportional to its vectorial angle. Ans. The hyperbolic or reciprocal spiral, pd = a. (c) the logarithm of its radius vector is proportional to its vectorial angle. Ans. The logarithmic spiral, logp = ad. (d) the square of its radius vector is inversely proportional to its vectorial angle. Ans. The lituus, p^d = a^. 101. Loci defined by a construction and a given curve. Many important loci are defined as the locus of a point obtained by a given construction from a given curve. The method of treatment of such loci is illustrated by Ex. 1. Find the locus of the middle points of the chords of the circle x2 + y2 -= 25 which pass through Pg (3, 4). Solution. Let Pi(Xi, yi) be any point on the circle. Then a point P (x, y) on the locus is obtained by bisecting P1P2. By the Corollary, p. 39, X = l(Xi + 3), 2/=l(yi + 4). (1) .-. a:i=:2x-3, y^ = 2y-4. Since Pi lies on the circle (Corollary, xi^ + 2/i2 = 25. Substituting from (1), (2 X - 3)2 + (2 ?y - 4)2 = 26, ic2 + 2/2_3a;_4y_o. p. 53), r> K ^ ^ ba: s m} ) / / / / / - / / \ <^ -?/: .Y .> X \ \ Wn g Vt) \ -'1 ^ ^ - y r' 250 ANALYTIC GEOMETRY As this equation expresses analytically that P(x, y) satisfies the given condition, it is the equation of the locus. The locus is easily seen to be a circle described on OP2 as a diameter, since its center is the point (|, 2) and its radius is f (Theorem I, p. 131). The method may evidently be expressed as follows : Rule for finding the equation of a locus defined by a construction and a given curve. First step. Find expressions for the coordinates of any point P\ (cci, 2/1) on the given curve in terms of a point P (x, y) on the required curve. Second step. Substitute the results of the first step for the coordi- nates in the equation of the given curve and simplify. The result is the required equation. This Eule may also be applied if polar coordinates are used instead of rectangular coordinates. Ex. 2. The witch. Find the equation of the locus of a point P constructed as follows : Let OA be a diameter of the circle x'^ -\-y^ — 2ay = and let any chord OPi of the circle meet the tangent at ^ in a point B. Lines drawn through Pi and B parallel respectively to OX and OY intersect at a point P on the required locus. Solution. First step. Let {x, y) be the coordinates of P and (xi, yi) of Pi. Then from the figure (1) 2/1 = y- From the similar triangles OCPi and PiPB we have xi yi (2) 00 _ CPi PiP ~ PB OM-OC=x- X — Xi 2a — y For OC = «i, P^P = OM- OC=x-x^, CP^= iji, P£ = MB- Solving (1) and (2) for Xi and ?/i, we obtain (3) MP=2a-y. xy 2a Second step. Substituting from (3) in the equation of the given circle x2 + ?/2 — 2 ay = 0, we get x2'i/2 + y'' or (4) 4a2 x2y = 4a2(2a 2 ay = 0, y)- LOCI 251 The locus of this equation is known as the witch of Agnesi. Discussion (p. 74). 1. The witch does not pass through the origin (Theo- rem VI, p. 73). 2. The witch is symmetrical with respect to the F-axis (Theorem V, p. 73). 3. Its intercept on the F-axis is 2 a, but the curve does not meet the X-axis (Rule, p. 73). 4. No values of x need be excluded, but all values of y must be excluded except 02a or if y <0, and hence these values must be excluded. 5. The witch extends indefinitely to the right and left and approaches nearer and nearer to the X-axis. For from the first of equations (5), as x increases without limit y decreases and approaches zero. Ex. 3, The conchoid. Find the locus of a point P constructed as follows : Through a fixed point A on the F-axis a line is drawn cutting the X-axis at Pi. On this line a point P is taken so that PiP = ± 6, where 6 is a constant. (5) Solution. First step. Use polar coordinates, taking A for the pole and A ¥ for the polar axis. Then it AO = a the equation of XX' is (6) p = a sec d. For the equation of a line perpendicular to the polar axis has the form (p. 156) C C pA cos + C = 0, or p = sec 6, and its intercept on the polar axis is A A If the coordinates of P are (/?, 6) and of Pi are {pi, 6i), then in any one of the figures we have by definition AP = ^Pi ± b. .'. 0i = d,pi = pTh. 252 ANALYTIC GEOMETRY Second step. Substituting in (6), we obtain (7) p = a sec d ±h. The locus of this equation is called the conchoid of Nicomedes. It has three distinct forms according as a rs greater, equal to, or less than 6. Discussion (p. 151). 1. The intercepts on the polar axis AY are a+ 6 and a — b. The pole also will lie on the conchoid if a sec 6±b = 0or9 = sec- 1 f ± -)• 2. The conchoid is symmetrical with respect to the polar axis AY. For sec (- 6) = sec 6 by 4, p. 19. 3. The conchoid recedes to infinity in the two opposite directions perpen- dicular to the polar axis A Y. For if = - or ^ — , sec = oo and hence p = oo . 2 2 j^p 4. If we transform to rectangular coordinates, using (2), p. 155, we get (x2. + 2/2)(x-a)2 = 62ic2. A is now the origin and J. Y the positive axis of X. To translate the axes 7t to and rotate them through we set (Theorem III, p. 164) x = y' -\- a, y — — x'. We thus obtain (8) x2?/2 = (2/ + a)2(62-2/2), which is the equation of the conchoid referred to OX and OY. From (8) it is easily seen that the conchoid approaches nearer and nearer to the X-axis as it recedes from the origin. PROBLEMS 1. Find the locus of a point whose ordinate is half the ordinate of a point on the circle x'^ ■\- y"^ = 64. Ans. The ellipse a:2 + 4 ?/2 = 64. 2. Find the locus of a point which cuts off a part of an ordinate of the circle x'^ + 2/2 = a2 whose ratio to the whole ordinate is h-.a. Ans. The ellipse 62^2 + aHj'^ = a^b^. 3. Find the locus of a point which divides an ordinate of (a) x^-\-y- = r^^ (b) y2 _ 2px, (c) 2xy = a^ into segments whose ratio is X. Ans. (a) X2x2 + (1 + X)2?/2 = x2r2 ; (b) (1 + \)^y2 = 2\^px; (c) 2 (1 + X) xy == Xa2. LOCI 253 4. Find the locus of the middle points of the chords of (a) an ellipse, (b) a parabola, (c) an hyperbola which pass through a fixed point P^ (X2, 2/2) on the curve. Ans. A conic of the same type for which the values of a and 6 or of p are half the values of those constants for the given conic. 5. Lines are drawn from the point (0, 4) to the hyperbola x^ — 4^/2 = 16, Find the locus of the points which divide these lines in the ratio 1 : 2. Ans. 9 x2 - 36 y2 + 192 y _ 272 = 0. 6. Lines drawn from the focus of the conic (II), p, 178, are extended their own lengths. Find the locus of their extremities. Ans. A conic with the same focus and eccentricity whose directrix is x = -2p. 7. Lines are drawn from a fixed point P2 {x^, 2/2) to (a) the line Ax + By + C = 0, (b) the parabola y'^ = 2px, (c) the central conic Ax^ + By^ + F=0. Find the locus of the points dividing these lines in the ratio X. Ans. (a) a straight line parallel to the given line ; (b) a parabola whose axis is parallel to that of the given parabola ; (c) a central conic whose axes are parallel to those of the given conic. 8. Find the locus of the middle points of chords of an ellipse which join the extremities of a pair of conjugate diameters. Ans. 2 62x2 + 2 a^y^ = a^l^. 9. A chord OPi of the circle x"^ + y^ -{■ ax = which passes through the origin is extended a distance PiP = a. Find the locus of P. Ans. The cardioia | (^' + ^' + <^)^ = ''Y + ^'); i,or p = a{l — cose cos 6). 10. A chord OPi of the circle x^ + 7/^ — 2ax = meets the line x = 2 a at a point A. Find the locus of a point P on the line OPi such that OP=PiA. f y^(2 a x) ^ x^ Ans. The cissoid of Diodes v « ' ■ t or p = 2asmd tan d. 11. Find the locus of the point P in problem 9 if PiP = b. Ans. The lima5on of Pascal, p = b — a cos d. The limaQon has three dis- tinct forms according as 6 = a. 102. Parametric equations of a curve. Equations (XY), p. 124, X = Xi-\- p cos a, y — y^-{- p COS (i, are called the parametric equations of the straight line because they give the values of the coordinates of any point {x, y) on the line in terms of a single variable parameter p. In general, if two 254 ANALYTIC GEOMETRY equations give the values of the coordinates of any point (x, y) on a curve in terms of a single variable parameter, those equations are called parametric equations of the curve. Ex. 1. Find parametric equations of the circle whose center is the origin and whose radius is r. Solution. Let P (», y) be any point on the circle and denote Z XOP by 6. Then from the figure (1) x = r cos e^ y — r sin d. These are the required equations. They possess two properties analogous to those of the equation of the locus (p. 53). 1. Correspgnding to any point P on the locus there is a value of d such that the values of x and y given by (1) are the coordinates of P. 2. Corresponding to every value of Q for which the values of x and y given by (1) are real numbers there is a point P (x, y) on the locus. The parameter in the parametric equations of a curve may be chosen in a great many ways, and hence the parametric equations of the same curve will often appear in very different forms. Thus in Ex. 1, if we had chosen for the parameter half the abscissa of P, denoting it by ^, then ^ = - ' and from the figure y - ±. V7*2 — x^, whence the parametric equation would have been x = 2 <, y = ± Vr^ — 4 1^. Rule to 'plot a curve whose parametric equations are given. First step. Assume values of the parameter and compute the corresponding values of x and y from the given equations. Second step. Plot the points whose coordinates are found in the first step. Third step. If the points are numerous enough to suggest the general shape of the locus, draw a smooth curve through the points. Ex. 2. Plot the curve whose parametric equations are (2) X = at^, y = aH\ Solution. Take a = ^. Then equations (2) become (3) x = it^ y = itK LOCI 255 First step. Assume values of t and compute x and y from (3) , For example, if t = 2, X = i22 = 2, ?/ = i 23 = 2. This gives the table : I FA t ic y 1 .5 .25 2 2 2 3 4.5 6.75 etc. etc. etc. - 1 .5 - .25 -2 2 -2 -3 4.5 -G.75 etc. etc. etc. Second step. Plot the points found. Third step. Draw a smooth curve through these points as in the figure. Rule to find the equation of a curve in rectangular coordinates whose parametric equations are given. Eliminate the parameter from the parametric equations. We shall justify the Kule for the examples in this section. In Ex. 1, if we square each of the equations (1) and add, we obtain (3, p. 19) x2 + y2 = r^ which is the equation of the given locus (Corollary, p. 58). In Ex. 2, if we cube the first of equations (2) and square the second, we get a;3 = aH% y^ = aH^. (4) .-. ?/2=aa;8. This is the equation of the semicubical parabola (p. 209) . To prove that (4) is the equation of the curve obtained in Ex. 2 we mi^st prove two things (p. 53) : 1. The coordinates of any point P\ {x\, yi) on the curve satisfy (4). ilf Pi (a;i, yi) is on (2), then (1, Ex. 1) there is a value ti such that (5) xi = atr^, yi = a'^tiK (6) .-. xi3 = aHi^, yi^ = aHi^. (7) .'.yi2=axiK Hence xi and yi satisfy (4) . 2. Ifxi and y\ satisfy (4), then Pi (xi, yi) is on the curve. For if (7) is true, then from the first of equations (5) we obtain a value ^i. Sub- stituting xi = ati^ in (7), we get yi = aHi^. Hence X\ and yi are given by (5), and 256 ANALYTIC GEOMETRY The parametric equations of a curve are important because it is sometimes easy to express the coordinates of a point on the locus in terms of a parameter when it is otherwise difficult to obtain the equation of the locus. Ex. 3. The cycloid. Find the parametric equations of the locus of a point P on a circle which rolls along the axis of x. 3T (8) Solution. Take for origin a point at which the moving point P touched the axis of x. Let a be the radius of the circle and denote the variable angle ABP by d. Then (p. 18) PC = a sin d, CB = a cos d. By definition, OA = arc^P = ad. For an arc of a circle equals Its radius times tlie subtended angle, from the definition of a radian (p. 19). Hence from the figure, if (x, y) are the coordinates of P, x = OD = OA-PC = ae-asmd, 2j = DP = AB-CB = a-acosd. 'X = aid -sin 6), y — a{l — cos^). These are the parametric equations of the cycloid. Discussion. 1. The cycloid passes through the origin, for if ^ = 0, x = yz=0. 2. The cycloid is symmetrical with respect to the Y-axis (Theorem IV, p. 72, and 4, p. 19). 3. Its intercepts on the X-axis are 2n7ta, where n is any positive or negative integer, or zero. For, from the second of equations (8), if y/ = 0, cos 9 = 1. .'. = 2mr ; and hence from the first of equations (8) x = a • 2mr. 4. The cycloid lies entirely between the lines y = and y = 2a, for - 1 0, A < ; = A = 0, 1)2-4 AF< 0. GENERAL EQUATION OF SECOND DEGREE 267 106. Degenerate conies of a system. Let the equations of two conies, degenerate or non-degenerate, be Ci : Aix^ -}- Bixy + Ci?/2 + DyX -h Eiy -{- Fi = and C2 : ^2^2 + Bzxy + C^y^ + Ax + Ezy + Fg = 0. (1) or (2) Then the equation Aix^ + Bixy + Ci?/2 + Dix + ^ly + Fi + A; (^2^2 + B^xy + C^y^ + DoX + ^22/ + 1^2) = 0, (^1 + kA2)x^ + (5, + ^J52)X2/ + (Ci + fcC2)y2 + (Z>i + JCD2) X + (^1 + A:^2) y + (i^i + kF2) = 0, where A: is an arbitrary constant, will represent a system of conic sections. If Ci and C2 intersect, all the conies of the system will pass through their points of intersection. This is proved as in the case of straight lines (Theorem XIII, p. 119) and circles (Theorem IV, p. 140). Ex. 1. Find the values of k for which the conies belonging to the sys- tem a;2 4. 2/2 - 4 + A: (x^ - 2/2 - 1) = are degenerate. Solution. The givep equation may he writ- ten in the form (3) (1 + A;) a:2 + (1 - k) y^ - (^ ^ k) = 0. Its discriminant is e = - 4:{1 + k) {1 - k) (4: + k). If the locus of (3) is degenerate, then (Theo- rem I, p. 266) e=^-4il + k)(l-k)i^-{-k) = 0. .: k = -l, 1, or -4. If A; = — 1, (3) becomes 21/2 — 3=0, or y = ± Vf. U k= 1, (3) becomes 2 a;2 — 5 = 0, or x = ± V|. If A; = - 4, (3) becomes 3x^-5y^ = 0, or y = ±V^x. \ i .- \ 1 1 / \ \ / ^ \ ^ T II / ■< s ^^ '«j A ^ Js fc— 2^\ {— k^ -1 / iS y^ \ A ' \ r ^\ \ k= -1 X ■i' ^^ J> vv / y ^ ^ / / V < / , \ \ / "v In each case the locus is a pair of lines. The figure shows the circle a:2 -|- 1/2 _ 4 = 0, the hyperbola a;2 — ?/2 and the three pairs of lines. 1 = 0. Theorem n. In mery system of conies whose equation has the form (1) there is at least one degenerate conic and., in general, there cannot be more than three. These are obtained by substituting for k, in the equation of the system, the roots of its discriminant 0. 268 ANALYTIC GEOMETRY Proof. The discriminant © of (1), when set equal to zero, gives an equa- tion of the third degree in k. For each term in © consists of the product of three of the coefficients of (2), and such products will contain the third power of k. The roots of this cubic equation will be the values of k giving the degen- erate conies of the system (Theorem I, p. 266). There are, therefore, not more than three values of k for which the locus is degenerate. In a special case, however, all of the coefficients in this cubic might be zero, in which case the locus of (2) is degenerate for all values of k (see problem 4, (c), p. 266). Two or all three of the roots might be equal, and hence there might be but two, or even but one, degenerate conic in the system. Two of the roots might be imaginary and hence could not be used. But one of them must be real,* and hence there is always at least one real value of k for which the locus of (1) is degenerate, t q.e.d. Systems of conies defined by equations of the form (1) are classified according to the nature of the common solutions of C^ and Cj. In Algebra it is shown that two equations of the second degree have, in general, four pairs of common solutions for x and y. Hence five cases arise : 1. Four distinct pairs of solutions. 2. Two pairs are identical and the other two pairs are distinct. 3. Three pairs are identical and the fourth pair is different. 4. Two pairs are identical and the other two pairs are also identical. 5. All four pairs are identical. If the four pairs of solutions are all real, then these five cases have the following geometrical interpretation. 1. Ci and Cj have four distinct points of intersection. All the conies of the system pass through these four points. There are three degenerate conies in the system [Ex. 1 and problem 1, (a)]. 2. Cj and Cg are tangent at one point and intersect in two other points. All the conies of the system are tangent at the first point and pass through the other two points. There are two degenerate conies in the system [problem 1, (b)]. 3. Cj and Cj are tangent at one point and intersect in a second point. All the conies of the system are tangent at the first point and pass through the second point. There is but one degenerate conic in the system [problem 1, (c)]. 4. Ci and C^ are bi-tangent, that is, tangent at two different points. All of the conies of the system are tangent at these two points. There are two degenerate conies in the system [problem 1, (d)]. 5. C-i and C^ are tangent at one point and do not intersect elsewhere. All of the conies of the system are tangent at this point. There is but one degenerate conic in the system [problem 1, (e)]. * In Algebra it is shown that the imaginary roots of an equation with real coefficients must enter in pairs. Hence if the degree is an odd number, one root, at least, must be real. t It is tacitly assumed, as is true, that not all of the roots of the discriminant, when substituted for k, give equations which have no locus. But this point is not essential for our further reasoning. GENERAL EQUATION OF SECOND DEGREE 269 PROBLEMS 1 . Find the values of k for the degenerate conies of the following systems. Plot Ci, C2, and the degenerate conies. (a) x2 + 2/2 _ 16 + A; (x2 + 92/2 _ 36) = 0. Ans. fc = - 1, -\, - f . (b) 4x2 + 2^2 _ i6x + A; (x2 ^y^-^x) = 0. Ans. A: = - 2, - 2, - 1. (c) x2 4. 2x2/ + 22/2 + 8x + 82/ + A:(x2 + 22/2 + 82/) = 0. Ans. fc = — 1, — 1, — 1. (d) x2 + 2/2 _ 36 + A; (x2 + 4 2/2 - 36) = 0. Ans. k = -l, - 1, - 1. (e) x"^ + y^ - 4:x + k{ix^ + y^ - 4:x) = 0. Ans. k = -l, - 1, - 1. 2. Find the points of intersection of Ci and C2 in problem 1. Ans. (a) (fV6,iVl0),(fV6,-iVl0),(-|y6,^Vi0),(-|V6,-|Vu;). (b) (0, 0), (0, 0), (I, I V2), (f , -fV2). (c) (0, - 4), (0, - 4), (0, - 4), (0, 0). (d) (6,0), (6,0), (-6,0), (-6,0). (e) (0, 0), (0, 0), (0, 0), (0, 0). 3. Discuss the following systems of conies. (a) x2 + 2/2 - 16 + A; (x2 - 4 y2 + 16) = 0. (b) x2 - 2 2/ + 4 + A; (x2 + 8 2/) = 0. (c) X2/ + 82/ + 8 + A:(X2/ + 8) = 0. (d) x2 + 2 2/2 - 8 + A; (x2 + 2/2 - 4) = 0. (e) x2 + 6 2/ + 9 + A; (x2 + 6 2/) = 0. (f) 2j^-4.x-\-k{y^ + ix) = 0. (g) x2 - 2/2 + 25 + A;(x2 -f 2/2) = 0. (h) x2 - 2/2 + A; (x2 + 2/2) = 0. (i) 2/2 - 4 X - 16 + fc(x2 - 2/2 + 8 X + 16) = 0. (j) x2 - 2/2 + A:(x2 - 42/2 - 3) = 0. 107. Invariants under a rotation of the axes. Lemma III. If the axes are rotated about the origin, then for any point whose old and new coordinates are respectively (x, y) and (x', y') we have x2 + 2/2 = x'-^ + 2/'2. Proof. To rotate the axes through an angle 6 we set (Theorem II, p. 162) J X =: x' cos ^ — 2/' sin ^, '^ \y = x' sin d -\- y' cos 6. Then x2 + y2 _ ^x' cos d - y' sin 9)^ + (x' sin 6 + y' cos 5)2 = x'2 (cos2 d -f sin2 5) + 2/'2 (siu2 5 + cos2 6) = x'2 + y-2^ (by 3, p. 19) Q.E.D. The lemma is evident geometrically since x^ + ?/2 and x'^ + if^ are the squares of the distance from the point to the origin [(JY), p. 31] in the new and old coordinates respectively. 270 ANALYTIC GEOMETRY "We are considering in this chapter the equation (2) Ax^ + Bxy + Cy^ -\- Dx -\- Ey + F = 0. If we substitute from (1) without simplifying the result, we obtain an equation of the form (3) A'x'-2 4- B'xY + Cy^ + D'x' + EY + F = 0, ^ which has the same constant term (Corollary, p. 170). Consider the system of conies (4) Ax^ + Bxy + Cy^ + Dx + Ey + F + k{x'^ -^ y^) = 0. If the axes be rotated by substituting from (1), the equation of the system becomes (6) ^ V2 + B'xY + CY^ + D'x' + EY -\-F+k (x'2 + y'^) = 0. For the left-hand member of (2) becomes the left-hand member of (3), and x^ + y' becomes a;'^ ^ y^2 \)y Lemma III. Denote the discriminants of (2), (3), (4), and (5) by 0, ©', ©i, and 0/ respectively. The locus of (4) is degenerate when and only when (Theorem I, p. 266) ei = i{A+k){C + k)F-\- BDE - {A + k)E-2 - {C -^ k) D'^ - FB^ = 0, or (6) AFk^-i-{4:AF-\-ACF-E^ -D'^)k + Q = Q* Similarly, the locus of (5) is degenerate when and only when (7) 4 Fk^ + (4 A'F + 4 C'F - E'^ - B'^) k ^ Q' = 0. The roots of (6) and (7) must be the same. For (4) and (5) are the equations of the same conic referred to different axes. Hence the locus of either equation is degenerate if the locus of the other is degenerate. Since the coefficients of A;^ in (6) and (7) are equal, the other coefficients must also be equal. Hence (8) 0' = and 4:A'F +^C'F- E'^ - D'^ = 4AF + 4CF - E^ - Z)2. An expression involving the coefficients A, B, C, D, E, and F whose value remains unchanged when the axes are changed is called an invariant of the general equation of the second degree under a transformation of coordinates. It is assumed in this definition that the equation in the new coordinates is not simplified by multiplying or dividing by a constant. An expression involv- ing the coordinates which remains unchanged when the equation in the new coordinates is simplified is called an absolute invariant. Hence, from (8), * This quadratic may be regarded as a cubic equation with one infinite root, by a theorem analogous to Theorem IV, p. 15. The locus of (4) for k=co is x^ + y^ = 0, which is one of the degenerate conies of the system. GENERAL EQUATION OF SECOND DEGREE 271 Theorem m. The discriminant of an equation of the second degree is invariant under a rotation of the axes. Corollary. The expression ^ = 4-42^ + 4 CF — E'^ — Ifi is invariant under a rotation of the axes. Lemma IV. An invariant of (9) Ax^ + Bxij + C?/2 + F = under a rotation of the axes which involves only A, J?, and C is also an invariant of (2). Proof. Substituting in (9) from (1), we obtain A'x'^ + B'xY + Cy^ -{-F = 0, where A', B\ and C have the same values as in (3). For when we substitute from (1) in Ax^ + Bxy + Cy^ we obtain only terms in x'^, x'y', and y'"^, and substituting in Dx + Ey in (2) we obtain only terms in x' and y'. Hence an expression involving only A^ B, and C will be an invariant of (2) if it is an invariant of (9). q.e.d. Theorem IV. The expressions A = B^-4tAC, 1I = A + C, are invariants of an equation of the second degree under a rotation of the axes. Proof. Consider the system (10) Ax^ -\-Bxy + Cy^ + F+k {x^ + 7/2) = o. Rotating the axes, this equation becomes (11) A'x'^ + B'xY + CY^ + F + A: (x'2 + y-i^ ^ q. Denote the discriminants of (10) and (11) by 0i and ©/. Then the locus of (10) is degenerate when and only when ■ ei = 4(A -\- k) (C + k) F - Fm = or (12) - 4 A;2 + 4 (^ + C) A: - (^ - 4 ^ C) = 0. Similarly, the locus of (11) is degenerate when and only when (13) 0/ = 4 A:2 + 4 {A' +C')k- {B'^ - 4 A'C) = 0. Since (10) and (11) have the same locus, (12) and (13) have the same roots. And since the coefficients of A;2 in (12) and (13) are equal, the remaining coefficients are equal. Hence B'^-iA'C' = B^-^AC and A'' + C = A -]- C. q.e.d. 272 ANALYTIC GEOMETRY Ex. 1. Transform x^-\-xy + x — 2y + 4: = 0'bj rotating the axes through — Compute A, H, and © for the given and required equations. Solution. In the given equation A=1,B = 1,C=0,D = 1,E = -2,F=4:. .'. H = 1 + = 1, A = 12 - 4 • 1 • = 1, 0=4-l-O-4 + l-l(-2)-l(-2)2-O-12-4-12 = -lO, To rotate the axes set (Theorem II, p. 162) X = X' COS — 4 -.^=^-f y = X' &in- + y'cos- = x' -\- y' V2 This gives, after removing parentheses but not clearing of fractions, (14) V2 3 ' + 4 = :0. Here A = l, B = -l, C-0, D = 1 E^- 3 V2' F = 4. .-. A = i, H = i, e = -io. Hence the values of A, H, and are unchanged. But if we clear fractions in (14) we obtain V2a;'2 _ V2xV - cc' - 3y' + 4 V2 =: 0. For this equation A = 2, H = V2, = - 20 V2. Hence A, H, and are not absolute invariants under a rotation of the axes. H2 H3 Theorem V. The expressions — and — are absolute invariants of an equa- A tion of the second degree under a rotation of the axes.* Proof. The given expressions are invariants because A, H, and are invariants. To show that they are absolute invariants we must prove that their values are unchanged when we multiply (15) Ax^ + Bxy + Cy2 i-Dx + Ey-h F=0 by a constant. Multiplying (15) by k, we get (16) kAx'^ + kBxy + kCy^ + kDx + kEy -\- kF = 0. Denote the invariants of (16) by A^-, H^-, and Qk- Then (17) Ax,. = k^B^ - 4 kAkC = k2{B^-^AC) = k^A. (18) Jlk = kA-\-kC = k{A + C) = kli. (19) Sk = k^ {iACF + BDE - AE^ - CD^ - FB^) = k^B. * The proof also holds for a translation of the axes after Theorems VI and VII are proved. GENERAL EQUATION OF SECOND DEGREE 273 Dividing the square of (18) by (17), Ak ~ A Dividing the cube of (18) by (19), H2 H3 Hence — and — are absolute invariants. q.e.d. A PROBLEMS H2 H3 1. Compute — and — for the equations in problem 2, p. 168, and also A for their answers. 2. The values of A\ B\ and C in (3), p. 270, are respectively the coefficients of x'2, x'y\ and y"^ in (4), p. 170, Compute the values of £'2 _ 4 vl'C and A' + C in terras of ^, B, and C rj 3. Show that -===:z^ is an invariant of the line Ax -i- By + C = ± V^2 + ^ under a rotation of the axes. 4. Show that — ^ - is an invariant of the line Ax-\-By -\- C = and the point Pi (iCi, ^i) under a rotation of the axes. 6. Show that V(xi - Xz)^ + (yi - 2/2)2 is an invariant of the points Pi{^u yi) and P2(X2, 2/2) under a rotation of the axes. 6. Show that ^ ^ ^— ? is an invariant of the lines Axx-\- Biy + Ci = AiA2-\- B1B2 and A2X + B^y + C2 = under a rotation of the axes. 7. Interpret geometrically the meaning of the invariants in problems 3 to 6. 108. Invariants under a translation of the axes. Theorem VI. The expressions A = B^-4AC, Ti=A-i-C are invariants of an equation of the second degree under a translation of the axes. Proof. If an equation of the second degree be transformed by translating the axes, the coefficients^, B, and C are unchanged (Corollary I, p. 171). Hence any expression involving these letters, as A or H, is an invariant. q.e.d. 274 ANALYTIC GEOMETRY Lemma V. If the axes are translated to the point {h, fc), then for any point P whose old and new coordinates are respectively (x, y) and (x', y') we have kx — hy = kx' — hy\ Proof. To translate the axes we set (Theorem I, p. 160) x = x' -\-h, y = y' -\-k. Then ' kx - hy :^k{x' + h) - h {7/ + k) = kx' — hy'. Q.E.D. The Lemma is evident geometrically since either kx - hy or kx^ - Tiy' is the area of the triangle whose vertices are P and the old and new origins [(VIII), p. 42]. Theorem VII. The discriminant of the equation (!) Ax^ + Bxy + Cy^+Dx-{-Ey +F=0 is an invariant under a translation of the axes (2) x = x' + h, y = y' + k. Proof. Consider the system (3) Ax^ + Bxy + Cy^ -\- Dx + Ey -{- F + k' {kx - hy) = 0. Substituting in (3) from (2), we obtain (4) Ax'^ + BxY + Cy'^ + D'x' + Wy' + F' + ¥ {k 0. From (6), if A<0, A' and C have From (6), if A>0, A' and C have the same signs and the locus belongs opposite signs and the locus belongs to to the elliptic type (p. 195). the hyperbolic type (p. 195). From (8), if 5^ 0, then F' ^0 and From (8), if 9^ 0, then F' ^0 and the locus is an ellipse if H and differ the locus is an hyperbola. in sign, or there is no locus if H and agree in sign. For A' and C" have the sign of H, from (7), and F' has the sign of 0, from (8). From (8), if = 0, then ii^= and From (8), if = 0, then i^=: and the locus is a point. the locus is a. pair of intersecting lines. The values of A^, C, and F^, if desired, may be found by solving (6), (7), and (8). Case II. A = and 0^0. The locus is a parabola (p. 180). Substituting from (3) in (5), we get C^ = H and — CD'- = ©, from wbicb the values of C and D' may be found if desired. , Case III. A = and = 0. Substituting from (4) in (5), we obtain the single equation C = H, which does not enable us to compare the signs of C" and F' in (III). But ^ = 4.AF +^CF -FT^ - D^ is invariant under a rotation of the axes, and when A = = 0, ^ is also an invariant under a translation of the axes. For, substituting the values of D', E', and F' given by (5), p. 170, and setting A' = A, B'=B, C'= C (Corollary I, p. 171) in ^' =iA'F' + ^ C'F' - E'-^ - Z>'2, weget ^'={4:CD-2BE)h + {^AE-2BD)k + ^. But if A = = 0, then 2 BD -4AE = 0, from (6), p. 265. Multiplying this by B and set- ting B2 = 'i AC {from A = 0), we have iACD - 2 ABE = 0, or 4 CD - 2BE= 0. Hence ^'= ^ For (III) we have ^ = 4 C'F\ and hence 4 C'F' = ^ Hence (p. 196) if ^ < 0, the locus is two parallel lines ; if ^ = 0, the locus is a single line ; if f > 0, there is no locus. The results of this section are embodied in GENERAL EQUATION OF SECOND DEGREE 277 Theorem IX. The nature of the locus of the equation Ax^ + Bxy + Cy^ -i- Dx + Ey + F .= depends upon the values of the invariants A = B^-4:AC, R = A-\-C, e = ^ACF-{- BDE - AE^ - CD^ - FB^, and ^ = 4:AF-\-^CF-E'2-D^, as indicated in the following table. 09^0 Conic. A<0 Ellipse, if H and differ in sign. No locus, if H and agree in sign. A = Parabola. A>0 - Hyperbola. = Degenerate Conic. A<0 Point. A = Two parallel lines, if ^ < 0. One line, if ^ = 0. No locus, if ^ > 0. A>0 Two intersecting lines. PROBLEMS 1 . Find the exact nature of the locus of (a) a^2 ^ 2xy -h 2y^ - 6x - 2y +9 = 0. (b) x^ - 2 xy -\- 2 y'2 - 4:y + 8 = 0. (c) x^-\-6xy + 9y^-j-2x-6y = 0. (d) x2 - 2 xy - 2/2 + 8 X - 6 = 0. (e) 4 x2 4- 9 2/2 + 4 X + 1 = 0. (f) 4x2 + 4x7/ + ?/2 + 4x4-2 2/ -48 = 0. (g) 4x2 -20x?/4- 25?/2 + 12x-30?/ + 9 = (h) 9x2 -12x?/ + 4?/2- 18x4-122/ + 34 = 0. (i) 3x2 - lOxy + 7 2/2 + 15x- 7 2/ -42 = 0. 2. Find a^ and b^, orp, for the following conies : (a) x2 - 2 X2/ + 2/2 - 8 X = 0. (b) 3x2 - 10x2/ + 32/3-8 = 0. (c) 5 x2 + 2 x?/ + 5 2/2 - 12 X - 12 2/ = 0. Hint. Compute the absolute invariants — and — for tlie given equation and for that A one of the typical forms (III), p. 179, (V) and (VI), p. 185, to which it may be reduced. Equate and solve for a^ and b^ or for jo. Ans. Ellipse. Ans. No locus. Ans. Parabola, Ahs. Hyperbola. Ans. Point. Ans. Two parallel lines. Ans. One line. Ans. No locus. Ans. Intersecting lines. Ans. p = V2. Ans. a2 = l, 62:^4. Ans. a2 = 3, 62 = 2. 278 ANALYTIC GEOMETllY 3. Show that A' and C in (I), p. 275, are the roots of the quadratic 4x2 — 4Hx — A = and show that they are always real. When will they also be equal ? 110. Equal conies. The object of this section is to determine when two conies whose equations are given are equal. The solution of this problem affords a further application of the theory of invariants. Theorem X. The axes of a non-degenerate central conic whose equation is Ax^ + Bxy + Cy^ -{- DX + Ey + F=0 H2 H3 are determined by the values of the absolute invariants — and — A Proof. The equation of a central conic may be reduced to the form [(11), P- 187] ^ + t = i a p The absolute invariants of this equation are H'2 \a /3/ (a + i8)2 A' - 4 - 4 a^ ' 0' - 4 - 4 a2^ a/3 a/3 Hence (Theorem VIII, p. 275) (a + ^)2 _ H2 {a + (8)3 _ H3 - 4 a^ A ' - 4 a2/32 ' (1) H2 H^ where — and — are known. These equations can be solved for a and /3, A and the values of the axes determined from them by 1 and 2, p. 187, and the definition of the axes (p. 185). q.e.d. Equations (1) may be solved as follows : Dividing the second by the first, (2) ^±1 = ^. Dividing the first of equations (1) by (2), (3) - ^™ Dividing (3) by (2), A2 4 ©2 Then, by Theorem I, p. 3, a and /3 are the roots of the quadratic equation 4K© 4 ©2 (4) x^ + ^^x-^-^ = 0, or A3a;2 + 4AH©a;-4©2 = 0. A2 A3 The roots of (4) are always real, for the discriminant is (4 AH©)2 - 4 A3 (- 4 ©2) = 16 A2©2 (H^ + A) = 16A2©2(^2 + 2^C+C2+52-4^0 = 16A202[(^-C)2+iB2], which is always positive when the coefiicients A, B, C, D, E, and F are real numbers. GENERAL EQUATION OF SECOND DEGREE 279 Theorem XI. The value of p for a parabola whose equation is Ax^ + Bxy +Cy^ + Dx-{-Ey + F=0 is determined by the value of the absolute invariant Proof For the parabola we have 7/2 = 2px H'3 IS 1 0' - 4l)2 4p2 Hence (Theorem VIII) 1 H3 4p2 - ' whence ^ = W-|3- Q.E.D. As the value of p is always a real number, © and H must have opposite signs. This may also be proved from the values of © and H by means of the condition A = 0. Theorem XII. Two non-degenerate conies C:Ax'^ + Bxy + Cy"^ + Dx + Ey + F = and C : A'x^ + B'xy + C'y^ + D'x + E'y + F' = are equal when and only when A' "* A ' 0' " " Proof If the conies are central conies, they are equal when and only when their axes are equal. But the axes of C and C are determined in the same JJ2 JJ3 JJ'2 JJ'3 manner from — and — and from — and — respectively (Theorem X). A A' 0' ^ "^ ^ ' Hence the axes are equal when and only when H'2 H2 ^ H'3 H3 — = — and — = A' A 0' If C and C" are parabolas, they are equal when and only when they have le same value of p, that is (Theorem XI), when and only when = Q.E.D. 0' 111. Conies determined by five conditions. The equation of any conic las the form (1) ^x2 + Bxy + Cy^-\- Dx + Ey + F=0, id the conic is completely determined if five of the coeflBcients are known terms of the sixth. Any geometrical condition which the curve must satisfy gives rise to an equation between one or more of the coeflBcients. Hence five conditions will determine the equation of a conic. The locus may be degenerate, or there may be no locus, which would mean that the five conditions are inconsistent. 280 ANALYTIC GEOMETRY Rule to determine the equation of a conic which satisfies five conditions. First step. Assume that the equation of the conic is Ax^ + Bxy + Cy"^ -{- Dx + Ey -\- F = 0. Second step. Find five equations between the coefficients, each of which expresses that the conic satisfies one of the given conditions. Third step. Solve these equations for five of the coefficients in terms of the Fourth step. Substitute the results of the third step in the equation in the first step and divide out the remaining coefficient. The result is the required equation. PROBLEMS 1. Show that the following pairs of conies are equal and determine the nature of the conies. (a) x^-iy'^-2x-16y -U = 0, Sx^ + 10xy -\- Sy^ - 2 = 0. (b) 9x2 + 24xy + 16y2_80x+602/=0, x^-2xy + y^-4:V2x-4:V2y=0. (e) x2 + 2/2 _ 2x - 8?/ - 8 = 0, x'^-\-y^ + 6x- lOy + 9 = 0. (d) 2 x2 + ?/2 - 12 X + 10 2/ + 41 = 0, 17 x^ -12xy -\- 22 y^ - 26 = 0. 2. Find the equations of the conies determined by the following condi- tions and determine the nature of the conic in each ease. (a) Passing through (0, 0), (2, 0), (0, 2), (4, 2), (2, 4). Ans. x^ — xy + y^ — 2x — 2y = 0. (b) Passing through (0, 0), (10, 0), (5, 3) and symmetrical to the A"-axis. Ans. 9x2 + 25^2 _ 90x = 0. * (c) Passing through (- 4, 0), (0, 4), (0, - 4), (5, 6) if A = 0. Ans. y^-4x-16 = 0. (d) Passing through (0, 5), (5, 0) and symmetrical with respect to both axes. Ans. x^ -\- y^ — 25 = 0. (e) Passing through (0, 0), (2, 1), (- 2, 4), (- 4, - 2), (2, - 4). Ans. 2 x2 - 3 X2/ - 2 ?/2 = 0. (f) Passing through (0, 2), (— 2, 0), (2, — 8) and symmetrical with respect to the origin. Ans. x^ -{■ ixy + y^ — i = 0. 3. Show that, in general, two parabolas may be constructed which pass through four given points. 4. Find the parabolas passing through the following points and construct the figures. (a) (0, 2), (0, -2), (4, 0), (-1, 0). Ans. x^ ±2xy -\- y^ - Sx - 4: = 0. (b) (2,0), (0,-8), (-2,0), (0,2). Ans. 4 x2 ± 4 xy + ?/2 + 6 y - 16 = 0. (c) (0,1), (0,-1), (2,0), (-1,0). Ans. x^±ixy + iy^ — x — 2y — 2 = 0. CHAPTER XIII EUCLIDEAN TRANSFORMATIONS WITH AN APPLICATION TO SIMILAR CONICS 112. An operation which replaces a given figure by a second figure in accordance with a given law is called a transformation. If a transformation replaces the points of one figure by the points of a second, it is called a point transformation. If a point transformation replaces P(x, y) by P'(x', y'), then the equations expressing x' and y' in terms of x and y, or conversely, are called the equations of the transformation. In this chapter we shall con- sider the transformations which replace a given figure by one equal or similar to it. They are called Euclidean transformations, because the properties of equal and similar figures are studied in the Elementary Geometry of Euclid. 113. Equal figures. Two figures whose corresponding lines and angles are equal may be brought into coinci- dence and are therefore equal. Equal figures in the same plane are said to be congruent if the corresponding parts are arranged in the same order, and sym- metrical if they are arranged in the opposite order. Thus the triangles ABC and A'B'C are congruent, and either is symmetrical to A"B"C'% because the directions established on the perimeters by the corresponding vertices are the same (clockwise) in the first case but are different in the second case. In Plane Geometry we do not study symmetrical figures as such. It is true that we study figures which are symmetrical with respect to a point or with respect to a line. But it should be noticed, as is seen from the figures, that figures which are symmetrical with respect to & point are congruent, while figures which are symmetrical with respect to a line are sym- metrical in the sense defined above. The essential distinction be- tween congruent and symmet- rical figures is this : two con- gruent figures may be brought into coincidence by moving them around in the plane, but before two symmetrical figures can be brought into coincidence one of them must be taken out of the plane and turned over. 281 282 ANALYTIC GEOMETRY 114. Translations. A translation is the transformation which moves all points of a figure through the same distance in the same direction. Hence if a translation replaces any point P by P', the projections of PP' on the axes will be constant. Theorem I. The equations of a translation through the directed length whose projections on the axes are respectively h and k are (I) 00^ = a^ -\- h, Proof. By Theorem III, p. 31, the projections of PP^ on the axes are respectively x' - X, y' - y. Then, by hypothesis, x'— x= h, y'— y = k. Solving for x' and y% we obtain (I). q.e.d. If we solve (I) for x and y and substitute their values in the equation of a curve, the result will X evidently be the equation of the curve after it has been translated. If P is the origin (0, 0), then P^ is the point (h, k). If we solve (I) for x and y, we obtain x = x'-h, y = y'-k. These may be regarded as the equations for translating the axes to a new origin {-h, -k) (Theorem I, p. 160). ^>^ o' ^.<^^^P) (1) ri (-hrk) (3) It is evident that the relative position of the new figure and the old axes (Fig. 1) is the same as that of the old figure and the new axes (Fig. 2). Hence it is. immaterial whether we regard equations (I) as the equations of a transla- tion of a figure in one 'direction or as the equations of a translation of the axes in the opposite direction. • 116. Rotations. The transformation which turns all points through the same angle about a given point is called a rotation. is called the center of the rotation. If a rotation replaces P by P', then OP' = OP. EUCLIDEAN TRANSFORMATIONS 283 Theorem n. The equations of a rotation about the origin through an angle e are = acco89 — y sin 9^ 05 sin ^ + y cos 0. (H) Proof. Let the polar coordinates of P be {p, + 6). Hence (Theorem I, p. 155) x' = p cos (0 + d) = p cos COS ^ — ^ sin sin 6 (by 10, p. 20) = X cos ^ — y sin ^, since [(I), p. 155] X = p cos ^, y = p sin 0. Similarly, y^ = xshid + y cosd. q.e.d. Q / / / Ad If we solve (II) for x and y, we get x = x' cos e + y^sine = x' cos (- 0) - y' sin (- 5), y = -x'9>m.Q ^y' cos Q — x' sin (- 0) + y' cos (- 0). (by 4, p. 19) (by 4, p. 19) These may be regarded (Theorem II, p. 162) as the equations for rotating the axes tlirough an angle — Q. Hence it is immaterial whether we regard equations (II) as the equations of a rotation of a, figure in one direction or of the axes in the opposite direction. This should be illustrated by figures analogous to Figs. 1 and 2, p. 282. PROBLEMS 1 . Plot the following curves, translate them through the directed length whose projections are given, and find the equations of the curves in their new positions. (a) y^ = Ax, h = -3,k = 2. Ans. ?/2-4x-4?/-8 = 0. (b) xy = 6,h = 2,k = - 2. Ans. xy + 2x - 2y - 2 = 0. (c) x2 + 92/2 = 25, /i = 0, A: = |. Ans. x^ + 9y^ - 30y = 0. 2. Plot the following curves, rotate them about the origin through the given angle, and find the equations of the curves in their new positions. Ans. y^ -x^ = 16. Ans. x^-^y'^+Sx + 12 = 0.. Ans. 4 x2 + 2/2 4. 18 2/ = 0. (a) xy = 8,d It ~ 4 (b) x2 + 2/2 - 8x + 12 = 0, ^ = - It. (c) «2 + 42,2 -18x = 0, d = - It ' 284 ANALYTIC GEOMETRY 3. Translate the locus of x^ + 4 2/ = through a distance whose projec- tions are A = 0, k = — 4 and then rotate it about the origin through an angle of-. Ans. 2/2-4x + li3 = 0. 2 4. Rotate the curve in problem 3 through the given angle and then trans- late it. Ans. 2/2 _ 4 X + 8 2/ + 16 = 0. 5. Prove from equations (II) that the origin is unchanged by a rotation, that is, that the origin is a fixed point. 6. Find the equations of the straight lines which are unchanged by the translation (I). Hint. Translate Ax + By + C=0 and then determine A, B, and C so that this line coincides with the line into which it is translated by Theorem III, p. 88. Ans. kx — hy — 0. 7. Find the equations of all circles which are unchanged by the rotation (II). Ans. x2 + 2/2 + F = 0. 8. Show that no straight lines are invariant under the rotation (II). Hint. See the hint, problem 6, and apply Theorem IV, p. 90. 9. Prove analytically that no points are unchanged by a translation unless all points are unchanged. 116. Displacements. A transformation which replaces any figure by one congruent to it (p. 281) is called a displacement. Hence a figure is displaced when it is moved in the plane from one position to another. This may evi- dently be accomplished in many different ways. Two displacements which move a figure from one position to the same second position are said to be equivalent. Lemma I. A displacement is equivalent to a translation or to a rotation followed by a translation. Proof. Let the given displacement replace any figure F by a figure F\ Then if corresponding lines in F and F^ are parallel and have the same direc- tion, F may be translated into F', and hence the displacement is equivalent to a translation. If this is not the case, then F may be rotated into a position F'' such that corresponding lines in F'' and F^ are parallel and have the same direction and then F'' may be translated into F'. Hence the given displacement -is equivalent to a rotation followed by a translation. q.e.d. EUCLIDEAN TRANSFORMATIONS 285 Theorem HI. The equations of any displacement have the form ajf = 05 cos ^ — 2/ sin ^ + /«,, Jaj' =iccosd — y sind -\- \y^ =z gc sin 9 -{■ y cos 6 + (III) 1 .w _ ^ „;« fl I ., COS ^ + A; ly^ere 6, h, and k are arbitrary constants. Proof. Let the given displacement replace any figure F by a congruent figure F\ Then by Lemma I it is equivalent to a translation whose equa- tions have the form (III) when ^ = (Theorem I, p. 282), or to a rotation which replaces F by a figure F'' followed by a translation which replaces F'' hy F\ By Theorem II, x'' = xcosd — y sin d, y'' = xsmd + y cos ^, and by Theorem I, x' = x" + h, y' — y" + A:. Substituting the values of x" and y" in these equations, we obtain (III). Q.E.D. If a point is unchanged by a transformation, it is called a fixed or an invariant point. Thus the center of a rotation is an invariant point. Theorem IV. If a displacement is not equivalent to a translation, there is one fixed point. Proof The point (x, y) will be a fixed point when and only when x' = x and y' = y. Substituting in (III) and transposing, we get J (1 — cos ^) X + sin ^ ■ ?/ = ^, ' ^ t - sin ^ • X +' (1 - cos d)y=h These equations can be solved, in general, for one pair of values of x and y (Theorem IV, p. 90), and hence there will be, in general, but one fixed point. -„ ^ .. 1 — cos sin d But if = , — sin ^ 1 — cos d or, reducing, cos ^ = 1 , there will be no solution, that is, there is no fixed point. If cos = 1, then sin ^ = (by 3, p. 19) and equations (III) become x' = x-\-h, y' = y-^k, which are the equations of a translation. Hence there is one fixed point unless the displacement is a translation. Q.E.D. There cannot be an infinite number of solutions of (1) unless h = k = 0. For if 1 - cos 9 sin h - sin " 1 - cos e~ Ic' then, as above, cos = 1 and sin = 0. Substituting in (1), we get A = and k = 0. In this case every point (x, y) is a fixed point, that is, there is no displacement. 286 ANALYTIC GEOMETRY Theorem V. Every displacement which is not equivalent to a translation is equivalent to a rotation. Proof. If the displacement is not equivalent to a translation, then it has a fixed point (Theorem IV). Let the fixed point be chosen as origin. Then if ic = and y == 0, we get x' = and y' — 0. Substituting in (III), we obtain ^ = 0, k = Q as the conditions that the origin is the fixed point. For these values of h and k equations (III) reduce to (II), p. 283, and hence the displacement is equivalent to a rotation. q.e.d. Corollary I. Any two congruent figures may he brought into coincidence by a rotation or a translation. Corollary n. The perpendicular bisectors of the lines joining corresponding points of two congruent figures pass through the same point or are parallel. For if the figures may be brought into coincidence by a rotation, they pass through the center of tlie rotation; and if the figures may be brouglit into coincidence by a translation, they are perpendicular to the direction of the translation. -, ^ PROBLEMS 1 . Show analytically that the angle between two lines is unchanged by a displacement. Hint. Show that the value of tan Q given by (X), p. 109, is an absolute invariant of the displacement (III). 2. Show analytically that the distance between two points is unchanged by a displacement. Hint. Show that the value of I given by (IV), p. 31, is an absolute invariant of (III). 3. Prove Corollary II geometrically and derive Theorem V from it. 4. Show that a rotation about the origin through an angle of 7t replaces any figure by the figure symmetrical to it with respect to the origin. 5. Find the equations of a rotation about the point (1, 4) through an angle of -. Ans. x' = i VSx -iy + 3 - i Vs, ?/' = ix + i V3y + | -2 V3. 6 6. Find the equations of a rotation about the point (3, — 2) through an angle of Ans. x' = y + 6, y' = — x ~\- 1. 7. Find the equations of a rotation about the point (xi, 2/1) through an angle 6. . Ans. x' = (x — Xi) cos 6 — {y — y\) sin ^ + Xi, y' = {x- Xi) sin d + {y - yi) cos d + ?/i. EUCLIDEAN TRANSFORMATIONS 287 117. The reflection in a line. A transformation which replaces any figure by one symmetrical to it (p. 281) is called a symmetry transformation. The simplest symmetry transformation is the reflection in a line, which replaces a point by the point symmetrical to it with respect to that line. Hence a reflection in a line replaces a figure by the figure which is symmetrical to it with respect to that line. Theorem VI. The equations of a reflection in ^^ the X-axis are (VI) W = -y' a C)- 118, Symmetry transformations. Lemma II. A symmetry transformation is equivalent to a reflection in any line followed by a displacement. Proof. Let the given transformation replace a figure F by a symmetrical figure F\ Let F be transformed into a figure F'' by a reflection in any line. Then since F' and F'^ are both symmetrical to F, they are congruent to each other. For the parts of F^ and F^^ are equal, since they are equal to the parts of F, and they are arranged in the same order, for they are in each case arranged in the opposite order to those of F. Hence F'' can be brought into coincidence with F' by a displacement, that is, F may be transformed into F' by a reflection in any line followed by a displacement. q.e.d. (VII) Theorem VII. The equations of any symmetry transformation have the form (oc' =aocosd -\- y sinO + h, \y' = 00 sin6 — y cosd -{- k, where 6, h, and k are arbitrary constants. Proof Let the given transformation replace any figure i^ by a symmetrical figure F'. Then by Lemma II it is equivalent to a reflection in the JT-axis which replaces F by a figure F'% followed by a displacement which replaces F'' by F\ By (VI), X- and by (III), p. 285, x' = x" cos d — y'' sin y h, y' = x'' sin ^ + y" q,q>%Q ■\- k. Substituting the values of x'' and y" in these equations, we get (VII). Q.E.D. 288 ANALYTIC GEOMETRY Theorem VIII. The line whose equation is X cos w 4- 2/ sin w — p = is transformed by (VII) into the line whose equation is X cos {6 — uj) -\- y sin {6 — w)— [p -\- h cos {6 — ca) + k sin {d — w)] = 0. This is proved by solving (VII) for x and y, substituting in the given equation, simpli- fying by 9 and 11, p. 20, and dropping primes. A line is said to be invariant under a transformation if it is transformed into itself by that transformation. Theorem IX. There is always one line which is invariant under the sym- metry transformation (VII), and if h cos le -\-ksmld = 0, then all of the lines perpendicular to that line are invariant. Proof. If the lines in Theorem VIII coincide, then (Theorem III, p. 88) cos w _ sin w _ p cos (d — w) sin {d — u}) p -\- h cos {d — w) -\- k sin {6 — w) From the first two ratios sin (6 — w) cos w — cos {9 — w) sin w = 0, or (9, p. 20) sin (6' - 2 w) = 0. Hence ^ — 2 w =: or tt, .: (a = Id or u = ^6 — ^ it. Case I. ia — \d — \Tt. Substituting this value of w in the last two ratios of (1) and simplifying by 4, p. 19, and 6, p. 20, we get — cos \Q _ p cos \Q p — h sin \d ■\- k cos \ 6 Solving for p, p = | (Ji sin \d — k cos i 6). Hence there is always one pair of values of w and p for which (1) is true, that is, there is always one line which is transformed into itself by (VII). Case II. w = i 6. Susbtituting this value of w in the last two ratios in (1), we get sinig p sin 1 ^ p -{- h cos \d -\- k^\n\d The first of these ratios equals 1, but the second is never equal to 1 unless (2) hcoQ\e -\-ksm\e = 0, in which case p may have any value. Hence there is, in general, but one invariant line. But if (2) is satisfied, all of the lines of a system of parallel lines are invariant. Since the values of w in Case I and Case II differ by — , the invariant system of parallel lines is perpendicular to the single invariant line. q.e.d. EUCLIDEAN TRANSFORMATIONS 289 Theorem X. If the invariant line of a symmetry transformation is the X-axis^ then the equations of the transformation are (X) (xf = cc -{■ h, \yl =-y. Proof. If the X-axis is invariant, then, if y = 0, we must have y' = for all values of x. Substituting y = and y' = in the second of equations (VII), we get X sin + A: = 0. This is true for all values of x when and only when sin ^ = and A: = 0. If sin ^ — 0, then cos ^ = ± 1- Substituting A; = 0, sin = 0, and cos ^ = 1 in (VII), we get (X). Substituting A: = 0, sin ^ = 0, and cos ^ = — 1 in (VII), we get x' = — x-\-h, y' = y. This transformation leaves all of the lines parallel to the JT-axis invariant, for if y = a, then y' = a. Hence the X-axis is not the single invariant line, so that this case is to be excluded ; that is, equations (VII) reduce to (X) if the X-axis is the invariant line in Case I of Theorem IX. q.e.d. Corollary I. A symmetry transformation is equivalent to a reflection in a line or to a reflection in a line followed by a translation parallel to it. For if ^ = 0, equations (X) reduce to equations (VI). If h ^ 0, equations (X) are equivalent to the two transformations f x" = x, -, x' = x" + ^, ■i. ,, and \y"=-y, y'=y'\ which are respectively a reflection in the X-axis and a translation parallel to it. Corollary 11. The middle points of the lines joining corresponding points of two symmetrical figures lie on a straight line. T ,'-' For let (X) be the equations of the symmetry transformation which transforms one figure into the other. The middle point of the line PP' is (Corollary, p. 39) iiix + x'), i(y + y')]. Substituting the values of «' and y^trom (X), this becomes (x + ^k, 0), which is a point on the JC-axis. Ix 290 ANALYTIC GEOMETRY PROBLEMS 1. Find the equations of the curves symmetrical to the following curves with respect to the X-axis and construct the figure.* (a) 2/2 - 4 ic = 0. (c) x2 + 4 2/2 - 4 X = 0. (b) x2 + 0:2/ - 2 2/2 = 0. (d) x3 - 8 2/ = 0. 2. Show analytically that the distance between two points is unchanged by (a) a reflection in a line, (b) any symmetry transformation. 3. Show analytically that the numerical value of the angle which one line makes with another is unchanged by (a) a reflection in a line, (b) any symmetry transformation, but that its sign is changed in both cases. 4. Find the equations of the invariant lines which are proved to exist in Theorem IX. 1 9 5. Find the equations of a reflection in the Y-axis. 6. Prove that a reflection in a line followed by a reflection in a line per- pendicular to the first is equivalent to a rotation through it. 7. Asymmetry transformation (VII) has, in general, no fixed points, but if ^(1 4- cos ^) + jfcsin ^ = 0, then all of the points of the line x(l— cos B) — y sin d = k are fixed points. 8. If /i (1 + cos e) -[-Icmid = 0, then (VII) is a reflection in a line. 9. Find the equations of a reflection in the line 3x + 42/ — 10 = 0. Ans. X' =i^x- §4 2/ + V, y' = - ||x - J~,y + ^f. Hint. The distances from the line to P{x, y) and P'{x\ y^) (Rule, p. 106) must be equal numerically with opposite signs, and the slope of PP' (Theorem V, p. 35) must be equal to the negative reciprocal of the slope of the given line (Theorem VI, p. 36). These conditions give two equations which may be solved for x' and y' in terms of x and y. 10. Find the equations of a reflection in the line 5x — 12 2/ — 27 = 0. Ans. X' = iifx + i||2/ + iih V' = Hf « - my - fu- ll. Find the equations of a reflection in the line Ax + By -{- C = 0. , &-A^ 2AB 2AC Ans. X = —- — ^ A^+ B^ A'^ + B^ A^-h B^ 2AB m-A^ 2BC y =z — X y • ^ ^2 + 2^2 ^2 + ^^ A^-\-B^ EUCLIDEAN TRANSFORMATIONS 291 119. Congruent and symmetrical conies. The conditions that two conies should be equal are given in Theorem XII, p. 279. We shall now prove Theorem XI. Two equal conies are both congruent and symmetrical. Proof. Since a conic is symmetrical with respect to its principal axis (p. 174), it is unchanged by a reflection in that axis. Let C and C be two congruent conies, and let D be the displacement which transforms C into C. Then C may be transformed into C by a reflection in its principal axis followed by the dis- placement D, that is (Lemma II, p. 287), by a symmetry transformation. Hence C and C are also symmetrical. Conversely, let C and C be two sym- metrical conies, and let S be the symmetry transformation which transforms C into C\ Then S is equivalent to a reflection in the principal axis of C followed by a displacement D. Since C is unchanged by a reflection in its principal axis it may be transformed into C" by the displacement D, and hence C and C" are congruent. Hence two equal conies are both congruent and symmetrical. q.e.d. In the figure C may be transformed into C by a rotation about or by a symmetry transformation consisting of (Corollary I, p. 289) a reflection in the line S which replaces C by C, followed by a translation parallel to S. 120. nomothetic transformations. Given a fixed point O, the transfor- mation which replaces a point P by a point P' on the line OP such that 0P'=\- OP, .--'-"A ^-^, ::^9^'-^^ X<0 where X is constant, is called a homothetic transformation. is called the center and X the ratio of the transformation. Corresponding figures are called homothetic figures. They may easily be proved similar, with the ratio of similitude (that is, the ratio of corresponding lines) equal to X. Homothetic figures are also similarly placed. 292 ANALYTIC GEOMETRY Theorem xn. The equations of a homothetic transformation whose center is {h, k) and whose ratio is \ are Proof. Let P and P' be two corresponding points. Then, by definition, ^>. > r OP'=\-OP. Projecting on the X-axis (Theorem III, p. 31), x' — h = \{x — li). > Hence x' = 'Kx + h{l— X). Similarly, y' = \y -\-k (1— X). q.e.d. X' Corollary. The equations of a homothetic transformation whose center is the origin and whose ratio is X are foe' =Ax, \2 \y' = Ay. 121. Similitude transformations. A transformation which replaces any figure by one similar to it is called a similitude transformation. It is said to be direct or inverse according as corresponding figures are directly or inversely similar, that is, according as the corresponding parts of the similar figures are in the same or opposite order. If F and F' are two similar figures whose ratio of similitude is X, then a homothetic transformation with any center and with the ratio X will transform F into a figure F'' which is equal to F'. F" may be transformed into F' either by a displacement or by a symmetry transformation according as F' and F" are congruent or symmetrical, that is, according as F and F' are directly or inversely similar. Hence Theorem XIII. A similitude transformation is equivalent to a homothetic transformation with any center and with its ratio equal to the ratio of simili- tude of corresponding figures, followed by a displacement or a symmetry transformation according as the similarity is direct or inverse. PROBLEMS Problems 1 to 4 and 5 to 10 are to be solved in order by using those preceding. 1. The equations of a transformation of direct similitude have the form x' = X (x cos6 — y sin d + h), y' = \ {x sind -\-y cos d + k). 2. A transformation of direct similitude has one fixed point. I EUCLIDEAN TRANSFORMATIONS 293 3. If the fixed point is the origin, the equations of a transformation of direct similitude have the form x'=\(xcosd — y sin 6), y' =\{xsme + y cos d). 4. A transformation of direct similitude is equivalent to a rotation fol- lowed by a homothetic transformation with the same center. 5. The equations of a transformation of inverse similitude have the form ^ x' =\{x cos 6 + ysind + h), y' = \ {x sind — y cos d + k). 6. The line a; cos w + y sin w — p = is transformed by a transformation of inverse similitude into the line X cos {d — (f) 2/2 = 2px, (X - /i)2 = 2p\y - k). 2. Show that the ellipses in Ex. 1, p. 200, are similar. 3. Show that the hyperbolas in Ex. 2, p. 201, for which k is positive or for which k is negative, are similar. 4. Show that the locus of Ax^ + Bxy + Cy^ = k is, in general, a system of similar conies. Discuss all possible special cases in which this statement is not exact. 5. Any homothetic transformation is equivalent to a homothetie trans- formation whose center is the origin followed by a translation. 6. By means of problem 4 prove that two conies are homothetic if the coefficients of the terms of the second degree are proportional. 7. Find the center and ratio of the homothetic transformation which transforms y^ = 2px into 2/^ = 2_p'x. ^^^^ /q q\ x = :^, 8. A homothetic transformation whose center is 0(0, 0) and whose ratio is X followed by a homothetic transformation whose center is 0'{a, 0) and whose ratio is X' is equivalent to a homothetic transformation whose center is ( » I , that is, a point on 00' and whose ratio is XX'. V 1 - XX' / 9. A circle may be transformed into any other circle by two homothetic transformations whose centers, called the centers of similitude of the circles, lie on the line of centers. Hint. Take the center of one circle for the origin and let the JT-axis pass through the center of the other circle. Substitute from (XII), p. 292, in the equation of the first circle and determine h, k, and A so that the result coincides with the second circle. 10. Given three circles, the line joining a center of similitude of one pair with a center of similitude of a second pair will pass through a center of similitude of the third pair. Hint. Apply problem 8. 1 1 . The six centers of similitude of three circles taken by pairs lie three by three on four straight lines. Hint. Apply problem 10. CHAPTER XIV INVERSION 123. Definition. Let be a given point and let P be any point of a figure F. Construct P' on OP such that OP' • OP = 1. By letting P assume different positions on P, P" will move on a figure F\ The operation or transformation which replaces P by P' is called an inversion, while F and F' are called inverse figures. is called the center of the inversion. \ / The figure has been accurately constructed and indi- ^v ^^"^ cates tliat the inverse of a triangle is a figure bounded "' ' by three curves. Hence we may expect to find that the properties of inverse figures are, in general, quite different from those of equal or similar figures. Two important properties of an inversion are immediately evident from the definition. 1. If P approaches the origin, Pi recedes to infinity, and conversely. For if OP approaches zero, then OP' must become infinite since OP' • OP = 1, and conversely. 2, The points of the circle of unit radius whose center is are fixed points. For if OP = 1, then from OP' . OP = 1 we get OP' = 1. Hence P' coincides with P, that is, P is a fixed point. This fact is useful in plotting inverse figures, for the points in which a figure cuts this circle will be points of the inverse figure. 1 24. Equations of an inversion. By the equations of an inversion we mean two equations involving the coordinates of two corresponding points P and P'. These equations must express the two conditions : 1. That P and P' lie on a line through the center. 2. That OP' 0P = 1. The first of these conditions is satisfied when the triangles 0PM and OP'M' are similar, whence X _y _ OP x'~y'~~dP'' (1) p'(:^',y') (2) The second condition may be written, by dividing by OP'^ OP _ 1 _ 1 OP'" OP'2"~x'2 + y'2' 297 (by (IV), p. 31) 298 ANALYTIC GEOMETKY From (1) and (2), V ^ 1 y' x'2 + 2/'2' ;:;' V y Hence we have Theorem I. The equations of an inversion whose center is the origin are (I) _ X' _ V Ex. 1. Find the inverse of the line 2a; + 4y — 1 = 0. Solution. Substitute the values of x and y given by (I) in the given equation. We thus obtain 2x' + 4?/' 1 = 0. Reducing and dropping primes, we get a;2 + 2/2 _ 2x — 4?/ = 0. This is the equation of a circle_ whose center is the X point (1, 2) and whose radius is Vs (Theorem I, p. 131). In the figure a number of inverse points are indicated by the dotted lines. Ex. 2. Find the inverse of the straight line Ax + By + C — 0. Solution. Substitute in the given equation the values of x and y given by (I). This gives Ax' + C7=0. Simplifying and dropping primes, Cx^ + Cy^ + Ax-\-By = 0. The locus of this equation is a circle (Theorem II, p. 132) which passes through the origin (Theorem VI, p. 73). If C = 0, the locus is the given line. Hence The inverse of a straight line which does not pass through the origin is a circle, and a line which passes through the origin is invariant under an inversion. Ex. 3. Find the inverse of the circle x^ + y"^ -{- Dx -\- Solution. Substituting from (I), we get 4- Dx' W + i^=0. + i^=0. (x'2 + l/'2)2 (a.-2 + y/2)2 cc'2 + |/'2 ^"^ ^ y'l Multiplying by x"^ + y"^ and dropping primes, (3) Fx'^ + i^?/2 + Dx + JE;?/ + 1 = 0. The locus is a circle (Theorem II, p. 132) unless J^ = 0, in which case (3) is an equation of the first degree and its locus is a straight line (Theorem II, p. 86) . Hence The inverse of a circle is, in general, a circle, but the inverse of a circle which passes through the origin is a straight line. INVERSION 299 PROBLEMS 1. If the origin is the center of inversion, find the inverse of each of the following curves. Construct the figure in each case. (a) 2x = l. (b) 42/ = 1. (c) X + y - 1 = 0. (d) x2 + (e) x2 + ?/ = 4. (f ) x2 + 2/2 - 2 X 4x = 0. 4 2/ + 1 = 0. (g) x2 + 2/2 + 4 X - (h) 3x-42/ = 0. (i) x^-y^ = 0. (j) 4x-32/ = l. (k) x2 + 2/2 + 2 2/ : (1) 2/2 = 4 X. 62/ 0. I 2. Find the inverse of the points (0, 2), (3, 0), (3, 4), (2, 1), (i, 0), (i, i), (a, 0), and (0, b). Plot the given and inverse points. 3. Prove by (I) that the points on the unit circle are fixed points. 4. Find the equation of all circles which are unchanged by an inversion whose center is the origin. Ans. x"^ -\- y^ + Dx -}- Ey + 1 = 0. 5. Show that the inverse of the center of a circle is not, in general, the center of the inverse circle. 6. Show that the center of the circle obtained in Ex. 2 lies on the perpen- dicular drawn from the origin to the given line. 7. Show that the inverse of a circle whose center is the center of inver- sion is a concentric circle. 125. Inversion of conic sections. In this section we shall discuss several curves which are obtained by inverting a conic section. These curves have been otherwise defined in Chapter XI. Theorem II. The inverse of the parabola is the cissoid if the vertex of the parabola is the center of inversion. Proof. If the vertex of the parabola is the origin, its equation is y^ = 2px. Then, from (I), p. 298, r 2px' (X'2 + 2/'2)2 x'2 + y'2 Reducing and dropping primes. This is the equation of the cissoid of Diodes (problem 10, p. 253). If we replace — by 2 a, 2p we obtain the form of the equation usually given, namely, (1) x^ = y^{2a-x). q.e.d. 300 ANALYTIC GEOMETRY A general discussion (p. 74) gives us the following properties of the cissoid. 1. The cissoid passes through the origin (Theorem VI, p. 73). 2. It is symmetrical with respect to the X-axis (Theorem V, p. 73). 3. Its intercepts on both axes are zero (Rule, p. 73). 4. The cissoid lies entirely between the F-axis and the line x = 2 a. For, solving (1) for y, (2) -~^<^ If X is negative, tlie numerator is negative and the denominator positive ; and if a; > 2 a, the numerator is positive and the denominator negative. In either case the fraction is negative and y is imaginary. 5. The cissoid recedes indefinitely from the X-axis and approaches the line x= 2 a. For as x approaches 2 a the fraction in (2) becomes larger and approaches infinity as a limit. This may also be seen by transforming (1) to polar coordinates, vrhich gives p= 2a sine tan as the polar equation of the cissoid ; and hence, if = - or — > p = oo. Theorem III. The inverse of the equilateral hyperbola is the lemniscate if the center of inversion is the center of the hyperbola. Proof. The equation of the equilateral hyperbola is (p. 186) x2 - 2/2 = a2. The equation of the inverse curve is (by (I), p. 298) Reducing and dropping primes, (X2 + 2/2 (X2-2/2). The locus is the lemniscate of Bernoulli (problem 1, (g), p. 248, and problem 4, p. 262) . Replacing by a'2, we get the form of the equation usually given, namely, (3) (X2 + 2jY = a'2 (x2 - 2/2). Q.E.D. A discussion of the equation of the lemniscate in polar coordinates is given in Ex. 2, p. 152. From (3) it is evident that the lemniscate is symmetrical with respect to both axes and the origin (Theorem V, p. 73). In the figure a< 1 and a' > 1. If a = a'=l, the lemniscate will be tangent to the hyperbola at its vertices. If a > 1 and a' < 1, the two curves will not intersect. INVERSION 301 Theorem IV. The inverse of the equilateral hyperbola is the strophoid if the center of inversion is a vertex of the hyperbola. The equation of the equilateral hyperbola, when the origin is the right- hand vertex, is x2 - 2/2 + 2 ax = 0. This is obtained from a;2 - ?/2 = a^ by setting (Theorem I, p. 160) x- x' + a,y = y', and dropping primes. The inverse curve, from (I), p. 298, is 2nx' (X^2 + y^2y2 (;^^2 + y^2y2 Reducing and dropping primes, 1 0. 0. X(x2 + ?/2) + -— (X2-2/2 2 a The locus of this equation is the strophoid (problem 9, p. 262). Repla- cing — by a' and solving for y^, we 2a get the form of the equation usually given, namely. (4) y ., a + X x^ a' — X Q.B.D. In the figure a'=2a = l. If a' >1 and 2a < 1, the left-hand branch of the hyperbola will intersect the loop of the strophoid. If a' < 1 and 2 a > 1, the left-hand branch of the hyperbola will not meet the strophoid. A general discussion of (4) gives us the following properties of the strophoid. 1. It passes through the origin (Theorem VI, p. 73). 2. It is symmetrical with respect to the X-axis. 3. Its intercepts are'?/ = or and x = — a', 0, or 0. Hence it passes twice through the origin. 4. The strophoid lies entirely between the lines x = a' and x = — a\ For, solving (4) for y, (5) Va' + X X /—- = ± Va'2 - x^ . a' — X a' - X The quadratic under the radical is negative for values of x not lying between the roots (Theorem III, p. 11), and for these values y is imaginary. 5. The strophoid recedes indefinitely from the X-axis and approaches the line X = a\ For, from (5), y becomes infinite when x approaches a'. 302 ANALYTIC GEOMETRY Theorem V. The inverse of a conic is a limaQon if the center of inversion is a focus of the conic. Proof. The equation of a conic whose focus is the origin is (Theorem II, p. 178) (1 - e2) x2 + 2/2 _ 2 e^px - e^p^ = 0. Substituting from (I), p. 298, the equation of the inverse curve is (l-e2)x'2 2 e'^px' - e2p2 = 0. (X'2 + 2/'2)2 (x'2 4. y^y a;'2 + y^2 Clearing of fractions, transposing, and dropping primes, e2p2 (x2 4- 2/2)2 4_ 2 e^x {x^ + 2/2) = (1 - e2) x^ + 2/2. Adding 6^x2 to both sides and dividing by e2_p2^ (x2 + 2/2 + ia:)2 = J-(x2 + y2). p e2p2 The locus of this equation is the lima9on (problem 11, p. 253). If we set — = a and — — = b^, we get the form of the equation usually given, namely, p e2p2 (6) (x2 + 2/2 + ax)2 = 62 (a;2 + yzy q e.d. The lima9on has three distinct forms corresponding to the three forms of conies, according as a is less than, equal to, or greater than b. If a= b, the lima9on is some- times called the cardioid (Ex. 2, p. 158). a >b A general discussion of (6) gives us the following properties of the limagon. 1. It passes through the origin (Theorem VI, p. 73). 2. It is symmetrical with respect to the X-axis (Theorem V, p. 73). 3. Its intercepts are x = 0, 0, — a — b, and — a + 6 and y = 0, 0, 6, and b. Hence the lima^on passes twice through the origin. 4. The lima9on is a closed curve. For, transforming to polar coordinates, (6) becomes (Theorem I, p. 155) (p2 + ap cos 0)2 = 62p2. Solving for p, p = 6 - a cos fl. Since — 1 ^ cos ^1, p cannot become infinite. INVERSION 303 PROBLEMS 1. Construct the following conies, find the equations of the inverse curves, and discuss and construct their loci. (a) y2 = a;, 2/2 = 8a;, x^ = iy. (b) x^ - y^ = i, x^ - y^ = I, x^ - ?/ = l,2xy = l. (c) x2 - 2/2 + V2 x = 0, a;2 _ 2/2 + X = 0, x2 - ?/2 + 4 X = 0. (d) 3x2 + 4?/2-4x = 4, 2/2 -4x =16, 3x2 - 2/2 + 16x + 16 = 0. 2. Find the inverse of the hyperbola Srx^ — ry^ + 2x = 0, and discuss its properties. Ans. The trisectrix of Maclaurin x (x2 + 2/^) = - (2/'"^ 3x2). 3. Prove that the inverse of (a) the cissoid is a parabola ; (b) the lemniscate is an equilateral hyperbola ; (c) the strophoid is an equilateral hyperbola ; (d) the limagon is a conic, if the origin is the center of inversion. 4. Prove analytically and geometrically that if a curve C inverts into C, then C inverts into C. 5. Show that the inverse of the locus of an equation of the second degree is, in general, a curve whose equation is of the fourth degree. In what combination of x and y will the terms of the fourth degree enter ? What will be the degree if the given locus passes through the origin ? 126. Angle formed by two circles. If radii be drawn to a point of intersection of two circles, the angle formed is equal to one of the angles formed by the tangents at that point, since their sides are respectively per- pendicular. That angle d is called the angle formed by two circles. Theorem VI. The angle 6 formed hy two intersecting circles Ci : x2 + 2/2 + 1^1^ + ^12/ + Fi = and Oa : x2 + 2/2 + DgX + ^22/ + -^^2 = is given hy (VI) COS e J>il>2 + E^E^ -^F^-^F^ Vl>i2 _|_ J5:^2 _ 4 p^ ^Jj2 _,_ jg;^2 _ 4 j^^ 304 ANALYTIC GEOMETRY Proof. By definition d equals the angle formed by tlie radii drawn to a point of intersection. Hence from tlie figure and 17, p. 20, (1) 2rir2 where ri and r^ are the radii of C\ and C2 respectively and d is the length of the line of centers. By Theorem I, p. 131, r2-iVD22 + ^22_4j.2, and the centers of C\ and C% are respec- «ve„(-f,-f)a„d(-f,-f). Hence (by (IV), p. 31) Substituting in (1) and reducing, we get (VI). ^ q.e.d. Corollary. Ci and C2 are orthogonal if BiD^ + J&i£'2 - 2 i^i - 2 i^2 = 0. 127. Angles invariant under inversion. Theorem VII. The angle between two circles is equal to the angle formed by the inverse circles. Proof. Let the equations of two circles be Oi : x2 + ?/2 + Dix + Eiy + F^ = and C2 : x^ + 2/2 + I)2X + E2y + F2 == 0. Then the equations of the inverse circles are respectively [(3), p. 298] J_ Fi 1 Ci':.2 + ,. + ^^ + ^l + and C2':x2 + y2 + :^a; F2 -F2 i'a By Theorem VI the angle formed by Ci and C2' is given by ^ _ 2 Fi F2 cos r = D1D2 . E\E2 F1F2 F1F2 Mhrn-rMhrnH D1D2 + -E^ijEJa - 2 i?'i - 2 F2 VDi2 + J?i2 cos ^. 4 Fi VDs^ + ^2^ - 4 i<^2 where 6 is the angle formed by Ci and C2. Since 6' and ^ are both less than 7t, we therefore have 6' = 6. Q.E.D. INVERSION 305 Corollary. The angles formed by two intersecting curves are equal to the angles formed by the inverse curves. For draw two circles respectively tangent to the given curves at a point of inter- section. The inverse circles will be tangent to the inverse curves at a point of intersec- tion. The angles formed by either pair of curves and the tangent circles are identical, and the angles formed by the two pairs of circles are equal. Hence the angles formed by the given curves and by the inverse curves are equal. PROBLEMS 1. Find the angles formed by the following pairs of curves and the angles formed by the inverse curves, and show that they are equal. (a) x-y = 0, x + 2y = 0. (b) x-\-3y -2 = 0, x-2y = 0. (c) x^ + y^-^ix-Sy = 0, x^-hy^-^x = 0. (d) x^+y^-4x + 12 = 0, x^ + y^ -8y = 0. (e) x2 -f 2/2_6x + 4y = 0, 6x-4y-l = 0. 2. Show that the circles found in problem 4, p. 299, are orthogonal to the circle x^ -{- y'^ = 1. 3. If P and P' are two inverse points, show that all of the circles which pass through P and are orthogonal to x^ + y^ = 1 will also pass through P\ 4. How may problem 3 be used to define an inversion ? * 5. Into what kind of a figure will three lines forming a triangle invert if the center of inversion is not on one of these lines ? 6. Into what kind of a figure will three circles which have a point in common invert if that point is the center of inversion ? 7. Three circles pass through a point and intersect each other in three other points. Show that the sum of the angles formed by the circles at these three points is two right angles. Hint. Invert the figure, using the point common to the three circles as the center of inversion. 8. Three circles pass through the same point. Show how to construct four circles tangent to the three given circles. Hint. Suppose the required circles constructed. Invert the figure, using the common point as the center of inversion and show how to construct the inverse of the required circles. Then invert the figure so constructed, using the same center of inversion. 9. Show that the sign of an angle is changed by an inversion. 306 ANALYTIC GEOMETRY 128. Inversion of systems of straight lines. Theorem "VIII. The inverse of a system of parallel lines is a system of tan- gent circles whose centers lie on a line perpendicular to the lines of the system. Proof Choose one of the lines of the system for the Y-axis. Then the equation of the system is x = a, where a. is an arbitrary constant. The inverse system is therefore (by (I), p. 298) x'2 + y's a, or, reducing and X' dropping primes, cc^ + ?/2 x = 0. This is the equation of a system of circles whose centers lie on the X-axis and which are tangent to each other at the origin (Theorem VIII, p. 144). q.e.d. Theorem IX. The inverse of a system of lines passing through a point is a system, of circles passing through the origin and through the inverse of that point. Proof Let the system of lines be y = mx + 6, where h is constant and m varies. By Theorem I, p. 298, the inverse of the system is, after reducing and dropping primes, x-^ + 2/2 + y = 0. This is the equation of a system of circles passing through the origin (Theorem VI, p. 73) and through (0, t ) (Corollary, p. 53), which is the inverse of (0, b) through which the lines pass, q.e.d. INVERSION 307 129. Inversion of a system of concentric circles. Theorem X. The inverse of a system of concentric circles is a system with two limiting points, one at the origin and the other at the inverse of the center of the concentric circles. Proof. The equation (1) ic2 -f 2/2 _ 2 /3?/ + (82 - r2 = represents a system of concentric circles if j8 is constant and r varies (Tlieorem II, p. 58). (2) The inverse of (1) is [(3), p. 298] 2^ X'^ + y-^ ^2 y + The locus of (2) is a system of circles with their centers on the F-axis (Corollary, p. 131). The radius of any one is (Theorem I, p. 131) r" 2 \\B2-r^/ Hence r'= if r = 0, and the locus of (2) is then the point-circle (0, - j » which is the inverse of (0, /3), the center of (1). If r = co, (2) becomes x'2 _j_ ^2 — 0, whose locus is the origin. Hence the system (2) has two limiting points (p. 144), at the origin and at the inverse of the center of (1). q.e.d. PROBLEMS 1. Why do we not consider tlie system of lines passing through the origin in proving Theorem IX ? 2. Why do we not take the origin for the center of the system of circles in proving Theorem X ? 3. Construct a number of lines of the system x = a and the inverse circles. 308 ANALYTIC GEOMETRY 4. Construct a number of lines of the system y = mx + I and the inverse circles. 5. Construct a number of circles of the system x^ -{- y^ — Sx i- 1 — r^ = and the inverse circles. 6. What is the inverse of a system of tangent circles if the point of tan- gency is the center of inversion ? 7. What is the inverse of a system of circles passing through two points one of which is the center of inversion ? 8. What is the inverse of a system of circles with two limiting points one of which is the center of inversion ? 9. The point Pi(xi, yi) may be regarded as a point-circle whose equa- tion is (x — Xi)2 + (y — ?/i)2 =: 0. Show that the system of circles repre- sented by (x — Xi)2 + (y — y\Y -f A; (x^ + y^ — 1) = has two limiting points, namely, Pi and the inverse of Pi. What is the nature of the system if Pi lies on the circle x^ + y^ = 1 ? 10, How may problem 9 be used to define an inversion ? 130. Orthogonal systems of circles. Two systems of circles are said to be orthogonal if each circle of one system is orthogonal (p. 143) to every circle of the other system. The preceding sections enable us to construct such systems with ease. Consider two systems of parallel lines such that the lines of one system are perpendicular to the lines of the other. If we invert these systems of lines, we get two systems of tangent circles whose centers lie respectively on two perpendicular lines (Theorem VIII, p. 306). Since angles are preserved by inversion (Corollary, p. 306) these systems are orthogonal. Hence i INVERSION 809 Theorem XI. Two systems of tangent circles are orthogonal if they have the same point of tangency and if their centers lie on perpendicular lines. It is also evident that all of the lines passing through the same point P and all of the circles having the center P intersect orthogonally. The inverse of the system of lines is the system of circles passing through the origin, and the inverse of P (Theorem IX^ p. 306), and the inverse of the system of concentric circles is the system of circles having the origin and the inverse of P as limiting points (Theorem X, p. 307). Hence Theorem XII. Two systems of circles are orthogonal if all the circles of one system pass through two points which are the limiting points of the other. MISCELLANEOUS PROBLEMS 1. Show that the degree of an equation is, in general, doubled by an inversion. Will this be true if the terms of the highest ^degree contain x2 + 2/2 as a factor ? 2. Construct a linkage consisting of a deformable rhombus APBP' and two bars of equal length OA and OB which are free to rotate about the fixed point 0. Show that P and P' describe inverse curves if is the center of inversion and OA^ — AP^ is the unit of length. 3. If P is that point of the rhombus in problem 2 which lies nearest to 0, then by adding a bar O'P, which is free to rotate about the fixed point 0', P will be constrained to move in a circle. How will P' move ? This linkage is known as Peaucellier's Inversor. 4. Show how to construct four circles passing through a given point and tangent to each of two given circles which do not intersect. Hint. Invert the figure, using the given point as the center of inversion. 5. Find the properties of the cissoid, lemniscate, strophoid, cardioid, and limaQon, which may be obtained from problems 3, 4, 5, 6, 9, 10, 12, and 13, p. 220, by inversion with a proper center. 6. Show that the angle which one line makes with a second equals the angle between the inverse circles, without using the Corollary on p. 305. CHAPTER XY POLES AND POLARS. POLAR RECIPROCATION 131. Pole and polar with respect to a circle. Let Pi(xi, yi) be any point and let the equation of a given circle C be (1) a;2 + 2/2 = r2. The line ii, whose equation is (2) a?iX + i/i2/ = ^^ is called the polar of Pi (xi, yi) with respect to C, and Pi is called the pole of Xi. Theorem I. The polar of a point on a circle is the tangent to the circle at that point The proof follows at once from the definition and from the fact that (2) has the same form as the equation of the tangent (Theorem T, p. 212). Theorem II. The polar of a point Pi vjith respect to a circle is perpendicular to the line passing through Pi and the center of the circle. Proof. Thp equation of the line passing through Pi and the origin, the center of the circle (1), is (Theorem VII, p. 97) yix - Xiy = 0. This line is perpendicular to (2), the polar of Pi (Corollary III, p. 87). Q.E.D. Corollary. The angle formed by the polar s of two points with respect to a circle is equal to the angle formed by the lines joining those points to the center of the circle. Theorem III. The polar of any point of a given line passes through the pole of that line. Proof Let Li be the given line and let Pi (-^1, yi) l)e its pole. Then the equation of ii is (3) xix + yiy = r^. Let P2 (X2, 2/2) he any point on ii; then (Corollary, p. 53) (4) X1X2 + yiyz = r2. The equation of the polar L2 of the point P2is X2X + 7/oy = r2. 310 POLES AND POLARS 311 This line passes through Pi, for if the coordinates of Pi be substituted for X and y, we obtain equation (4), which is known to be true. Corollary. The pole of any line is the point of intersection of the polars of any two of its points. Theorem IV. The pole of any line passing through a given point lies on the polar of that point. Proof. Let Pi (xi, yi) be the given point. Its polar is the line (5) Xi : Xix + yiy = r^. Let P2(iC2, 2^2) l>e the pole of a line L2 which passes through Pi.' The equation of L^ is then X2X + 2/22/ = r^- Since L2 passes through Pi we have (Corollary, p. 53) (6) xgxi + y^yi = r2. Then P2 lies on ii, for when the coordinates of P2 are substituted in (5) for X and y, we obtain equation (6), which is known to be true. q.e.d. Corollary. The polar of any point is the line passing through the poles of any two lines which pass through the given point. 132. Construction of poles and polars. Construction I. To construct the polar of a point P outside of a circle, draw the tangents to the circle which pass through P. The line joining the points of contact of these tangents is the polar of P. Proof. Let Xi and Lo be the tan- gents to 0, and let Pi and P2 be their points of tangency. Then the polars of Pi and P2 are the lines ii and L2 (Theorem I). Since ii and L2 pass through P, the polar of P is the line L passing through Pi and P2 (Corollary to Theorem IV). q.e.d. In like manner the following con- structions are proved. Construction II. To construct the pole >f a line L which cuts the circle, draw le tangents at the points at which L itersects the circle. The point of inter- action of these tangents is the pole of L (Corollary to Theorem III). 312 ANALYTIC GEOMETKY Construction III. To construct the polar of a point P within a circle, construct the poles Pi and P2 of two lines Xi and L^ passing through P (Construction II). The line join- ing Pi and P2 is the polar of P (Corollary to Theorem IV). Construction IV. To construct the pole of a line L which does not cut the circle, construct the polars Xi and L2 of two points Pi and P^ on L (Construction I) . The inter- section of ii and L2 is the pole of L (Corollary to Theorem III). Y' - \ X' { ^ X 1 ^ rC^ \ ^ \ I r' -T-- PROBLEMS 19 1. Find the equation of the polar of each of the following points with respect to the given circle and construct the figure. (a) (3, _ 4), x2 + 2/2 = 4. (d) (3, 4), x^ + ?/2 = 36. (b) (- 1, 2), x2 + y2 ^ 25. (e) (5, 0), x^ + y^ = 49. (C) (7, - 2), X2 + 2/2 ^ 9. (f) (_ 3, 4), x2 + 2/2 = 25. 2. Find the pole of each of the following lines with respect to the given circle and construct the figure. (a) 3 x + ?/ = 25, £c2 + y2 ^ 25. Ans. (3, 1). (b) 3x - 2?/ = 18, x2 + 2/2 ^ 36. Ans. (6, - 4) (c) x-4y + 8 = 0, x2 + ?/2 = 16. Ans. (-2,8) (d) 2x-y = 64, x2 + 2/^ = 64. Ans. (2, - 1) (e) x-32/ + 16 = 0, x2 + 2/2 = 16. Ans. (-1,3) if) x-^y = 18, x2 + 2/2 ^ 9. Ans. (i, - |) (g) Ax + By + C = ^, x2 + 2/2 = r^. Ans. (^-^ J5r2\ Hint. Let Pi(«i, ?/i) be the pole of the given line and write down the equation of the polar of Pt with respect to the given circle. From the conditions that this line shall coincide with the given line (Theorem III, p. 88) determine x^ and y^. 3. Find the distance from the origin to the polar of Pi with respect to x2 + 2/2 = r2. Ans. Vxi2 + yi' 4. By problem 3 show that (a) if Pi approaches the origin its polar recedes to infinity; (b) if Pi recedes to infinity its polar approaches the origin. POLES AND POLARS 813 6. By problem 2, (g), show that if a line recedes to infinity its pole approaches the origin, and if the line approaches the origin its pole recedes to infinity. 6. Find the pole of the line joining Pi {xi, y\) and P2 (X2, 2/2) and prove that it is the point of intersection of the polars of Pi and P2. \X1y2 - iczz/i' Xiy2 - x^yj 7. Find the polar of -the point of intersection of AiX -f Biy + Ci = and A where x' = 4 Xi y' = 2 2/1 Xi Hence Xi=- 4 x'' yi = 2 2/' X' Substituting in (1), y'-2 = -4a' > Vs — -1^ — h ^ ^ V s. ^1 >^ x ' ^ ^ •' sc 7 ^ >* \ X ^ X ^ X ^s / S. y k ^ y' S X ^ 1 -1- ^ 1 This is the equation of the locus of P', that is, of the polar reciprocal of the given parabola. The polar reciprocal is therefore a parabola of the same size, turned to the left instead of to the right. The method consists in finding the pole P' of the tangent to the given curve at Pi, expressing xi and 2/1 in terms of x' and y^, and substituting in the given equation. POLES AND POLARS 315 PROBLEMS 1. Find the polar reciprocal of each of the following circles with respect to the circle x^ -^ y^ = 4. (a) x2 + 2/2 _ 4 x = 0. Ans. y^ + 4x = ^. (b) x2 + 2/2 - 2 X - 3 = 0. Ans. 3 x2 + 4 y2 + g x - 16 = 0. (c) x2 + 2/2 - 6 X + 5 = 0. Ans. 5 x2 - 4 2/2 - 24 x + 16 = 0. 2. Find the polar reciprocal of each of the following curves with respect to the given circle. (a) x2 + 4 2/2 = 16, x2 + 2/2 = 1. Ans. 16 x2 + 4 2/2 = 1. (b) 2/^ = 2x -6, x2 + 2/^ = 9. Ans. 6x2 - 2/2 - 18x = 0. (c) 4x2 + 2/2 = 8x, x2 + 2/2 = 4. Ans. 2/^ + 2x - 4 = 0. 3. Verify the answers to problems 1 and 2 by finding the polar recipro- cals of the curves given in the answers and applying Theorem VI. 4. Show that the equilateral hyperbola 2 X2/ = 9 is transformed into itself by a polar reciprocation with respect to the circle x^ + y^ = 9. 5. Show that the locus of'x2 — y"^ = a'^ is transformed into itself by a polar reciprocation with respect to the circle x^ -{- y^ = a^. 134. Pole and polar with respect to the locus of any equation of the second degree. Let Pi (xi, 2/1) be any point and let any equation of the second degree be (1) ^x2 + Bxy ■i-Cy^-h Dx + Ey + F=0. The line Xi, whose equation has the same form as the tangent, namely (Theorem II, p. 212), (2) A...+B^i^^^ + Cp.y+1>'^ + B^ + F=0, is called the polar of the point Pi with respect to the locus of (1)^ Pi is called the pole of Li. In what follows we speak always of poles and polars with respect to the locus of (1) unless the contrary is stated. The following theorems are generalizations of the theorems indicated, and are proved in the same way by using (1) and (2) of this section instead of (1) and (2) of section 131, p. 310. Theorem VII. (Generalization of Theorem I.) The polar of a point on the locus of (1) is the tangent at that point. Theorem VIII. (Generalization of Theorems III and IV.) The polar of any point on a given line passes through the pole of that line. Conversely., the pole of any line passing through a given point lies on the polar of that point. 316 ANALYTIC GEOMETRY Corollary I. The pole of any line is the point of intersection of the polars of any two of its points. Corollary II. The polar of any point is the line passing through the poles of any two lines which pass through the given point. The constructions on pp. 311 and 312 enable us to construct poles and polars with respect to (1), for the theorems by which the constructions are proved have been generalized for the locus of (1). A good idea of the direction of the polar of a point with respect to a conic is afforded by Theorem IX. The polar of a point Pi with respect to a conic is parallel to the tangent to the conic at the point where the diameter through Pi cuts the conic. Proof. The proof is separated into two cases according as the conic is a central conic or a parabola. Case I. Central conic. If the center is the origin, its equation may be written Ax'^ -{■ Cy2 -{- F = 0. The equation of the polar of Pi is \P (3) Axix + Cyiy + F = 0. Let the diameter through Pi cut the conic at P2. The equation of the tangent at P2 is (4) Ax2X + Cyip + F = 0. Since Pi and P2 are on a line through the origin (Corollary, p. 242), xi_yi^ and hence the lines (3) and (4) are parallel (Corollary II, p. 87). , Case II. Parabola. Its equation is y^ = 2px. The equation of the polar of Pi is (5) yiy^p{x-\-^\)- Let the diameter through Pi cut the parabola at P2. The equation of the tangent at P2 is (6) 2/22/=p(« + iC2). Since (Theorem X, p. 242) y^ = ?/2, the lines (5) and (6) are parallel. q.e.d. PROBLEMS 1 . Find the equations of the polars of the following points with respect to the given conies and construct the figures. (a) (3, 4), 9x2 + 4y2 = 36. (e) (- 1, 3), a;2 + a;?/ - 6y + 4 = 0. (b) (2, - 1), 16 x2 - ?/2 = 64. (f ) (4, 5), xy -{- ^x - Qy -^ = 0. (c) (3, 6), x2 + 4 y = 0. (g) (2,-6), x'^ + 'lxy ^y^ + x -y = 0. (d) (2, - 4), xy -\Q = 0. (h) (3, 2), 5x2 + 6a:?/ + 5?/2 - 12 = 0. POLES AND POLARS 317 2. Find the poles of the following lines with respect to the given conies and construct the figures. (a) 9x + 4y = 36, 9x2 + ^/^ = 36. Ans. (1, 4). (b) 2x-3y + 4 = 0, 2/2 = 4x. Ans. {2, - S). (c) x-2y = 16, xy = 8. Ans. (- 2, 1). (d) 14x + 2/ = 8, 4x2 - ?/2 = 16. Ans. (7, - 2). (e) 2x - y + 13 = 0, x2 4- 4?/ = 16. Ans. (- 4, - 6). (f ) X + 4 = 0, x2 + 4 xy + y2 + 2 X + 4 = 0. Ans. (0, 0). (g) llx + 2y + 18 = 0, 17x2-12x?/ + 82/2-68x + 24y - 12 = 0. Ans. (0, -2). 3. Tangents are drawn from the point (8, 4) to the ellipse x2 + 4^/2 = 16. Find the equation of the line joining their points of tangency. Ans. x + 2?/-2 = 0. 4. Tangents are drawn to the hyperbola 16x2 _ y2 — 64 at the points of intersection of the hyperbola and the line 8x + 3?/ + 32 = 0. Find the coordinates of their point of intersection. Ans. (—1,^). 5. How does the polar of a point with respect to a central conic behave if the point approaches the center ? if the point recedes to infinity ? 6. The polar of the focus of any conic with respect to that conic is the corresponding directrix. 7. The polar of any point on the directrix of a conic passes through the corresponding focus. 8. The polars of a point with respect to conjugate hyperbolas are parallel. 9. The polar of a focus of an ellipse with respect to the major auxiliary circle is the corresponding directrix. 10. What is the locus of a point which lies on its polar with respect to a given conic ? 1 1 . That part of the diameter of a parabola included between any point on it and its polar is bisected by the point of contact. 135. Polar reciprocation with respect to the locus of any equation of the second degree. Let (1) ^x2 + Bxy + Cy^ + Dx + Ey + F=0 be any equation of the second degree. Let C be any curve and let C be the locus of the poles of the tangents to C with respect to the locus of (1). C is called the polar reciprocal of C with respect to (1). Theorem X. (Generalization of Theorem VI.) If C is the polar reciprocal of C with respect to (1), then C is the polar reciprocal of C. The proof is identical with that of Theorem VI, p. 314. For the theorems on poles and polars with respect to a circle, used in proving that theorem, have been extended to the locus of (1). 318 ANALYTIC GEOMETRY Corollary. The polar reciprocal of C is a curve C whose tangents are the polar s of the points of C. The polar reciprocal of a curve C with respect to (1) may therefore be regarded in either one of two ways : 1. As the locus of the poles of the tangents to C. 2. As the curve whose tangents are the polars of the points of C. In either case the fact to be observed is that to a point of one figure corre- sponds a straight line of the other figure and vice versa. The transformation which replaces C by C is called a polar reciprocation with respect to (1). Analytically the polar reciprocation with respect to (1) is completely defined by the equation 2 -"2 2 For, in the first place, the locus of (2) gives us at once the polar of Pi i^u yi)' In the second place, the pole of any line (3) A'x -\- B'y -{■ C = > -> is found from (2) as follows. Let Pi (xi, 2/1) be the pole of (3). Then since (2) and (3) are the equations of the polar of the same point Pi, their loci coincide. Hence (Theorem III, p. 88) ^Xi + -^1 + - -x,+ Cy, + - D E ^ -Xi + -yi + P A' B' C These equations can, in general, be solved for Xi and yi (Theorem IV, p. 90). The .method of finding the equation of the polar reciprocal of a given curve C is illustrated in the following example. Ex. 1. Find the equation of the polar reciprocal of the ellipse (7:4x2 + 92/2-1 = with respect to the ellipse (4) a;2 + 4?/2 + 2x = 0. Solution. Let Pi (xi, yi) be any point on C. Then (5) 4a;i2 + 9?/i2-l = 0. The equation of the tan- gent to C at Pi is (Theorem III, p. 214) (6) 4:xix + 9 yiy -1 = 0. Let P'{x\ y') be the pole of (6) with respect to (4) . The equation of the polar of P' is (7) {x'-\-l)x + ^y'y + x'=0. s». y ^ ZP x. y ^. y r 25" \ t > ^ V 3^-. r t i --— z^ ^ V Z S .^ "S. ^'^ ^s,^ -r- — ^;: POLES AND POLARS 319 Since (6) and (7) have the same locus (Theorem III, p. 88), ' 4a;i ^^yi^ -1, x' + l 4 y' x' Solving for xi and yi, we obtain Substituting in (5) , we have the required equation Reducing and dropping primes, we obtain 27x2 _64y2_i8x- 9 = 0, whose locus is an hyperbola. In the figure three divisions are taken for unity. PROBLEMS 1. Find the polar reciprocal of the first of the following curves with respect to the second. Construct the figure in each case. (a) y2 _ 4 a; = 0, x^ + iy = 0. Ans. xy -2 = 0. (b) ic2 + ?/2 = 1, x^-y^ = 4. Ans. x^ -{- y^ = 16. (c) x2 + 4?/2 = 4, 4x2 + y2 = 4. j^y^s. 64x2 + 2/2 = 16. (d) x2 -42/2 = 16, x2 + 4^2:^ 2x. Ans. 16x2 - 64 2/2 - 32x + 16 = 0. (e) xy - 4 = 0, x^ - y^ = 16. Ans. xy + 16 = 0. (f) 8 2/ - x3 = 0, x2 - 2/2 = 4. Ans. 2 x^ = 27 y. 2. Verify the answers to problem 1 by showing that the polar reciprocals of the curves in the answers are the given curves. 3. Show that either of the following curves is unchanged by a polar recip- rocation with respect to the other. (a) 62x2 + a22/2 = aW, hH'^ - a^y'^ = a^"^. (b) &2x2 _ a^y2 - c^252^ 12x2 _ a^y2 ^ _ ^252. (c) 2/2 - 2px = 0, 2/2 + 2px = 0. 4. If the vertices of one triangle are the poles of the sides of a second triangle, then the vertices of the second are the poles of the sides of the first. Two triangles such that the vertices of either are the poles of the sides of the other are called conjugate triangles. If the vertices of a triangle are the poles of the opposite sides, the triangle is said to be self-conjugate. 5. Show that (2, 1), (4, 4), and (3, 2) are the vertices of a self-conjugate triangle with respect to the hyperbola x^ — y^ = 4. 6. Show how to construct a self -con jugate triangle with respect to a given conic if one vertex is given. How many may be constructed ? 320 ANALYTIC GEOMETRY 7. Show that if we reciprocate the figure which is given or implied in one of the following statements, we obtain the corresponding statement. (a) Two points determine the line on which they lie. (b) Three points on the same line. (c) Three points at the vertices of a triangle. (d) n points at the vertices of a poly- gon. (e) An infinite number of points lying on a curve. (f) A line intersecting a curve in n points. (g) A curve passing twice through the same point. (h) A conic section. (i) A conic may be constructed which passes through five given points. (j) Two conies intersect in general in four points. Two lines determine the point in which they intersect. Three lines through the same point. Three lines forming a triangle. n lines forming the sides of a poly- gon. An infinite number of lines tangent to a curve. A point through which pass n lines tangent to a curve. A curve tangent twice to the same line. A conic section. A conic may be constructed which is tangent to five given lines. Two conies have in general four common tangents. 136. Polar reciprocation of a circle with respect to a circle. The equa- tions of any two circles C and Oi may be put in the forms C : a:2 + y2 ^ r2 and Ci : x2 + ?/2 + Zte + i?' = by taking the center of C as origin and the line of centers of C and C\ as the X-axis, We shall now find the polar reciprocal of C with respect to C\. Let Pi (xi, 2/i) be any point on C. Then (Corollary, p. 53) (1) Xi2 + yi2 = r\ and the equation of the tangent to C at Pi is (Theorem I, p. 212) (2) xix + yiy - r2 = 0. Let P' (x', y") be the pole of (2) with respect to Ci ; then the polar of P' is X'X ^y'y^B ?^^ + F OX (3) x' + -)x + 2/'2/ + -x' + P 0, 0. Since (2) and (3) have the same locus (Theorem III, p. xx _yi_ -y^ ^ y' Solving for Xi and yi, we obtain r2 (2 x' + D) x' + F Xi = Bx' + 2F y\ = 2r^y' Dx' -\-2F POLES AND POLARS 321 Substituting these values in (1), reducing, and dropping primes, we have the equation of C\ namely, C : (4r2 - 2)2) x2 + 4r22/2 + 4i)(r2 -F)x-\- {r^B^ - 42^2) _ o. The discriminant of C is (p. 265) 0' = 16 r2 (4 r2 - 2)2) (rW^ -iF^)- 64 rW^ (r^ -F)^ = -167^ (2)2 - 4 2?')2. As ^ v2)2 - 4 2^ is the radius of Ci (Theorem I, p. 131), it follows that 0' is not zero if the radii of C and Ci are not zero. Hence (Theorem I, p. 266) Theorem XI. The polar reciprocal of the circle C : x^ + y^ = r^ with respect to the circle Ci : x^ + y^ + Dx -\- F = is the non-degenerate conic C whose equation is (XI) (4r2 - 2)2)a;2 + 4r22/2 + 42)(r2 - F)x + (r22)2 - 42^2) = q. The nature of the conic C depends upon the sign of A' = _ 4 . 4 r2 (4 r2 - 2)2). It is evident that 2)2 A' < if 4 r2 - 2)2 > 0, or r2 > — ; 4 2)2 A' > if 4 r2 - 2)2 < 0, or r2 < — ; 4 A' = if 4 r2 - 2)2 = 0, or r2 = — . Hence (Theorem IX, p. 277) 2)2 the conic C is an ellipse if r2 > — ; 4 the conic C is an hyperbola \ir^ <—^ the conic C" is a parabola if r2 4 4 2)2 / 2) \ But — is the square of the distance from the origin to ( — — , ) , the 4 \ 2 / center of Ci (Theorem I, p. 131), and therefore 2)2 the center of Ci is inside of C if r2 > — ; 4 2)2 the center of Ci is outside of C if r2 < — ; 4 and the center of Ci is on C 2)2 ifr2=— . 4 322 ANALYTIC GEOMETRY Hence Theorem XII. The polar reciprocal of a circle C with respect to a circle C\ is an ellipse, hyperbola, or parabola according as the center of Ci is inside of, outside of, or on the circle C. PROBLEMS 1. Find the polar reciprocal of the circle x^ -\- y^ = i with respect to each of the following circles and construct the figure. (a) x2 + 2/2-4x-5 = 0. Ans. 4?/2-36iK-9 = 0. (b) x2 + 2/2_2a;-;3:=0. Ans. 3x2 + 42/2 -Ux- 5 = 0. (c) x'^ + y^-6x = 0. Ans. 5x2 _ 42/2 + 24x- 36 = 0. 2. Show that the center of Ci (Theorem XI) is a focus of (XI) and that the corresponding directrix is the polar of the center of C with respect to Ci. Hint. Transform (XI) by moving the origin to the center of C^, find the focus and directrix by comparison with (II), p. 178, and transform to the old coordinates. 3. If Pi and P2 are two points whose polars with respect to a circle Ci are ii and L2, then — = — , where Zi and I2 are the distances from the center of di di C\ to Pi and P2, d\ is the distance from L^ to Pi, and ^2 from Li to P2. Hint. The center of C^ may be taken as the origin. Apply (IV), p. 31, and the Rule, p. 106. 4. Prove Theorem XII and problem 2 by means of problem 3 and the defi- nition of a conic (p. 173). Hint. Let P^ of problem 3 be the center of C. 6. The angles which two lines L\ and L^. (Fig., p. 311), which are tan- gent to a circle (7, make with the polar L of their point of intersection are evidently equal. If we reciprocate the figure with respect to a circle Ci, what will be the corresponding theorem in the new figure ? Hint. The polar reciprocal of C is a conic whose focus is the center of C^ (problem 2). To i, and L^ correspond two points on the conic, and to their points of contact correspond the tangents to the conic at these points. To L corresponds the point of intersection of these tangents. Draw lines from the focus to the points of contact of the tangents and to their point of intersection, and apply the Corollary to Theorem II, p. 310. Ans. If two tangents be drawn to a conic, the line joining the focus to their point of intersection bisects the angle between the focal radii drawn to the point of contact. POLES AND POLARS 323 6, Obtain the following theorems in the right-hand column from those in the left-hand by means of a polar reciprocation with respect to a circle. (a) Any tangent to a circle is per- The lines from a focus to any point pendicular to the radius drawn to the on a conic and to the point where the point of contact. tangent at that point meets the directrix are perpendicular. (b) The angle formed by two tan- The angle formed by the focal radii gents to a circle is bisected by the line of a conic drawn to its points of inter- drawn from the center to their point 'Section with any line is bisected by the of intersection. line joining the focus to the intersec- tion of that line and the directrix. (c) The points of intersection of Chords of a conic which subtend tangents to a circle which intersect at equal angles at the focus are tangent a constant angle lie on a concentric to a conic with the same focus and circle. directrix. 137. Correlations. Any transformation which makes the points of one figure correspond to the lines of a second figure is called a correlation. Polar reciprocations with respect to conies are the most important correlations. A correlation is completely determined when we are able to find 1. The equation of the line corresponding to a given point. 2. The coordinates of the point corresponding to a given line. We shall now see that a correlation is defined by an equation of the form (1) {aiXi -f biyi + Ci) X + (a2Xi + 622/1 + 02)2/+ {a^Xi + 632/1 + Cs) = 0, which is of the first degree in x and y and in Xi and yi. The locus of (1) is the line corresponding to a given point Pi{xi, 2/1). To find the point corresponding to a given line (2) Ax + By + C = 0, we suppose that Pi (cci, yi) is the required point. The equation of the line jorresponding to Pi is (1). Hence (1) and (2) have the same locus and lerefore aiXi + 6iyi + ci _ ^23:1 + 62^1 + C2 _ «3a^i + 63^1 + Cs ' Z ~ B ~ C ' These equations may, in general, be solved for Xi and 2/1. As far as defining the line corresponding to a given point is concerned, the parenthe- in (1) might be any complicated expressions in x^ and y^. But if the expressions in lose parentheses were not of the first degree, then the equations (3) would have more lan one pair of solutions for x^ and y^, and hence there would be more than one point corresponding to a given line. In general the point Pi will not lie upon the locus of (1). The condition that Pi should lie on the locus of (1) is (Corollary, p. 63) (aiXi + 6i?/i + ci) xi + {a2Xi + 622/1 + Cg) 2/1 + (as^i + 632/1 + a^) = 0, or aiXi2 + (61 + 02) xiyi + 622/1^ + (ci + as) Xi + (C2 + 63) yi + as = 0. 324 ANALYTIC GEOMETRY This is also the condition that Pi shall lie upon the locus of the equation (4) aix2 + (61 + a^) xy + &22/^ + (ci + as) x + (cg + 63) y + as = 0. The manner in which the conic sections enter into the theory of correlations is thus given by Theorem XIII. The locus of the points which lie upon the lines corresponding to them in the correlation defined by (1) is the conic or degenerate conic whose equation is (4). • It should be noticed that the correlation defined by (1) is not, in general, a polar reciprocation in the curve (4), for (1) is not the equation of the polar of Pi(xi, yi) with respect to (4). Suppose, however, that 61 = a^, Ci = as, and Cg = &3- Then (4) becomes (5) ttix^ + 2 a^xy + hy^ + 2 asx + 2 b^y + as = 0, and (1) becomes (aiXi + azyi + az)x-\- (a2Xi + 622/1 + 63) 2/ + (as^i + 632/1 + C3) = 0, or (6) aixix + a2 {yiX + Xiy) + 622/12/ + as (x + Xi) + 63 (y + 2/1) + C3 = 0. The locus of (6) is the polar of Pi(Xi, 2/1) with respect to (5). Hence we have Theorem XIV. Ifbi = ao, Ci = as, and c^ = b^, then the correlation defined by (1) is a polar reciprocation with respect to the locus of (5). CHAPTER XYI CARTESIAN COORDINATES IN SPACE 138. Cartesian coordinates. The foundation of Plane Analytic Geometry lies in the possibility of determining a point in the plane by a pair of real numbers (x, y) (p. 25). The study of Solid Analytic Geometry is based on the determination of a point in space by a set of three real numbers x, y, and z. This deter- mination is accomplished as follows : Let there be given three mutually perpendicular planes inter- secting in the lines XX\ YY\ and ZZ' which will also be mutually perpendicular. These three planes are called the coordinate planes and may be distin- guished as the ZF-plane, the FZ-plane, and the ZX-plane. Their lines of intersection are called the axes of coordi- nates, and the positive direc- tions on them are indicated by the arrowheads.* The point of intersection of the coordinate planes is called the origin. Let P be any point in space and let three planes be drawn through P parallel to the coordinate planes and cutting the axes 2it A, B, and C. Then the three numbers OA =>: x, OB = y, and OC = z are called the rectangular coordinates of P. * XX' and ZZ' are supposed to be in the plane of the paper, the positive direction on XX' being to the right, that on ZZ' being upward. YY' is supposed to be perpendicular to the plane of the paper, the positive direction being in front of the paper, that is, from the plane of the paper toward the reader. 325 326 ANALYTIC GEOMETRY Any point P in space determines three numbers, the coordinates of P. Conversely, given any three real numbers x, y, and z, a point P in space may always be constructed whose coordinates are x, y, and z. For if we lay off OA = x, OB = y, and OC = z, and draw planes through A, B, and C parallel to the coordinate planes, they will intersect in such a point P. Hence Every point determines three real numbers, and conversely, three real numbers determine a point. The coordinates of P are written (x, y, z), and the symbol P(x, y, z) is to be read, "The point P whose coordinates are x, y, and zJ' The coordinate planes divide all space into eight parts called octants, designated hj.O-XYZ, 0-X'YZ, etc. The signs of the coordinates of a point in any octant may be determined by the Rule for signs. X is positive or negative according as P lies to the right or left of the YZ-plane. y is positive or negative according as P lies in front or in back of the ZX-plane. z is positive or negative according as P lies above or below the XY-plane. If the coordinate planes are not mutually perpendicular, we still have an analogous system of coordinates called oblique coordinates. In this system the coordinates of a point ^ are its distances from the coordi- ^ nate planes measured parallel to the axes instead of perpendicular to the planes. We shall confine ourselves to the use of rectangular coordinates. Points in space may be conveniently plotted by marking the same scale on XX' and ZZ' and a somewhat smaller scale on YY\ Then to plot any point, for example (7, 6, 10), we lay ofi OA — 1 on OX, draw ylQ parallel to OF and equal to 6 units on r, and QP parallel to OZ and equal to 10 units on OZ. CARTESIAN COORDINATES IN SPACE 327 PROBLEMS 1. What are the coordinates of the origin ? 2. Plot the following sets of points. (a) (8, 0,2), (-3, 4, 7), (0,0, 5). (b) (4, -3,6), (-4, 6,0), (0,8, 0). (c) (10, 3, -4), (-4, 0,0), (0,8,4). (d) (3, -4, -8), (-5, -6,4), (8,6, 0). (e) (-4, -8, -6), (3, 0,7), (6, -4,2). (f) (-6,4, -4), (0,-4, 6), (9, 7, -2). 3. Where can a point move ifcc = 0? if y = 0? if z = 0? 4. Where can a point move if ic = and y = 0? it y = and z = ? if 2; = and X = ? 5. Show that the points {x, y, z) and (— x, y, z) are symmetrical with respect to the FZ-plane ; (x, y, z) and (x, — y, z) with respect to the ZX- plane ; (x, y, z) and ^, y, — z) with respect to the XZ-plane. 6. Show that the points (x, y, z) and (— x, - y, z) are symmetrical with respect to ZZ' ; (x, y, z) and (x, — y, —z) with respect to XX' ; (x, y, z) and (- X, y, —z) with respect to YY' ; (x, y, z) and (- x, — y, - z) with respect to the origin. 7. What is the value of z if P (x, y, z) is in the JTF-plane ? of x if P is in the FZ-plane ? of ?/ if P is in the ZA"-plane ? 8. What are the values of y and z if P (x, y, z) is on the JT-axis ? of z and X if P is on the F-axis ? of x and ?/ if P is on the Z-axis ? 9. A rectangular parallelopiped lies in the octant 0-XYZ with three faces in the coordinate planes. If its dimensions are a, 6, and c, what are the coordinates of its vertices ? 139. Orthogonal projections. To extend the first theorem of projection (p. 30) we define the angle between two directed lines in space which do not intersect to be the angle between two intersecting directed lines (p. 28) drawn parallel to the given lines and having their positive directions agreeing with those of the given lines. The definitions of the orthogonal projection (p. 29) of a point upon a line and of a directed length AB upon a directed line ^hold when the points and lines lie in space instead of in the plane. It is evident that the projection of a point upon a line 328 ANALYTIC GEOMETRY may also be regarded as the point of intersection of the line and the plane passed through the point perpendicular to the line. As two parallel planes are equidistant, then the projections of a directed length AB upon two parallel lines whose positive directions agree are equal. Theorem I. First theorem of projection. If A and B are points upon a directed line making an angle of y ivith a directed line CD, then the projection of the length AB upon CD = AB cos y. Proof Draw CD' through A parallel to CD. Then by defini- D' tion the angle between AB and ^ CD' equals y. Since CD' and AB ^ intersect we may apply the first theorem of projection in the plane (p. 30), and hence the projection of the length AB upon C'D' = AB cos y. Since the projection of AB on CD equals the projection of AB upon CD' we get (I). q.e.d. Theorem II. Second theorem of projection. If each segment of a broken line in space he given the direction determined in passing continuously from one extremity to the other, then the algebraic sum of the projections of the segments upon any directed line equals the projection of the closing line. The proof given on p. 48 holds whether the broken line lies in the plane or in space. Corollary I. The projections on the axes of coordinates of the line joining the origin to any point P are respectively the coordi- nates of P. For the projection of OP (Fig., p. 325) upon OX equals the sum of the projec- tions of OA, AQ, and QP, which are respectively equal to x, 0, and [by (I)]. Similarly for the projections on OTand OZ. CARTESIAN COORDINATES IN SPACE 329 Corollary II. Given any two points P^ {x^, y^, z^) and P^ {x^, 2/2? ^i)i then a?2 — a?! = projection of JP^P^, upon XX', y^ — y^ = projection of ^Pi-Pg upon YY^, z^ — z^z= projection of -Pi 1*2 upon ZZK For if we project P\OP-i-l) 7. Find two expressions for the projections upon the axes of the line drawn from the origin to the point P{x, y, z) if the length of the line is p and the angles between the line and the axes are a, /3, and y. 8. Find the projections of the coordinates of P(x, y, z) upon the line drawn from the origin to P if the angles between that line and the axes are a, j3, and y. Ans. x cos a, y cos j3, z cos 7. 140. Direction cosines ^of a line. The angles a, /3, and y between a directed line and the axes of coordinates are called the direction angles of the line. If the line does not intersect the axes, then by definition (p. 327) a, j3, and y are the angles between the axes and a line drawn through the origin parallel to the given line and agreeing with it in direction. The cosines of the direction angles of a line are called the direction cosines of the line. Reversing the direction of a line changes the signs of the direc- tion cosines of the line. For reversing the direction of a line changes a, /3, and 7 into (p. 28) it — a, TT — /3, and it — y respectively, and (5, p. 20) cos (jt — x) = — cos x. Theorem III. If a., ft, and y are the direction angles of a line, then (III) cos'* a + cos^^ + cos^* 7 = 1- That is, the sum of the squares of the direction cosines of a line is unity. Proof. Let ^S be a line whose direction angles are a, ji, and y. Through draw OP parallel to AB and let OP = p. By definition (p. 327) Z XOP = a, Z Y0P = (3, ZZOP=y. Projecting OP on the axes, we get by Corollary I, p. 328, and Theorem I, p. 328, p cos (3, z = p cos y. CARTESIAN COORDINATES IN SPACE 331 Projecting OP and OCQP on OP, we get (Theorems I and II) (2) p = xcosa -^ y cosft -\- z cos y. Substituting from (1) in (2) and dividing by p, we obtain (HI). Q.E.D. ^. cosa COS/8 cosy Corollary, if = — r-^ = } then a ' b cos a = J cos £j = ± Va2 + 62 _^ c^ ± Va^ + &2 + c2 c COS y = ^ That is, if the direction cosines of a line are proportional to three numbers, they are respectively equal to these numbers each divided by the square root of the sum of their squares. For if r denotes the common value of the given ratios, then (3) cos a = ar, cos j3 = br, cos y = cr. Squaring, adding, and applying (III) , I=:r2(a2 + 62 + 02). ± Va2 + 62 _|. c2 Substituting in (3), we get the values of cos a, cos /3, and cos y to be derived. ' If a line cuts the XF-plane, it will be directed upward or downward according as cos 7 is positive or negative. If a line is parallel to the XF-plane, cos 7 — and it will be directed in front or in hack of the ZX-plane according as cos j8 \& positive or negative. If a line is parallel to the X-axis, cos /S = cos 7 = 0, and its positive direction will agree or disagree with that of the X-axis according as cos or = 1 or — 1 . These considerations enable us to choose the sign of the radical in the Corollary so that the positive direction on the line shall be that given in advance. 141. Lengths. Theorem IV. The length I of the line joining two points Pi(xi, 1/1, ^1) «^^ A (^2, 1/2, ^2) ^'-^ ff^^en by (IV) I = V(^, - ^.,y + (y, - y,y +. (z, - z,y. 332 ANALYTIC GEOMETRY Proof. Let the direction angles of the line PiPg he a, ^, and y. Projecting P^P^ on the axes, we get, by Theorem I, p. 328, and Corollary II, p. 329, (1) I cos a = X2 — Xi, I COS /3 = 1/2 — ^Jl) I cos y = ^2 — ^1- Squaring and adding, Z^ (cos^ a + cos^^ /? + cos^ y) = {x^ — x^y-\- (2/2 — ViY + («2 — ^l)^ = (^1 - ^lY + (i/i - 2/2)' + («i - ^2)'. Applying (III), p. 330, and taking the square root, we get (IV). Q.E.D. Corollary. The dh'ection cosines of the line drawn from P^ to P2 are proportional to the projections of P1P2 on the axes. cos a _ cos /3 For, from (1), X2 — Xl -2/1 cos 7 ^2 - 21 ' since each ratio equals -y and the denominators are the projections of P1P2 on the axes (Corollary II, p, 329). ^t V rj;E=n'-i /I ~ 1/ V If we construct a rectangular parallelepiped by passing planes through P^ and P2 parallel to the coordinate planes, its edges will be paral- lel to the axes and equal numerically to the projections of P1P2 upon the axes. P1P2 will be a diagonal of this parallelepiped, and hence / ^ l^ will equal the sum of the squares of its three dimensions. AVe have thus a second method of deriving (IV). PROBLEMS 1. Find the length and the direction cosines of the line drawn from (a) Pi (4, 3, -2) to P2(-2, 1, —5). (b) Pi (4, 7, -2) to P2(3, 5, -4). (c) Pi (3, - 8, 6) to P2 (6, - 4, 6). Ans. 7, - f, - f, - f. Ans. 3, -i, - I, - f. Ans. 5. f, f, 0. 2. Find the direction cosines of a line directed upward if they are propor- tional to (a) 3, 6, and 2 ; (b) 2, 1, and - 4 ; (c) 1,-2, and 3. Ans. (a) f, f, f; (b) ^ ^ -2 ^ ^ •(c)J-, , V21' -V2T' + V21' V14' VTi' Vi4 CARTESIAN COORDmATES IN SPACE 333 3. Find the lengths and direction cosines of the sides of the triangles whose vertices are the following points ; then find the projections of the sides upon the axes by Theorem I, p. 328, and verify by Corollary III, p. 329. (a) (0, 0, 3), (4, 0, 0), (8, 0, 0). (b) (3,2,0), (-2,5,7), (1, -3, -5). (c) (-4,0, 6), (8, 2,-1), (2, 4, 6). (d) (3, - 3, - 3), (4, 2, 7), (- 1, - 2, - 5). 4. In what octant {0-XYZ, 0-X'YZ, etc) will the positive part of a line through lie if (a) cos a > 0, cos ^3 > 0, cos 7 > ? (e) cos a < 0, cos j8 > 0, cos 7 > ? (b) cosa>0, cos/3>0, cos7<0? (f) cosa<0, cos/3<0, cos7>0? (c) cos a: > 0, cos j3 < 0, cos 7 < ? (g) cos a < 0, cos /3 < 0, cos 7 < ? (d) cos a > 0, cos /3 < 0, cos 7 > ? (h) cos or < 0, cos /3 > 0, cos 7 < ? 5. What is the direction of a line if cos a = 0? cosjQ = ? cos 7 = 0? cos a = cos /3 = ? cos /3 = cos 7 = 0? cos 7 = cos ex = 0? 6. Find the projection of the line drawn from the origin to Pi (5, — 7, 6) upon a line whose direction cosines are f, — 7, and f. Ans. 9. Hint. The projection of OP^ on any line equals the projection of a broken line whose segments equal the coordinates of P^. 7. Find the projection of the line drawn from the origin to Pi (xi, ?/i, Zi) upon a line whose direction angles are a, p, and 7. Ans. Xi cos a + yi cos ^ + Zi cos 7. 8. Show that the points (- 3, 2, - 7), (2, 2, - 3), and (- 3, 6, - 2) are the vertices of an isosceles triangle. 9. Show that the points (4, 3, - 4), (- 2, 9, - 4), and (- 2, 3, 2) are the vertices of an equilateral triangle. 10. Show that the points (-4, 0, 2), (-1, 3 V3, 2), (2, 0, 2), and (-1, V3, 2 + 2 Vq) are the vertices of a regular tetraedron. 11. What does formula (IV) become if Pi and P2 lie in the XF-plane ? in a plane parallel to the XF-plane ? 12. Show that the direction cosines of the lines joining each of the points (4, - 8, 6) and (- 2, 4, - 3) to the point (12, - 24, 18) are the same. How are the three points situated ? 13. Show by means of direction cosines that the three points (3, — 2, 7), (6, 4, — 2), and (5, 2, 1) lie on a straight line. 14. What are the direction cosines of a line parallel to the X-axis ? to the F-axis ? to the Z-axis ? 334 ANALYTIC GEOMETRY 15. What is the value of one of the direction cosines of a line parallel to the XY-plane ? the YZ-plane ? the ZX-plane ? What relation exists between the other two ? 16. Show that the point (—1, — 2, — 1) is on the line joining the points (4, — 7, 3) and (— 6, 3, — 5) and is equally distant from them. 7t It 17. If two of the direction angles of a line are — and - , what is the third ? 3 * , TT 27r Ans. —or 3 3 18. Find the direction angles of a line which is equally inclined to the three coordinate axes. Ans. a = p = y = cos-i ^^S. 19. Find the length of a line whose projections on the axes are respectively (a) 6, - 3, and 2. Ans. 7. (b) 12, 4, and - 3. Ans. 13. (c) - 2, - 1, and 2. Ans. 3. 142. Angle between two directed lines. Theorem V. If a, ^, y and a', /3', y' are the direction angles of two directed lines, then the angle 6 between them is given by (V) cos ^ = cos a cos a' + cos p cos jff' + cos y cos y'. Proof Draw OP and OP' parallel to the given lines and let OP == p. Then by definition, p. 327, Z POP' = 6. Project OP and OABP on OP'. ^ Then by Theorem I, p. 328, and Theorem II, p. 328, (1) p cos 6 = X cos a'-\- y cos ^'+ ^ COS y'. Projecting OP on the axes (Corollary I, p. 328, and Theorem I), (2). X = p cos a, y = p COS ^, z = p cos y. Substituting in (1) from (2) and dividing by p, we obtain (V). Q.E.D. 'AX CARTESIAN COORDINATES IN SPACE 336 Theorem VI. If a, ^, y and a', p\ y' are the direction angles of two lines, then the lines are (a) parallel and in the same direction* when and only when a=a',/3 = /3',y = y'', (b) perpendicular "f ivhen and only ivhen cos a cos a'-\- COS P COS /8' H- cos y COS y' = 0. That is, two lines are parallel and in the same direction when and only luhen their direction angles are equal, and perpendicular when and only when the sum of the products of their direction cosines is zero. Proof The condition for parallelism follows from the fact that both lines will be parallel to and agree in direction with the same line through the origin when and only when their direction angles are equal. The condition for perpendicularity follows from (Y), for if IT = —} then cos ^ == 0, and conversely. q.e.d. Li Corollary. If the direction cosines of the lines are proportional to a, h, c and a\ b', c', then the conditions for parallelism and perpendicularity are respectively ^ b C , T 7 I t r. — = - = -, aa' + bb' -\- cc' = 0. a' b c' 143. Point of division. Theorem VII. The coordinates (x, y, z) of the point of division P on the line joining Pi(xi, yi, Zi) and P2(^2> y2} ^2) such that the ratio of the segments ^s are given by the formulas (VII) «= = ^Vtx'' ^ = ^ttx'' " = ttx- This is proved as on p. 39. * They will he parallel and differ in direction when and only when the direction angles are supplementary. t Two lines in space are said to he perpendicular when the angle hetween them is — » but the lines do not necessarily intersect. 336 ANALYTIC GEOMETRY Corollary. The coordinates (x, y, z) of the middle point P of the line joining P\{xi, yi, z^) and ^2(^2? 1/21 ^2) ^^^ 03 = i(Xi + a?2), 2/ = i(2/i + Vi), « = 1(^1 + «2)- • PROBLEMS 1. Find the angle between two lines whose direction cosines are respectively (a) f, f, -f and a, - |, f Ans. |. (b) f, - 1, I and - T?3, r%, {h ^ns. cos-iif. (c) f, - f, 1 and f, f, f. ^ns. cos-i(- f^). 3. Show that the lines joining the following pairs of points are either parallel or perpendicular, -^ (a) (3, 2, 7), (1, 4, 6) and (7, - 5, 9), (5, - 3, 8). (b) (13, 4, 9), (1, 7, 13) and (7, 16, - 6), (3, 4, - 9). (c) (- 6, 4, - 3), (1, 2, 7) and (8, - 6, 10), (15, - 7, 20). 4. Find the coordinates of the point dividing the line joining the follow- ing points in the ratio given. (a) (3, 4, 2), (7, - 6, 4), \ = \. Ans. (-V-, |, |). (b) (- 1, 4, - 6), (2, 3, - 7), X =3 - 3. Ans. (|, |, - ^i). (c) (8, 4, 2), (3, 9, 6), \ = - i. Ans. (V, f , 0). (d) (7, 3, 9), (2, 1, 2), X = 4. Ans. (3, |, -U). 5. Show that the points (7, 3, 4), (1, 0, 6), and (4, 5, — 2) are the vertices of a right triangle. 6. Showthatthepoints(-6, 3, 2), (3, - 2, 4), (5, 7, 3), and (- 13, 17,-1) are the vertices of a trapezoid. 7. Show that the points (3, 7, 2), (4, 3, 1), (1, 6, 3), and (2, 2, 2) are the vertices of a parallelogram. 8. Show that the points (6, 7, 3), (3, 11, 1), (0, 3, 4), and (- 3, 7, 2) are the vertices of a rectangle. 9. Show that the points (6, - 6, 0), (3, - 4, 4), (2, - 9, 2), and (- 1, — 7, 6) are the vertices of a rhombus. 10. Show that the points (7, 2, 4), (4, - 4, 2), (9, - 1, 10), and (6, - 7, 8) are the vertices of a square. CARTESIAN COORDINATES IN SPACE 337 11. Show that each of the following sets of points lies on a straight line, and find the ratio of the segments in which the third divides the line joining the first to the second. (a) (4, 13, 3), (3, 6, 4), and (2, - 1, 5). Ans. - 2. (b) (4, - 5, - 12), (- 2, 4, 6), and (2, - 2, - 6). Ans. ^ (c) (- 3, 4, 2), (7, - 2, 6), and (2, 1, 4). Ans. 1. ' 12. Find the lengths of the medians of the triangle whose vertices are the points (3, 4, - 2), (7, 0, 8), and (- 6, 4, 6). Ans. VTlS, V89, 2V29. 13. Show that the lines joining the middle points of the opposite sides of the quadrilaterals whose vertices are the following points bisect each other. (a) (8, 4, 2), (0, 2, 5), (- 3, 2, 4), and (8, 0, -6). (b) (0, 0, 9), (2, 6, 8), (- 8, 0, 4), and (0, - 8, 6). (c) Pi(xi, yi, zi), P2{X2, 2/2, 22), Psixs, 2/3, Zs), P^ix^, 2/4, Z4). 14. Show that the lines joining successively the middle points of the sides of any quadrilateral form a parallelogram. 15. Find the projection of the line drawn from Pi (3, 2, — 6) to P2(— 3, 5, — 4) upon a line directed upward whose direction cosines are proportional to 2, 1, and - 2. Ans. 4i. 16. Find the projection of the line djawn from Pi (6, 3, 2) to P2(4, 2, 0) upon the line drawn from P3(7, - 6/6) to P4(- 5, - 2, 3). Ans. ff. 17. Find the coordinates of the point of intersection of the medians of the triangle whose vertices are (3, 6, — 2), (7, — 4, 3), and (— 1, 4, — 7). Ans. (3, 2, - 2). 18. Find the coordinates of the point of intersection of the medians of the triangle whose vertices are any three points Pi, P2, and P3. Ans. [i (xi + X2 + X3), i {yi + 2/2 + 2/3), i (^i + Z2 + 23)]. 19. The three lines joining the middle points of the opposite edges of a tetraedron pass through the same point and are bisected at that point. 20. The four lines drawn from the vertices of_any tetraedron to the point of intersection of the medians of the opposite face meet in a point which is three fourths of the distance from each vertex to the opposite face (the center of gravity of the tetraedron). CHAPTER XYII SURFACES, CURVES, AND EQUATIONS 144. Loci in space. In Solid Geometry it is necessary to con- sider two kinds of loci : 1. The locus of a point in space which satisfies one given con- dition is, in general, a surface. Thus the locus of a point at a given distance from a fixed point is a sphere, and the locus of a point equidistant from two fixed points is the plane which is , perpendicular to the line joining the given points at its middle point. 2. The locus of a point in space which satisfies two conditions * is, in general, a curve. For the locus of a point which satisfies either condition is a surface, and hence the points which satisfy both conditions lie on two surfaces, that is, on their curve of intersection. Thus the locus of a point which is at a given distance r from a fixed point Pi and is equally distant from two fixed points P^. and P3 is the circle in which the sphere whose center is Pi and whose radius is r intersects the plane which is perpendicular to P2P3 at its middle point. These two kinds of loci must be carefully distinguished. 145. Equation of a surface. First fundamental problem. If any point P which lies on a given surface be given the coordinates (x, y, z), then the condition which defines the surface as a locus will lead to an equation involving the variables x, y, and z. The equation of a surface is an equation in the variables x, y, and z representing coordinates such that : 1. The coordinates of every point on the surface will satisfy the equation. 2. Every point whose coordinates satisfy the equation will lie upon the surface. * The number of conditions must be counted carefully. Thus if a point is to be equi- distant from three fixed points P^, P^, and P3, it satisfies two conditions, namely, of being equidistant from P^ and P^ and from P^ and P3. 338 SURFACES, CURVES, AND EQUATIONS 339 If the surface is defined as the locus of a point satisfying one condition, its equation may be found in many cases by a Rule analogous to that on p. 53. Ex. 1. Find the equation of the locus of a point whose distance from Pi (3, 0, - 2) is 4. Solution. Let P {x, y, z) be any point on the locus. The given condition may be written p p _ 4 By (IV), p. 331, PiP = V{x - 3)2 + y2 + ^^ 4. 2)2. .-. V(x - 3)2 + 2/2 + (z + 2)2 = 4. Simplifying, we obtain as the required equation x2 + 2/2 + 22 _ 6x + 42; - 3 = 0. That this is indeed the equation of the locus should be verified as in Ex. 1, 52, and Ex. 1, p. 53. PROBLEMS 1 . Find the equation of the locus of a point which is (a) 3 units above the XF-plane. (b) 4 units to the right of the FZ-plane. (c) 5 units below the XF-plane. (d) 10 units back of the ZX-plane.-*- (e) 7 units to the left of the FZ-plane. (f) 2 units in front of the ZX-plane. 2. Find the equation of the plane which is parallel to (a) the XF-plane and 4 units above it. (b) the XF-plane and 5 units below it. (c) the ZX-plane and 3 units in front of it. (d) the FZ-plane and 7 units to the left of it. (e) the ZX-plane and 2 units back of it. (f) the FZ-plane and 4 units to the right of it. 3. Find the equation of the sphere whose center is the point (a) (3, 0, 4) and whose radius is 5. Ans. ic2 + 2/2 + 2=2 - 6x - 8 2; = 0. (b) (—3, 2, 1) and whose radius is 4. Ans. ic2 4-2/2 + z2 + 6x-42/-22;-2=0. (c) (6, 4, 0) and whose radius is 7. Ans. x2 -f 2/2 + z2 - 12 X - 8 2/ -f- 3 = 0. (d) {a, /3, 7) and whose radius is r. Ans. x2 + y2 + 22 _ 2 ax - 2 /??/ - 2 72; + ^2 + ^2 + ^2 _ y2 = 0. 340 ANALYTIC GEOMETRY 4. What are the equations of the coordinate planes ? 5. What is the form of the equation of a plane which is parallel to the XY-plane ? the FZ-plane ? the Z X-plane ? 6. Find the equation of the locus of a point which is equally distant from the points (a) (3, 2, - 1) and (4, - 3, 0). ' Ans. 2x -lOy -\- 2 z -11 = 0. (b) (4, - 3, 6) and (2, - 4, 2). Ans. 4x + 2?/ + 82 - 37 = 0. (c) (1, 3, 2) and (4, - 1, 1). Ans. Sx - 4:y - z - 2 = 0. (d) (4, - 6, - 8) and (- 2, 7, 9). Ans. 6x - 13?/ - 172 + 9 = 0. 7. Find the equations of the six planes drawn through the middle points of the edges of the tetraedron whose vertices are the points (5, 4, 0), (2, —5, —4), (1, 7, —5), and (—4, 3, 4) which are perpendicular to the edges, and show that they all pass through the point (—1, 1, — 2). 8. What are the equations of the faces of the rectangular parallelepiped which has one vertex at the origin, three edges lying along the coordinate axes, and one vertex at the point (3, 5, 7) ? 9. Find the equation of the sphere whose center is the point (6, 2, 3) which passes through the origin. Ans. x^ -]- y'^ -{■ z^ — 12 x — 4y — 6z = 0. 10. Find the equation of the locus of a point which is three times as far from the point (2, 6, 8) as from (4, — 2, 4) and determine the nature of the locus by comparison with the answer to problem 3, (d). 11. Find the equation of the locus of a point the sum of the squares of whose distances from (1, 3, — 2) and (6, — 4, 2) is 50 and determine the nature of the locus by comparison with the answer to problem 3, (d). 146. Planes parallel to the coordinate planes. We may easily prove Theorem I. The equation of a plane which is parallel to the XY-plane has the form z = constant; parallel to the YZ -plane has the form x = constant; parallel to the ZX-plane has the form y = consta7it. 147. Equations of a curve. First fundamental problem. If any point P which lies on a given curve be given the coordinates {x, y, z), then the two conditions which define the curve as a locus will lead to two equations involving the variables x, y, and z. SURFACES, CURVES, AND EQUATIONS 341 The equations of a curve are two equations in the variables X, y, and z representing coordinates such that: 1. The coordinates of every point on the curve will satisfy- both equations. 2. Every point whose coordinates satisfy both equations will lie on the curve. If the curve is defined as the locus of a point satisfying two conditions, the equations of the surfaces defined by each condi- tion separately may be found in many cases by a Rule analogous to that on p. 53. These equations will be the equations of the curve. Ex. 1. Find the equations of the locus of a point whose distance from the origin is 4 and which is equally distant from the points Pi (8, 0, 0) and A (0,8,0). ^^ Solution. First step. Let P{x, y, z) be any point on the locus. Second step. The given conditions are (1) P0 = 4, PPi = PP2. Third step. By (IV), p. 331, Vx2 + ?/2 + Z^, PO PPl = V(X - 8)2 + 2/2 + z% PP2 = Vx2 + (y - 8)2 + z2. Substituting in (1), we get (8,0,0) ^ Vx2 + y2 + 2;2 = 4, V(a; - 8)2 + 2/2 + ^2 = Vx"^ + {y - 8)2 + z^ Squaring and reducing, we have the required equations, namely, x2 + 2/2 + 22 = 16, x-y = 0. These equations should be verified as in Ex, 1, p, 52. Ex. 2. Find the equations of the circle lying in the XF-plane whose center is the origin and whose radius is 5. Solution. In Plane Geometry the equation of the circle is (Corollary, p. 68) (2) x^ + y^ = 25. Regarded as a problem in Solid Geometry we must have two equations which the coordinates of any point P(x, y, z) which lies on the circle must satisfy. Since P lies in the XF-plane, (3) 2 = 0. Hence equations (2) and (3) together express that the point P lies^m the XF-plane and on the given circle. The equations of the circle are therefore x2 + ?y2 = 25, z = 0. 342 ANALYTIC GEOMETRY The reasoning in Ex. 2 is general. Hence If the equation of a curve in the XY-jplane is known, then the equations of that curve regarded as a curve in space are the given equation and « = 0. An analogous statement evidently applies to the equations of a curve lying in one of the other coordinate planes. From Theorem I, p. 340, we have at once Theorem II. The equations of a line ivhich is parallel to the X-axis have the form y = constant, z = constant; the Y-axis have the form z = constant, x = constant; the Z-axis have the form x = constant, y = constant. PROBLEMS 1. Find the equations of the locus of a point which is (a) 3 units above the JTF-plane and 4 units to the right of the FZ-plane. • (b) 5 units to the left of the YZ-plan.e and 2 units in front of the ZX-plane. (c) 4 units back of the ZX-plane and 7 units to the left of the FZ-plane. (d) 9 units below the XY-plane and 4 units to the right of the rZ-plane. 2. Find the equations of the straight line which is (a) 5 units above the XY-plane and 2 units in front of the ZX-plane. (b) 2 units to the left of the YZ-plane and 8 units below the JTF-plane. (c) 3 units to the right of the FZ-plane and 5 units from the Z-axis. (d) 13 units from the X-axis and 5 units back of the ZX-plane. (e) parallel to the F-axis and passing through (3, 7, — 5). (f) parallel to the Z-axis and passing through (—4, 7, 6). 3. Find the equations of the locus of a point which is (a) 5 units above the XF-plane and 3 units from (3, 7, 1). Ans. z = 5, x^ + y^ + z^-Gx-Uy -2z + 60 = 0. (b) 2 units from (3, 7, 6) and 4 units from (2, 5, 4). Ans. x2 -h ?/2 + ^2 - 6 X - 14 i/ - 12 2; + 90 = 0, x2 -f- 2/2 -I- 22 _ 4x _ 10 ?/ - 8 2; + 29 ^ 0. (c) 5 units from the origin and equidistant from (3, 7, 2) and ( — 3, - 7, —2). Arts. x2 -f- ?/2 -}- z2 _ 25 = 0, Sx + 7 y -\- 2z = 0. (d) equidistant from (3, 5, — 4) and (— 7, 1, 6), and also from (4, — 6, 3) and (- 2, 8, 5). Ans. 6x -\- 2y - 5z + 11 =0, 3x-7y-z + 8 = 0. (e) equidistant from (2, 3, 7), (3, - 4, 6), and (4, 3, - 2). Ans. 2x-Uy -2z-{-l=:0, x+ly-Sz + 16 = 0. SURFACES, CURVES, AND EQUATIONS 343 4 . What are the equations of the edges of a rectangular parallelepiped whose dimensions are a, b, and c, if three of its faces coincide with the coordinate planes and one vertex lies in 0-XYZ ? in 0-XY'Z ? in 0-X'Y'Z ? 5. What are the equations of the axes of coordinates ? 6. The following equations are the equations of curves lying in one of the coordinate planes. What are the equations of the same curves regarded as curves in space? (a) 2/2 = 4a;, (e) x2 + 42 + 6x = 0. (b) x^ + z^ = 16. (f) y2-z^-4y = 0. (c) 8x2 - 2/2 = 64. (g) yz^-\-z^-Qy = 0. (d) 4 22+ 9 2/2 = 36. (h) z2_4a;2 + 82 = 0. 7. Find the equations of the locus of a point which is equally distant from the points (6, 4, 3) and (6, 4, 9), and also from (— 5, 8, 3) and (— 5, 0, 3), and determine the nature of the locus. Ans. 2 = 6, y = 4. 8. Find the equations of the locus of a point which is equally distant from the points (3, 7, — 4), (— 5, 7, — 4), and (— 6, 1, — 4), and determine the nature of the locus. Ans. x = — 1, y = ^. 148. Locus of one equation. Second fundamental problem. The locus of one equation in three variables (one or two may be lacking) representing coordinates in space is the surface passing through all points whose coordinates satisfy that equation and through such points only. The coordinates of points on the surface may be obtained as follows : Solve the equation for one of the vai'iables, say z, assume pairs of values of a; and y, and compute the corresponding values of z. A rough model of the surface might then be constructed by taking a thin board for the XF-plane, sticking needles into it at the assumed points (x, y) Avhose lengths are the computed values of z, and stretching a sheet of rubber over their extremities. The second fundamental problem, namely, of constructing the locus, is usually discarded in space on account of the mechanical difficulties involved. 149. Locus of two equations. Second fundamental problem. The locus of two equations in three variables representing coor- dinates in space is the curve passing through all points whose coordinates satisfy both equations and through such points only. The coordinates of points on the curve may be obtained as follows : Solve the equations for two of the variables, say x and y, in terms of the third, z, assume values for z, and compute the corresponding values of x and y. 344 ANALYTIC GEOMETRY 150. Discussion of the equations of a curve. Third funda- mental problem. The discussion of curves in Elementary Ana- lytic Geometry is largely confined to curves which lie entirely in a plane which is usually parallel to one of the coordinate planes. Such a curve is defined as the intersection of a given surface with a plane parallel to one of the coordinate planes. The method of determining its nature is illustrated in Ex. 1. Determine the nature of the curve in which the plane 2 = 4 inter sects the surface whose equation is y^ + z^ = 4 x. Solution. The equations of the curve are, by definition, (1) 2/2 + z2 = 4 X, 2; = 4. Eliminate z by substituting from the second equation in the first. This gives (2) *?/'^-4x + 16 = 0, z = 4. Equations (2) are also the equations of the curve. ' For every set of values of {x, y, z) which satisfy both of ecLuations (1) will evidently satisfy both of equations (2), and conversely. If we take as axes in the plane 2 = 4 the lines O'X' and O'Y' in which the plane cuts the ZX- and YZ-planes, then the equation of the curve when referred to these axes is the first of equations (2), namely, (3) y^ -4tx^\Q = 0. For the second of equations (2) is satisfied by all points in the plane of X^, O', and Y\ and the first of equations (2) is satisfied by the points in that plane lying on %e curve (3), because the values of the first two coordinates of a point are evidently the same when referred to the axes O'X', O' Y', and O'Z as when referred to the axes OX, O Y, and OZ. The locus of (3) is a parabola (Rule, p. 197) whose vertex, in the plane z = A, m the point (4, 0) for which p = 2. SURFACES, CUKVES, AND EQUATIONS 345 The method employed in Ex. 1 enables us to state the Rule to determine the nature of the curve in which a plane par- allel to one of the coordinate planes cuts a given surface. First step. Eliminate the variable occun^ng in the equation of the plane from the equations of the plane and surface. The result is the equation of the curve referred to the lines in which the given plane cuts the other two coordinate planes as axes. Second step. Determine the nature of the curve obtained in the second step by the methods of Plane Analytic Geometry. PROBLEMS 1. Determine the nature of the following curves and construct their loci. (a) x2 - 4 ?/2 = 8 z, z = 8. (e) x2 + 4 2/2 + 9 z2 zr 36, ?/ = 1. (b) x2 + 9?/2 = 9z* z = 2. (f) a;2 - 4?/2 + ^2 = 25, x = - 3. (c) x2-4?/2 = 4z, 2/ = -2. (g) x2-2/2-4z2 + 6x = 0, x=2. (d) x2 + ?/2 + z2 = 25, X = 3. (h) 2/2 + z2 - 4 X + 8 = 0, y = 4. 2. Construct the curves in which each of the following surfaces intersect the coordinate planes. (a) x2 + 4 2/2 + 16 z2 = 64. (d) x2 + 92/2 = 10 z. (b) x2 + 4 2/2-16z2 = 64. (e) x2-9 2/2=10z. (c) x2 - 4 2/2 - 16 z2 = 64. (f) x2 + 4 2/2 - 16 z2 = 0. 3. Show that the curves of intersection of each of the surfaces in problem 2 with a system of planes parallel to one of the coordinate planes are similar conies. In what cases must' this statement be modified ? 4. Determine the nature of the intersection of the surface x2 + 2/2 + 4 z2 = 64 with the plane z = k. How does the curve change as k increas^^ irom 0- to 4? from — 4 to 0? What idea of the appearance of tiie surface is thus obtained ? 6. Determine the nature of the intersection of the surface 4x — 2 2/ = 4 with the plane y = k; with the plane z = k'. How does the intersection change as k or ¥ changes ? What idea of the form of the surface is obtained ? 151. Discussion of the equation of a surface. Third funda- mental problem. Theorem III. The locus of an algebraic equation passes through the origin if there is no constant term in the equation. The proof is analogous to that of Theorem VI, p. 73. 346 ANALYTIC GEOMETRY Theorem IV. If the locus of an equation is unaffected by chang- ing the sign of one variable throughout its equation, then the locus is symmetrical with resj^ect to the coordinate plane from which that variable is measured. If the locus is unaffected by changing the signs of two variables throughout its equation, it is symmetrical with respect to the axis along which the third variable is measured. If the locus is unaffected by changing the signs of all three variables throughout its equation, it is symmetrical with respect to the origin. The proof is analogous to tliat of Theorem IV, p. 72. Rule to find the intercepts of a surface on the axes of coordinates. Set each pair of variables equal to zero and solve for real values of the third. The curves in which a surface intersects the coordinate planes are called its traces on the coordinate planes. From the first step of the Rule, p. 345, it is seen that The equations of the traces of a surface are obtained by succes- sively setting x = 0, y = 0, and z = in the equation of the surface. By these means we can determine some properties of the surface. The general appearance of a surface is determined by considering the curves in which it is cut by a system of planes parallel to each of the coordinate planes (Eule, p. 345). This also enables us to determine whether the sur- face is closed or recedes to infinity. Ex. 1. Discuss the locus of the equation y^ 4- z'^ = iz. Solution. 1. Tha surface passes through the origin since there is no constant term in its equation. 2. The surface is sym- metrical with respect to the XF-plane, the ZX-plane, and the X-axis. For.the locus of the given equation is unaffected by changing the sign of z, of y, or of both together. SURFACES, CURVES, AND EQUATIONS 347 3. It cuts the axes at the origin only. 4. Its traces are respectively the point-circle y^-\-z^ = and the parabolas z2 = 4 cc and y^ = 4:X. 5. It intersects the plane x = A: in the curve (Rule, p. 345) 2/2 + 2;2 = 4 A;. This curve is a circle whose center is the origin, that is, is on the X-axis, and whose radius is 2 Vfc if A; > 0, but there is no locus if k<0. Hence the surface lies entirely to the right of the FZ-plane. If k increases from zero to infinity, the radius of the circle increases from zero to infinity while the plane x = k recedes from the FZ-plane. The intersection of a plane z = k or y=k', parallel to the XY- or ZX-plane, is seen (Rule, p. 345) to be a parabola whose equation is (compare Ex. 1, p. 344) y2 = 4x-fe2 or z^ = 4:X-k'^. These parabolas are found to have the same value of p, namely, p = 2, and their vertices recede from the YZ- or ZX-plane as k or k' increases numerically. PROBLEMS 1 . Discuss the loci of the following equations. (a) x2 + 22 = 4 a;. (f ) X2 + 2/2 - 22 = 0. (b) x2-fy2 + 422 = 16. (g) X2 - 2/2 _ 22 ^ 9. (C) X2 + ?/2 - 4 Z2 = 16. (h) x2 + 2/2-2;2 + 2xy = (d) 6x + 4?/+3z = 12. (i) x + y-6z = e. (e) Sx + 2y + z = 12. (j) 2/2 + 22 = 25. 2. Show that the locus of Ax + By + Cz + D — is a. plane by considering its traces on the coordinate planes and the sections made by a system of planes parallel to one of the coordinate planes. 3. Find the equation of the locus of a point which is equally distant from the point (2, 0, 0) and the FZ-plane and discuss the locus. Ans. 2/2 -1- 22 _ 4 X + 4 = 0. 14. Find the equation of the locus of a point whose distance from the point (0, 0, 3) is twice its distance from the XF-plane and discuss the locus. Ans. x2 + 2/2- 3 22-62 + 9 = 0. [ 5. Finj^ the equation of the locus of a point whose distance from the point (0, 4, 0) is three fifths its distance from the ZX-plane and discuss the locus. Ans. 25x2 + 16 2/2 + 2522 -200?/ + 400 = 0. CHAPTER XVIII THE PLANE AND THE GENERAL EQUATION OF THE FIRST DEGREE IN THREE VARIABLES 152. The normal form of the equation of the plane. Let ABC be any plane, and let ON be drawn from the origin per- pendicular to ABC at D. Let the positive direction on ON he from O toward N, that is, /ro7^ the origin toward the plane, and denote the directed length OD by p and the direction angles of ON (p. 330) by a, (3, and y. Then the position of any j^^ci.ne is determined by given positive values of p, a, (3, and y. Conversely, a given plane determines a single set of positive values of p, a, ^, and 7 unless p = 0. lip= 0, the positive direction on ON becomes meaningless. Ifp — 0, we shall suppose that ON is directed upward, and hence cos 7 > since It 7 < — • If the plane passes through OZ, then ON lies in the XF-plane and 2 It cos 7 = 0; in this case we shall suppose ON so directed that /3 < 17 and hence cos /3 > 0. Finally, if the plane coincides with the TZ-plane, the positive direction on ON shall be that on OX. 348 THE PLANE 349 Theorem I. Normal form. TJie equation of a plane is (I) X cos a + 2/ cos ^ + 5! cos y — jp = O, where p is the perpendicular distance from the origin to the plane, and a, ^, and y are the direction cosines of that perpendicular. Proof Let P(x, y, z) be any point on the given plane ABC. Project OEFP and OP on the line ON. By Theorem II, p. 328, proj. of OE + proj. of EF + proj. of FP = proj. of OP. Then by Theorem I, p. 328, and by the definition, p. 29, X cos a + y cos ^ + z cos y = p. Transposing, we obtain (I). q.e.d. Corollary. The equation of any plane is of the first degree in X, y, and z. 153. The general equation of the first degree, Aoc + By + Cz + 1> = 0. Theorem II. (Converse of the Corollary.) The locus of the gen- eral equation of the first degree in x, y, and z, (II) Aic-\- By -\-Cz + D = 0, is a plane. Proof We shall prove the theorem by showing that (II) may' be reduced to the form (I) by multiplying by a proper constant. To determine this constant, multiply (II) by k, which gives (1) kAx + kBy + kCz + kD = 0. Equating corresponding coefficients of (1) and (I), we get (2) kA = cos a, kB = cos (3, kC = cos y, kD = —p. Squaring the first three of equations (2) and adding, k^(A^ -\-B^ -{- C^) = cos^a + cos^^ + cos^y = 1. (by (III), p. 330) (3) .\ k= ^ - 350 ANALYTIC QEOMETRY From the last of equations (2) we see that the sign of the radical must be opposite to that of D in order that p shall be positive. If Z) = 0, then p = 0; and from the third of equations (2) the sign of the radical must be the same as that of C, since when p = cos 7 > 0. If D = and C = 0, then p—0 and cos 7=0 ; and from the second of equations (2) the sign of the radical must be the same as that of B, since when p — and cos 7 = cos /3 > 0. Substituting from (3) in (2), we get COS « = , ) COS B = i C -D COS y We have thus determined values of a, /?, y, and p such that (I) and (II) have the same locus. Hence the locus of (II) is a plane. q.e.d. Corollary I. The direction cosines of a normal to the plane (II) are respectively A, B, and C each divided hy ± Vvl^ -\- B'^ -\- C^. The sign of the radical is opposite to that of D, the same as that of C if D = 0, the same as that of B if C = D = 0, or the same as that of A ifB = C = D = 0. Corollary II. To reduce the equation of a plane to the normal form divide its equation hy zt V^^ -\- B'^ -\- C^, choosing the sign of the radical as in Corollary I. Corollary III. Two planes whose equations are Ax + By + Cz + D = 0, A'x + B'y -i- C'z -[- D' = are parallel when and only when the coefficients of x, y, and z are proportional, that is, A__B__C^ A'~B'~C'' For from Corollary I the direction cosines of a normal to (II) are proportional to A, B, and C, and two planes are evidently parallel when and only when their normals are parallel (Corollary, p. 335) . Corollary IV. Two planes are perpendicular when and only when AA' ^BB^ ^- CC = 0. This follows from Corollary I by the Corollary on p. 335, since two planes are perpendicular when and only when their normals are perpendicular. THE PLANE 351 Corollary V. A plane whose equation has the form Ax -{- By -\- D = is perpendicular to the XY-plane; By -\- Cz -{- D z= is perpendicular to the YZ -plane; Ax-\- Cz -\- D = (i is perpendicular to the ZX-plane. That is, 4f one variable is lacking, the plane is perpendicular to the coordinate plane corresponding to the two variables which occur in the equation. For these planes are respectively perpendicular to' the planes 2; = 0, x = 0, and y = by Corollary IV. Corollary VI. A plane whose equation has the form Ax -^ D = is perpendicular to the axis of x; By + D = is perpendicular to the axis of y ; Cz -{- D = is perpendicular to the axis of z. That is, if two variables are lacking, the plane is perpendicular to the axis corresponding to the variable which occurs in the equation. For by Corollary I two of the direction cosines of the normal to the plane are zero and hence the normal is parallel to one of the axes and the plane is therefore perpendicular to that axis. PROBLEMS 1. Eind the intercepts on the axes and the traces on the coordinate planes of each of the following planes and construct the figures. (a) 2ic + 3?/ + 4z - 24 = 0. (e) 5a; - 7y - 35 = 0. Xb) Tx-Sy + 2-21=0. (f) 4X + 32; +36 = 0. ■^^ (c) 9x-7y-9z + 6S = 0. (g) 5?/ - 82 - 40 = 0. (d) 6x + 4y -2+ 12 = 0. (h) 3x + 52 + 45 = 0. 2. Find the equations of the planes and construct them by drawing their traces, for which Ans. V2 X + 2/ + z - 12 = 0. Ans. x + V2y-2; + 16 = 0. Ans. 6x-2y + Sz-2Sz=0. Ans. 2x + y + 22 + 6 = 0. ,p = e (b)a = ?^,^ = 5^,.= It = 3'^ cos a cos/3 C0S7 ^'^ 6 - -2 3 ' 1> = 4, ,,, cos or cosjS cos 7 (^) _2 = -l = -2' p = 2, 352 ANALYTIC GEOMETRY 3. Find the equation of the plane such that the foot of the perpendicular from the origin to the plane is the point (a) (-3,- 2, a). Ans. 3x - 2 ?/ - 62 + 49 = 0. (b) (4, 3, - 12). Ans. 4 x + 3 ?/ - 12 z - 169 = 0. (c) (2, 2, - 1). Ans. 2x + 2y -z-9 = 0. 4. Reduce the following equations to the normal form and find a-, j8, 7, and p. (a) 6x-Sy + 2z-7 = 0. Ans. cos-if, cos-i(- f), cos-if, 1. (b) x-V2y + z + 8 = 0. Ans. ?^, j, ^, 4. o 4 o (c) 2x-2y -z + 12 = 0. Ans. cos-i(- |), cos-if, cos-ii, 4. (d) y-z + 10 = 0. Ans. -, — , -, 5V2. (e) Sx-{-2y -6z = 0. Ans. cos-i(- f), cos-i(-f), cos-if, 0. 5 . Find the distance from the origin to the plane 12x — 4?/ + 3z — 39 = 0. Ans. 3. 6. Find the distance between the parallel planes 6x + 2y — Sz — 6Pi = and6x + 22/-32; + 49 = 0. Ans. 16. 7. What may be said of the position of the plane (I) if (a) cos a = 0? (c) cos 7 = ? (e) cos j3 = cos 7 = ? (b) cos /3 = ? (d) cos « = cos/3 = ? (f ) cos 7 = cos or = ? 8. What are the equations of the traces on the coordinate planes of the plane Ax + By + Cz -{■ D = 0? 9. Show that the following pairs of planes are either parallel or perpen- dicular. J2x-hBy-ez + S = 0, fex-Sy + 2z-1 =0, W \6x+l^y-lSz-6 = 0. ^^' \sx + 2y-6z-\-28 = 0. 3x-5?y-4z + 7 = 0, ,^,^ (Ux - 7 y - 21 z - 50 = 0, 3z + 12 = 0. fSx-6y-4z + 7 = 0, ri4x-7; ^ ^ \6x + 2y + 2z-7 = 0. ^ ^ \ 2x-y 10. For what values of a, )3, 7, and p will the locus of (I) be parallel to the XF-plane ? the YZ-plane ? the ZX-plane ? coincide with each of these planes ? 11. For what values of a, ^, 7, and p will the locus of (I) pass through the X-axis ? the F-axis ? the Z-axis ? 12. Show that the coordinates of the point of intersection of three planes may be found by solving their equations simultaneously for x, y, and z. THE PLANE 353 13. Find the coordinates of the point of intersection of the planes x-{-2y -{■ z = 0, X -2y -8 = 0, and x + y-\-z-S = 0. Ans. (2, - 3, 4). 14. Show that the plane x + 2y — 2z — 9 = passes through the point of intersection of the planes x + y-^z — 1 = 0, x—y — z — l = 0, and 2xi-Sy -8 = 0. 15. Show that the four planes x + y + 2z — 2 = 0, x + y — 2z -\- 2 = 0, X — y + 8 = 0, and 3x — y-2z + 18 = pass through the same point. 16. Show that the planes 2x — y-\-z + S = 0, x — ?/ + 4z = 0, Sx + y -2z + 8 = 0,^x-2y + 2z-5 = 0,9x + Sy-6z-T=0,a,nd7x-7y -\- 28z — 6 = bound a parallelopiped. 17. Show that the planes Qx — Sy + 2z = i, Sx + 2y -6z =10, 2x + 6y + 32 = 9, Sx + 2y -6z = 0, 12x + 36y + 18z-ll = 0, and I2x -6y + 42 — 17 =0 bound a rectangular parallelopiped. 18. Show that the planes x+2y — z = 0, y+1 z—2=0, Xj2)f — z—A=0, 2x + y — 8 = 0, and 3x + 3y — z — 8 = bound a quadranaj^lar pyramid. 19. Derive the conditions for parallelism of two plane^'from the fact that two planes are parallel if their traces are parallel lines. / / 154. Planes determined by three conditions. If three of the coefficients of (1) . Ax+By -{-Cz^D = are known in terms of the fourth, then the plane is completely determined, for if their values be substituted in (1), the equation may be divided by the fourth coefficient. Three conditions which the plane satisfies will lead to three equations in the coefficients which may be solved for three of the coefficients in terms of the fourth. Hence a plane is, in general, determined by three con- ditions. Its equation may be obtained by a Rule analogous to that on p. 93, using equation (1) in the first step. Thus to find the equation of a plane passing through three points we proceed as in Ex. 1, p. 93, using equation (1) in the first step. In the second step three equations involving A, B, C, and 7) are obtained, which may be solved for three of these coefl&cients in terms of the fourth. 354 ANALYTIC GEOMETRY Ex. 1. Find the equation of the plane which passes through the point Pi (2, -7, I) and is parallel to the plane 21 ic - 12 ?/ + 28 z - 84 = 0. Solution. Let the equation of the required plane be (2) Ax + By-{-Cz + D = 0. Since Pi lies on (2), (3) 2A-7B + IC + D = 0, and since (2) is parallel to the given plane (Corollary III, p. 350), A _ Ji^ _ C^ 21 ~ -12" 28* (4) Solving (3) and (4) for J., J5, and D in terms of C, we get A = ^C, B = -^C, D=-6C.- Substituting in (2), we obtain lCx-^Cy + Cz-6C = 0. Clearing of fractions and dividing by C, 21 X - 12 2/ + 28 z - 168 = 0. PROBLEMS 1. Find the equation of the plane which passes through the points (2, 3, 0), (-2, -3, 4), and (0, 6, 0). Ans. Sx + 2y -{- 6z -12 = 0. \y 2. Find the eqiiation of the plane which passes through the points " 1, i, -1), (-2, -2, 2), and (1, -1, 2). Ans. x-Sy -2z = 0. 3. Find the equation of the plane which passes through the point (3, — 3, 2) and is parallel to the plane Sx — y + z — 6 = 0. Ans. 3x — y-\-z — 14: = 0. \) THE PLANE 355 4. Find the equation of the plane which passes through the points (0, 3, 0) and (4, 0, 0) and is perpendicular to the plane 4x*— 6y — z = 12. Ans. Sx + iy -12z-l2 = 0. 5. Find the equation of the plane which passes through the point (0, 0, 4) and is perpendicular to each of the planes 2x — Sy = b and x-iz = 3. Ans. 12a: + 82/ + 32 -12 = 0. 6. Find the equation of the plane whose intercepts on the axes are 3, 5, and 4. Ans. 20x + 12?/ + 15 2 - 60 = 0. \/t. Find the equation of the plane which passes through the point (2, — 1, 6) and is parallel to the plane a>— 2^ — 32 + 4 = 0. Ans. x-2y-3z + 14: = 0. 8, Find the equation of the plane which passes through the points (2, —1, 6) and (1, -2, 4) and is perpendicular to the plane x — 2?/ — 2 2 + 9=0. Ans. 2x + 4y-32; + 18 = 0. 9. Find the equation of the plane w^ose intercepts are —1, —1, and 4. Ans. 4x + 42/-2 + 4=0. 10. Find the equation of the plane which passes through the point (4, —2, 0) and is perpendicular to the planes x+y—z=0 and 2x— 4y + z=5. Ans. x-\-y + 2z-2 = 0. 11. Show that the four points (2, -3,4), (1,0,2), (2, -1,2), and (1, — 1, 3) lie in a plane. 12. Show that the four points (1, 0, -1), (3, 4, -3), (8, -2, 6), and (2, 2, — 2) lie in a plane. 13. Find the equation of the plane which is perpendicnlai" to the line joining (3, 4, — 1) to (5, 2, 7) at its middle point. 4iP^ '* \ . Ans. x ^' t ^ 2 -^ 13 = 0. ^'^ i \ 14. Find the equations of the faces of the tetraedraiBf^hose v^tices^rfe' the points (0, 3, 1), (2, - 7, 1), (0, 5, - 4), and (2, 0, l).'' Ans. 25x + 52/ + 22 = 17, 5x- 22 = 8, 2 = 1, 15x + lOy + 42 = 34. 15. The equations of three faces of a parallelepiped are x — 4?/ = 3, 2x-2/ + 2 = 3, and 3x + y-22 = 0, and one vertex is the point (3, 7, - 2). What are the equations of the other three faces ? ' Ans. x-42/ + 25 = 0, 2x-?/ + 2 + 3 = 0, 3x + ?/-22 = 20. 16. Find the equation of the plane whose intercepts are a, 6, c y b Ans. - + - + - V- 356 ANALYTIC GEOMETRY 17. What are the equations of the traces of the plane in problem 16? How might these equations have been anticipated from Plane Analytic Geometry ? 18. Find the equation of the plane which passes through the point -Pi {^ii 2/i5 Zi) and is parallel to the plane AiX + Biy + dz + Di = 0. Ans. Ai (X - Xi) + Bx{y - yi) + Ci (2 - Zi) = 0. 19. Find the equation of the plane which passes through the origin and -Pi i^u 2/1? ^1) a-nd is perpendicular to the plane AiX + Biy + CiZ + Di = 0. Ans. {Bizi - Ciyi)x + (CiXi - AiZi) y + {AiVi - BiXi) z = 0. 155. The equation of a plane in terms of its intercepts. Theorem III. If a, b, and c are the intercepts of a plane on the X-j Y-, and Z-axes respectively, then the equation of the plane is (III) ^ + ^ + ^ = 1. ? Proof By Theorem II the equation of any plane has the form (1) Ax-\-Bij -\-Cz^D = 0. By the Rule, p. 346, we get D ^ D whence A= 5 B= — —i C= a G Substituting in (1), dividing by — D, and transposing, we obtain (III)- q.e.d. Equation (III) should be compared with (VI), p. 96. 156. The distance from a plane to a point. The positive direc- tion on any line perpendicular to a plane is assumed to agree with that on the line drawn through the origin perpendicular to the plane (p. 348). Hence the distance from a plane to the point Pi is positive or negative according as Pi and the origin are on oppo- site sides of the plane or not. If the plane passes through the origin, the sign of the distance from the plane to Pi must be determined by the conventions for the special cases on p. 348. D — F a "-T THE PLANE 357 Theorem IV. The distance d from the plane X cos cc -\- y cos y8 + ^ cos y — p = to the point P^ (x-i, yi, z-^) is given by ^■ (IV) ci = Xi cos a + 2/1 cos p + z^cosy — p. Proof. Projecting OP^ on ON, we evidently get jo + d. Projecting OE, EF, and FP^ on ON, we get respectively (Theo- rem I, p. 328) Xi cos a, 2/1 cos (3, and Zi cos y. Then by Theorem II, p. 328, p -\- d z= Xi cos a + 2/1 cos y8 H- ^1 cos y. .-. d = XiGOSa -\- 2/1 cos ft -\- Zi cos y —p. Q.E.D. Prom Theorem lY we have at once the Rule to find the distance from a given plane to a given point. First step. Reduce the equation of the plane to the normal form {Corollary II, p. S50). Second step. Substitiite the coordinates of the given point in the left-hand side of the equation. The result is the required distance. 157. The angle between two planes. The plane angle of one pair of diedral angles formed by two intersecting planes is evi- dently equal to the angle between the positive directions of the normals to the planes. That angle is called the angle between the 358 ANALYTIC GEOMETRY Theorem V. The angle 6 hetiveen two planes is given by ^7ie si(7W5 o/ ^Ae radicals being chosen as in Corollary I, p. 350. Proof. By definition the angle between the planes is the angle between their normals. By (4), p. 350, the direction cosines of the normals to the planes are cos oTi = , "'' ? cos cr. cos Pi = . ^ ? cos ^2 = ^ii ±^A, + c\' ±Va{ ' + B,' c. + Ci^ ±vi; ' + B, '^ci ±Va, '+B^' '^c^ C, cos yi = , ? COS 72 . By (V)^ P- 334, we have COS ^ = COS «i COS CTs + COS ^1 COS ^2 + COS yi COS ya. Substituting the values of the direction cosines of the normals, we obtain (V). q.e.d. PROBLEMS 1. Find the distance from the plane (a) 6x - 3 ?/ + 2 2: - 10 = to the point (4, 2, 10). Ans. 4. (b) X + 2 2/ - 2 z - 12 = to the point (1, - 2, 3). Ans. - 7. (c) 4 X 4- 3 ?/ + 12 2; + 6 = to the point (9, -1,0). Am. - 3. (d) 2x - 5?/ + 3 2; - 4 = to the point (-2, 1, 7). Ans. y% V38. 2. Do the origin and the point (3, 5, — 2) lie on the same side of the plane 7x — 2/ — 32 + 6 = 0? An&. Yes. 3. Find the distance from the plane ^x + % + Cz + D = to the point -Pi(a^i, 2/1, ^i)- ^^^ Axx + Byx + Czx + D THE PLANE ' 359 4. Find the locus of points which are equally distant from the planes 2x-y -2z-3 = 0and6x-Sy + 2z-\-4i = 0. Ans, 32x-16y-82-9 = 0. 5. Find the length of the altitude of the tetraedron whose vertices are (0, 3, 1), (2, - 7, 1) (0, 5, - 4), and (2, 0, 1) which is drawn from the first vertex. Ans. ^^V29. 6. Find the volume of the tetraedron whose vertices are (3, 4, 0), (4, - 1, 0), (1, 2, 0), and (6, - 1, 4). Ans. 8. 7. Find the angles between the following pairs of planes. (a) 2x + y -2z-9 = 0,x-2y -\-2z = 0. Ans. cos-i(-f). (b) x-\-y-4.z = 0,Sy-3z + 7 = 0. Ans. cos- 1 1. (c) ^x + 2y -{-4z-7 = 0, 3x~4y = 0. Ans. cos-i(-^2^). {d)2x-y-^z = 7,x-\-y + 2z = ll. . Ans. -. o 8. Show that the angle given by (V) is that angle formed by the planes which does not contain the origin. 9. Find the vertex and the diedral angles of that triedral angle formed by the planes x + y + z = 2^x — y — 2z=^^ and 2x-\-y — z = 2\n which the origialies. ^^ ^^_ _^^ 2), cos-.iVi, ^, cos->(-l%^). o o \ 3 / ^ 10. Find the equation of the plane which passes through the points (0, —1, 0) and (0, 0, — 1) and which makes an angle of — with the plane y + z = l. r- ^ Ans. ±V6x + y + 2; + l = 0. 11. Find the locus of a point which is 3 times as far from the plane Zx — Qy — 2z = Q as from the plane 2x — y + 2z = ^. Ans. 17x-13y + 12z-63 = 0. 158. Systems of planes. The equation of a plane which satis- fies two conditions will, in general, contain an arbitrary constant, for it takes three conditions to determine a plane. Such an equa- tion therefore represents a system of planes. Systems of planes are used to find the equation of a plane satisfying three conditions in the same manner that systems of lines are used to find the equation of a line satisfying two condi- tions (Eule, p. 114). 360 ANALYTIC GEOMETRY Theorem VI. The system of planes parallel to a given plane Ax+Bij -^Cz + D^^) is represented by (VI) Ax + By -\-Cz-]-h = 0, where k is an arbitrary constant. Hint. Show that all of the planes (VI) are parallel to the given plane by Corollary III, p. 350, and that every plane parallel to the given plane is represented by (VI), by finding a value of k for which (VI) passes through a given point P^. Theorem VII. The system of planes passing through the line of intersection of two given planes A^x + B^y + C^z + Z)i = 0, A^x + B^ + Cg^ + A = is represented by (VII) A^oc + Biu + C^z + I>i + A; {A^x + B^y + C^z + D^) = O, where k is an arbitrary constant. Hint. Show that (VII) passes through any point on the intersection of the given planes, and find a value of k for which (VII) passes through any point not on the intersection. Theorem VIII. If the equations of the planes in Theorem VII are in normal form, then — k is the ratio of the distances from those planes to any point in (^VII). Hint. Let P-^ (a;,, y^, z^) be any point on the plane X cos a^ 4- 2/ cos Pi + s cos y^- Pi + k (x cos a^ + y cos ^2 + ^ cos yg - ^2)= 0. Then x^ cos a^ + y^ cos ^^ + z^ cos yi - i?i + k (x-^ cos a^ + 2/1 cos p^ + ^1 cos yj — Pz) = ^^ Solve for A; and interpret the result by Theorem IV, p. 357. Corollary. The equations of the planes bisecting the angles for 7ned by two giveii planes are found by reducing their equations to the normal form and adding and subtracting them. The plane (VII) will lie in the external or internal angles (p. 121) formed by the given planes according as k is positive or negative. The equation of a system of planes which satisfy a single con- dition must contain two arbitrary constants. One of the most important systems of this sort is given in THE PLANE 361 Theorem IX. The system of planes passing through a given point Pi (xi, )/i, Zi) is represented by (IX) A(oo-aOi) + B(y-yi)-\-C{z-Zi)=0. Proof. Equation (IX) is the equation of a plane which'passes through Pi, for the coordinates of Pi satisfy (IX). If any plane whose equation is Ax -\- By -{- Cz -\- D = passes through Pj, then Axi + %i + C^, + p = 0. Subtracting, we get (IX). Hence (IX) represents all planes passing through Pi. q.e.d. Equation (IX) contains two arbitrary constants, namely, the ratio of any two coefficients to the third. PROBLEMS 1. Determine the vakie of k such that the plane x-\-ky — 2z — ^ = shall (a) pass through the point (5, — 4, — 6). Ans. 2. (b) be parallel to the plane 6a; — 2?/ — 12 2 = 7, Ans. — \. (c) be perpendicular to the plane 2x — 42/4-2 = 3. Ans. 0. (d) be 3 units from the origin. Ans. ± 2. (e) make an angle of — with the plane 2x — 2y -\- z — 0. Ans. — f V35. o 2. Find the equation of the plane which passes through the point (3, 2, — 1) and is parallel to the plane 7 x — y + z = 14. Ans. 7x — y-^z — IS = 0. 3. Find the equation of the plane which passes through the intersection of the planes 2x + y — i = and 2/ + 2 z = and which (a) passes through the point (2, — 1, 1); (b) is perpendicular to the plane Sx -\- 2y — Sz = 6. Ans. (a) cc + y + 2 -2 = 0; (b) 2x -f 3?/ + 42 - 4 = 0. 4. Find the equations of the planes which bisect the angles formed by the planes 2x — y + 2z = and x-\-2y — 2z = 6. ^ Ans. 3x + 2/-6 = 0, x-3y + 4z + 6 = 0. 5. Find the equations of the planes passing through the intersection of the planes 2x + y — z = 4: and x — y -\-2z = which are perpendicular to the coordinate planes. Ans. 5ic + y = 8, 3x + z = 4, 3?/ — 52=4, 362 ANALYTIC GEOMETKY 6. Find the equations of the planes which bisect the angles formed by the planes 6x — 2y — 3z = and ix-\-Sy — ISz = 10, and verify by means of (V). 7. Find the equation of the plane passing through the intersection of the planes AiX + Biy + C12; + Di = and A^x + ^22/ + C22; + A = which passes through the origin. Arts. (JliDg - A2B1) X + {B1B2 - A^Bi) y + (CiA - C2B1) z = 0. 8. Find the equations of the planes which bisect the angles formed by the planes ^ix + Biy + Ciz + Di = and A^x + Biy + C^z + D2 = 0. Aix + Biy + Ciz + Di _ A^x + Biy + C^z + B^ -a.7lS. . = — ± — • V^i2 + 5^2 + Ci^ VvlaS + ^a^ + C22 9. Find the equations of the planes passing through the intersection of the planes AiX + B^y 4- C\Z + Di = and A^x + B^y + C2Z + I>2 = which are perpendicular to the coordinate planes. Ans. (^iBg - AiBx) y - (Ci^2 - C^Ai) z + A^B^ - A^Bi = 0, {A^Bi - A2B{)x - {B1C2 - B2C1) z - {B1B2 - B2B1) = 0, {C1A2 - C2Ai)x - {B1C2 - B2Ci)y + C1B2 - C2B1 = 0. 10. Find the equation of the plane which passes through Pi (xi, yi, Zi) and is perpendicular to the planes Aix + Biy 4- Ciz + X)i = and A2X + -B22/ + C^z + D2 = 0. Ans. {BiG2-B2Ci){x-Xi)+{CiA2-C2Ai){y-yi) + {AiB2-A2Bi){z-Zi)=0. CHAPTER XIX THE STRAIGHT LINE IN SPACE 159. General equations of the straight line. A straight line may be regarded as the • intersection of any two planes which pass through it. The equations of the planes regarded as simul- taneous are the equations of the line of intersection, and hence (Corollary, p. 349) Theorem I. The equations of the straight line are of the first degree in x, y, and z. Conversely, the locus of two equations of the first degree is a straight line unless the planes which are the loci of the separate equations are parallel. Hence, by Corollary III, p. 350, we have Theorem II. The locus of two equations of the first degree^ ^ ^ \A^x^B^y-\-C^z + D^ = 0, is a straight line unless the coefficients ofx, y, and z are proportional. To plot a straight line we need to know only the coordinates of two points on the line. The easiest points to obtain are usually those lying in the coordinate planes, which we get by setting one of the variables equal to zero and solving for the other two. If a line cuts but one of the coordinate planes, we get only one point in this way, and to plot the line we draw a line through that point parallel to the axis which is perpendicular to that plane. The direction of a line is known when its direction cosines are known. The method of obtaining these is illustrated in Ex. 1. Find the direction cosines of the line whose equations are 3x + 2y-2;-l=:0, 2x-y + 22;-3 = 0. Solution. Let the direction cosines of the line be cos a, cos j9, and cos y. The direction cosines of the normals to the planes in which the line lies are respectively (Corollary I, p. 350) 3 2 _ _^ ^^^2 _ 1 2 Vli' Vli VTi 3 3 3 363 364 ANALYTIC GEOMETRY Since the intersection of the two planes is perpendicular to the normals to both, we have (Theorem VI, p. 335) o 2 1 2 12 — izz COS a + —^= cos j3 =z cos 7 = 0, - cos (^^ — ~ cos i3 + - cos 7 = 0. Vl4 Vl4 Vl4 3 3 3 Solving for cos /3 and cos 7 in terms of cos a, we get cos j8 = — I cos a, cos 7 = — | cos a, 3cos/3 3 cos 7 and hence cos a = — = — • Dividing by 3, the least common multiple of the numerators, we get cos a _ cos /3 _ cos 7 3 ~ - 8 ~ -7 * Then by the Corollary, p. 331, 3^-8 -7 cos p = ; — J cos 7 ±Vl22 ±Vl22 ±Vl22 The line will be directed downward or upward according as the positive or negative sign of the radical is chosen. The method is general and may be formulated as the Rule to find the direction cosines of a line whose equations are given. First step. Find the direction cosines of the 7iormals to the planes in which the line lies (^Corollary I, p. 350). Second step. Find the conditions that the given line is perpen- dicular to the normals in the first step (^Theorem VI, p. S35) and solve for two of the direction cosines of the line in terms of the third. Third step. Express the results of the third step as a continued proportion and apply the Corollary, p. 331. Ex. 2. Find the direction cosines of the line whose equations are 4x + 3z-10 = 0, 4a;-2y + 3z-l = 0. Solution. First step. The direction. cosines of the normals to the given planes are 4 3,4 2 3 0, - and — =, --=^ —=■ 5 5 V29 V29 V29 Second step. If the direction cosines of the line are cos a, cos jS, and cos 7, then - cos a + - cos 7 = 0, — — : cos a —=. cos p -\ — ~= cos 7 = 0, 5 5 V29 ■V29 V29 and hence cos 7 = — | cos a, cos ^ = 0. THE STRAIGHT LINE IN SPACE 365 COS Ct COS 'V Third step. From these equations — — - = , cos /3 = 0, and hence 3 ■ — 4 cos a, cos 13, and cos 7 are proportional to 3, 0, and —4. Then (Corollary, P- ^'^^^' cosa: = ±|, cosi8 = 0, cos7 = :f|. The line will be directed downward or upward according as the upper or lower signs are used. Theorem III. If a, ^, and y are the direction cosines of the line {II), then cos a cos p cos y This is proved by the ahove Rule without carrying out the last part of the third step. PROBLEMS 1. Find the points in which the following lines pierce the coordinate planes and construct the lines. (a) 2a: + 2/-z = 2, x-y + 2z = 4. (c) x + 2y = 8, 2a;-4y = 7. (b) 4x + 3?/-62 = 12, 4x-3y = 2. (d) y + z = 4, x - ?/ + 2z = 10. 2. Find the direction cosines of the following lines. (a) 2x-2/ + 2z = 0, x+ 22/-22; = 4. _ Ans. ± 6^3 V65, T g'^V^, T rV^65. (b) x + 2/ + z = 5, x-2/ + z = 3. Ans. ± i V2, 0, T i ^2. (c) 3x + 2y-z = 4, x-2y-22; = 5. Ans. ± /^ V5, :f_l V5, ± ^s^ V5. (d) X + 2/ - 32; = 6, 2 X - 2/ + Sz = 3. Ans. 0, ± tV VlO, ± yV VlO. (e) X + ?/ = 6, 2 X - 3 z = 5. Ans. ± ^V V22, T ^h^/^, ± jV^^- (f) y + 3z = 4, 32/-5z = l. Ans. ±1,0,0." (g) 2x-3?/ + z = 0, 2x-3i/-2z = 6. _ Ans. ± x^a Vl3, ± fV Vi3, 0. (h) 5x- 14z- 7 =0, 2x + 7z = 19. ^ns. 0, ± 1, 0. 3. Show that the following pairs of lines are parallel and construct the lines. (a) 2?/ + 2: = 0, 3?/ - 4z = 7 and 5?/ - 2z = 8, 4?/ + 11 z = 44. (b) x + 2?/-z = 7, 2/ + 2;- 2x = 6 and 3x + 6?/ - 3z = 8, 2x- ?/ - z = 0, (c) 3x + z = 4, 2/ + 2z=:9 and 6x-2/ = 7, 3?/+.6z = l. 4. Show that the following pairs of lines meet in a point and are perpendicular. (a) x4-22/ = l, 2y — z = l and x — y = l,x — 2z = 3. (b) 4x-fy-3z + 24 = 0, z = 5 and x + y + 3 = 0, x + 2 = 0. (c) 3x + ?/-z = l, 2x-z = 2 and 2x-2/+2z = 4, x-?/ + 2z = 3. 866 ANALYTIC GEOMETRY 6, Find the angles between the following lines, assuming that they are directed upward or in front of the ZJT-plane. (a) X (b) X y — z = 0, y-^z = and x — y = l, x — Sy + z = 0. Ans. 2y + 2z = l,x-2z = l and 4x + 3y z-\-l=0,2x+Sy = 0. Ans. cos-i^f. (c) x-2y -\-z = 2,2y-z = l Siud x - 2y -^ z'= 2, x - 2y + 2 z = 4. Ans. cos-ii. 6. Find the equations of the planes through the line x-\-y — z = 0, 2x — y-\-Sz = 6 which are perpendicular to the coordinate planes. Ans. 3x + 2z = 5, 3y^5z-{-6 = 0, 5x + 2y = 5. 7. Show analytically that the intersections of the planes x — 2y — z = Z and 2x — 4y — 2z = 6 with the plane x + y — Sz = are parallel lines. 8. Verify analytically that the intersections of any two parallel planes with a third plane are parallel lines. 160. The projecting planes of a line. The three planes passing throngh a given line and perpendicular to the coordinate planes are called the projecting planes of the line. If the line is perpendicular to one of the coordinate planes, any plane con- taining the line is perpendicular to that plane. In this case we speak of but two projecting planes, namely, those drawn through the line perpendicular to the other coordinate planes. If the line is parallel to one of the coordinate planes, two of the projecting ~j^ planes coincide. By Theorem VII, p. 360, the equation of any plane through the line (1) A^x + B^y + C^z 4- Z)i = 0, A^x -f B^y -^ C^z + D^ = has the form (2) (Ai + kA^) ^ + (5i + kB^) 2/ 4- (Ci + A^Cs) ^ + (A + kD^) = 0. If (2) is to be perpendicular to the ZF-plane, z = 0, then (Corollary IV, p. 350) C, + kC^ = 0, whence k = --^- Substi- tuting in (2) and reducing, we get (3) (C1J2 - C^A,)x - (B,C, - B,C{)7j + CiA - C2A = 0. THE STRAIGHT LINE IN SPACE 367 Similarly, if (2) is perpendicular to the YZ- or ZX-plane, it becomes (4) (yliZ^a - A^B^) y - (C^A^ - C^A{) z + AJ)^ - A^D, = 0, (5) (A,B^ - A^B,) X - (B,C^ - B^C^) z - (B^D^ - B^D,) = 0. Equations (3), (4), and (5) are the equations of the projecting planes of the line (1), and any two of them may be used as the equations of the line. If ^1^2 - A^Bi-^ 0, that is, if the -^ line is not parallel to the ZF-plane (Theorem III), equations (5) and (4) may be written in the forms X = mz -\- a, y = nz -\- b. n, / If A^B^ — J 2^1 = and B^C^ — B^C^^^ 0, that is, if the line is parallel to the ZT-plane but is not parallel to the F-axis, equations (5) and (3) may be written in the forms z = a, ?/ = mx + b. If A^B^ — A^Bi = and BiC^ — B2Ci = 0, that is, if the line is parallel to the F-axis, equations (4) and (3) may be written in the forms z = a, X = b. Hence we have Theorem IV. The equations of a line which pierces the XY-plane, or which is parallel to the XY-plane but not to the Y-axis, or which is parallel to the Y-axis, may be put in the following forms respectively : (oc = mz-\-a, (z = a, ( z = a, ^ ^ \y=nz-\-b, \y = rnoc-\-b, \qc = b. 368 ANALYTIC GEOMETRY To find the equations of the projecting planes of a given line we may proceed as above by considering the system of planes which pass through the given line (Theorem VII, p. 3G0) and determining the parameter k so that the plane shall be perpendicular to each of the coordinate planes in turn. These equations may also be found by eliminating 2;, x, and ij in turn from the equations of the line. To reduce the equations of a given line to one of the forms (IV) we solve them for X and y in terms of z. If there is no solution for x and y (Theorem IV, p. 90), we solve for y and z. Finally, if there is no solution for y and z, we solve them for z and x. PROBLEMS 1. Find the equations of the projecting planes of the following lines. (a) 2x + y-2; = 0, x-2/ + 22; = 3. Ans. 5x '+?/ = 3, 3x + 2; = 3, 3^-52 + 6-0. (b) X + ?/ + z = 6, X - ?/ - 2 z = 2. Ans. 3x + y = 14, 2x-z = 8, 2y+3z = 4. (c) 2x + 2/-z = l, x-?/ + z = 2. Ans. x = l, 2/-z + l = 0. (d)x4-2/-4z = l, 2x + 22/ + z = 0. Ans. 9x + 92/ = l, 9z + 2 = 0. (e) 2 2/ + 3 z = 6, 2 1/ - 3 z = 18. Ans. y = Q, z=-2. (f) 2x-2/ + z = 0, 4x + 3y + 2z = 6. Ans. 5?/ = 6, lOx + 5z = C. (g) X + z = 1, X — z = 3. Ans. x = 2, z = — 1. 2. Reduce the equations of the following lines to one of the forms (IV) and construct the lines. (a) x + y — 2z = 0, X — ?/ + z = 4. Ans. x = i z + 2, y = | z — 2. (b)x + 2y-z = 2, 2x + 42/ + 2z = 5. Ans. z = i, y = - ^ x + |. (c) X — 2?/ + z = 4, x + 2?/ — z = 6. Ans. x = 5, y = iz-\-^. (d) x + 3z = 6, 2x + 5z = 8. Ans. z =4,x=-6. (e)x + 22/ — 2z = 2, 2x + ?/ — 4z=:l. Ans. x = 2z,y = l. (f) x-2/ + z = 3, 3x-3?/ + 2z=:6. Ans. z = S, y = x. 3. Interpret geometrically the meaning of the constants in each of equa- tions (IV) by determining numbers proportional to the direction cosines of each line and the point in which the first line cuts the JTY-plane, the second the FZ-plane, and the third the ZX-plane. 4. Interpret the geometric significance of the constants in equations (IV) by considering the traces of the planes which are the loci of those equations taken separately. 5. Show that a straight line in space is determined by four conditions, and formulate a rule by which to find its equations. 6. Find the equations of the line passing through the points (—2, 2, 1) and (- 8, 5, - 2). Ans. x — 2 z - 4, y = — z -\- S. THE STRAIGHT LINE IN SPACE 369 7. Find the equations of the projection of the line a; = z + 2, 2/ = 2z — 4 upon the plane x + y — z = 0. Ans. x = ^ z + Y, 2/ = 1 2; — Y-. 8. Find the equations of the projection of the line z = 2, y = x — 2 upon the plane X — 2y — 3z = 4. Ans. x=:— 6z + 4, y = — 4^;. 9. Show that the equations of a line may be written in one of the forms fy = mx + a, fx = a, Jx — a, \z= nx +b, \z = my -]-b, \y = b, according as it pierces the FZ-plane, is parallel to the FZ-plane, or is parallel to the Z-axis. 10. Show that the condition that the line x = mz -\- a, y = nz -\- b should , ,. , , , ^, . a — a' b —b' mtersect the Ime x — m'z + a, y = nz + 0' is = -• m — m n — n 161. Various forms of the equations of a straight line. Theorem V. Parametric form. The coordinates of any point P(x, 1/, z) on the line through a given point Pxioc\, yi, z-^ whose direction angles are a, /3, and y are given by (V) oc = oci-\- p cos a, y = y^-\- pcos p, z = z^ + p cos y, where p denotes the variable directed length PiP. Proof. The projections of PiP on the axes are respectively (Corollary II, p. 329) x-x^, y-yi, « - «i, or (Theorem I, p. 328) p cos a, p COS /3, p cos y. Hence X — Xi = p cos ct, y — yi = p cos p, z — Zi = p cos y. Solving for x, y, and z, we obtain (Y). q.e.d. Theorem VI. Symmetric form. The equations of the line passing through the point P^ (xi, y^, z^) whose direction angles are a, ^, and y have the form ^ ^ cos a cos p cos y Hint. Solve each of equations (V) for p and equate the values obtained. 370 ANALYTIC GEOMETRY ^ ,, ^. cosa cos/8 cosy Corollary. If = — y~- — -y then the sijmmetnc equations of the line may he written in the form X — Xj_ y — Vi z — Zi a h c Theorem VII. Two-point form. The equations of the straight line passing through Pi (iCj, y^, z^ and Pg (^2? 2/2? --2) <^^^ ^ ^ i»2 — «i 2/2 — 2/1 «2 — «i" Froof. The line (VI) passes through. Pj. If it also passes through P2, then cos a cos ^ COS y Dividing (VI) by this equation, we obtain (VII). q.e.d. Equations (VI) and (VII) each involve three equations, namely, those obtained by neglecting in turn each of the three ratios. These equations are, in different form, the equations of the projecting planes, since one variable is lacking in each (Corollary V, p. 351). Any two of the three equations are independent and may be used as the equations of the line, but all three are usually retained for the sake of their symmetry. PROBLEMS 1. Find the equations of the lines which pass through the following pairs of points, reduce them to one of the forms (IV), p. 367, and construct the lines. (a) (3, 2, - 1), (2, - 3, 4). Ans. x = -\z^ V, 2/ = - 2 + 1- (b) (1, 6, 3), (3, 2, 3).* An». 2=^3, y = -2x + 8. (c) (1, - 4, 2), (3, 0, 3). Ans. x = 2z - S, y = 4z - 12. (d) (2, - 2, - 1), (3, 1, - 1). Ans. z = -l,y = Sx-8. (e) (2, 3,' 5), (2, - 7, 5). Ans. z = 6,x = 2. 2. Show that the two-point form of the equations of a line become ^~^^ = y ~y^ , z = Zi, if zi = Z2. What do they become if 2/1 = 2/2? X2 -xi 2/2 - 2/1 if xi = X2 ? * From (VII), '^^ = ^-^ = ^-^- • The value of the last ratio is infinite unless 2-3=0. 3-1 2-6 3-3 If 3-3 = 0, then the last ratio may have any value and may be equal to the first two. Hence the equations of the line become ^-^ = , z = 3. Geometrically, it is evident 2 -4 that the two points lie in the plane z = 3, and hence the line joining them also lies in that plane. THE STRAIGHT LINE IN SPACE 371 3. What do the two-point equations of a line become if Xi = X2 and 2/1 = 2/2? if yi = 2/2 and zi = Z2? if Zi = Z2 and xi — x^? 4. Do the following sets of points lie on straight lines ? (a) (3, 2, - 4), (5, 4, - 6), and (9, 8, - 10). Ans. Yes. (b) (3, 0, 1), (0, - 3, 2), and (6, 3, 0). Ans. Yes. (c) (2, 5, 7), (- 3, 8, 1), and (0, 0, 3). Ans. No. 5. Show that the conditions that the three points Pi (iCi.yi, 21), P2(iC2, 2/2, 2^2), and Ps (X3, 2/3, 23) should lie on a straight line are ^LH^ = Vs - Vi ^ Zs - Zi «2 -xi 2/2 - 2/1 22 - 2l 6. Find the equations of the line passing through the point (2, -1, - 3) whose direction cosines are proportional to 3, 2, and 7, and reduce them to the form (IV), p. 367. Ans. x = f 2 + V, y = ^z- ^. 7. Find the equations of the line passing through the point (0, — 3, 2) which is parallel to the line joining the points (3, 4, 7) and (2, 7, 5). X 2/ + 3 z -2 Ans. 1-3 2 8. Show that the lines ^^ = ^^^ .. " and ^±i = ^^ = ^±^ are parallel. ^ -2 4 -3 2 -4 9. Find the equations of the line through the point (—2, 4, 0) which is jjj ■?/ -I- 2 z 4 parallel to the line - = = , and reduce them to the form (IV), p. 367. ^ ~^ Ans. x = -iz-2,y = -3z + 4. ■tn ax, *u^^i,T x + 2 y — S z — 1 ,x — 3 y z + 3 10. Show that the hues — — = - — — = and = -^ = — ■ — are perpendicular. ^ ~^ ^ 2 6 3 11. Find the angle between the lines = - — ~ and 2 1-1 cc + 22/-72;.-,^, J. ,j ^ ^ 27r — :; — = = - if both are directed upward. Ans. 12 1 ^ • 3 12. Find the parametric equations of the line passing through the point (2, — 3, 4) whose direction cosines are proportional to 1, — 2, and 2. Ans. x = 2 + ip, 2/ = - 3 - f /), 2 = 4 + f p. 13. Construct the lines whose parametric equations are (a) X = 2 + fp, 2/ = 4 - ip, z = 6 + f p. (b) X = - 3 - f /), ?/ = 6 - f p, z = 4 + f p. 14. Find the distance, measured along the line x = 2 — ^gp, 2/ = 4 + j|/), 2 = — 3 + j-**3 p, from the point (2, 4, — 3) to the intersection of the line with the plane 4x — 2/ — 2z = 6. Ans. If. 16. Show that the symmetric equations of the straight line become ^ = - — — » 2 = 2i if cos 7 = 0. What do they become if cos a = ? cos or cos/3 if cos/3 = 0? 372 ANALYTIC GEOMETRY 16. Show that the symmetric equations of the straight line become z = zi, x = Xi if cos 7 = cos a = 0. What do they become if cos a = cos j3 = ? if cos /3 = cos 7=0? 17. Reduce the equations of the following lines to the symmetric form. (a) x-2y + z = 8, 2x-3?/ = 13. Ans. ^^I^ = ^-±-? = - . O JJ 1. (b) 4x-5?/ + 3z = 3, 4x-5y + z + 9 = 0. Ans. - = ^-^ , z = Q. 5 4 (c) 2x + z + 5 = 0, x + 32;-5=0. Ans. 2 = 3, x = - 4. (d)x + 22/ + 62 = 5, 3x-2y-10z = 7. Ans. ?-^l- = ^^^ = 1 2 — 7 2 /;p 3 2 (e)3x-2/-22; = 0, 6x-3?/-42; + 9 = 0. ^ns. = -,2/ = 9. 2 o (f) 3 X - 4 y = 7, X + 3 2/ = 11. ^ns. x = 5, y = 2. (g) 2x + y + 22 = 7, x + 3y+62; = ll. Ans. ^— — = — - , x = 2. A — i (h)2x-3?/ + z = 4, 4x-6y-z = 5. Ans. -=y^^, z =1. o 2 (i) 3 2 + ?/ = 1, 4 2; - 3 y = 10. Ans. y = -2,z = l. ^ X — a y — h z (]) X = mz -}- a, y = nz -{- b. Ans. = = -■ m n 1 Hint. Find the coordinates of a point on the line hy assuming a value of one variahl* and solving the equations of the line for the other two variables. In the answers thi point is the point in which the line pierces the X T-plane, or the point in which it pierce the FZ-plane if it is parallel to the XF-plane, or the point in which it pierces the ZX- plane if it is parallel to the T-axis. Find the direction cosines of the line by the Rule, p. 364 (or numbers proportional to them by Theorem III, p. 365), and substitute in the symmetric equations of the line (oi in the form given in the Corollary to Theorem VI). If one or two of the direction cosines are zero, the symmetric equations take the fori given in problems 15 and 16. 18. Find the equations of the line passing through the point ^2, 0, —2) which is perpendicular to each of the lines = - = and - = — - = — — 2 1 2 o — 1 2 Ans. l^ = ?^ = i±?. 4 2 -5 19. Find the equations of the line passing through the point (3, —1,2) which Ls perpendicular to each of the lines x = 2z — 1, y = z + B, and - = - = -. ^ o 4 . x-3 y+1 z-2 Ans. — -— = — = — - — THE STRAIGHT LINE IN SPACE 373 20. Find the equations of the line through Pi(xi, 2/1, Zi) parallel to , . x-Xi y -Vi Z-Z2 (a) = — 7 — - = a b c (b) X = mz + a, y = nz + h. (c) z = a, y = mx + b. (d) Aix + Biy + Ciz + Di = 0, A^x + B^y + C^z A X-Xi Ans. A X-Xi Ans. - = a y. -yi b Z-Zi c . X — X\ Ans. i = m y. -yi n Z-Zx 1 X — Xi Ans. = 1 L ~yi, m Z = Zi. B^y + C2Z + A J = 0. y-yi z -zx BIC2 — B2CI CIA2 — A2CI AIB2 — A2BI 21. Find the equations of the line passing through Pi(Xi, 2/1, Zi) which is perpendicular to each of the lines X — X2 _ y — z/2 _ z - Z2 , x-xs _ y - ys _ z - zz a% &2 C2 Ct3 63 C3 -xx y -y\ z-zi Ans. &2«3 — bsCz Citts — Cza2 ^263 — «3&2 162. Relative positions of a Hne and plane. If the equations of a line have the general form (II), p. 363, then the line will lie in a given plane if a value of k in (VII), p. 360, may be found such that the locus of that equation is the given plane. If the equations of the line have the form (IV), we substitute the values of two of the variables given by (IV) in the equation of the plane and see whether the result is true for all values of the third variable. If such is the case, the line lies in the plane. An analogous procedure may be followed if the equations of the line have the form (V), (VI), or (VII). Theorem VIII. A line ivhose direction angles are a, ^, and y and the plane Ax -{- By -\- Cz -\- D = are (a) parallel when and only when Acqs a + B cos p -\- C cosy = 0\ (h) perpendicular when and only when A B C cos a ~ COS jff ~ cos y Proof. The direction cosines of the normal to the plane are (Corollary I, p. 350) A B C 374 ANALYTIC GEOMETRY The line and plane are parallel when and only when the line is perpendicular to the normal to the plane, "* that is (Theorem VI, p. 335), when and only when A Gos a -{- Bcos 13 + C cos y _ Multiplying by the radical, we get the condition for parallelism. The line and plane are perpendicular when and only when the line is parallel to the normal to the plane, that is (Theorem VI, p. 335), when and only when cos a = — — , COS /? = ±V^2 + i>'' + C' ±-^A-' + B-' + C^ C cos V = ± Va^ -^B^ + C^ Dividing these equations by A, B, and C respectively and inverting, we at once obtain the conditions for perpendicularity. Q.E.D. 163. Geometric interpretation of the solution of three equa- tions of the first degree. The coordinates of a point which lies on each of three planes will satisfy the equations of the three planes, and hence to each point common to three planes there will correspond a solution of their equations. Hence we have the following correspondence between the relative positions of three planes and the number of solutions of their equations. Position of planes Number of solutions of equations forming a triedral angle. One solution. Forming a prismatic surface, t No solution. Passing through the same line.$ A singly infinite number. § Three parallel planes. $ No solution. Three coincident planes. A doubly infinite number. || * If the line is perpendicular to the normal to the plane, it may, in a special case, lie in the plane. t Two of the planes may be parallel in a special case. X Two of the planes may coincide in a special case. § The solution contains one arbitrary constant. II The solution contains iwo arbitrary constants. THE STRAIGHT LINE IN SPACE 375 If the three planes form a triedral angle, the point of intersection is found without difficulty by solving their equations. If the three planes form a prismatic surface, their lines of intersection are par- allel. Whether this is the case or not may be determined by Theorem III, p. 365, and the Corollary, p. 331. If the three planes pass through the same line, the intersection of two planes lies in the third. Whether this is the case or not may be determined by the method on p. 373. To solve their equations set one variable equal to k and solve tioo of the equations for the remaining variables. The results will be solutions for all values of k. Whether the three planes are parallel or not may be determined by Corollary III, p. 350. If the three planes coincide, all of their coefficients are proportional. To solve their equations set two of the variables equal to k\ and k-z, and solve one of the equations for the remaining variable. The results will be solutions for all values of k\ and ki. PROBLEMS jc _j. 3 y 4 z 1 . Show that the line = = - is parallel to the plane 4 x + 2 y + 22 = 9. ^ -7 3 2. Show that the line - = - = - is perpendicular to the plane 3x + 2?/ + 7z = 8. 3 2 7 3. Show that the line x = 2; — 4, 2/ = 2z— 3 lies in the plane 2x — Zy + 42-1 = 0. 4. Show that the line x=-2 + f/3, y = -|p, z = 6 + ip lies in the plane x-2y-6z + 38 = 0. 5. Find the coordinates of the points of intersection of the following planes and determine the relative positions of the planes. (a) 2x + y-22; = ll,x-?/ + 2; = 0, jc + 2?/-2; = 7. Ans. (3, 1, — 2); planes form a triedral angle. (b)2x + 42/ + 22 = 3, 3x + 3?/ + 2; = 0, 3x-6y-52;==8. Ans. None; planes form a prismatic surface, (c) X — ?/ — 32 = 1, x + y + 2;=:2, 3x — y — 5-2; = 4. Ans. (I + A:, I — 2 fc, A:); planes pass through a line. (d)3x-2/ + 52; = 0, 21x-7y + 35z = 8, 22/-10z-6x = 4. Ans. None ; planes are parallel. (e)2x-3y + 42 = 3, 6y-4x-8z + 6 = 0, 6x-9y + 12 z = 9. Ans. [fci, ^2, 1(3 — 2 A;i + 3 ^2)] ; planes coincide. 6. Show that the line ^-^- = ^ = — - — lies in the plane 2 x + 2 2/ -2 + 3 = 0. ^ -^ ^ 376 ANALYTIC GEOMETRY 7. Find the equations of the line passing through the point (3, 2, — 6) which is perpendicular to the plane ix — y + Sz = d. x-3 y -2 2+6 Ans. = = 4-13 8. Find the equations of the line passing through the point (4, — 6, 2) which is perpendicular to the plane x-{- 2y — Sz = 8. . ic-4 y + 6 2-2 1 2 -3 9. Find the equations of the line passing through the point (—2, 3, 2) which is parallel to each of the planes Bx — y -i- z = and x — z = 0. x + 2 y - S z - 2 Ans. 1 4 1 10. Find the equation of the plane passing through the point (1,3, — 2) which is perpendicular to the line = = Ans. 2x + 5y — z = 19. 11. Find the equation of the plane passing through the point (2, — 2, 0) which is perpendicular to the line z = 3, y = 2x — i. Ans. x-\-2y +2 = 0. 12. Find the equation of the plane passing through the line x + 2 z = 4, 3j 3 ■j/_l4 2; 7 y — z = 8 which is parallel to the line = = . 2 3 4 Ans. x + lOy -8z-84: = 0. 13. Find the equation of the plane passing through the point (3, 6, — 12) which is parallel to each of the lines — — = ^ ~ = ^-^ and ^ ~ _z+2 3-13 2 -—^'2/ -3. Ans. 2x-\-3y -z = S6. 14. Find the equations of the line passing through the point (3, 1, — 2) which is perpendicular to the plane 2x — y — 5z = 6. Ans. x = -|z + -U, 2/ = i2; + |. 15. Show that the lines ^^ ^ ^±i ^ ^ ^^d ^^ = ^±i ^ ^ 3 4-2 -1 32 intersect, and find the equation of the plane determined by them. Ans. 14 X - 4 2/ + 13 z = 32. />. 2 y I 3 16. Find the equation of the plane determined by the line =: z-1 2-2 1 and the point (0, 3, - 4). Ans. x-^2y + 2z-^2 = 0. 17. Find the equation of the plane determined by the parallel lines x-^l _ y -2 _z X -S _ y + 4 _z -1 3 2 13 2 1 Ans. 8x + y-26z-\-6 = 0. THE STRAIGHT LINE IN SPACE 377 18. Find the equations of the line passing through Pi(xi, ?/i, Zi) which is perpendicular to the plane Ax -{- By i- Cz + D = 0. Ans. ^^1^ = 2^1-1 = ^^111. ABC 19. Find the equation of the plane passing through the point Pi (cci, yi, zi) which is perpendicular to the line = — ; — - = . a c Ans. a{x-Xi)-\-b{y- yi) -\- c{z-Zi) = 0. 20. Find the angle 6 between the line = — = and the plane ^x + ^y + C2; + D = 0. "* . ^ .,. ^^ , . ^ Aa + Bb + Cc Ans. sm 9 = — Hint. The angle between a line and a plane is the acute angle between the line and its projection on the plane. This angle equals ~ increased or decreased by the angle between the line and the normal to the plane. 21. Find the equation of the plane passing through Ps{xs, ys? Zs) which 11 w u r *i T x-xi y -y\ z-zi .x-x^ v -yi IS parallel to each of the lines = — - — = and = f__^ ai &i ci ai 02 Z — Z2 Ans. (61C2-&2C1) {x-Xz) + {cia2-a2C\) (2/-2/3) + (ai?>2-a2&i) {z-Zz)=Q. 22. Find the condition that the plane A^x + B\y + C\Z + Di = should be parallel to the line A^x + B^y + C^z + D2 = 0, A^x + B^y + C^z +1)3 = 0. Ans. Ai{B2Cs -^3C2) + 5i(C2^3-C3^2)+Ci(^2^3-^3-B2)= 0. 23. Find the equation of the plane determined by the point Pi {xi, yi, Zi) and the line Aix -{■ Biy + CiZ +Di = 0, A2X +B2y+ C^z +D2 = 0. Ans. [A^xi + B^yi + C^Zi +D2) {A-LX+B^y+ C^z +i)i) = (^ixi + jBi2/i + Cizi +Di) {A2X +B2y + C^z +D2). 24. Find the equation of the plane determined by the intersecting lines x-xi _ y-yi _ z -zi ^^^ x - xi _ y - yi _ z - zi _ ai 6i ci a2 62 C2 Ans. (61C2 - &2C1) (x - Xi) + (cia2 - C2ai) {y - yi) + (ai62 — aa&i) {z - Zi) = 0. 25. Find the equation of the plane determined by the parallel lines X — xi _y — yi _z — zi , x — X2 _ y — 1/2 _ z — Z2 ^ a b c a b c Ans. [(2/1 - 2/2) c - (zi - Z2) b]x + [(zi - Z2) a - (xi - X2) c] y + [(xi - X2) & - (^1 - 2/2) a] z + (yiZ2 - y2Zi) a + (2:1X2 - Z2X1) b -1- (Xi?/2 - iC22/l) c = 0. 378 ' ANALYTIC GEOMETRY 26. Find the conditions that the line x = mz -{- a, y = nz +h should lie in the plane Ax + By -\- Cz + D = 0. Ans. Aa + Bb-\-D = 0, Am + Bn + C = 0. 27. Find the equation of the plane passing through the line 6i C\ a^ 62 C2 Ans. (61C2 — 62C1) (x — cci) + (cia2 — C2ai) (2/ — 2/1) + (ai&2 — a2&i) (2 — Zi) = 0. i;^ CHAPTER XX SPECIAL SURFACES 164. In this chapter we shall consider spheres, cylinders, and cones ^ (surfaces considered in Elementary Geometry) and sur- faces which may be generated by revolving a curve about one of the coordinate axes or by moving a straight line. 165. The sphere. Theorem I. The equation of the sphere whose center is the point (tr, /?, y) and whose radius is r is (a^-ay + (v-l3y + (z-Yy = A or (I)_ 3c''-^y^ + z^-2ax-- Q^y - 2 ys; + a^ +)8'^ + y^ - r-^ = O. Froof. Let P (x, y, z) be any point on the sphere, and denote the center of the sphere by C. Then, by definition, PC = r. Substituting the value of PC given by (IV), p. 331, and squar- ing, we obtain (I). q.e.d. Theorem II. The locus of an equation of the form (II) ic'-^tf + z' + GQC-\- Hy -^Iz + KzzzO is determined as follows : (a) When G'^ -\- H"^ -\- P — ^ K > 0, the locus is a sphere whose center is ( — —^ — ^j "~ o ) ^^^ whose radius is {h) When G'^ + H^ + I^ — 4:K= Q, the locus is the point-spheret (g_ _H _/ V 2' 2' 2, (c) When G^ + H^ -^ I^ - 4:K < 0, there is no locus. * In Analytic Geometry the terms sphere, cylinder, and cone are usually used to denote the spherical surface, cylindrical surface, and conical surface of Elementary Geometry, and not the solids bounded wholly or in part by such surfaces. + That is, a point or sphere of radius zero. 379 380 ANALYTIC GEOMETRY Proof. Comparing (II) with (I), we obtain whence G ^ H I Hence, if G^ + H^ -{■ I^ - 4: K > 0, the locus is a sphere. To determine the general appearance of the locus of (II) when G^ -{- H^ -\- I^ — 4:K^0, we consider the section formed by the plane z = k, whose equation is (Eule, p. 345) (1) x'' + 2f-{-Gx-{-Hi/-^k'' + Ik-^K= 0. The discriminant of (1) is (p. 131) © = G'^ + 7/2 _ 4 A;2 _ 4 Ik - 4 k'^^ = - 4 7^2 - 4 //i: + G^2 ^ iy2 _ 4 K The discriminant of this quadratic in k is (p. 2) A = 16 /2 + 16 (9^ + 16 ^2 - 64 ii: = 16(G'2 + //2_^/2_4 7r). In discussing the locus of (1) three cases arise which depend upon the sign of © (Theorem I, p. 131). (a) If (92 + iy2 + 72 _ 4 7^ > 0, is positive for values of k lying between the roots of © (Theorem III, p. 11), and the section (1) formed by th,e plane ^ = A; is a circle. Equation (II) has a locus, as we have seen. (b) If (92 + 7^2 4- 72 - 4 7: = 0, © is negative for all real values of k (Theorem III, p. 11) except the roots, which are real and equal (Theorem II, p. 3), and for this single value of k the locus of (1) is a point-circle. As but one plane, z = k, intersects the locus of (II), and as this intersection is a point-circle, the locus is a point which may be regarded as a sphere of zero radius. (c) If (92 -f TT^ + 72 - 4 A^ < 0, © is negative for all real values of k (Theorem III, p. 11). Hence (1) has no locus whatever the value of k may be, and therefore (II) has no locus. q.e.d. SPECIAL SURFACES 381 Theorem III. The locus of the general equation of the second degree in three vanables (III) Ax^-\- Bi/ + Cz^ + Dgz H- Ezx + Fxij + Gx -\- Hg -\-Iz + 7i = {5 a sphere when and only when A =z B = C, D = E = F = Oj and G^ + !{'-{- 1^-4: AK . —^ is positive. This is proved by comparing (III) with (II). PROBLEMS 1 . Find the equation of the sphere whose center is the point (a) {a, 0, 0) and whose radius is a. Ans. x^ -{- y^ + z^ — 2 ax = 0. (b) (0, iS, 0) and whose radius is /3. Ans. x^ -\- y^ -\- z^ - 2 ^y =0. (c) (0, 0, 7) and whose radius is 7. Ans. x^ -{- y^ + z^ — 2yz =0. 2. Determine the nature of the loci of the following equations and find the center and radius if the locus is a sphere, or the coordinates of the point- sphere if the locus is a point-sphere. (a) x2 -f 2/^ -H 22 - 6 X + 4 z = 0. (c) x^ -{- y"^ + z^ + ix - z + 7 = 0. (b) x2-|-2/2 + 22 + 2a;-4y-5 = 0. (d) x'^ -\- y^ -{- z^ - 12x + 6y + 4: z = 0. 3. Where will the center of (II) lie if (a) G = 0? (c)7=0? (e) H = I=0? {h) H = 0? {d) G=R=0? {t) I=G = 0? 4. Show that a sphere is determined by four conditions and formulate a rule by which to find its equation. 5. Find the equation of the sphere which (a) has the center (3, 0, — 2) and passes through (1, 6, — 5). Ans. x2 + y^ + z^-Qx-\-4z-3G = 0. (b) passes through the points (0, 0, 0), (0, 2, 0), (4, 0, 0), and (0, 0, - 6). Ans. x^ -]-y^ + z^ -4:X-2y + 6z = 0. (c) has its center on the F-axis and passes through the points (0, 2, 2) and (4, 0, 0). Ans. x2 -f y2 _|. 22 _[. 4 y _ 16 = 0. (d) passes through the points (1, 1, 0), (0, 1, 1), and (1, 0, 1) and whose radius is 11. Ans. x2 4- y2 _}_ ^2 _ 14 x - 14 2/ - 14 z + 26 = 0. (e) has the line joining (4, - 6, 5) and (2, 0, 2) as a diameter. Ans. x2 + ?/2-^22_6x + 62/-7z-Hl8 = a 382 ANALYTIC GEOMETRY 6. Given two spheres Si : x"^ + ij^ + z"- -^ GiX + Hiy + Iiz + -^i = and >S2 : x2 + ?/2 -f 2;2 + G2X + HiV + hz + 7^2 = ; show that the locus of Sk : x2 + 2/2 + z2 + Gix + Ihy + hz + K^ + /b (x2 + 2/2 + 2;2 + G2X + H^y + J22; + X2) = is a circle except when k = — 1. In this case the locus is a plane called the radical plane of Si and S^. 7. The center of the sphere Sk in problem 6 lies on the line of centers of Si and S-i and divides it into segments whose ratio is equal to k. 8. The equation of the radical plane of Si and ^2 (problem 6) is ((?i - G2)x + {Hi-H2)y-\- {Ii-h)z + {K1-K2) = 0. 9. The radical plane of two spheres is perpendicular to their line of centers. 10, The radical planes of three spheres taken by pairs intersect in a line perpendicular to their plane of centers which is called the radical axis of the spheres, 11, The radical planes of four spheres taken by pairs intersect in a point Called the radical center of the spheres. 12, When two spheres Si and S2 (problem 6) intersect, the system Sk con- sists of all spheres passing through their circle of intersection, 13, When the spheres Si and S2 (problem 6) are tangent, the system Sk consists of all spheres tangent to Si and -82 at their point of tangency, 14, The equation of the system Sk (problem 6) may be written in the form x^-\-y^ + z^ + k'x + K= 0, where ¥ is an arbitrary constant, if the X-axis is chosen as the line of centers and the YZ-plane as the radical plane of ^1 and ^2- 15, The spheres of the system in problem 14 have their centers on the X-axis and (a) pass through the circle y^ + z^ + E = 0, x = ii K <0. (b)- are tangent to each other at the origin if K = 0. (c) are orthogonal to the sphere x^ -}- y^ + z^ = K it K> 0, 16, The product of a secant of a sphere drawn from a fixed point and its external segment is constant. 17, Find the square of the length of a tangent from a point Pi (xi, 2/1, 2:1) to the sphere x^ -{- y^ + z"^ -\- Gx -{- Hy -\- Iz + K = 0. Ans. Xi2 + 2/i2 + zi^ + Gxi + Hyi + Izi + K. 18, Show that the equations of an inversion (p. 297) in space are x' y' ^ z' y- X'2 + /2 + 2'2 x'2 + ?/'2 + 2;'2 x'2 + ?/'2 + 2'2 SPECIAL SURFACES 383 19. Show that the inverse of a plane is a sphere unless the plane passes through the origin, and that in this case the plane is invariant. 20. Show that the inverse of a sphere is a sphere unless it passes through the origin, when the inverse is a plane. 166. Cylinders. Ex. 1. Determine the nature of the locus ofy^ = ix. Solution. The intersection of the surface with a plane parallel to the FZ-plane, x = k, are the lines (Rule, p. 345) (1) X = fc, y = ±2Vk, which are parallel to the Z-axis (Theorem II, p. 342) . If A; > 0, the locus of equations (1) is a pair of lines ; if A; = 0, it is a single line (the Z-axis) ; and if Aj < 0, equations (1) have no locus. Similarly, the intersection with a plane parallel to the ZX-plane, y = k, is a, straight line whose equations are (Rule, p. 345) X = ^ A;2, y = k, and which is therefore parallel to the Z-axis. The intersection with a plane parallel to the JTZ-plane is the parabola z = k, 2/2 = 4 X. For different values of k these parabolas are equal and placed one above another. It is therefore evident that the surface is a cylinder whose elements are parallel to the Z-axis and intersect the parabola in the XF-plane 2/2 _ 4 a;, z = 0. It is evident from Ex. 1 that the locus of any equation which contains but two of the variables x, y, and z will intersect planes parallel to two of the coordinate planes in one or more straight lines parallel to one of the axes and planes parallel to the third coordinate plane in equal curves. Such a surface is evidently a cylinder. Hence Theorem IV. The locus of an equation in which one variable is lacking is a cylinder whose elements are parallel to the axis along which that variable is measured. 88-i ANALYTIC GEOMETKY 167. The projecting cylinders of a curve. The cylinders whose elements intersect a given curve and are parallel to one of the coordinate axes are called the projecting cylinders of the curve. Their equations may be found by elimiDating in turn each of the variables x, y, and z from the equations of the curve; for if we eliminate z, for example, the result is the equation of a cylinder (Theorem IV) which passes through the curve, since values of x, y, and z which satisfy each of two equations satisfy an equation obtained from them by eliminating one variable. The equations of two of the pi:ojecting cylinders may be con- veniently used as the equations of the curve.* The figure shows the curve whose equations are 2if- + z'^-\-^x--=^z, ?/2 + 322_8a; = i2z. ^ Eliminating x, y, and z in turn, we obtain the equations of the projecting cylinders 4 2, 4x= 4z, + 4X: The figure shows the first and third of these cylinders. If the curve lies in a plane parallel to one of the coordinate planes, then two of these cylinders coincide with the plane of the curve, or part of it. * In general, the equations of a curve may be replaced by any two independent equa- tions to which they are equivalent, that is, by two independent equations which are satisfied by all values of x, y, and z satisfying the equations of the curve, and only by such values. SPECIAL SURFACES 385 The projecting cylinders of a straight line are evidently planes. The equations of a line in terms of its projecting cylinders or planes have already been given (Theorem IV, p. 367). 168. Cones. Ex. 1. Determine the nature of the locus of the equation 16 x^ + y2 _ z^=0. Solution. Let Pi {xi, ?/i, Zi) be a point on a curve C in which the locus intersects any plane, for example z = k. Then (1) 16 xi2 + yi2 - zi^ = 0, zi = k. The origin lies on the surface (Theorem III, p. 345). We shall- show that the line OPi lies entirely on the surface. The direction cosines of OPi are (Corollaries, Xi pp. 332 and 331) — , — , and -, where Pi Pi Pi Pi2 ^ xi"^ + 2/i2 + zi"^ = 0Pi2. Hence the coordi- nates of any point on OPi are (Theorem V, p. 369) Xi x = — p, PI Pi z = -p. PI Substituting these values of cc, y, and z in the given equation, we obtain (2) 16 a^lV^ yi2^2 2;i2p2 + 0. pr pr pi'' This is true for all values of p since it may be obtained from the first of equations (1) by multiplying by — • Hence every point on OPi lies on the surface, that is, the entire line lies on the surface. Hence the surface is a cone whose vertex is the origin. The essential thing in the solution of Ex. 1 is that (2) may be obtained from the first of equations (1) by multiplying by a power of — • This may be done whenever the equation of the surface is homogeneous* in the variables x, y, and z. Hence Theorem V. The locus of an equation which is homogeneous in the variables ic, y, and z is a cone whose vertex is the origin. * An equation is homogeneous in x, y, and z when all the terms in the equation are of the same degree (footnote, p. 17). 386 ANALYTIC GEOMETRY PROBLEMS 1 . Determine the nature of the following loci ; discuss and construct them. (a) a;2 ^ y% = 36. (e) a;2 _ 2/2 + 36 22 = 0. (b) x2 + 2/2 = 22. (f ) 2/2 - 16 x2 + 4 z2 = 0. (c) ?/2 + 4 2;2 = 0. (g) x2 + 16?/2 - 4x = 0. (d) X2 - Z2 3:: 16. (h) x2 + yz 2. Find the equations of the cylinders whose directrices are the following curves and whose elements are parallel to one of the axes. (a) 2/2 + 22 _ 4 y = 0, a; = 0. (c) Wx'^ - a'^y'^ = a%^, z = 0. (b) z2 _|. 2x = 8, 2/ = 0. (d) 2/2 + 2pz = 0, x = 0. 3. Find the equations of the projecting cylinders of the following curves and construct the curve as the intersection of two of these cylinders. (a) x2 + y2 _^2;2 = 25, x2 + 4 2/2 - z2 = 0. (b) x2 + 42/2 - 22 = 16, 4x2 + 2/2 + z2 =: ^c) x2 + 2/2 = 4 z, x2 - 2/2 = 8 z. (d) x2 + 2 2/2 + 4 22 = 32, x2 + 4 2/2 = 4z. (e) 2/2 + zx = 0, 2/2 + 2 X + 2/ - z = 0. 16. v 4. Discuss the following loci. (a) x2 + 2/2 = 22tan2 7. (b) 2/2 + z2 = ic2tan2a:. (c) 22 + x2 = 2/2tan2^. (d) x2 + 2/2 = r2. (e) 2/2 + 22 rz r2. (f) z2 + x2 = r2. 169. Surfaces of revolution. The surface generated by revolv- ing a curve about a line lying in its plane is called a surface of revolution. Ex. 1.- Find the equation of the surface of revolution generated by revolv- ing the ellipse x2 + 4 2/2 — 12 x = 0, z = about the JT-axis. Solution. Let P (x, 2/, z) be any point on the surface. Pass a plane through P and OX which cuts the surface along one position of the ellipse, and in this plane draw OF' perpendicular to OX. Referred to OX and OY' as axes, the equation of the ellipse is evidently x2 + 4 2/'2 - 12 X = 0. But from the right triangle PAB we get y'^ = y^ + z''. Substituting in (l), we get (2) x2 + 42/2 + 4z2-12x=:0. SPECIAL SURFACES 387 This equation expresses the relation which any point on the surface must satisfy, and it is easily shown that any point whose coordinates satisfy equa- tion (2) lies on the surface. It is therefore the equation of the surface. The method of the solution enables us to state the Rule to find the equation of the surface generated by revolving a curve in one of the coordinate planes about one of the axes in that plane. Substitute in thejequation of the curve the square root of the sum of the squares of the two variables not measured along the axis of revolution for that one of these two variables which occurs in the equation of the curve. If the intersections of a surface with all planes parallel to one of the coordinate planes are circles, then the surface is evidently a surface of revolution whose axis is the coordinate axis perpen- dicular to the planes of the circular sections. This enables us to determine whether or not a given surface is gaseurface of revolu- tion whose axis is one of the coordinate axes. 170. Ruled surfaces. A surface generated by a moving straight line is called a ruled surface. If the equations of a straight line involve an arbitrary constant, then the equations represent a sys- tem of lines which form a ruled surface. If we eliminate the parameter from the equations of the line, the result will be the equation of the ruled surface. For if («!, yi, Z\) satisfy the given equations for some vakie of the parameter, they will satisfy the equation obtained by eliminating the parameter, that is, the coordinates of every point on every line of that system satisfy that equation. Cylinders and cones are the simplest ruled surfaces. Ex. 1. Find the equation of the surface generated by the line whose equations are -^ x-{-y = kz, x-y = ~z. Ic Solution. We may eliminate k from these equations of the line hy multi- plying them. This gives (1) x2 - y2 ^ z^. This is the equation of a cone (Theorem V, p. 385) whose vertex is the origin. As the sections made hy the planes x = k are circles, it is a cone of revolution whose axis is the X-axis. 388 ANALYTIC GEOMETRY We may verify that the given line lies on the surface (1) for all values of A; as follows : Solving the equations of the line for x and y in terms of z, we get 2\ k Substituting in (1), we obtain ^-\i.--> -K-D^- 4^ k) 4V k) an equation which is true for all values of k and z, as is seen by removing the parentheses. Hence every point on any line of the system lies on (1), since its coordinates satisfy (1). Ex. 2. Determine the nature of the surface z^ — 3 zx + 8 y = 0. Solution. The intersection of the surface with the plane z = k is the straight line (Rule, p. 345) k^-Skx-\-8y = 0, z = k. ^ Hence the surface is the ruled surface generated by this line as k varies. To construct the surface consider the intersections with the planes x = and jc = 8 whose equations are respectively x = 0, 8 y + z3 = and X = 8, Sy -24z + z^ = 0. Joining the points on these curves which have the same value of z gives the lines generating the surface. SPECIAL SURFACES 389 PROBLEMS 1 . Find the equations of the surfaces of revolution generated by revolving the following curves about the axes indicated, and construct the figures. (a) ?/2 = 4a;-16, JT-axis. Ans. y^ -{- z^ = ix - 16. (b) x2 + 4 y2 = iG, r-axis. Ans. x^ + i y^ -{- z^ = 16. (c) x2 — 4 2, Z-axis. Ans. x"^ -\-y^ = iz. (d) x2 - ?/2 _ 16^ r-axis. Ans. x^ - y^ + z^ = 16. (e) x2 - 2/2 = 16, X-axis. Ans. x^ - y^ - z^ = 16. (f) 2/2 -^ 22 _ 25, Z-axis. Ans. x^ + y^ + z^ - 25. (g) 2/2 = 2j9z, Z-axis. Ans. A paraboloid of revolution, x^ + 2/2 = 2pz. X2 ?/2 X^ ?/2 z2 (h) 1 — =1, JT-axis. Ans. An ellipsoid of revolution, h — -\ — = 1. a2 62 (j2 52 52 a;2 1/2 — (i) -^ - ^ = 1, r-axis. ^ ^ a;2 y^ z2 -4 ns. An hyperboloid of revolution of one sheet, — 1 = 1. X^ y^ ^' ^' «^ ^ ° iC2 ^2 22* ^ns. An hyperboloid of revolution of two sheets, =1. a2 52 52 2. Show that the following loci are either surfaces of revolution or ruled surfaces whose generators are parallel to one of the coordinate planes. Con- struct and discuss the loci. (a) 2/2 + z2 = 4 X. (e) 4 x2 + 4 2/2 - 2;2 = 16. (b) 0:2 - 4 2/2 + 2:2 = 0. (f ) r,fiy _ 22 = 0. (c) z^-zx + y = 0. (g) x2 -h z2 = 4. (d) x-hf + xz = y. (h) {x^ + z'^)y = ia'^{2a - y). 3. Verify analytically that a sphere is generated by revolving a circle about a diameter. 4. Show that the systems of spheres in problem 15, p. 382, may be gen- erated by revolving the systems of circles in Theorem VIII, p. 144, about the X-axis. 6. Find the equation of the surface of revolution generated by revolving the circle ic2 + 2/2 — 2 ax + a2 - r2 = about the T-axis. Discuss the sur- face when a>r^ ix = r^ and a r the surface is called an anchor ring or torus. 6. Find the equations of the ruled surfaces whose generators are the following systems of lines, and discuss the surfaces. (a) x + y = k,k{x-y) = a'^. Ans. x'^ - y"^ - a2. (b) 4x-2 2/ = A:z, A:(4x-f-2 2/) = z. Ans. 16x2-4 2/2 = ^2. (c) x-22/ = 4^2, A;(x-22/) = 4. Ans. x^-4y^ = lQz. (d) X -1- A;2/ -I- 4 2 = 4 A;, fcx - 2/ - 4 fcz = 4. Ans. x^ + y^ - 16 z^ = 16. 390 ANALYTIC GEOMETRY 7. Find the equation of the cone whose vertex is the origin and whose elements cut the circle x'-^ + y'^ = W, z = 2. Ans. x^ + ^/^ — 4 2^ = 0. 8. Find the equations of the cones of revolution whose axes are the coordinate axes and whose elements make an angle of with the axis of revolution. Ans. y^ + z'^ = x^ tan2 ; z^ -\- x^ = y^ tan2 ; x"^ + y^ = z^ tan2 0. 9. Find the equations of the cylinders of revolution whose axes are the coordinate axes and whose radii equal r. Ans. y^ -{- z"^ = r^ ; z^ + x'^ = r'^; x^ + y^ = r^. IP CHAPTER XXI TRANSFORMATION OF COORDINATES. DIFFERENT SYSTEMS OF COORDINATES 171. Translation of the axes. Theorem I. The equations for translating the axes to a new origin 0' (h, k, I) are (I) x = x'-\-7i, y = yf+k, z = z*+l. Proof. Let the coordinates of any point before and after the translation of the axes be (cc, y, z) and {x\ y\ z') respec- tively. Projecting OP and OO'P on each of the axes (Theorem II, p. 328), we get equations (I). q.e.d. 172. Rotation of the axes. Theorem II. If a^, fi^, y^ a^, ft, ys, a'^id a^, ft, yg are respec- tively the direction angles of three mutually ijerpendicular lines 0X\ OY', and OZ', then the equations for rotating the axes to the position 0-X'Y'Z' are (ijc = x^ cos tti + 2/' cos ttg + z^ cos as, (II) J 1/ = cc' cos ^1 + y^ cos ft + 2!' cos ft, [ s = a?' cos yi + y^ cos y^ + z^ cos yg.^* * By Theorem III, p. 330, and Theorem VI, p. 335, we see that the direction cosines of OX', O Y', and OZ' satisfy the six equations cos2 a, + cos2 j3i + cos2 Yi = 1, cos Oj cos Oj + cos /3i cos ^^ + cos y, cos 72 = 0. cos2 02 + C082 (8j + cos2 V2 = 1 > ^^s o^ COS flg + COS 182 COS ^3 + cos 72 «os Vs = 0, C082 flg + COS2 Pg 4- COS2 73 = 1, cos Og COS O^ + COS /Bg COS Pi + COS 73 COS Yj = 0. Hence only three of the nine constants in (II) are independent. 391 392 ANALYTIC GEOMETRY Proof. Let the coordinates of any point P before and after the rotation of the axes be respec- tively {x, y, z) and (x\ y\ z'). Pro- jecting OP and OA'B'P on each of the axes OX, Y, and OZ, we get, by Corollary I, p. 328, and Theorems I and II, p. 328, equations (II). q.e.d. Theorem III. The degree of an equa- tion is unchanged hy a transforma- tion of coordinates. Hint. Show that any transformation of coordinates may be effected by applying Theorems I and II successively, then that the degree cannot be raised by changing to new coordinates, and finally that it cannot be lowered. Z' ^ \ P ) / / y A . / ^^ B PROBLEMS V 1 . Transform the equation x'^-\-y^ — ^x-\-2y — Az-\-l=0 by trans- lating the origin to the point (2, — 1, — 1). Ans. x'^ + y"^ — ^z = 0. 2. Transform the equation bx^ ^Sy"^ + 5z^ — 4:yz + Szx-\-4:Xy — ix + 2y -f 4 2 = by rotating the axes to a position in which their direction cosines are respectively f , f^ i . i, _ 2^ i ; 2^ _ i, _ i. Ans. Sx^ + Sy^ = 2 z. 3. Formulate a rule by which to simplify a given equation (a) by trans- lating the axes, (b) by rotating the axes. How many terms may, in general, be removed from a given equation by a general transformation of coordinates ? 4. Derive the equations for rotating the axes through an angle 6 about (a) the Z-axis, (b) the X-axis, (c) the F-axis. rx = x' cosd — y' sin d, Ans. (a) < y = x' sin Q -\- y' cos ^, Vz — z'. 5. Simplify the following equations by translating the axes or by rotating them about one of the coordinate axes. (a) x2 + 2/^ - 2;2 _ 6 x - 8 ^ -h 10 z = 0. Ans. (b) 3x2- 8x?/ -32/2 _ 5^2 + 5 :=0. Ans. (c) ?/2 + 4 2;2 _ 16 x - 6 ?/ + 16 z -}- 9 = 0. Ans. (d) 2x2 - 5?/2 - 5^2 - 62/z = 0. Ans. (e) 9x2-25y2_i_i6z2_242x-80x-60z = 0. Ans. x2 - 2/2 = 4 z. 6. Show that Ax -\- By -\- Cz + D =z may be reduced to the form x = by a transformation of coordinates. Hint. Remove the constant term by translating the axes, then remove the z-term by rotating the axes about the T-axis, and finally remove the y-term by rotating about the ^-axis. y 22 = 0. x^ -y^ + z^ = 1. ?/2 + 4z2=:16x. X2 _ 4 y2 - 2;2 =r 0. TRANSFORMATION OF COORDINATES 393 7. Show that the xy-term may always be removed from the equation Ax^ + By^ + Cz^ + Fxy -\-K = by a rotation of the axes. 8. Show that the y2;-term may always be removed from the equation Ax^ + By^ + Cz'^ + Dyz + K = by rotating the axes, 9. What are the direction cosines of OX, OZ, and OZ (Fig., p. 392) referred to 0X% 0Y\ and OZ'? What six equations do they satisfy? 10. Show that the six equations obtained in problem 9 are equivalent to the six equations in the footnote, p. 391. 11. If (x, y, z) and {x% y', z') are respectively the coordinates of a point before and after a rotation of the axes, show that x2 + y2 + 2;2 ^ X'2 -f- 2/^2 + 2/2, 173. Polar coordinates. The line OP drawn from the origin to any point P is called the radius vector of P. Any point P determines four numbers, its radius vector p and the direction angles of OP, namely, a, yS, and y, which are called the polar coordinates of P. These numbers are not all independent since a, jS, and 7 satisfy (III), p. 330. If two are known, the third may then he found, hut all three are retained for the sake of symmetry. Conversely, any set of values of p, a, /?, and y which satisfy (III), p. 330, determine a point whose polar coordinates are p, or, /8, and y. / Projecting OP on each of the axes, we get, by Corollary I, p. 328, and Theorem I, p. 328, Theorem IV. The equations of transformation from rectangular to polar coordinates are (IV) Qc = pcosa, y = pcosfi, z = p cos y. From Theorem (IV), p. 331, we obtain (1) p' = Gc' + y^ + z\ which expresses the radius vector in terms of x, y, and ». 394 ANALYTIC GEOMETRY 174. Spherical coordinates. Any point P determines three numbers, namely, its radius vector p, the angle 6 between the radius vector and the Z-axis, and the angle <^ between the pro- jection of its radius vector on the JTF-plane and the Z-axis. These num- bers are called the spherical coordinates of P. 6 is called the colatitude and <^ the longitude. ^ Conversely, given values of p, 6, and <^ -^ determine a point P whose spherical coordinates are (p, 0, cf>). Projecting OP on OA, we get OM = p sin Oj and then projecting OP and OMP on each of the axes, we obtain Theorem V. The equations of transformation from rectangular to spherical coordinates are (V) a; = /> sin ^ cos ^, y = p sin ^ sin ^, « = /) cos 9. The equations of transformation from spherical to rectangular coordinates may be obtained by solving (V) for p, 6, and <^. Z< P 175. Cylindrical coordinates. Any point P (x, y, z) determines three numbers, its distance z from the XF-plane and the polar coordinates (r, <^) of its projection (x, y, 0) on the ZF-plane. These three numbers are called the cylin- drical coordinates of P. Conversely, three values of r, <^, and z determine a point whose cylindrical coordinates are (r, , z). From Theorem I, p. 155, we have at once ^^ Theorem VI. The equations of transformation from rectangular to cylindrical coordinates are (VI) Qc = r cos , -y = r sin ^, z = z. The equations of transformation from cylindrical to rectangu- lar coordinates may be obtained by solving (VI) for r, <^, and z. TRANSFORMATION OF COORDINATES 395 PROBLEMS 1. What is meant by the "locus of an equation " in the polar coordinates P, a, ^, and 7 ? in tlie spherical coordinates p, 6, and ? in the cylindrical coordinates r, 0, and z ? 2. Show that the locus of an equation in polar coordinates is symmet- rical with respect to the pole if only the form of the equation is changed when p is replaced by — p ; with respect to one of* the coordinate planes if only the form of the equation is changed when a is replaced by ;r — or, /3 by ;r — /8, or 7 by TT — 7. Under what conditions will it be symmetrical with respect to each of the rectangular axes ? 3. Find rules by which to determine when the locus of an equation in spherical or cylindrical coordinates is symmetrical with respect to the origin, each of the rectangular axes, and each of the coordinate planes. 4. How may the intercepts of a surface on the rectangular axes be found if its equation in polar coordinates is given? if its equation in spherical coordinates is given ? if its equation in cylindrical coordinates is given ? 5. Transform the following equations into polar coordinates. (a) x2 + 2/2 + 22 ^ 25. Ans. p = 5. (b) x2 + y2 _ ^2 = 0. Ans. 7 = -• 4 (c) 2x2-2/2-22:^0. Ans. a = cos-iiV3. 6. Transform the following equations into spherical coordinates, (a) x2 + y^ + z'^ = 16. Ans. p = 4. (b)2x + 3?/ = 0. Ans. = tan-i(-|). (c) 3 x2 + 3 2/2 =: 7 2;2. Ans. 6 = tan-i i V2T. 7. Transform the following equations into cylindrical coordinates. (a) 5x — ?/ = 0. An^. 0=rtan-^5. (b) x2 + 2/2 = 4. Ans. r = 2. 8. Find the equation in polar coordinates of (a) a sphere whose center is the pole. (b) a cone of revolution whose axis is one of the coordinate axes. Ans. (a) p = constant ; (b) a = constant, /3 = constant, or 7 = constant. 9. Find the equation in spherical coordinates of (a) a sphere whose center is the origin. (b) a plane through the Z-axis. (c) a cone of revolution whose axis is the Z-axis. Ans. (a) p = constant ; (b) = constant ; (c) 6 = constant. 396 ANALYTIC GEOMETRY 10. Find the equation in cylindrical coordinates of (a) a plane parallel to the XY-plane. (b) a plane through the Z-axis. (c) a cylinder of revolution whose axis is the Z-axis. Ans. (a) z = constant; (b) = constant; (c) r = constant. 11. In rectangular coordinates a point is determined as the intersection of three mutually perpendicular planes (p. 326). Show that (a) in polar coordinates a point is regarded as the intersection of a sphere and three cones of revolution which have an element in common. (b) in spherical coordinates a point is regarded as the intersection of a sphere, a plane, and a cone of revolution which are mutually orthogonal. (c) in cylindrical coordinates a point is regarded as the intersection of two planes and a cylinder of revolution which are mutually orthogonal. 12. Show that the square of the distance between two points whose polar coordinates are (pi, cri, j8i, 71) and {p2, a^^ ^21 72) is r2 = p-^ + p2^ _ 2 P1P2 (cos ai cos a^ + cos /3i cos ^2. + cos 71 cos 72). 13. Find the general equation of a plane in polar coordinates. Ans. p {A cos a -\- B cos jS + C cos 7) + D = 0. 14. Find the general equation of a sphere in polar coordinates. Ans. /)2 + yo (G^ cos a + JBTcos /3 + I cos 7) + ii = 0. CHAPTER XXII QUADRIC SURFACES AND EQUATIONS OF THE SECOND DEGREE IN THREE VARIABLES 176. Quadric surfaces. The locus of an equation of the second degree, of which the most general form is (1) Ax'^-^Bi/^Cz'^-^Dyz+Ezx+Fxy-{-Gx-\-Hy+rz4K = 0, is called a quadric surface or conicoid. Theorem I. The intersection of a quadric with any plane is a conic or a degenerate conic. Proof. By a transformation of coordinates any plane may be taken as the ZF-plane, z = 0, and referred to any axes the equa- tion of a quadric has the form (1) (Theorem III, p. 392). Then the equation of the curve of intersection referred to axes in its plane is (Kule, p. 345) Ax^ + Fxy 4- By^ -\- Gx -{- Hy -j- K = 0, and the locus is therefore a conic or a degenerate conic (Theo- rem XIII, p. 196). Q.E.D. Corollary. The intersection of a cone of revolution with a plane is an ellipse, hyperbola, or parabola according as the plane cuts all of the elements, is parallel to two elements (cutting some on one side of the vertex and some on the other), or is parallel to one element (cutting all the others on the same side of the vertex). Theorem II. The intersections of a quadric with a system of par- allel planes are, in general, similar conies. Proof. By a transformation of coordinates one of the planes of the system may be taken as the ZF-plane, and hence the equa- tion of the system is z — k^ while that of the quadric has the 397 398 ANALYTIC GEOMETRY form (1) (Theorem III, p. 392). Hence the equation of the curve in which the plane z = k intersects the quadric is (Kule, p. 345) (2) Ax^-\-Fxy+By'' + {Ek+G)x+{Dk+H)y + Ck'^+Ik+K=(). For different values of k this equation represents a system of similar conies^ (Corollary I, p. 295). q.e.d. 177. Simplification of the general equation of the second degree in three variables. If equation (1) be transformed by rotating the axes (Theorem II, p. 391), it can be shown that the new axes may be chosen so that the terms in yz, zx, and xy drop out and hence (1) reduces to the form A '^2 -f Bhf 4- C'^2 _^ Qt^ j^ jj^y _|_ p^ ^ j^^ ^ Q^ Transforming this equation by translating the axes (Theorem I, p. 391), it can be shown that the axes may be chosen so that the transformed equation has either the forni (1) A V + B^'if + C"^2 _p j^v ^ Q or the forin (2) A"x^ + B'Y -\- I"z = 0. If all of the coefficients in (l)'and (2) are different from zero, (1) and (2) may, with a change in notation, be respectively written in the forms (3) ±$±$±$ = 1, (4) S^S=2- * If the invariants of (2) (Theorem YIII, p. 275) for two different values of k are respec- tively A, H, © and A', H', ©' then in order that the conies be similar the value of A given by = — — must be a real number. ©' A2 © All the sections will belong to the same type because A will have the same sign for all valvies of k. If the sections are ellipses, H and © have opposite signs (Theorem IX, p. 277) and A will be real. The same is true if the sections arcparabolas (p. 279). If the sections are hyperbolas, then, in general, for values of A; between certain limits the hyperbolas will be similar, and for the remaining values of k, exclusive of the limits, the sections will also be similar (compare problem 3, p. 296). QUADRIC SURFACES 399 The purpose of the following sections is to discuss the loci of these equations,* which are called central and non-central quadrics respectively. If one or more of the coefficients in (1) or (2) are zero, the locus is called a degenerate quadric. If K" = 0, the locus of (1) is a cone (Theorem V, p. 385) unless the signs of A'% B", and C" are the same, in which case the locus is a point, namely, the origin. If one of the coefficients A", B", and C" is zero, the locus is a cylinder (Theo- rem IV, p. ;^3) whose elements are parallel to one of the axes and whose directrix is a conic of the elliptic or hyperbolic type (p. 195). If K" = 0, the locus will be a pair of intersecting planes or a line. If tioo of the coefficients J.'', B'\ and C" are zero, the locus is a, pair of parallel planes (coincident if K'^ = 0) or there is no locus. If one of the coefficients in (2) is zero, the locus is a cylinder (Theorem IV, p. 383) whose directrix is a parabola or a degenerate central conic. If two of the coefficients are zero, the locus is a pair of coincident planes. {A" and B'^ cannot be zero simultaneously, as the equation would cease to be of the second degree.) PROBLEMS 1. Construct and discuss the loci of the following equations. (a) 9x2 _ 362/2 + 4^2 ^ q. (e) 4y2 _ 25 = 0. (b) 16x2 _ 4^/2 - z2 = 0. (f) 3y2 + 7 22 = 0. (c) 4x2 + 22 - 10 = 0. (g) 8 ^2 ^. 25 z = 0. (d) 2/2 - 9 z2 + 36 zz 0. (h) z^-\-lQ = 0. 3.2 y2 ^2 2. Discuss the locus of the equation ± — ± — ± — = (a) if all the a2 52 c2 signs agree ; (b) if two signs are positive. When will the locus be a cone of revolution about the X-axis ? the F-axis ? the Z-axis ? 3. Show geometrically by means of Theorem I that the sections of a cylinder whose equation is of the second degree made by planes cutting all of the elements are conies of the same type. Show also that the orthogonal projection on a plane of an ellipse is an ellipse ; of an hyperbola is an hyper- bola ; and of a parabola is a parabola. 4. Show how to find the equations of the projections of a curve upon the coordinate planes by means of their projecting cylinders. 5. Prove the Corollary to Theorem I by determining the nature of the intersection of the cone x2 + y2 _ tan2 7 • z"^ with the plane x = tan /3 • z + &• 6. Prove the Corollary to Theorem I by transforming x"^ + y^ = tan2 7 . z^ by rotating the axes about OY through an angle and considering the sec- tions formed by the plane z' = fc if ^ = 7- * There is a locus unless all of the coefficients of (3) are negative, when there is no locus. 400 ANALYTIC GEOMETRY 178. The ellipsoid _ + f- + _ = i. If all of the coefficients in a? W c^ (3), p. 398, are positive, the locus is called an ellipsoid. A discus- sion of its equation gives us the following properties. 1. The ellipsoid is symmetrical with respect to each of the coordinate planes and axes and the origin (Theorem IV, p. 346). These planes of symmetry are called the principal planes of the ellipsoid. 2. Its intercepts on the axes are respectively (Eule, p. 346) X —±. a, y =±b, z =:k c. The lines AA' = 2 a, BB' =2b,CC' = 2G are called the axes of the ellipsoid. 3. Its traces on the principal planes are the ellipses ABA'B', BCB'C, and ACA'C, whose equations are (p. 346>' x^ z' = \. 4. The equation of the curve in which a plane parallel to the ZF-plane, z — k, intersects the ellipsoid is (Rule, p. 345) (1) x"^ ?/ _ k^ -2 + ^ = 1-7 or + r (C2 -k"") IP- = 1. X The locus of this equa- tion is an ellipse, and for different values of k the ellipses are similar. If k increases from to c, or decreases from to — c, the plane recedes from the J^F-plane, and the axes of the ellipse decrease from 2 a and 2 h respectively to when the ellipse degenerates (p. 195). \ik> c oy k < — c, there is no locus, and hence the ellipsoid lies entirely between the planes ;^ = ± c. /C^ Z-i:).^;r^ r^^^^^~-— \/r _:>A X'-4'W^ i/'jo ; ^ ^^..^r7 \ i '' _---^ ^-m IP QUADRIC SURFACES 401 In like manner the sections parallel to the YZ- and ZX-planes are similar ellipses whose axes decrease as the planes recede,' and the ellipsoid lies entirely between the planes x = ±a and y = ±b. Hence the ellipsoid is a closed surface. If a = bj the section (1) is a circle for values of k such that — c a' 6^ c^ QUADRIC SURFACES 403 1. The hyperboloid is symmetrical with respect to each of the coordinate planes and axes and the origin (Theorem IV, p. 346). 2. Its intercepts on the X-axis ai'e x =±.a, but it does not cut the Y- and Z-axes. 3. Its traces on the XY- and ZZ-planes (p. 346) are respec- tively the hyperbolas r 1, which have the same transverse axis AA' = 2a, but it does not cut the FZ-plane. 4. The equation of the curve in which a plane parallel to the yz-plane, x = k, intersects the hyperboloid of one sheet is (Rule, p. 345) y or r Xk-'-a^) 1. a" ^ ■ a^^ ^ This equation has no locus \i — a < k < d. li k =±a, the locus is a degenerate ellipse, and as k increases from a to oo, or decreases from —a to — oo, the locus is an ellipse whose axes increase indefinitely. Hence the surface consists of two branches or sheets which recede indefi- nitely from the FZ-plane and from the X-axis. In like manner the sections, formed by all planes parallel to the XY- and ZZ-planes are hyperbolas whose axes increase indefinitely as their planes recede from the coordinate planes. The hyperboloid (1) is said to " lie along the Z-axis." The equations (2) r + are the equations of hyperboloids of two sheets which lie along the Y- and Z-axes respectively. 404 ANALYTIC GEOMETRY If h = Cj c = a, or a = h, the hyperboloids (1) and (2) are respectively hyperboloids of revolution. It should be noticed that the locus of (3), p. 398, is an ellipsoid if all the terms on the left are positive, an hyperboloid of one sheet if but one term is negative, and an hyperboloid of two sheets if two terms are negative. If all the terms on the left are negative, there is no locus. If the locus is an hyperboloid, it will lie along the axis corresponding to the term whose sign differs from that of the other two terms. PROBLEMS 1. Discuss and construct the loci of the following equations. (a) 4x2 + 97/2 + I6z2 ^ 144. (e) 9x2 _ ^/S + 9^2 ^ 35. (b) 4x2 + 9^,2 _ i6;22 ^ 144. (f) 22 _ 4x2 _ 4^/2 = \q, (c) 4x2 _ 9^/2 _ i6z2 := 144. (g) i6a;2 j^yij^iQ^'^^ 64. (d) x2 + 16 2/2 + z2 ^ 64. (h) x2 + y2 _ ^2 = 26. 2. For what values of k or ¥ will the sections of the hyperboloid of one ^2 y2 2;2 sheet, — \-~ = 1, formed by the planes x = k or y — V be similar a' ^"^ ^' x2 2/2 z2 hyperbolas ? the hyperboloid of two sheets 1 — =1? a2 62 q1 3. Show analytically that the intersection of an ellipsoid with any plane is a conic of the elliptic type. 4. Show analytically that the section of an hyperboloid of (a) one sheet, (b) two sheets formed by a plane passing through the axis along which the hyperboloid lies, is an hyperbola. rffh 2/2 ^2 5. Show that 1 f- — = {Ax -\- By ■\- Czf is the equation of the cone a2 62 Q% whose vertex is the origin which passes through the intersection of the r^ y1 2;2 ellipsoid f- — H — = 1 and the plane J.x + JBy + C2; = 1. a2 62 c2 6. Show that xsf i - i^ + 2/2('i - -") + ^2/1 _ i\ = is the equa- \a2 r2/ \62 r2/ \c2 r'^) ^ tion of the cone whose vertex is the origin which passes through the intersection of the ellipsoid and the sphere x2 + 2/2 + 22 = r^. 7. If, in problem 6, a>b>c and r = 6, show that the cone degenerates into a pair of planes whose intersections with the ellipsoid are circles. What is the nature of the cone if r=:a? ifr = c? 8. Find the equations of the planes whose intersections with the ellipsoid 9 x2 -I- 25 2/2 + 169 z2 = 1 are circles. Ans. 4 x = ± 12 z + ^. QUADRIC SURFACES 405 9. Find the equation of the cone whose vertex is the origin which 2;2 yl ^2 through the intersection of (a) the hyperboloid of one sheet 1- ^ = 1. « o o a^ ^' c2 2j2 y1 ^2 (b) the hyperboloid of two sheets -^ — rr — ^ = 1 with the sphere x^ + y^ + 2^ = r2. For what value of r will the cone degenerate into a pair of planes whose intersections with the hyperboloid are circles ? An.. (a)x^(l-i) + 2,^(l-I)-.^(i + l) = 0;r=aif«>6. 10. Find the equations of the two systems of planes whose intersections with (a) an ellipsoid, (b) an hyperboloid of one sheet, (c) an hyperboloid of two sheets, are circles. 2 2 181. The elliptic paraboloid ^ + ^ = 2 cs;. If the coefficient a o of y^ in (4), p. 398, is positive, the locus is called an elliptic paraboloid. A discussion of its equation gives us the following properties. 1. The elliptic paraboloid is symmetrical with respect to the YZr and ZZ-planes and the Z-axis (Theorem IV, p. 346). 2. It passes through the origin (Theorem III, p. 345) but does not intersect the axes elsewhere (Rule, p. 346). 3. Its traces on the coordinate planes (p. 346) are respectively the conies „ „ „ 0, 2r^ ^ 2cz, of which the first is a degenerate ellipse (p. 195) and the others are parabolas. 4. The equation of the curve in which a plane parallel to the AT-plane, z = k, cuts the paraboloid is (Rule, p. 345) or ir ^^^'' '' 2^k + ?r x: 2b'ck 1. " i- <5) 406 ANALYTIC GEOMETRY The curve is an ellipse if c and k have the same sign, but there is no locus if c and k have opposite signs. Hence, if c is positive, the surface lies entirely above the XF-plane. If k increases from to oc, the plane recedes from the ZF-plane and the axes of the ellipse increase indefinitely. Hence the surface recedes indefinitely from the ZF-plane and from the Z-axis. In like manner the sections parallel to the YZ- and ZX-planes are parabolas whose vertices recede from the ZF-plane as their planes recede from the coordinate planes. The loci of the equations (1) |! + S = 2-. 5 + S = 2*^ are elliptic paraboloids which lie along the X- and F-axes respectively. If a = h, the first surface considered is a paraboloid of revolu- tion whose axis is the Z-axis ; and ii b = c and a = c, the parab- oloids (1) are surfaces of revolution whose axes are respectively the X- and T-axes. An elliptic paraboloid lies along the axis corresponding to the term of the first degree in its equation, and in the positive or negative direction of the axis according as that term is positive or negative. 182. The hyperbolic paraboloid —-^ = ^cz. If the coeffi- cient of y"^ in (4), p. 398, is negative, the locus is called an hyperbolic paraboloid. 1. The hyperbolic paraboloid is symmetrical with respect to the YZ- and ZA-planes and the Z-axis (Theorem IV, p.' 346). 2. It passes through the origin (Theorem III, p. 345) but does not cut the axes elsewhere (Rule, p. 346). 3. Its traces on the coordinate planes (p. 346) are respectively the conies of which the first is a degenerate hyperbola (p. 195) and the others are parabolas. QUADRIC SURFACES 407 4. The equation of the curve in which a plane parallel to the ZF-plane, z = kj cuts the paraboloid is (Kule, p. 345) —„ — T7: = 2ch or 2a''ck 2b''ck If c is positive, the transverse axis The locus is an hyperbola, of the hyperbola is parallel to the .Y- or F-axis according as k is posi- tive or negative. If k increases from to 00, or decreases from to — 00, the plane recedes from the XF-plane and the axes of the hyperbolas increase indefinitely. Hence the surface recedes indefi- nitely from the ZF-plane and the Z-axis. The surface has approximately the shape of a saddle. In like manner the sections parallel to the other coordinate planes are parabolas whose vertices recede from the XF-plane as their planes recede from the coordinate planes. The loci of the equations ..--. = ^by, f,- 2 ax x" Z' are hyperbolic paraboloids lying along the F- and Z-axes respectively. An hyperbolic paraboloid also lies along the axis which corresponds to the term of the first degree in its equation. PROBLEMS 1. Discuss and construct the following loci. (a) ?/2 + z2^4a;. (b) y2_22^4a;. (c) 9z2_4a;2 (d) 16 x2 + z2 : : 288 y. 64 y. 2. Prove that the parabolas of the systems obtained by cutting (a) an elliptic paraboloid, (b) an hyperbolic paraboloid by planes parallel to one of the coordinate planes, are all equal. 3. Show analytically that any plane parallel to the axis along which (a) an elliptic paraboloid, (b) an hyperbolic paraboloid lies, intersects the surface in a parabola. 408 ANALYTIC GEOMETRY 4. Show analytically that any plane not parallel to the axis of an elliptic paraboloid intersects the surface in an ellipse. 5. Show analytically that any plane not parallel to the axis of an hyper- bolic paraboloid intersects the surface in an hyperbola. 6. Find the equation of the cone whose vertex is the origin which passes through the intersection of the paraboloid 1- — = 2 cz and the sphere x2 + 2/2 + ^2 = 2 rz. ^^ ^ns. x2(^^-c) + y2(^^-c)-cz2 = o. 7. By means of problem 6 find the equations of two systems of planes whose intersections with the paraboloid are circles. 183. Rectilinear generators. The equation of the hyperboloid of one sheet (p. 401) may be written in the form ^^ W" C" h'' ]9 As this equation is the result of eliminating h from the equa- tions of the system of lines a G I b J a G ky b the hyperboloid is a ruled surface (p. 387). Equation (1) is also the result of eliminating k from the equations of the system of lines / \ -, / a G ^\ bj a c k\ b and the hyperboloid may therefore be regarded in two ways as a ruled surface. In like manner the hyperbolic paraboloid contains the two systems of lines a b a b k T X , y . X y 2g and - -\- ^ — kz, 7 = -^^ a b a b k These lines are called the rectilinear generators of these surfaces. Hence Theorem III. The hyperboloid of one sheet and the hyperbolic paraboloid have two systems of reetilinear generators, that is, they may be regarded in tivo ways as mled surfaces. T^LATK IT Elliptic Paraboloid Hyperbolic Paraboloid Non-Central Quadrics Hyperboloid of one sheet Hyperbolic Paraboloid Riled Quadrics V QUADRIC SURFACES 409 MISCELLANEOUS PROBLEMS 1. Construct the following surfaces and shade that part of the first inter- cepted by the second. (a) ^2 + 4 y2 _|. 9 2;2 = 36, x2 + y^ + 2;2 = 16. (b) x2 + 2/2 _|. 2;2 ^ 64, x2 + 2/2 - 8x = 0. (c) 4 x2 + 2/2 - 4 z = 0, x2 + 4 2/2 - 22 = 0. 2. Construct the solids bounded by the surfaces (a) x2 + 2/2 = a2, z = mx, 2 = ; (b) x2 + 2/2 = az, x2 + 2/2 = 2 ox, z = 0. 3. Show that two rectilinear generators of (a) an hyperbolic paraboloid, (b) an hyperboloid of one sheet, pass through each point of the surface. 4. If a plane passes through a rectilinear generator of a quadric, show that it will also pass through a second generator and that these generators do not belong to the same system. 5. The equation of the hyperboloid of one sheet (p. 401) may be written in the form = 1 By treating this equation as we treated equa- 62 c2 a2 tion (1), p. 408, we obtain the equations of two systems of lines on the sur-' face. Show that these systems of lines are identical with those already obtained. 6. Show that a quadric may, in general, be passed through any nine points. 7. If a > 6 > c, what is the nature of the locus of x2 2/^ j 2^ ^-, a2 - X 62 _ X c2 - \ ifX>a2? ifa2>X>62? if62>X>c2? ifX> a^ + l = "'' and the line (2) X = Xi + p cos a, 2/ = 2/1 + p cos i3, z = Zi + p cos 7. Substituting from (2) in (1), we obtain the equation in p (p. 410) If Pi(xi, 2/1, zi) is to lie on (1), and (2) is to be tangent to (1) at Pi, both roots of (3) must be zero, and hence (Case III, p. 6) scicosa 2/1 cos /3 . Xi* , 2/1^ „ ^ (4) .J__ + ?^-ccos7 = 0, _ + _-2«i = 0. Solving (2) for the direction cosines, we get X -Xi ^ y -Vi z — z\ (5) cos a — -, cos /3 = , cos 7 = • 412 ANALYTIC GEOMETRY Substituting from (5) in the first of equations (4), we get (Q) ^i(^-'gi) , y{y- yi) ^ c (g - ^i) as the condition that P{x, y, z) should lie on a line tangent to (1) at Pi. Simplifying (6) by means of the second of equations (4), we obtain (') f + f^ -(--)• This is the equation of a plane (Theorem II, p. 349). Hence all of the lines tangent to (1) at Pi lie in a plane which is called the tangent plane. This method may be summed up in the Rule to derive the equation of the plane which is tangent to a quadric at a given point Pi(xi, y^ ^i)- First step. Derive the equation in.p and set the coefficient of p and the constant term equal to zero. Second step. Solve the parametric equations of the line for its direction cosines and substitute in the first equation obtained in the first step. Third step. Simplify the equation obtained in the second step by means of the second equation obtained in the first step. The result is the required equation. By means of this Rule we obtain Theorem I. The equation of the plane which is tangent at Pi (xi, ?/i, Zi) to the central quadric ± _ ± |- ± _ = 1 is ± ^ ± ^ ± -J^ = 1; x2 7/2 sc.ac y^y non-central quadric —± — = ^cz is — — 4- — — = c{z -\- Zj). a^ b^ a b Theorem n. The equation of the plane which is tangent to any quadric at Pii^h 2/ii ^i) *s found by substituting X\X, yiy, and ZiZ for x^, y^, and z^; i {yix + xiy), i {ziy + yiz), and i {xiz + zix) for xy, yz, and zx; and i (x + Xi), i (y + 2/i)» ttw<^ H^ + ^i) f^'"' ^» y^ ^"•^ g ^^ ^^^ equation of the quadric. 186. Polar planes. If Pi is a point on a quadric, the equation of the tangent plane at Pi may be found by Theorem II. If Pi is not on the quad- ric, the plane found by Theorem II is called the polar plane of Pi, and Pi is called the pole of that plane. In particular, the polar plane of a point on a quadric is the plane tangent to the quadric at that point, and the pole of a tangent plane is the point of tangency. 187. Circumscribed cones. All of the lines passing through a point not on a given quadric which are tangent to the surface form a cone which is said to be circumscribed about the quadric. LINE AND QUADIUC 413 Ex. 1. Find the equation of the cone circumscribed about the ellipsoid a;2 + 3 ^2 + 3 ^2 = 9 whose vertex is the point P^ (4, — 2, 4) . Solution. The parametric equations of any line through P^ are (Theorem V, p. 369) (1) a; = 4 + /) cos a, y = - 2 + pcosjS, 2; = 4 + /) cos 7. Substituting these values of x, y, and z in the equation of the ellipsoid, we obtain the equation in p (2) (cos2 a + 3 cos2/3 + 3 cos2 7) p2 _^ (g cos or - 12 cos /3 + 24 cos 7) /o + 67 = 0. If (1) is tangent to the ellipsoid, then [(6), p. 410] (3) (8 cos a - 12 cos iS + 24 cos 7)2 - 4 • 67 (cos2 a + 3 cos2/3 + 3 cos2 7) = 0. Solving (1) for the direction cosines, substituting in (3), and multiplying by p^, we get (4) [8(a;-4)-12(y + 2) + 24(z-4)]2-268[(x-4)2 + 3(y + 2)2 + 3(z-4)2] = as the condition that P {x, y, z) should lie on a line passing through P^ which is tangent to the ellipsoid. Hence (4) is the equation of the required cone. That the locus of (4) is really a cone whose vertex is Pi is easily seen by moving the origin to Pi and applying Theorem V, p. 385. In constructing the figure, two divisions on each axis were taken for the unit. The reasoning employed in the solution of Ex. 1 justifies the Rule to find the equation of the cone whose vertex is Pi (xi, yi, Zi) which circumscribes a given quadric. First step. Derive the equation in p and set its discriminant equal to zero. Second step. In the result of the first step substitute the values of the direc- tion cosines of a line through Pi obtained from the parametric equations of the line. The result is the required equation. 414 ANALYTIC GEOMETRY PROBLEMS , 1. Prove that the plane of the two rectilinear generators which pass through any point on a ruled quadric is the tangent plane at that point. 2. Prove that every plane which passes through a rectilinear generator of a ruled quadric is tangent to the quadric at some point of that generator. 3. Prove analytically that every plane tangent to a cone passes through the vertex. 4. Prove that the polar plane of any point in a given plane passes through the pole of that plane. 5. Prove that the pole of any plane which passes through a given point lies in the polar plane of that point. 6. Prove that the curve of contact of a cone circumscribed ahout a quadric lies in the polar plane of the vertex. 7. Show how to construct (a) the polar plane of a point outside of a quadric, (b) the pole of a plane which cuts the quadric, (c) the polar plane of a point within a quadric, (d) the pole of a plane which does not meet the quadric. 8. Show that the polar plane of a point Pi with respect to a sphere is perpendicular to the line drawn from the center to Pi- 9. Show analytically that the polar plane of a point Pi with respect to a central quadric recedes from the center as Pi approaches the center, and conversely. 10. Show that the distances from two points to the center of a sphere are proportional to the distances of each of these points from the polar plane of the other. 11. Show how the ideas of "polar reciprocal curves" and "polar recip- rocation" with respect to a conic may be generalized to "polar reciprocal surfaces" and "polar reciprocation" with respect to a quadric. 12. What is the polar reciprocal of a cone or cylinder with respect to a sphere 2 of a plane curve ? 13. Generalize problem 7, p. 320, for polar reciprocation with respect to a quadric. •' 14. Prove that the distance p from the origin to the plane which is tangent to the ellipsoid + |- + ^ 1 at Pi is given by -^ = :^ + ^^ + ^. a2 62 (.2 p2 (j4 ^4 c* 15. Prove that the plane Ax + By -\- Cz -\- D = is tangent to the ellipsoid X2 w2 2;2 - + ^ + 1 = lif ^2^2 + ^252 4. C2c2 = X>2. a2 o2 c2 LINE AND QUADRIC 415 16. The locus of the point of intersection of three mutually perpendicular tangent planes to an ellipsoid is a sphere whose radius is Va^ + b^ -\^^. Hint. From problem 15 we get the equations of three tangent planes. Square and add these equations, making use of the conditions that the planes shall be mutually perpendicular. 17. Show that the plane Ax + By + Cz + D = is tangent to the parabo- loid — ±y^ = 2czit A^a^c ± B^b^c = 2 CD. 18. Show that the locus of the point of intersection of three mutually perpendicular tangent planes to a paraboloid is a plane. The line perpendicular to a plane which is tangent to a surface at the point of tangency is called the normal to the surface at that point. 19. Find the equation of the normal to each of the quadrics at a point Pi. 20. If the normal to an ellipsoid at Pi meets the principal planes in ^, B, and C, then Pi^, PiB, and PiC are in a constant ratio. 21. Find the equation of the cone circumscribing a paraboloid whose vertex is Pi (xi, ?/i, Zi). 22. Find the equation of the cylinder circumscribing an ellipsoid if the direction angles of the elements of the cylinder are a, j8, and y. 188. Asymptotic directions and cones. If the coefficient of p"^ in the equation in p for any quadric is zero, one root is infinite (Theorem IV, p. 15), and the line meets the quadric in one point which is at an infinite distance from Pi. The direction of such a line is called an asymptotic direction. It is evident that a line having an asijmptotic direction of a quadric meets the quadric in but one point in the finite part of space. It is easily proved that the coefficient of p2 is formed by substituting cos a, cos jS, and cos y for x, y, and z in the terms of the second degree in the equation of the quadric (compare the footnote, p. 236). Hence the direction cosines of the asymptotic directions of the non-degenerate quadrics ±^±^±?^.l, ^^i^^ = 2c. respectively satisfy the equations cos^a cos2/3 COS27 cos^a cos^/S (1) "^-aT^-^^"^^^' ~ar^~^=^' 416 ANALYTIC GEOMETRY By considering the number of sets of real numbers satisfying these equations for the various combinations of signs we obtain Theorem m. The hyperboloids and the hyperbolic paraboloid have an infinite number of asymptotic directions, the elliptic paraboloid has one, and the ellipsoid has none. The lines passing through a given point Pi(xi, yi, Zi) which have the asymptotic directions of a quadric will, in general, form a cone. The equation of this cone for the hyperboloid of one sheet a^2 y2 2;2 ^ ' a2 62 c2 is found as follows. The direction cosines of an asymptotic direction satisfy the equation (3) c^^c^_c<|.^„^ ^^^^^^^ If the equations of a line through Pi are (4) x = Xi + p cos a, 2/ = 2/1 + P cos /5, z — Zi + p cos 7, then X — Xi ^ y — y\ z — Zi (5) cos a = , cos /3 = — , cos 7 = . P P P Substituting in (3) and multiplying by p^, we get ^ ' a2 52 c2 as the condition that P (x, y, z) should lie on a line through Pi which has an asymptotic direction of (2). Hence (6) is the equation of the cone whose vertex is Pi and whose elements have the asymptotic directions of (2). That (6) is really the equation of a cone is verified by translating the origin to P^. In general, we have the Rule to find the equation of the cone of asymptotic directions of a quadric whose vertex is a given point. Set the coefficient of p^ in the equation in p equal to zero, and substitute the values of the direction cosines derived from the parametric equations of the line. If the coefficients of p^ and p in the equation in p are both zero, then both roots are infinite * (Theorem IV, p. 15) and the line is called an asymptotic line. * This assumes that the constant term is not zero. If the constant term is zero, Pi lies on the quadric, and when the coefficients of p^ and p are both zero, any number is a root and the line lies entirely on the quadric. LINE AND QUADRIC 417 Let Pi be any point not on the hyperboloid (2) and let us seek the condi- tions that a, j8, and 7 must satisfy if the line (4) is an asymptote. The equation in p for the hyperboloid is cos^a cos2/3 COS27 \ a2 62 y+2( ccicosa 2/1 cos ^ 2^ COS 7 • + \ a2 62 + 62 C2 0. (7) If (4) is an asymptote, then, by definition, COS27 cos2a cos2/3 a2 "^ 62 c2 = 0, Xi COS a yi cos /S a2 62 Zi COS 7 C2 = 0. These are therefore the conditions which a, /3, and 7 must satisfy. Equa- tions (7) can be solved for cos a and cos /3 in terms of cos 7 and there will be two solutions which may be real and unequal, real and equal, or imaginary, and from these we can determine two sets of numbers to which cos a:, cos/3, and cos 7 are propor- tional. Hence there will pass through Pi either two asymptotes, one, or none. But if xi = yi = zi = 0, that is, if Pi is the center of the hyperboloid, the second of equations (7) is true for all values of a, jS, and 7 ; and as the first of equations (7) is identical with (3), we see that the elements of the cone of asymptotic directions whose vertex is the center (0, 0, 0) are all asymp- totic lines. From (6) the equation of this cone, which is called the asymptotic cone, is seen to be a2 62 c2 Hence we have Theorem IV. The equation of the asymptotic cone of the hyperboloid of one sheet 62 a^ Ir &• The figure shows the hyperboloid (2) in outline and its asymptotic cone which lies entirely within the surface. As the hyperboloid recedes to infinity it approaches closer and closer to its asymptotic cone in the same way that an hyperbola approaches its asymptotes (Theorem IX, p. 190). 418 ANALYTIC GEOMETRY In like manner we may prove the following theorem, P Theorem V. The equation of the asymptotic cone of the ,2 ^,2 ^2 hyperboloid of two sheets hyperbolic paraboloid 2cz 1 is O; The latter cone degenerates into a pair of intersecting planes. PROBLEMS 1. Show that a plane perpendicular to the axis of an hyperbolic parab- oloid intersects the surface in an hyperbola whose asymptotes form the intersection of the plane with the asymptotic cone. 2. Show that a plane passing through the axis of an hyperboloid inter- sects the surface in an hyperbola whose asymptotes form the intersection of the plane with the asymptotic cone. Hint. Rotate the axes about the axis of the hyperboloid. 3. Show that the asymptotic directions of any quadric are detern ined by the locus of the equation obtained by setting the terms of the second degree equal to zero. 4. Show that a plane passing through the center and a generator of an hyperboloid of one sheet is tangent to the asymptotic cone. 6. Show that any plane parallel to an element of the asymptotic cone of an hyperboloid intersects the hyperboloid in a parabola. 6. Show that a plane tangent to the asymptotic cone of an hyperboloid cuts the hyperboloid in two parallel lines. 7. Show that every asymptotic line of an hyperboloid is parallel to an element of the asymptotic cone and lies in the plane tangent to the cone along that element. LINK AND QUADRIC 419 8. By means of problem 7 show how to construct the asymptotic lines of an hyperboloid which pass through any point Pi other than the center. Show that there will be two, one, or no asymptotic lines through Pi according as Pi is outside of, on, or inside of the asymptotic cone. x^ ?/2 z2 ^.2 yl ^2 9. Show that the hyperboloids = 1 and h — H — = 1 have the same asymptotic cone. How are they situated relative to this cone ? 10. Show that two asymptotes of an hyperbolic paraboloid pass through every point not on the asymptotic cone, and that each of these lines is par- allel to one of the planes which fortn the cone. 189. Centers. A point Pi(xi, yi, Z\) is a center of symmetry of a quadric if it is the middle point of every chord passing through it. In order that Pi shall be the middle point of a chord, the roots of the equation in p must be equal numerically with opposite signs, and hence (Case II, p. 4) the coefficient of p must be zero. The coefficient of p in the equation in p for the general equation of the second degree is easily seen to be (2 Axx + Fyi + Ezi + G) cos a + (PiCi + 2 %i + Dzx + H) cos iS + {Ex^ + Dyi + 2 Czi + I) cos 7. This is zero for all lines passing through Pi, that is, for all values of cos a, cos j3, and cos 7, when and only when the three parentheses are zero. Setting these 'parentheses equal to zero and solving for Xi, 2/1, and Zi, we obtain the coordinates of the center. By means of the discussion in § 163, p. 374, we see that a quadric may have a single center, that there may be no center, or that all of the points of a line or of a plane may be centers. 190. Diametral planes. The locus of the middle points of a system of parallel chords of a quadric is found to be a plane which is called a diametral planp. Consider the ellipsoid (1) • t + yl + ^ = i ^' a'^ b^ c^ and the system of parallel lines (2) x = xi + p cos a, y = yi + p cos jS, z = zi + p cos 7. These equations represent a system of parallel lines if x^,yi, and Zj are arbitrary while a, /3, and y are constant. The equation in p for (1) is /cosVr cos2/3 cos27 \ ^ / Xicoscr yi Cos/3 2i_cos7\ ^^ V a2 "^ 62 + c2 r V a2 "^ 62 c^ J^ ^ V a2 ^ 62 ^ c2 / 420 ANALYTIC GEOMETRY If Pi is the middle point of the chord of (1) formed by the line (2), then the roots of (3) must be numerically equal with opposite signs ; and hence (Case II, p. 4) xi cos a yicos^ ziC0S7 _ ^ ^ ^ a^ Ifi c^ is the condition that Pi shall be the middle point of the chord. But (4) is the condition that Pi should lie in the plane X cos a y cos /3 z cos 7 _ ^ ^2 + 62 + c2 "" ' and this is therefore the equation of the locus of the middle points of all chords whose direction angles are or, /3, and 7. By proceeding in this manner with the other quadrics we obtain Theorem VI. The equation of the diametral plane bisecting all chords whose direction angles are a, /3, and 7 of the x2 7/2 ^2 Oleosa ycosB zcosy central quadric ± - ± _ ± - = Hs ± _^— i — ^ +_ ^ = O; x^ v^ ^ occosa ycosB non-central quadric — ± -- = 2cz is r-i- = c COS v. PROBLEMS 1. Determine geometrically the number of centers of each of the types of quadrics and degenerate quadrics. 2. Find the equation of the diametral plane of the locus of the general equation of the second degi-ee bisecting all chords whose direction angles are a, /3, 7. From the form of the equation prove that the plane passes through the center of the quadric if there is a center. 3. Prove that every plane through the center of a central quadric or parallel to the axis of a paraboloid is a diametral plane, and find the direc- tion cosines of the chords which it bisects. 4. The line of intersection of two diametral planes is called a diameter. Show that a central quadric has three diameters such that the plane of any two bisects all chords parallel to the third. Such lines are called conjugate diameters, and the plane of any two is said to be conjugate to the third. 5. Find the equation of the plane which bisects all chords of (a) a cen- tral quadric, (b) a paraboloid, which are. parallel to the diameter passing through a point Pi on the quadric. 6. The planes tangent to a quadric at the extremities of a diameter are parallel to the conjugate diametral plane. LINE AND QUADRIC 421 7. The sum of the squares of the projections of three conjugate semi- diameters of an ellipsoid on each of the axes of the ellipsoid is constant. Hint. Let P^, P^, and P^ be the extremities of three conjugate diameters. Find tlie conditions that tliese points are on the ellipsoid and that any two are on the plane conjugate to the diameter through the third. Then show that a^ b^ c^ a' b ^ c' a ' b ^ c are the direction cosines of three mutually perpendicular lines, and that if these lines be chosen as axes, then a^ a^ a^ b^ b ' b ' c ' c ' c are also the direction cosines of three lines. Then apply Theorem III, p. 330. 8. By means of problem 7 show that the sum of the squares of three conjugate semi-diameters of an ellipsoid is equal to a^ + 6^ 4. c2. IP INDEX Abscissa, 24 Absolute invariant, 270 Algebraic equation, 17 ; curve, 72 Anchor ring, 389 Arbitrary constant, 1 Auxiliary circle, 206 Euclidean transformation, 281 External angle, 121 Fixed point, 285 Focal radii of conies, 193, Four-leaved rose, 263 194 Cardioid, 158, 253, 302 Center of similitude of circles, 296 Central conic, 183 ; quadric, 399 Cissoid of Diodes, 253, 263, 299 Complete quadrilateral, 123 Conchoid of Nicomedes, 251 Condition for tangency, 230 Confocal conies, 203 Congruent figures, 281 Conicoid, 397 Conjugate diameters of quadrics, 420 Cubical parabola, 72 Curtate cycloid, 259 Cycloid, 256 Degenerate ellipse, 195; hyperbola, 195; parabola, 196; quadric, 399 Direction cosines of a line, 123, 330, 364 Director circle, 261 Discriminant of the equation of a circle, 131 ; of the general equation of the second degree, 265; of a quadratic, 2 Graph of an equation, 83 nomothetic figures, 291 * Hyperbolic spiral, 249 Hypocycloid, 259 ; of four cusps, 257 Intercepts, 73 Internal angle, 121 Invariant, 270 ; line, 288 ; point, 285 Involute of a circle, 259 Latus rectum, 181 Lemniscate of Bernoulli, 153, 248, 262, 300 Lima^on of Pascal, 253, 262, 302 Limiting points of a system of circles, 144 Normal to a curve, 210; length of, 214 ; to a surface, 415 Ordinate, 24 Orthogonal circles, 143; systems of circles, 308 Epicycloid, 259 Parabolic spiral, 263 Equations of a transformation, 281 Parameter, 1 423 424 INDEX Peaucellier's Inversor, 309 Point-circle, 131 Polar reciprocal curves, 313, 317 Prolate cycloid, 259 Radian, 19 Radical axis, 137, 382 Reciprocal spiral, 249 Subnormal, 214 Subtangent, 214 Torus, 389 Traces of a surface, 346 Transcendental equation, 17 Transformation, 281 Trisectrix of Maclaurin, 303 Self-conjugate triangles, 319 Semicubical parabola, 209 Spiral of Archimedes, 249 Strophoid, 262, 263, 301 Vertex of a conic, 174, 175 Witch of Agnesi, 250 ^ OF THE UNIVERSITY IP 14 DAY USE In RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below or on the date to which renewed. ' Renewed books are subject to immediate recall 28rei)'59Ai rh:c'd ld ILRJSJBm. ^^ !-_28Mat:60GM REC'D LD m MAR 1 1 193 f*^ec.'S8% ~~\-SLs:GinjjQ_ 4*i«A-44J3ui____ =M5-U:i ;^-jj|t-3-49aa ts^epeeiff — r>tTr*^r. . —, JAWrC LD 21A-50w-8,'57 (C8481sl0)476B . General Library University of California Berkeley X /9(> ^ ^ f THE UNIVERSITY OF CALIFORNIA LIBRARY 'k'v't^" N ': 'r w