GIFT OF INTRODUCTION TO ANALYTIC GEOMETRY BY PEECEY R SMITH, PH.D. N PROFESSOR OF MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL YALE UNIVERSITY AND AKTHUB SULLIVAN GALE, PH.D. ASSISTANT PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF ROCHESTER GINN & COMPANY BOSTON NEW YORK CHICAGO LONDON COPYRIGHT, 1904, 1905, BY ARTHUR SULLIVAN GALE ALL BIGHTS RESERVED 65.8 GINN & COMPANY PRO- PRIETORS BOSTON U.S.A. PEE FACE In preparing this volume the authors have endeavored to write a drill book for beginners which presents, in a manner conform- ing with modern ideas, the fundamental concepts of the subject. The subject-matter is slightly more than the minimum required for the calculus, but only as much more as is necessary to permit of some choice on the part of the teacher. It is believed that the text is complete for students finishing their study of mathematics with a course in Analytic Geometry. The authors have intentionally avoided giving the book the form of a treatise on conic sections. Conic sections naturally appear, but chiefly as illustrative of general analytic methods. Attention is called to the method of treatment. The subject is developed after the Euclidean method of definition and theorem, 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 object of the two short chapters on Solid Analytic Geom- etry is merely to acquaint the student with coordinates in space iii 20206fi iv PREFACE and with the relations between surfaces, curves, and equations in three variables. Acknowledgments are due to Dr. W. A. Granville for many helpful suggestions, and to Professor E. H. Lockwood for sugges- tions regarding some of the drawings. NEW HAVEN, CONNECTICUT January, 1905 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 . . 5 6. Variables 9 7. Equations in several variables . . . . . . . 9 8. Functions of an angle in a right triangle ..... 11 9. Angles in general 11 10. Formulas and theorems from Trigonometry .... 12 11. Natural values of trigonometric functions . . . . . 14 12. Rules for signs ..." 15 13. Greek alphabet 15 CHAPTER II CARTESIAN COORDINATES 14. Directed line 16 15. Cartesian coordinates 17 16. Rectangular coordinates 18 17. Angles 21 18. Orthogonal projection 22 19. Lengths 24 20. Inclination and slope 27 21. Point of division 31 22. Areas . 35 23. Second theorem of projection 40 vi CONTENTS CHAPTER III THE CURVE AND THE EQUATION SECTION PAGE 24. Locus of a point satisfying a given condition 44 25. Equation of the locus of a point satisfying a given condition . 44 26. First fundamental problem 46 27. General equations of the straight line and circle . . . 50 28. Locus of an equation ^ 52 29. Second fundamental problem 53 30. Principle of comparison ......... 55 31. Third fundamental problem. Discussion of an equation . . 60 32. Symmetry . . . . .65 33. Further discussion 66 34. Directions for discussing an equation . . . . . .67 35. Points of intersection t 69 36. Transcendental curves 72 CHAPTER IV THE STRAIGHT LINE AND THE GENERAL EQUATION OF THE FIRST DEGREE 37. Introduction . . . 76 38. The degree of the equation of a straight line .... 76 39. The general equation of the first degree, Ax -f By + C = . . 77 40. Geometric interpretation of the solution of two equations of the first degree 80 41. Straight lines determined by two conditions 83 42. The equation of the straight line in terms of its slope and the coordi- nates of any point on the line 86 43. The equation 91 the straight line in terms of its intercepts . . 87 44. The equation/of the straight line passing through two given points 88 45. The normal form of the equation of the straight line ... 92 46. The distance from a line to a point .... 96 47. The angle which a line makes with a second line .... 100 48. Systems of straight lines . . 104 49. The system of lines parallel to a given line 107 50. The system of lines perpendicular to a given line . . . 108 61. The system of lines passing through the intersection of two given lines HO CONTENTS vii CHAPTER V THE CIRCLE AND THE EQUATION a; 2 + y 2 + Dx + Ey + F = SECTION PAGE 52. The general equation of the circle ... .115 53. Circles determined by three conditions 116 54. Systems of circles 120 CHAPTER VI POLAR COORDINATES 55. Polar coordinates 125 56. Locus of an equation 126 67. Transformation from rectangular to polar coordinates . . . 130 58. Applications 132 59. Equation of a locus 133 CHAPTER VII TRANSFORMATION OF COORDINATES 60. Introduction . ... . 136 61. Translation of the axes 136 62. Rotation of the axes . . . 138 63. General transformation of coordinates 139 64. Classification of loci- 140 65. Simplification of equations by transformation of coordinates . 141 66. Application to equations of the first and second degrees . . 144 CHAPTER VIII CONIC SECTIONS AND EQUATIONS OF THE SECOND DEGREE 67. Equation in polar coordinates . . . . . . . 149 68. Transformation to rectangular coordinates .... 154 69. Simplification and discussion of the equation in rectangular coordi- nates. The parabola, e = 1 . . . . : . .154 70. Simplification and discussion of the equation in rectangular coordi- nates. Central conies, e ^ 1 158 71. Conjugate hyperbolas and asymptotes 165. 72. The equilateral hyperbola referred to its asymptotes . . . 167 viii CONTENTS SECTION PAGE 73. Focal property of central conies 168 74. Mechanical construction of conies 168 75. Types of loci of equations of the second degree .... 170 76. Construction of the locus of an equation of the second degree . 173 77. Systems of conies 176 CHAPTER IX TANGENTS AND NORMALS 78. The slope of the tangent ........ 180 79. Equations of tangents and normals 182 80. Equations of tangents and normals to the conic sections . . 184 81. Tangents to a curve from a point not on the curve . . . 186 82. Properties of tangents and normals to conies .... 188 y 83. Tangent in terms of its slope 192 CHAPTER X CARTESIAN COORDINATES IN SPACE 84. Cartesian coordinates 196 85. Orthogonal projections. Lengths 198 CHAPTER XI SURFACES, CURVES, AND EQUATIONS 86. Loci in space 201 87. Equation of a surface. First fundamental problem . . . 201 88. Equations of a curve. First fundamental problem . . . 202 89. Locus of one equation. Second fundamental problem . . 205 90. Locus of two equations. Second fundamental problem . . 205 91. Discussion of the equations of a curve. Third fundamental problem 206 92. Discussion of the equation of a surface. Third fundamental __ problem 207 93. Plane and straight line 210 94. The sphere 211 95. Cylinders . . . . 213 96. Cones 214 97. Non-degenerate quadrics . . . . . . . .215 UNIVERSITY', OF ANALYTIC 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 b V 1, where b is a real number, and (V I) 2 =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^T) 2 = a 2 - b 2 + 2 ab V^l, 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 of 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"), of' (read "a double prime or a second"), a'" (read "a third"), are all different constants. Using subscripts, &! (read " 6 one"), 6 a (read " 6 two"), are different constants. Using both, c/(read "c one prime"), c 8 " (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 2 + 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 2 4 A C 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 to satisfy that equation. REVIEW OF ALGEBRA AND TRIGONOMETRY In Algebra it is shown that (2) or (1) has two roots, Xi and obtained by solving (1), namely, (3) Adding these values, we have B ^Vw* \& (4) x -f- x = Multiplying gives (5) XiX z = 'A Hence Theorem I. T^e sum of the roots of a quadratic is equal to the coefficient of the first power of the unknown with its sign changed divided by the coefficient of th& second power. The product of the roots equals the constant term divided by the coefficient of the second power. The quadratic (2) may be written in the form (6) Ax* + Bx + C =*A(x - x,} (x - * 2 ), as may be readily shown by multiplying out the right-hand member and substituting from (4) and (5). For example, since the roots of 3 a; 2 4 a; + 1 = are 1 and , w$ have iden- tically 3 32-43 + 1 = 3*(x - 1) (x - *). The character of the roots x l and x z as numbers ( 1) when the coefficients A, B, C are real numbers evidently depends entirely upon the discriminant. This dependence is stated in Theorem II. If the coefficients of a quadratic are real numbers, and if the discriminant be denoted by A, then when A is positive the roots are real and unequal; 4/i when A is zero the roots are real and equal ; when A is negative the roots are imaginary. *The sign = is read "is identical with," and means that the two expressions connected by this sign differ only inform. 4 ANALYTIC GEOMETRY In the three cases distinguished by Theorem II the quadratic may be written in three forms in which only real numbers appear. These are ' Ax 2 + Bx -f- C = A (xXi) (x x z ), from (6), if A is positive ; Ax 2 + Bx + C = A(xx 1 ) 2 , from (6), if A is zero ; [-/ B \2 AC-B 2 ~] Ax* + Bx + C = A\ (* + o~r) H 7-7* Y if ^ is negative. L L\ 2>A ) ^ A J The last identity is proved thus : adding and subtracting - within the parenthesis. 4. Special quadratics. If one or both of the coefficients B and C in (1), p. 2, is zero, the quadratic is said to be special. CASE I. C = 0. Equation (1) now becomes, by factoring, (1) Ax z + Bx = x(Ax + B) = 0. r> Hence the roots are x 1 = 0, x 2 = Therefore one root of yl a quadratic equation is zero if the constant term of that equation is zero. And conversely, if zero is a root of a quadratic, the con- stant term must disappear. For if x = satisfies (1), p. 2, by substitution we have C = 0. CASE II. B = 0. Equation (1), p. 2, now becomes (2) Ax 2 + C = 0. From Theorem I, p. 3, x^ -f x 2 = 0, that is, (3) !=-* REVIEW OF ALGEBRA AND TRIGONOMETRY 5 Therefore, if the coefficient of the first power of the unknown in a quadratic equation is zero, the roots are equal numerically but have opposite signs. Conversely, if the roots of a quadratic equation are numerically equal but opposite in sign, then the coefficient of the first power of the unknown must disappear. For, since the sum of the roots is zero, we must have, by Theorem I, B = 0. CASE III. B = C = 0. Equation (1), p. 2, now becomes (4) Ax 2 = 0. Hence the roots are both equal to zero, since this equation requires that x 2 = 0, the coefficient A bei v ng, by hypothesis, always different from zero. 5. Cases when the roots of a quadratic are not independent. If a relation exists between the roots Xi and x z of the Typical :Form Ax 9 + Ex + C = 0, then this relation imposes a condition upon the coefficients A, B, and C, which is expressed by an equation involving these constants. For example, if the roots are equal, that is, if Xi = x 2 , then B 2 -4: AC = 0, by Theorem II, p. 3. Again, if one root is zero, then x& 2 = ; hence C = 0, by Theorem I, p. 3. This correspondence may be stated in parallel columns thus : Quadratic in Typical Form Relation between the Equation of condition satisfied roots by the coefficients In many problems the coefficients involve one or more arbitrary constants, and it is often required to find the equation of condi- tion satisfied by the latter when a given relation exists between the roots. Several examples of this kind will now be worked out. 6 ANALYTIC GEOMETRY Ex. 1. What must be the value of the parameter k if zero is a root of the equation (1) 2x 2 -6z + fc 2 -3fc-4=0? Solution. Here A = 2, B = - 6, C = k' 2 - 3 k - 4. By Case I, p. 4, zero is a root when, and only when, C = 0. Solving, k = 4 or 1. Ex. 2. For what values of k are the roots of the equation kx 2 + 2kx-4x = 2-3k real and equal ? Solution. Writing the equation in the Typical Form, we have (2) fcx 2 + (2 k - 4)x + (3 k - 2) = 0. Hence, in this case, A = k, B = 2k-4, C = 3k-2. Calculating the discriminant A, we get A = (2 k - 4) 2 - 4 k (3 k - 2) = _8fc 2 -8A;-f 16 = -8(A: 2 + & - 2). By Theorem II, p. 3, the roots are real and equal when, and only when, A = ' .-. k* + k - 2 = 0. Solving, k= 2 or 1. Ans. Verifying by substituting these answers in the given equation (2) : when fc=-2, the equation (2) becomes 2x 2 -8x-8=0, or -2(x-f 2) 2 =0; when fc= 1, the equation (2) becomes x 2 2 +1=0, or (x 1) 2 =0. "$Hence, for these values of fc, the left-hand member of (2) may be trans- formed as in (7), p. 4. ^ Ex. 3. What equation of condition must be satisfied by the constants a^6, fc, and m if the roots of the equation (3) (62 + a 2 m 2) y2 + 2a? kmy + a 2fc2 _ a 2 & 2 = Q are equal ? Solution. The equation (3) is already in the Typical Form ; hence A = 6 2 + a 2 m 2 , B = 2 a 2 km, C = a?k 2 - a 2 6 2 . By Theorem II, p. 3, the discriminant A must vanish ; hence A = 4 a 4 fc 2 m 2 - 4 (6 2 + a 2 m 2 ) (a 2 & 2 - a 2 6 2 ) = 0. Multiplying out and reducing, a 2 & 2 (fc 2 - a 2 m 2 - 6 2 ) = 0. Ans. REVIEW OF ALGEBRA AND TRIGONOMETRY 7 Ex. 4. For what values of k do the common solutions of the simultaneous equations (4) 3x + 4y = fc, (5) x 2 + y 2 = 25 become identical ? Solution. Solving (4) for y, we have (6) y = t(*-3x), Substituting in (5) and arranging in the Typical Form gives (7) 25 x 2 - 6 kx +~k* - 400 = 0. Let the roots of (7) be x\ and x 2 . Then substituting in (6) will give the corresponding values y\ and y z of y, namely, (8) yi = i(& and we shall have two common solutions (xi, y\) and (x 2 , y 2 ) of (4) and (5). But, by the condition of the problem, these solutions must be identical. Hence we must have (9) xi = X 2 and y l = y 2 . If, however, the first of these is true (xi = x 2 ), then from (8) y\ and y^ will also be equal. Therefore the two common solutions of (4) and (5) become identical when-, and only when, the roots of the equation (7) are equal; that is, when the dis-. criminant A of (7) vanishes (Theorem II, p. 3). .-. A = 36 A; 2 - 100 (fc* - 400) = 0. Solving, A* = 625, k = 25 or 25. Ans. Verification. Substituting each value of k in (7), when A; =25, the equation (7) becomes x 2 6x + 9=0, or (x 3) 2 =0 ; .. x=3 ; when k= -25, the equation (7) becomes x 2 + 6x + 9=0, or (x-f 3) 2 =0; .-.x= 3. Then from (6), substituting corresponding values of k and x, when k = 25 and x = 3, we have y = ^(25 9)=4; when k = 25 and x = 3, we have y = $ ( 25 -f 9) = 4. Therefore the two common solutions of (4) and (5) are identical for each of these values of fc, namely, if k = 25, the common solutions reduce to x = 3, y = 4 ; if k = 26, the common solutions reduce to x = 3, y = 4. Q.E.D. 8 ANALYTIC GEOMETRY PROBLEMS 1. Calculate the discriminant of each of the following quadratics, deter- mine the sum, the product, and the character of the roots, and write each quadratic in one of the forms (7), p. 4. (a) 2x2 _ 6x + 4. (d) 4x 2 - 4x + 1. (b) x 2 - 9x - 10. (e) 5x 2 + lOx + 5. (c) 1 - x - x 2 . (f) 3x 2 - 5x - 22. 2. For what real values of the parameter k will one root of each of the following equations be zero ?. (a) 6x 2 '+ 5 kx - 3 k 2 + 3 = 0. Ans. k = l. (b) 2k - 3x 2 + 6x - & 2 + 3 = 0. Ans. k = - I or 3. 3. For what real values of the parameter are the roots of the following equations equal ? Verify your answers. (a) fcx 2 -3x-l = 0. Ans. k = - f . (b) x 2 - kx + 9 = 0. Ans. k = 6. (c) 2 kx 2 + 3 kx + 12 = 0. Ans. k = ^-. (d) 2 x 2 + kx - 1 = 0. Ans. None. (e) 5x 2 - 3x + 5fc 2 = 0. Ans. k = &. (f) x 2 -f kx + k 2 + 2 = 0. Ans. None. (g) x 2 - 2 kx - k - I = 0. Ans. k=-$. 4. Derive the equation of condition in order that the roots of the following equations may be equal. (a) m 2 x 2 + 2 kmx - 2px = - k 2 . Ans. p (p - 2 km) = 0. (b) x 2 + 2 mpx + 2 6p = 0. Ans. p (m 2 p - 2 6) = 0. (c) 2 mx 2 + 2 bx + a 2 = 0. Ans. fr 2 = 2 a 2 m. 5. For what real values of the parameter do the common solutions of the following pairs of simultaneous equations become identical ? (a) x + 2 y = k, x 2 + y 2 = 5. -4ns. k = 5. (b) y = mx 1, x 2 = 4 y. Ans. m = 1. (c) 2 x - 3y = 6, x 2 + 2 x = 3 y. -4ns. 6 = 0. (d) y = mx + 10, x 2 + y 2 = 10. .Ans. m = 3. (e) Ix + y - 2 = 0, x 2 - 8 y = 0. Ans. None. (f) x + 4 y = c, x 2 + 2 y 2 = 9. ^.ns. c = 9. (g) a 2 + y 2 - x 2 y = 0, x + 2 y = c. Ans. c = or 5. (h) x 2 + 4 y 2 8 x = 0, mx - y - 2 m = 0. ^.ris. None. 6. If the common solutions of the following pairs of simultaneous equations are to become identical, what is the corresponding equation of condition ? (a) bx + ay = a&, y 2 = 2px. Ans. ap (2 6 2 + ap) = 0. (b) y = mx + 6, .Ax 2 + .By = 0. . ^Ins. B (m z B - 4 bA) = 0. (c) y = m(x- a), By 2 + Dx = 0. -Ans. D(4 am 2 - D) = 0. REVIEW OF ALGEBRA AND TRIGONOMETRY 6. Variables. A variable is a quantity to which. jri jjfr investigation, an unlimited number of values can be assigned. In a particular problem the variable may,< in general, assume any value within certain limits imposed by the nature of th$ problem^ It is convenient to indicate these limits by inequalities. For example, if the variable x can assume any value between 2 and 5, that is, if x must be greater* than 2 and less than 5, the simultaneous inequalities *>-2, x<5, are written in the more compact form -2 5 or x = 5 are abbreviated to x ^ 2 and x ^5. Write inequalities to express that the variable (a) x has any value from to 5 inclusive. (b) y has any value less than 2 or greater than 1. (c) x has any value not less than 8 nor greater than 2. 7. 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 = 3 satisfy the equation 2x2 + 32,2=35, since 2(2)2 + 3(- 3)2 = 35. Similarly, x = 1, y = 0, z = 4 satisfy the equation 2z2_3 y 2 + 2 ;2_i 8 = o, since 2 ( 1)2 3 X + ( 4) 2 18 = 0. * The meaning of greater and less for real numbers ( 1) is defined as follows : a is greater than b when a - b is a positive number, and a is less than b when a - b 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 6. Thus 3 < 5, but - 3 > - 5. Therefore changing signs throughout an inequality reverses the inequality sign. 10 ANALYTIC GEOMETRY An equation is said to be algebraic in any number of variables, for example x, y, 2, if it can be transformed into an equation each of whose members is a sum of terms of the form ax m y n z p , where a is a constant and m, n, p are positive integers or zero. Thus the equations x* + x 2 y 2 - z + 2 a: 5 = 0, are algebraic. The equation x* + y* = a? is algebraic. For, squaring, we get x + 2 x$y* + y=a. Transposing, 2 x*y* = a x y. Squaring, 4 xy a 2 + x 2 + y z 2 ax 2 ay + 2 xy. Transposing, x 2 + y 2 2 xy 2 ax 2 ay + a 2 = 0. Q.E.D. The degree of an algebraic 'equation is equal to the highest degree of any of its terms.* An algebraic equation is said to Ibe 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' 2 + 3y' + 6x'-2x'y'-2 + z /3 = x' *if - y'* with respect to the variables z', y', we transpose and rewrite the terms in the order x '3 _ yfvtf + 2 X '2 2x'y' + y' 2 + 6a/ + 3y' 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 = sinx, y = 2*, logy=3x. PROBLEMS 1. Show that each of the following equations is algebraic; arrange the terms according to the variables x, y, or x, y, z, and determine the degree. (e) y = 2 + Vx 2 -2x-5. * The degree of any term is the sum of the exponents of the variables in that term, REVIEW OF ALGEBRA AND TRIGONOMETRY 11 (f) y .= x + 6 + V2x 2 -6x W*=-|j> + B + (h) y = ^ix + B + Vix' J 4- JMTx + N. 2. Show that the homogeneous quadratic * Ax 2 + Bxy + Cy* may be written in one of the three forms below analogous to (7), p. 4, if the discriminant A = B 2 4 AC satisfies the condition given : CASE I. Ax 2 + Bxy + Cy 2 = A(x- hy] (x - I 2 y), if A > ; CASE II. .Ax 2 + Bxy + Cy 2 = A(x- ky) 2 , if A = ; ', if A<0. CASE III. ^1x2 + Bxy + Cy* = A [(x + 8. 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 : opposite side (/ sin A = -** , / hypotenuse adjacent side cos A = -=-* , - hypotenuse esc A = sec A hypotenuse ^ opposite side' hypotenuse tan A = _ opposite side adjacent side adjacent side' adjacent side cot .4 = J - ^-. opposite side From the above the theorem is easily derived : B 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- nuse and the cosine of the angle adjacent to that side. A b C 9. Angles in general. In Trigonometry 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 clockwise. s *The coefficients A, B, Cand the numbers fj, I 2 are supposed real. 12 ANALYTIC GEOMETRY 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 ^J^ 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 = IT radians (IT = 3.14159 ). Or also, by solving this, 1 degree = ^ = .0174 radians, loU 1 80 1 radian = = 57.29 degrees. 7T 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. 10. Formulas and theorems from Trigonometry. 1. cotx = ; secx = ; cscx = tanx cosx smx sinx cosx 2. tanx = ; cotx = cosx sin x 3. sin 2 x + cos 2 x = 1 ; 1 + tan 2 x = sec 2 x ; 1 + cot 2 x = csc a z. 4. sin (- x) = - sin x ; esc (- x) = esc x ; cos( x) = cosx; sec( x) = secx; tan(- x) = tanx ; cot (- x) = cotx. REVIEW OF ALGEBRA AND TRIGONOMETRY 13 5. sin (if - x) = sinx ; sin (?r + x) = - sinx ; cos (x - x) = - cosx ; cosjtf ._&.= -', 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 M' 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 determines a point. The imaginary numbers of Algebra have no place in this repre- sentation, and for this reason elementary Analytic Geometry is concerned only with the real numbers of Algebra. 16. Rectangular coordinates. A rectangular system of coordi- nates is determined when the axes X'X and Y'Y are perpendicular X' -4} to each other. This is the usual case, and will be assumed unless otherwise stated. CARTESIAN COORDINATES 19 The work of plotting points in a rectangular system is much simplified by the use of coordinate or plotting paper, constructed by ruling off the plane into equal squares, the sides being parallel to the axes. In the figure, p. 18, several points arb plotted, the unit of length being assumed equal to one division on each axis. The method is simply this : Count off from along X'X 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 First off above or below the axis of x. (+ * +} ., Rectangular axes divide the plane into four * portions called quadrants ; these are numbered as in the figure, in which the proper signs of Second X' Third r Fourth 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), ( VI, 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 6 are positive numbers: (-a, 6)? (-a, -6)? (6, -a)? (a, 6)? 5. To what quadrants is a point limited if its abscissa is positive? nega- tive ? its ordinate is positive ? negative ? -. Plot the triangle whose vertices are (2, - 1), (- 2, 5), (- 8, - 4). 7. Plot the triangle whose vertices are ( 2, 0), (5 Vs - 2, 5), ( 2, 10). 8. Plot the quadrilateral whose vertices are (0, - 2), (4, 2), (0, 6), (-4,2). 20 ANALYTIC GEOMETRY 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 when its abscissa is zero? Where? Can it move when its ordinate is zero ? Where ? Can it move if both abscissa and ordi- nate are zero ? Where will it be ? 11. Where may a point be found if its abscissa is 2? if its ordinate is -3? 12. Where do all those points lie whose abscissas and ordinates are equal ? 13. Two sides of a rectangle of lengths a and b 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, y) and ( x, y) with respect to Y'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 VS), (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 origin ? CARTESIAN COORDINATES 21 17. Angles. The angle between two intersecting directed lines is defined to be the angle made by their positive directions. In the figures the angle between the directed lines is the angle marked 0. 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. Reversing the direction of either directed line changes to the supplement TT 0. If both directions are reversed, the angle is unchanged. 0=0 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). T Y r ^ \. B X' jS/K y y-' \0 "^^ | -^ -^*- \ /3 r' x x' o \A Y' X A (1) (2) (3) Theorem I. If a and ft are the angles between a line directed upward and the rectangular axes OX and OY, then (I) cos ft = sin a. Proof. The figures are typical of all possible cases. In Fig. 1, and hence /=!-, cos ft = cos / a \ = sin a. (by 6, p. 13) 22 ANALYTIC GEOMETRY In Fig. 2, ft = a - 1 9 I -A and hence cos J3 = cos ( a -- 1 = sin a. (by 4 and 6, p. 12) In Fig. 3, a = ^, ft = 0. .'. cos ft = 1 = sin a. 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. Hence for such a line a = 0, ft = > and the relation (I) still holds, since 7T cos 8 = cos = = sin = sin a. * PROBLEMS 1. Show that for lines directed downward cos /3 = sin a. 2. What are the values of a and 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 '_ a = 5 ; p + p = ?L. 2 2 18. Orthogonal projection. The orthogonal projection of a point upon a line is the foot of the perpendicular let fall from the point upon the line. Thus in the figure M is the orthogonal projection of P on X'X', M N is the orthogonal projection of P on FT; X' J/ - P f is the orthogonal projection of P' on X'X. If A and B are two points of a directed line, and 7kf and N their projections upon a second directed line CD, then MN is called the .projection of AB upon <7Z>. CARTESIAN COORDINATES 23 Theorem II. First theorem of projection. If A and B are points upon a directed line making an angle y with a second directed line CD, then the (II) projection of the length AB upon CD = AB cos y. Proof. In the figures let a = the numerical length of AB, I = the numerical length of AS or BT\ then a and I are 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. 11). A A B - T A > B A > B r Ns N _^\M $\~}MD ! C\M BK^JTIX^ ! c ^CT i . i i t ! "i> 5 r M N M N (1) (2) (3) (4) (5) C6) [M In Fig. 1, In Fig. 2, In Fig. 3, In Fig. 4, In Fig. 5, Hence In Fig. 6, Hence I = a cos BA S = a cos y, '. MN AB COS y. = E> 24 ANALYTIC GEOMETRY 19. Lengths. Consider any two given points Then in the figure MiMt = projection of PiP 2 on X'X, NiNz = projection of PiP a on YY. Y #i M z X' But by (1), p. 17, o Mi . ** (III) Hence Theorem HI. Given any two points P l (cc 1? y^), P 2 (# 2 , y 2 ) ; 2 &! = projection of JP^ F t on ! 1/1 = projection of P^f z on We may now easily prove the important Theorem IV. The length I of the line Y' joining two points P! (a^, 3^), P 2 (x 2 , y 2 ) ^ 2 is given by the formula ^ (IV) 1 = Proof. Draw lines through P! and x> PZ parallel to the axes to form the right triangle PiSP 2 . r Then SP, = (x\,yi t x PzS = yi - and hence (by III) (by III) Q.E.D. CARTESIAN COORDINATES 25 The method used in deriving (IV) for any positions of PI and P 2 is the following : Construct a right triangle by drawing lines parallel to the axes through P! and P 2 . The sides of this triangle are equal to the projections of the length P^P Z 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 line joining the points (1, 3) and ( 5, 6). Solution. Call (1, 3) P lt and (- 6, 5)P 2 . Then i = 1, y\ = 3, and Xa = - 5, y 2 = 5 ; and substituting in (IV), we have I = V(l + 5) 2 + (3 - 5) 2 = V40 = 2 VlO. 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 (IV) are true for all positions of the points P l and P 2 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 be 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 6, 7; length = V74. (b) (- V2, V3) and (\/3, V2). -4ns. Projections Vjj + \/2, V2 V ; length = VlO. 26 ANALYTIC GEOMETRY (c) (0, 0) and (-, - -V Ans. Projections -, - V3; length = a. \2 2 / 22 (d) (a + 6, c + a) and (c + a, 6 + c). Ans. Projections c - 6, b - a; length = v(6 - c) 2 +(a - b) 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, - 5), (- 1, - 8). (c) (a, 6), (6, c), (c, d). 3. Find the letfgths of the sides of the triangles in problem 2. 4. Work out formulas (III) and (IV), (a) if Xi = x z ; (b) if y\ y 2 . 5. Find the lengths of the sides 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 V3, - 2 Vjj) 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 lelogram. Also find the lengths of the diagonals. 10. Show that (11, 2), (6, - 10), (-6, - 5), (- 1, 7) are a square. Find its area. 11. Show that .the points (1, 3), (2, Ve), (2, - V6) the origin, that is, show that they lie on a circle with its cen and its radius VlO. 12. Show that the diagonals of any rectangle are equal. v 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). 15. 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 ? CARTESIAN COORDINATES 27 17. What equation expresses algebraically the fact that the point (z, y) is equidistant from the points (2, 3) and (4, 5)? ^18. If the angle XOY (Fig., p. 17) equals o>, show that the length of tha line joining PI(ZI, yi) and P 2 (x 2 , ?/ 2 ) is given by I = V(xi - z 2 ) 2 + (y l - y 2 ) 2 + 2 (X! - x 2 ) (yi - y 2 ) cos w, "^19. If w = -, find distance between the points ( 3, 3) and (4, 2). Ans. V39. 20. If w = , find the perimeter of the triangle whose vertices are (1, 3), 3 (2, 7), (-4, -4). Ans. 21. If w = -, find the perimeter of triangle (1, 2), (- 2, - 4), (3, - 5). 6 Ans. 3 Vs + 2 V3 + V26 - 5 V3 + Vs3 - 14 Vg. 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 (\/3, -x/2), (- \/2, V). Ans. VlO + V2. 24. Show that the sum of the projections of the sides of a polygon upon eitlfll axis is zero if each side is given a direction established by passing luously around the perimeter. ion and slope. The inclination of a line is the angle is of x and the line when the latter is given the upward direction (p. 21). The slope of a line is the tangent of its inclination. The inclination of a line will be denoted by a, a^ a zj a', etc. ; its slope > by w, m 1? m 2 , m', etc., so that m = tan a, m l tan a l9 etc. The inclination may be any angle from to TT inclusive (p. 21). The 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. 28 ANALYTIC GEOMETRY 00 Theorem V. The slope m of the line passing through two points ( x ii yi)> PI ( x iu 2/2) ** given by Vi Vi Proof. (1) Similarly, (2) = x 2 x l = PiP 2 cos or. cos a = # 2 Xi. &!&* = ^2 2/i = PjP 2 COS /3. COS /3 = 2/2 2/1- cos /? = sin a. But Hence, from (2), (3) PiPa sin a = 2/2 - 2/i- Dividing (3) by (1), tan a = m = g ~~ ! Remark. Formula (V) may be verified by constructing a right triangle whose hypot- enuse is PiP 2 , as on p. 24, whence tan a (= tan Z SPiP 2 ) is found directly as the ratio of the opposite side, SP 2 = y z y w to the adjacent side, PjS = x 2 x lt * (by (III), p. 24) (by (II), p. 23) (by (III), p. 24) (by (II), p. 23) (by (I), p. 21) Q.E.D. * To construct a line passing through a given point P^ whose slope is a positive frac- tion ^ , we mark a point S b units to the right of P^ and a point P, a units above S, and draw P,P 2 . If the slope is a negative fraction, | , then either S must lie to the left of P t or P, must lie below S. CARTESIAN COORDINATES 29 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 2 / 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, 6), (-4, 2) are the vertices of a rectangle, and hence, by (IV), of a square. ". . 30 ANALYTIC GEOMETRY 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, 5), (5, 5), (- 5, 5) are the vertices of a trapezoid. 11. Show that the line joining (a, 6) and (c, d) is parallel to the line joining (a, 6) and ( c, d). 12. Show that the line joining the origin to (a, 6) 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. 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) ? Ans. Ans. f- 18. What is the inclination of the line joining (3, 0) and (2, V3) ? Ans. ^- 19. What is the inclination of the line joining (0, - 4) and (- V, - 5) ? Ans. - 20. What is the inclination of the line joining (0, 0) and (- V, 1) ? 5* 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, b + c), (6, c + a), and (c, a + b) 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). CARTESIAN COORDINATES 31 21. Point of division. Let P x and P 2 be two fixed points on a directed line. Any third point on the line, as P or P', is said .PI P P 2 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 2 . The position of the point of division depends upon the ratio of its distances from Pj and P 2 . 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 P! and P 2 , then P is said to divide PiP 2 into the segments P X P P P and PP 2 . The ratio of division is the value of the ratio* - We shall denote this ratio by X } that is, If the division is internal, P X P and PP 2 agree in direction and therefore in sign, and A. is therefore positive. In external divi- sion X is negative. , The sign of X therefore indicates whether the point of division P is within or without the segment PiP 2 ; and the numerical value determines whether P lies nearer P^ or P 2 . The distribution of X is indicated in the figure. & ' ' -1<\<0 Ac=0 \>0 X = oo -oo / y l / f P f T -4ns. Hence Pis (- 2, - 8). Ex. 2. Find the coordinates of the point of intersection of the medians of a triangle whose vertices are (x l5 ?/i), (x 2 , y 2 ), (x s , y s ). Solution. By Plane Geometry we have to find the point P on the median AD such that AP = f AD, that is, AP : PD : : 2 : 1, or \ = 2. By the Corollary, D is [i (x 2 + x 3 ), \ (y 2 + ys)]. ^f *jyJ To find P, apply (VII), remembering that A corre- sponds to (xi, y\) and D to (x 2 , y 2 ). This gives x = y = 1 + 2 1 + 2 x = x 2 + X 3 ), y = i (2/1 + yz + 2/s)- 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. 1 and 2 and 3 d and 5 PROBLEMS Find the coordinates of the middle point of the line joining (4, 6) (- 2, - 4). Ans. (1, - 5). Find the coordinates of the middle point of the line joining (a + 6, c + d) (a - 6, d c). Ans* (a, d). Find the middle points of the sides of the triangle whose vertices ai ), (4, - 6), and (-3, 6) ; also find the lengths of the medians. Find the coordinates of the point which divides the line joining (-1,4) (-5, - 8) in the ratio 1:3. . Ans. (- 2, 1). Find the coordinates of the point which divides the line joining !, - 5) and (6, 9) in the ratio 2 : 6. Ana. (- $, - 1). 34 ANALYTIC GEOMETRY 6. Find the coordinates of the point which divides the line joining (2, 6) and ( 4, 8) into segments whose ratio is f . 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), (f, 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. 1. 16. In what ratio does the point (16, 3) divide the line joining the points (-5,0) and (2, 1)? Ans. - f. 17. Given the triangle whose vertices are ( 5, 3), (1, 3), (7, 5); 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). D.' The middle point of a line is (6, 4), and one end of the line is (5, 7). What are the coordinates of the other end ? Ans. (7, 1). > 21. The vertices of a triangle are (2, 3), (4, - 5), (- 3, - 6). Find the 4 coordinates of the point where the medians intersect (center of gravity). CARTESIAN COORDINATES 35 22. Find the area of the isosceles triangle whose vertices are (1, 5), (5, 1), (_ o ? _ 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 XOT may- have any value. 25. How far is the point bisecting the line joining the points (5, 5) and (3, 7) from the origin ? What is the slope of this last line ? Ans. 2 Vl3, f . 22. 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, PI(X I} y-^), and P 2 (# 2 , y?) is given by the formula (VIII) Area of triangle OP i P 2 = |(x l2 / 2 - a? 22/1 ). Proof. In the figure let = Z. XOP 2 , (1) By 18, p. 13, (2) Area A 1 . OP 2 sin (3) But in the figure OP l . OP 2 (sin ft cos a - cos /3 sin a). (by 9, p. 13) sin a = OP 2 OP 2 OP 1 OP 2 OP 2 , cos a = -- - 1 OP l OP l Substituting in (3) and reducing, we obtain Area A OP^ = % ( Xl y 2 - x&J. Q.E.D. 36 ANALYTIC GEOMETRY 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 P 2 . Then Substituting in (VIII), Then Area = 11 unit squares. If, however, the formula (VIII) is applied by denoting ( 2, 4) by P 2 , 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 19 P 2 , if the area is on the left, as in Fig. 1, then (VIII) gives a posi- tive result; if the area is on the right, as in Fig. 2, then (VIII) gives a negative result. Proof. In the formula (4) Area A OP^ = OP l - OP 2 sin $ the angle 6 is measured from OP l to OP 2 within the triangle. Hence is positive when the area lies P to the left in passing around the per- imeter 0, P 19 P 2 , as in Fig. 1, since is V then measured counter-clockwise (p. 11). But in Fig. 2, is measured clockwise. Hence is negative and sin in (4) is also negative. 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. (2) CARTESIAN COORDINATES 37 Theorem X. The area of a triangle whose vertices are PI(X I} y^, -Pa( 2/ 2 ), ^a (*3, 2/a) w given by (X) Area A P^P^P^ = \ (a^y, - # 2 2/i + # 2 2/3 - a? 3 y 2 + x z y v - o^y,). 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^P^. 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 Pff^ = A OP^ + A OP 2 P 3 + A OPf v since these areas all have the same sign. Fig. 2, origin without the triangle. By inspection, (6) Area A P,P 2 P 3 = A OPJ> % + A OP 2 P 3 + A OP 3 P since OP^P^ OP Z P 1 have the same sign, but OP 2 P 3 the opposite sign, the algebraic sum giving the desired area. By (VIII), A OP,P 2 = (XM - x&J, A OP 2 P 3 = J (^ 2 7/ 3 - x 3 y 2 ), A OPgPi = i (a?^ - ,y 8 ). Substituting in (5) and (6), we have (X). Also in (5) the area is positive, in (6) negative. Q.E.D. 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, i y\ abscissas in one, ordinates in the other, repeating the x 2 y coordinates of the first vertex. 1 y\ Second step. Multiply each abscissa by the ordinate of the next row, and add results. This gives x^y z -f X 2 y 8 4- x^. Third step. Multiply each ordinate by the abscissa of the next row, and add results. This gives y^x^ + yx 8 -f- y z 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). 38 ANALYTIC GEOMETRY 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. Second step. Multiply each abscissa by the ordinate of the next row, and add. This gives 1 x 3 + (- 1 x - 4) + (- 3 x - 2) + 2 x 6 = 25. Third step. Multiply each ordinate by the abscissa of the next row and add. This gives 6 x -1 + 3 x - 3 + (- 4 x 2) + (- 2 x 1) = - 25. Fourth step. Subtract the result of the third step from the result of the second step, and divide by 2. .. Area = = 25 unit squares. -4ns. 2 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. i. 2. Find the area of the triangle whsSe" 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, 6). Ans. ab. 6. Find the area of the triangle whose vertices are (0, 0), (xi, y\), (x 2 , yz). Ans. CARTESIAN COORDINATES 39 6. Find the area of the triangle whose vertices are (a, 1), (0, 6), (c, 1). 7. Find the area of the triangle whose vertices are (a, 6), (6, a), (c, c). Ans. 8. Find the area of the triangle whose vertices are (3, 0) , (0, 3 V), (6, 3 V3). Ans. 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, b -f c), (6, 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. 40 ANALYTIC GEOMETRY 23. Second theorem of projection. Lemma I. If M l} M 2 , M z are any three points on a directed line, then in all cases Proof. Let be the origin. By (1), p. 17, M^Mt = OM 2 - OM lf M Z M 3 = OM 8 - Adding, M^M Z -f M Z M 3 == OM Z But by (1), p. 17, J^A/g = OM 3 .'. M^M Z = MJI* -f Q.B.D. This result is easily extended to prove Lemma II. If MI, M z , M 8 , - - , M H _ 19 M n are any n points on a directed line, then in all cases 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. O M (1) \2) Thus in Fig, 1 the closing line is PiP 8 ; in Fig. 2 the closing line is PiP 6 - CARTESIAN COORDINATES 41 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. The proof results immediately from the Lemmas. For in Fig. 1 M^MI = projection of PiP 2 > M Z M 3 = projection of P 2 P 3 ; M^MS = projection of closing line PiP 8 . But by Lemma I Q.E.D. and the theorem follows. Similarly in Fig. 2. 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 the 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 A B = proj. of OM + proj. of MP = OMcos - + MP cos - 4 4 (by first theorem of projection, p. 23) Ana. r/ \ H / X X M, \ / \ \ / \ \ ^ \ x s / r \ ^ \ A s ^ / (-51 0) O M 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. 42 ANALYTIC GEOMETRY Ex. 2. Find the perpendicular distance from the line passing through o _, (4, 0), whose inclination is , to the point (10, 2). Solution. In the figure draw OC perpendicular to the given line AB. = , or 120. = 30, Z ^OF = 60. Required the perpendicular dis- tance RP. Project the broken line OMP upon 0(7. Then, by the second theorem of projection, {\ \ \ \ / ^ N^ f . j? \ , ,,' \ no 2) 7T 2 \ _^ - -^r" - ' -', f ^ _ - ^-~- _- 5 sf A V' TT, .!/ lY tf> \ ! proj. of OP = proj. of OM + proj. of MP = OMcos Z XOS + MP cos Z SOY P) = 1 + 5 VS. .But in the figure proj. of OP = OS + ST = OA cos JTOS + (2) = 4 ! V3 + RP. From (1) and (2), RP + 2 V = 1 + 5 V|. EP = 1 + 3 Vs. Ans. PROBLEMS 1. Four points lie on the axis of abscissas at distances of 1, 3, 6, and 10 respectively from the origin. Find P\P 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 a line passing through the point (1, 1) whose inclination is . Ans. 3 V3 CARTESIAN COORDINATES 43 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 it. Ans. V2. 7. Find the projection of the broken line joining the points (1, 4), (3, 6), and (5, 0) upon a line whose inclination is Verify your result by finding the projection of the closing line. Ans. Vi. 8. Find the projection of the broken line joining (0, 0), (4, 2), and (6, 3) 2 TIT A Q *\XQ upon a line whose inclination is ^- Ans ~ _. 8 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 through the point (4, 0) with an inclination of Ans. = 4 V2 11. Find the perpendicular distance from the point (5, 1) to a line passing through the point (6, 0) and having an inclination of it. Ans. 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 24. 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, f Analytic Geometry is peculiarly adapted to the solution of both parts of a locus problem. 25. 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 " will hereafter signify any continuous 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 i3 relatively unimportant. 44 THE CURVE AND THE EQUATION 45 Ex. 1. Find the equation in x and y if the point whose locus is required ,11 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. 24, (2) f ^ PJ. = V(x + 2) a + (y-Q) 8 , P = V(x + 3) 2 + (y - 8)2. Substituting in (1), V(x + 2)2 + (y - 0)2 = V(x + 3)* + (y - 8)2. Squaring and reducing, (3) 2 x - 16 y + 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, 7/1) be any point on the locus. Then P\A = Pi-B, by defi- nition. Hence, by formula IV, p. 24, (4) V(zi + 2)2 + yi a = V(zi + 3)2 + O/i - 8)2, or, squaring and reducing, (5) 2x 1 -16 2 / 1 + 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, 2/1) 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 of A and B, namely, z = -2i, y = 4 (Corollary, p. 32), satisfy (3), since 2 (-2) -16 x 4 + 69 = 0. This example illustrates the following correspondence 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 locus). and y representing coordinates (satisfied by the coordinates of every point on the locus). 46 ANALYTIC GEOMETRY 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 curve when and only when its coordinates satisfy the equation of the curve. 26. First fundamental problem. To find the equation of a curve which is defined as the locus of a point satisfying a given 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 P! (4, 1) and having an inclination of Solution. First step. Assume P(z, y) any point on the line. Second step. The given condition, since the incli- nation a. is i may be written 4- 1 (1) Slope of PiP = tan a = ll Third step. From (V), p. 28, NS \ 1 sQ w -)- \ 3 O \ <% i) PJ \ (2) Slope of PiP = tan a = 3/1 -2/2 Xi x* x 4 [By substituting (x, y) for (x lf y x ), and (4, Ti for (x 2 , THE CURVE AND THE EQUATION from (1), 47 x + y 3 = 0. Ans. To prove that (3) is the required equation : 1. The coordinates (x 1? ?/i) of any point on the line will satisfy (3), for the line joins (xi, 2/1) and (4, - 1), and its slope is 1 ; hence, by (V), p. 28, substituting (4, 1) for (x 2 , 2/2), or 3 = 0, and therefore Xi and 2/1 satisfy the equation (3). 2. Conversely, any point whose coordinates satisfy (3) is a point on the straight line. For if (xi, 2/1) is any such point, that is, if Xi + 2/1 3 = 0, then also 1 = - is true, and (xi, 2/1) is a point on the line passing through (4, 1) and having an inclination equal to r- m Q.E.D. Ex. 2. Find the equation of a straight line parallel to the axis of y and at a distance of 6 units to the right. Solution. First step. Assume that P(x, y) is any point on the line, and draw NP perpendicular to OF. Second step. The given condition may be written (4) NP = 6. Third step. Since NP = OM = x, (4) becomes (5) x = G. Ans. tt (x y) M X The equation (5) is the required equation : 1. The coordinates of every point satisfying the given condition may be substituted in (5). For if PI (xi, yi) is any such point, then by the given condition x\ = 6, that is, (xi, j/i) satisfies (5). 2. Conversely, if the coordinates (xi, yi) satisfy (5), then Xi = 6, and Pi(*i> 2/i) is at a distance of six units to the right of YY'. Q.E.D. The method above illustrated of proving that the derived equation has the two characteristic properties of the equation of the locus should be carefully studied and applied to each of the following examples. 48 ANALYTIC GEOMETRY PC = V(x + I) 2 + (y - 2)2. Substituting in (6), V(x + I) 2 + (y - 2)2 = 4. Squaring and reducing, 1.2) Ex. 3. Find the equation of the locus of a point whose distance from (1, 2) is always equal to 4. Solution. First step. Assume that P(x,y) is any point on the locus. Second step. Denoting ( 1, 2) by C, the given condition is (6) PC = 4. Third step. By formula (IV), p. 24, (7) 2, 2x _4y- 11 = 0. !/) X This is the required equation, namely, the equation of the circle whose center is (1, 2) and radius equals 4. The method of proof is the same as that of the preceding examples. PROBLEMS 1. Find the equation of a line parallel to OY and (a) at a distance of 4 units to the right. (b) at a distance of 7 units to the left. (c) at a distance of 2 units to the right of (3, 2). (d) at a distance of 5 units to the left of (2, 2). 2. What is the equation of a line parallel to OF and a 6 units from it ? How does this line lie relative toOI r ifa>&>0? if > 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' ? 5. 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 - b lie if a >'b > 0? if 6 > a > 0? 8. What is the equation of the axis of x ? of the axis of y ? THE CURVE AND THE EQUATION 49 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 = 5 ? 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 = 5 and x = 9 ? equidistant from y = 3 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 m = 3 ? Ans. 3x + y 3 = 0. (b) PI is ( 4, 2) and m \ ? Ans. x-3y 2 = 0. (c) P! is (- 2, 3) andm = -? Ans. V *y = *~8. (i) x = 2/2 + 2?y - 3. < w ) 4x = ^ + 8 ' (3)4x^7/3. ( X)?y== _^ (k) 4x = 2/ 3 - 1. l + x ' 2 (m) y = x3 - x. 1 + y 2 (n) 7y = x3-x2-5. ,, g= 2 (o) x 2 + y 2 = 4. 1 + 2/ 2 2. Show that the following equations have no locus (footnote, p. 52). (a) x 2 + 2/2 + i _ o. (f ) x 2 + 2/2 + 2 x + 2 y + 3 = 0. (b) 2x3 + 32/ 2 = - 8. (g) 4x 2 + 2/ 2 + 8x + 5 = 0. (c) x 2 + 4 = 0. (h) y 4 + 2 x 2 + 4 = 0. (d) x 4 + 2/' 2 + 8 = 0. (i) 9x 2 + 42/ 2 +18x + 8i/+15=0. (e) (x -H I) 2 + 2,2 + 4-0. (j) x 2 + xy + ?/ 2 + 3 = 0. Hint. Write each equation in the form of a sum of squares and reason as in the foot- x note on p. 52. 30. Principle of comparison. In Ex. 1, p. 53, and Ex. 3, p. 54, we can determine the nature of the locus, that is, discuss the equa- tion, by making use of the formulas (I) and (II), p. 51. The method is important, and is known as the principle of comparison. 56 ANALYTIC GEOMETRY The nature of the locus of a given equation may be 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 tfye second step for the values^ of the coefficients of the general equation. Ex. 1. Show that 2x 3y+6=0 is the equation of a straight line (Fig., p. 53). Solution. First step. Compare with the general equation (I), p. 61, (1) * y = mx + 6. 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 coefficients 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. 27, the slope of a line may have any real value whatever, and of course the ordinate b of the point (0, 6) 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 f . 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. THE CURVE AND THE EQUATION 57 Ex. 2. Show that the locus of (5) x 2 + y 2 + 6x-16 = is a circle (Fig., p. 54). Solution. First steg. Compare with the general equation (II), p. 51, (6) x 2 + 2/ 2/ - 2 ax - 20y + a 2 + p - r 2 = 0. The right-hand members of (5) and (6) agree, and also the first two terms, X 2 + y 2 . Second step. Equating coefficients of x, (7) - 2 a = Q. Equating coefficients of y, (8) -2 = 0. Equating constant terms, (9) 2 + 2 - r 2 = - 16. Third step. From (7) and (8), a = - 3, /3 = 0. Substituting these values in (9) and solving for r, we find r 2 = 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) 2 x -f y 6 = 0. Ans. m = 2, 6 = 6. (b) x - 3 y + 8 = 0. Ans. m = |, b = 2g. (c) x-f 2y = q -4ns. m = - i, 6 = 0. M T (d) 5 x 6 y 5 = 0. Ans. m = ', 6 = - $. (e) i x - f y - I = 0. .Ana. m = f, b =- ^. (f) ? - y - - 1 = 0. Ana. m = f , 6 = - 6. 5 o (g) 7 x - 8 y = 0. Ans. m = |, 6 = 0. (h) f x - f y - | = 0. Ans. m = f, 6 = - 1-&. 58 ANALYTIC GEOMETRY 2. Plot the locus of each of the equations following, and prove that the locus is a circle, finding the center (a, /3) and the radius r in each case. (a) x 2 + 2/2 _ i 6 = o. Ans. (a, 0) = (0, 0); r = 4. (b) x 2 + 2/ 2 - 49 = 0. Ans. (a, ft = (0, 0); r = 7. (c) x 2 + y 2 - 25 = 0. .4ns. (a, ) = (0, 0) ; r = 5. (d) x 2 + 2/ 2 + 4x = 0. .4ns. (a, /3) = (- 2, 0); r = 2. (e) x 2 + y 2 - 8 y = 0. 4ns. (a, 0) = (0, 4); r = 4. (f) x 2 + y 2 + 4x - 8y = 0. 4ns. (or, 0) = (- 2, 4); r = V20. (g) x 2 + y 2 - 6x + 4y - 12 = 0. Ans. (a, 0) = (3, - 2); r = 5. (h) x 2 + y 2 - 4x + 9y - | = 0. Ans. (a, /3) = (2, - f); r = 5. (i) 3x 2 + 3?/ 2 - 6x - 8y = 0. 4ns. (a, /3) = (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 . 2t -v 'Su. Ans. 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. -?vo 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 + 4j/ + 3 = 0, 6x + 4y-13=0. 6. A point moves so as to be 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 x 4 = and y + 5 = 0. Find the equation of the locus and plot. Ans. The perpendicular straight lines x-y-9 = 0, x + y + I = Q. 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 x 2 + y 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 x 2 + y 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 3x 2 + 32/ 2 -24x-4z/ = 0. I THE CURVE AND THE EQUATION 59 In the proofs of the following theorems the choice of the axes coordinates 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 is 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 he written (a, 0), (- a, 0), and if the "constant difference " be denoted by k, we find for the locus 4 ax = k or 4 ax = k. 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* 15. Draw the locus of 4 x 2 - 9 y* = 0. Solution. Factoring, (1) (2x Then, by the theorem, the locus consists of the straight lines (2) 2x-3y = 0, (3) 2x + 3y = 0. Proof. 1. The coordinates of any point (xi, yi) which satisfy (1) witt satisfy either (2) or (3). For if (xi, yi) satisfies (1), (4) 60 ANALYTIC GEOMETRY This product can vanish only when one of the factors is zero. Hence either 2x l -8y 1 = 0, and therefore (xi, ?/i) satisfies (2) ; or 2 xi -f 3 yi = 0, and therefore (xi, y\) satisfies (3). 2. A point (xi, ?/i) on either of the lines defined by (2) and (3) will also lie on the locus of (I). For if (xi, ?/i) is on the line 2 x 3 y = 0, then (Corollary, p. 46) (5) 2xi-32/i = 0. Hence the product (2 Xi 3 y{) (2 Xi -f 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) x 2 - y 2 = 0. (j) 3 x 2 + xy - 2 y* + 6 x - 4 y = 0. (b) 9 X* - y' 2 = 0. (k) x 2 - y 2 + x + y = 0. (c) x 2 = 9y 2 . (1) x 2 - xy + 5x - 5y = 0. (d) x- - 4x - 5 = 0. (m) x 2 -2xy + ?/ 2 + 6x - Gy = 0. (e) y i-Qy = 7. (n) x 2 - 4 y 2 + 5x -f 10 y = 0. (f) ?/ 2 -5x?/ + 6?/ = 0. (o) x 2 + 4x?/ + 4?/ 2 + 5x + Wy + 6 = 0. (g) xy - 2x 2 - 3x = 0. (p) x 2 + Zxy + 2?/ 2 + x + ?/ = 0. (h) xy - 2x = 0. (q) x 2 - 4x?/ - 5y 2 + 2x - 10 y = 0. (i) x?/ = 0. (r) 3 x 2 - 2 x?/ - ?/ -f 5x - 5 y = 0. 17. Show that the locus of -4.x 2 + Bx + C = is a pair of parallel lines, a single line, or that there is no locus according asA^-B 2 4 AC is positive, zero, or negative. 18. Show that the locus of ^.x 2 -f Bxy + Cy 2 = is a pair of intersecting lines, a single line, or a point according &s A = B 2 4 AC is positive, zero, or negative. . 31. 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 THE CURVE AND THE EQUATION Gl made in this way, however, since the nature of the curve between any two successive points plotted is not determined. This objection 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 (1) x2 + 42/ 2 = 16. Discuss the equation. Solution. First step. Solving for x, (2) x = 2 V4 - 2/2. 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 1* 2.7 -H 2 -2 imag. 3 imag. -3 if 3- {x. Discussion. 1. Equation (1) shows that neither x nor y can be indefi- nitely great, since x 2 and 4 y 2 are positive for all real values and their sum must equal 16. Therefore neither x 2 nor 4y 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 y 2 beneath the radical must not be negative. (2) also shows that x has values only from 4 to 4 inclusive. 62 ANALYTIC GEOMETRY 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)* + 4 ( - y) a = 16, replacing (x, y} by (x, - y) ; (4) (- x)2 + 4 (yy = 16, replacing (x, y) by (- x, y) ; (5) (- x)* + 4 (- 7/)2 = 16, replacing (x, y) 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 III, p. 52). 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 l$cus is symmetrical with respect to the axis of y, and from (5), the locus is symmetrical with respect to the origin. The locus is called an ellipse. Ex. 2. Plot the locus of (6) y2_4 x + i 5 = 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 = 0/ 2 + 15). X y 3f 4 1 4| 6 2 3 7! 10 4 5 12f 6 etc. etc. w -y A" Second step. Assume values for y and compute x. THE CURVE AND THE EQUATION 63 Iince y* only appears in the equation, positive and negative values of y the same value of x. The calculation gives the table on p. 62. j?'or example, if y = 3, then x = 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 <_,)- 4 <*) + 16 = 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 - 2 y - 4 = 0. Solution. First step. Solving for y, Second step. Compute y, assuming values for x. When x = 2, y = $ = oo. 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 x = 2 + -, y it is evident that x approaches 2 as y increases indefinitely. Hence the locus extends both upward and downward indefinitely far, approaching in each case the line x = 2. X y X y -2 -2 1 -4 -1 -1 If -8 -2 i It -16 -4 -f 2 CO -5 4 2.f 16 : 2 1 8 -10 -1 3 4 etc. etc. 4 2 5 f 6 1 12 0.4 etc. etc. 64 ANALYTIC GEOMETRY 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, - y), 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. This curve is called an hyperbola. Ex. 4. Draw the locus of the equation (10) 4y = x 3 . Solution. First step. Solving for y, X y x y 1 i -1 -i H H -H -H 2 2 -2 -2 2* 3ft -2* -8ff 3 6f -3 -6f 8* lOff -3| -lOff 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 = i = J = 3, etc. THE CURVE AND THE EQUATION 65 Ek ^ Third step. Plot the points thus found. Fourth step. The points determine the curve of the figure. Discussion. 1. From the given equation (10), a; 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 replacing (x, y) by (- x, y). Hence the locus is symmetrical with respect to the origin. The locus is called a cubical parabola. 32. 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, b) 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 by 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. 10). The locus of an algebraic equation in the variables x and y is called an algebraic curve. Then from Theorem IV follows 66 ANALYTIC GEOMETRY 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. 33. Further discussion. In this section we treat of three uidre 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. 46). 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 ofy 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 YY'. 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. THE CURVE AND THE EQUATION 67 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 ? 34. Directions for discussing an equation. Given an equation, ;he following questions should be answered in order before plot- ing the locus. 1. Is the origin on the locus ? (Theorem VI}. 2. Is the locus symmetrical -with respect to the axes or the ifjin? (Theorems IV and V). 3. What are the intercepts? (Rule, p. 66). 4. What values of x and y must be excluded? (Rule, p. 66). 5. Is the curve closed or does it pass off indefinitely far? ( 31, >. 61). Answering these questions constitutes what is called a general Iscussion of the given equation. Ex. 1. Give a general discussion of the equation (1) x 2 Draw the locus. (04) 1. Since the equation contains no constant term, 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 y = 0, we find x = 0, the intercept on the axis of x. Putting x = 0, we find y = and 4, the intercepts on the axis of y. 4. Solving for x, (2) " 68 ANALYTIC GEOMETRY Hence all values of y between and 4 must be excluded, since for such a value y 2 4 y is negative. Solving for y, (3) y = 2 $ Vx 2 + 16. Hence no value of x is excluded, since x 2 + 16 is always positive. 6. 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. (a) x 2 _ 4 y = o. (n) Qy 2 _ x s = o. < (b) 2/ 2 - 4x + 3 = 0. (o) 92/2 + x 8 = 0. (c) x 2 + 4 1/ 2 _ 16 = 0. (p) 2xy + 3x - 4 = 0. (d) 9x 2 + y 2 _ 18 = 0. (q) x2 - xy + 8 = 0. (e) x 2 - 4 ?/2 _ 16 = 0. (r) & + xy - 4 = 0. (f) x 2 -4y 2 + 16 = 0. (s) x 2 + 2xy -3y = 0. (g) x 2 - 2/2 + 4 = o. (t) 2xy - y* + 4x = 0. (h) x 2 - y + x = 0. (u) 3x 2 - y + x = 0. -4(i) xy - 4 = 0. (v) 4y 2 - 2x - y = 0. x (j) 9y + x 3 = 0. (w) x 2 - ?/2 + Q X = o. (k) 4x - y s = 0. J (x) x 2 + 4 y 2 -f 8 y = 0. (l)6x-y* = 0. (y) 9x2 + ?/ 2 + 18x- 6y = 0. (m) 5x - ?/ + 2/ 3 = 0. (z) 9x 2 - 2/ 2 + 18x + 6y = 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) j/2 = 2 mx. (f ) x 2 - y* = a 2 . (b) x 2 - 2 my = m 2 . (g) x 2 + y 2 = r 2 . x 2 y 2 _ (h) x 2 + y 2 = 2rx. W ^+52-- (i) x2 + 2/ 2 = 2r2/. (d) 2xy = a 2 . (j) & -f y 2 = 2ox + 26y. X 2 y2 (k) a2/ 2 - a* (e) &-&- (1) a^ = ^ 3 - * For example, in (a) and (b) m= is a special value. In fact, in all these examples zero ia a special value for any constant. I THE CURVE AND THE EQUATION 69 3. Draw the locus of the equation y = (x-a)(*-6)(x-c), (a) when a < 6 < c. (c) when a < 6, b = c. (b) when a = 6 < c. (d) when a = 6 = c. The loci of the equations (a) to (f ) in problem 2 are all of the class known as conies, or conic sections , curves following straight lines and circles in the matter of their simplicity. A conic section is the locus of a point whose distances from a fixed point and a fixed line are in a constant ratio. 4. Show that every conic is represented by an equation of the second degree in x and y. Hint. Take Y Y' to coincide with the fixed line, and draw XX' through the fixed point. Denote the fixed point by (p, 0) and the constant ratio by e. Ans. (1 - e 2 )x 2 + y 2 - 2px + p* = 0. 5. Discuss and plot the locus of the equation of problem 4, (a) when e = 1. The conic is now called a parabola (see p. 63). (b) when e < 1. The conic is now called an ellipse (see p. 62). (c) when e > 1. The conic is now called an hyperbola (see p. 64). 6. Plot each of the following. a** -6 = 0. = - = - A rf'i (b) x*y-y + 2x = 0. (f) . (c) xz/2 - 4x + 6 = 0. (g) y = ^f?. (k) 4x = x + 1 y2 - 9 35. 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. 46). 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 whose equa- tions are given. 70 ANALYTIC GEOMETRY 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 be 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) z-72/ + 25 = 0, (2) z2 + 2/2 = 25. Solution. First step. Solving (1) for x, (3) x = 1y- 25. . Substituting in (2), (7 y- 25)2 + ?/2 = 25. Eeducing, y 2 7 y + 12 = 0. /. y = 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 the 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) 2 x 2 + 32/2 = 35, (5) 3 x 2 4 y = 0. Solution. First step. Solving (5) for x 2 , (6) x 2 = f2A Substituting in (4) and reducing, 9 2/2 -f 8 y 105 = 0. .-. y = 3 and - - 3 /. Substituting in (6) and solving, x = 2 and i V-210. 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). THE CURVE AND THE EQUATION 71 PROBLEMS Find the points of intersection of the following loci. 2. . Ans. (6,1). - ' <0. ).(-* -I)- 8- xyt: 20 ^ nS< ( 5 ' 4) ' ( 4 ' 5) 10. I . For what values of 6 are the curves tangent ? y o X T- J G9\ /i 9\ ir = ^* j x 2 = 4 ay ^1 13. y= 8a3 L. Ans. (2 a, a), (-2a, a). 14. 2 9x, ^Irw. (8, 6), (8, - 6). " 2 (a, CM- 0,0, 72 ANALYTIC GEOMETRY X 2 V 2 X 2 V 2 17. The two loci - -- = I and -- 1- = 4 intersect in four points. Find the lengths of the sides and of the diagonals of the quadrilateral formed by these points. Ans. Points, ( VlO, f Ve). Sides, 2 VlO, 3 Ve. 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-5y + 34 = 0, 3x-2y + 1 = 0. Ans. 36. 19. x + 2y = 5, 2x + y = 7, y = x + 1. 4ns. f. 20. x + y = a, x-2y = 4a, y-x + 7a = 0. Ans. 12 a 2 . 21. x = 0, y = 0, x = 4, y =-6. 4ns. 24. 22. x-y = 0, x + y = 0, x-y = a, x + y = 6. 4ns. . 23. y = 3x-9, y = 3x + 6, 2y = x - 6, 2y = x + 14. 4ns. 56. 24. Find the distance between the points of intersection of the curves 3x-2y + 6 = 0, x 2 + y 2 = 9. 4ns. 25. Does the locus of y 2 = 4x intersect the locus of 2x + 3y + 2 = 0? 4ns. Yes. 26 . For what value of a will the three lines 3x + y 2 = 0, ax + 2y-3 = 0, 2x y 3 = meet in a point ? 4ns. a = 5. 27. Find the length of the common chord of x 2 -f y 2 = 13 and y 2 = 3 x + 3. 4ns. 6. 28. If the equations of the sides of a triangle are x + 7y-fll = 0, 3x + y 7 = 0, x 3y-fl = 0, find the length of each of the medians. 4ns. 2 V5, | V2, 1 VnO. Show that the following loci intersect in two coincident points, that is, are tangent to each other. 29. yt-Wx- 6y-31 = 0, 2y-10x = 47. 30. 9x 2 -4y 2 + 54x-16y + 29 = 0, 15 x - 8y + 11 = 0. 36. Transcendental curves. The equations thus far consid- ered have been algebraic in x and y, since powers alone of the variables have appeared. We shall now see how to plot certain so-called transcendental curves, in which the variables appeal- otherwise than in powers. The Eule, p. 53, will be followed. THE CURVE AND THE EQUATION 73 Ex. 1. Draw the locus of (1) y = logio x. 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 = 10". Hence values may also be assumed for y, and x computed by (2). This is done in the table. In plotting, unit length on XX' 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. X y X y 1 .1 - 1 3.1 * .01 -2 10 1 .001 -3 100 2 .0001 -4 etc. etc. etc. etc. 3. In (1), putting x = 0, y = log = oo = intercept on YY'. In (2), putting y = 0, x = 10 = 1 = intercept on XX'. 4. From (2), since logarithms of negative numbers do not exist, all nega- tive values of x are 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 ; BO we see that the curve extends down indefinitely and approaches nearer and nearer to YY'. 74 ANALYTIC GEOMETRY Ex. 2. Draw the locus of (3) y = sin x if the abscissa x is the circular measure of an angle (Chapter I, p. 12). Solution. Assuming values for x and finding the corresponding number of degrees, we may compute y by the table of Natural Sines, p. 14. 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 it in the table. For example, if X y x y Tt 6 ".50 x Tt ~ 6 -.50 Tt .86 Tt -.86 8 3 Tt 1.00 7T -1.00 2 ~2 27T .86 2tf -.86 3 3 5_7r .50 57r -.50 6 6 Tt - Tt Tt x= s' = sin- = sin 60 = .86. 3 2 Tt 2 Tt 9 7T" x= - , y = sin - _= -sin ~ (4, p. 12) = - sin 120= -sin 60 (5, p. 13) In plotting, three divisions being taken as the unit of length, lay off AO- OB = Tt = 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 Tt + x) = sin x. Hence y = sinx = sin(2^r + x), that is, the curve is unchanged if x + 2 it be substituted for x. This means, however, that every point is moved a distance 2 Tt to the right. Hence the arc Y. (0 ,1 ',> y =7 V ^ ^ 1 ^N s ^ --_: _-7T rl 7T, fl "1 . F ~0 7 / % ^ * i \ ^ y^ f/ ' ^ ^ \ < i / 7T j 7- J 7T 5J r 1 <^ TT * \ \ i 1 / s ^ K, ^ ^ / Ss ^ P //= -1 ^? APO may be moved parallel to XX' until -4 falls on J5, that is, into the position BBC, and ii will also be a part of the curve in its new position. THE CURVE AND THE EQUATION 75 Also, the arc OQB may be displaced parallel' to XX' until O 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), y = sin x, or ?/ = sm(-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. 65). 3. In (3), if x = 0, y = sin = = intercept on the axis of y. Solving (3) for x, (4) x = sin- 1 y. In (4), if y=0, x = sin- ! = nit, 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 it from one another. 4. In (3), x may have any value, since any number is the circular measure 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 -f 1 inclusive. 5. The curve extends out indefinitely along XX' in both directions, but is contained entirely between the lines y = + 1, y = 1. The locus is called the wave curve, from its shape, or the sinusoid, from its equation (3). PROBLEMS Plot the loci of the following equations. 1. y = cosx. 7. y = 21ogiox. i 2. y = tan x. 8. y = (1 + x)* 3. = secx. 9. y = sin2x. 4. y = sin- 1 x. 10. y = tan-- 5. y = tan~ 1 x. 11. y = 2cosx. 6. y = '2 x . / 12. y = sinx + cosx CHAPTER IV THE STRAIGHT LINE AND THE GENERAL EQUATION OF THE FIRST DEGREE 37. 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. 38. The degree of the equation of a straight line. It was shown in Chapter III (Theorem I, p. 51) that (1) y = mm + b is the equation of the straight line whose slope is m and whose intercept on the F-axis is b ; m and b may have any values, positive, negative, or zero (p. 27). 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 equation of any straight line is of the first degree in the coordinates x and y. 76 THE STRAIGHT LINE 77 39. The general equation of the first degree, Avc+By+C=Q. The equation (1) Ax+By + C = 0, where A 9 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 = mx + 6 may be written mx y + 6 = 0, which is of the form (l)itA = m,B = l, C= 6; and the equation x = constant may be written x constant = 0, which is of the form (1) if A = 1, B = 0, C = constant. Theorem II. (Converse of Theorem I.) The locus of the general equation of the first degree Ax + By -h C = is a straight line. Proof. Solving (1) for y, we obtain This equation has the same locus as (1) (Theorem III, p. 52). By Theorem I, p. 51, the locus of (2) is the straight line whose A C slope is m = and whose intercept on the F-axis is b = B B If, however, B = 0, it is impossible to write (1) in the form (2). But if B = 0, (1) becomes Ax + C = 0, or The locus of this equation is a straight line parallel to the Faxis (1, p. 50). Hence in all cases the locus of-(i) 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. 78 ANALYTIC GEOMETRY Corollary II. The lines Ax + By + C = am? A 'x + '?/ + C' = are parallel when and only when the coefficients of x and y are proportional; that is, A B For two lines are parallel when and only when their slopes are equal (Theorem VI, p. 29) ; that is, when and only when _A__A_ B ~ B'' Changing the signs and applying alternation, we obtain = A'~ B'' Corollary III. The lines Ax + By + C = and A'x + B'y + C" = are perpendicular when and only when AA' + B& = Q. 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. 29) ; that is, A _B' ~B-^>' or AA' + BB' = 0. Corollary IV. The intercepts of the line Ax + By -f C = o?i the X- and Y-axes are respectively C C a = -- - ana o = A .> For the intercept on the X-axis is found (p. 66) by setting y == and solving for x, and the intercept on the Y-axis has been found in the above proof. Corollaries I and IV are given chiefly for purposes of reference. In a numerical example the intercepts are found most simply by applying the general rule already given (p. 66) ; and the slope is found by reducing the equation to the form y = mx + 6, when the coefficient of x will be the slope. THE STRAIGHT LINE 79 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 Rule on p. 53. Theorem III. When two equations of the first degree, (3) Ax + By + C = and (4) A'x + B'y -\- C 1 = 0, have the same locus, then the corresponding coefficients are propor- tional; that is, A \ T>f /~v I 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, - = > (Corollary I, p. 77) and since they have the same intercept on the F-axis, I = j,> (CoroUary IV, p. 78) by alternation we obtain A B AB and hence 80 ANALYTIC GEOMETRY Ex. 1. Find the values of a and b for which the equations 2 ox + 2y 5 = and 4z-3?/ + 7& = will represent the same straight fine. Solution, These two equations will represent the same straight line if (Theorem III) . = 3'' there is no solution unless A T> /~y when there is an infinite number of solutions. The proof follows at once from Corollary II, p. 78, and Theorem III. PROBLEMS 1 . Find the intercepts of the following lines and plot the lines. (a) 2 x + 3 y = 6. Ans. 3, 2. ( b ) 7; + 7 = 1- -* Y + !f ' ^ ^** 2 4 - x v * y ->\ ~_ i / "' q ^ ~ o ( d ) T + -^: = !- ^ ns - 4 - 2 - (c) - | = 1. ^ns. 3, - 5. o 2. Plot the following lines. (a) 2z-32/ + 5 = 0. (c) ? + | = 1. J o (b) ? y- 5-4z = 0. (d) |-|=-1- 3. Find the equations, and reduce them to the general form, of the lines for which (a) TO = 2, b = - 3. Ans. 2 x y 3 = 0. (b) w =-.5,6= f. -4ns. z + 2y-3 = 0. (c) m = 1 , 6 = - f . .4ns. 4 x - 10 y 25 = 0. (d) a = -, 6 = - 2. ^ns. x - y - 2 = 0. 4 O ,_* (e) or = ^ , 6 = 3. -4ns. x + y - 3 = 0. Hint. Substitute in ?/ = mx + b. 82 ANALYTIC GEOMETRY 4. Find the number of solutions of the following pairs of equations and plot the loci of the equations. < a > {4 x + 6 1 + 9 = Q! An8 ' N S0luti n - (b) ( X ~ y = }' Ans. One. ' \x + y = l. (c) -I ~-4 ^ )IS ' An infinite number. 20 = 0. . Nosol tl on. 5. Plot the lines 2x-3y + 6 = and x - y = 0. Also plot the locus of (2x-3y + 6) + k(x - y) = for k = 0, 1, 2. 6. Select pairs of parallel and perpendicular lines from the following. 4. (b) - i 2 : 8x + y + 1 = 0. Ans. LI [L 3 :9x-3y + 2 = 0. (c) i 2 : 5 y + 2 x = 8. -4ns. L 2 JL L 3 . 7. Show that the quadrilateral whose sides are 2z 3y + 4 = 0, 3x-y-2 = 0, 4x-6yj-9 = 0, and 6a 2y + 4 = 0isa parallelogram. i 8. Find the equation of the line whose slope is 2 which passes through the point of intersection of y = 3x + 4 and y = x + 4. -4ns. 2x + y 4 = 0. 9. What is the locus of y = mx + 6 if 6 is constant and m arbitrary ? if m is constant and 6 arbitrary ? 10. Write an equation which will represent all lines parallel to the line (a) y = 2x + 7. (c) y - 3x - 4 = 0. (b) y = - x + 9. (d) 2 y - 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 3y = whose intercept on the F-axis is - 2. Ans. 2x-3y-6 = 0. 13. What is the locus of Ax -f By + C if B and C are constant and A arbitrary ? if A and B are constant and C arbitrary ? THE STRAIGHT LINE 83 41. 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. 77) (1) Ax + By + 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, if A = 2 B and C = 3 B, equation (1) becomes 2Bx + By 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. 66) ; or if the slope is to be 3, then = 3 (Corollary I, p. 77). 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. 81). 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 9 ^,'and C to which they give rise. 84 ANALYTIC GEOMETRY Rule to determine the equation of a straight line which satisfies two conditions. First step. Assume that the equation of the line is 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 terms of the third. Fourth step. Substitute the results of the third step in the equa- 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 P 2 (2, -2). Solution. First step. Let the required equation be (1) Ax + By 4- C = 0. Second step. Since PI lies on the locus of (1) (Corollary, p. 46), (2) 5 A B -f C = 0; and since P 2 lies on the line, (3) 2^-2^+07 = 0. Third step. Solving (2) and (3) for A and B in terms of C, we obtain A 1C 1 7? 3 f 1 A. = C, -D -5 O. Fourth step. Substituting in (1), - i Cx 4- f Cy + C = 0. Dividing by C and simplifying, the required equation is x 3y - 8 = 0. Ex. 2. Find the equation of the line passing through Pj. (3, - 2) whose slope is i. Solution. First step. Let the re- quired equation be (4) Ax + By 4- C = 0. Second step. Since PI lies on (4), (5) 3^4-2^4-0 = 0; and since the slope is J, PCS,- (6) 11 THE STRAIGHT LINE 85 Third step. Solving (5) and (6) for 4 and C in terms of B, we obtain Fourth step. Substituting in (4), or PROBLEMS * - w $" ' 1. Find the equation of the line satisfying the following conditions and plot the lines. (a) Passing through (0, 0) and (8, 2). Ans. x 4 y = 0. (b) Passing through ( 1, 1) and (-3, 1). Ans. y I = 0. Passing through (3, 1) and slope = 2. Ans. 2x 7/ + 7 = 0. Having the intercepts a = 3 and fr = - 2._ Ans. 2 x 3 y - G = 0. (e) Slope = 3, intercept on X-axis = 4. Ans. 3 x -f ?/ 12 = 0. (f ) Intercepts a = 3 and b = - 4. Ans. 4x + 3y + 12 = 0. (g) Passing through (2, 3) and ( 2, 3). Ans. 3x - 2 y = 0. (h) Passing through (3, 4) and (- 4, 3). Ans. x y + 1 = 0. (i) Passing through (2, 3) and slope = 2. 4ns. 2 x + y 7 = 0. (j) Having the intercepts 2 and 5. 4ns. - - = 1. 2 5 2. Find the equation of the line passing through the origin parallel to the line 2 x 3 y 4. Ans. 2 x - 3 y = 0. 3. Find the equation of the line passing through the origin perpendicular to the line 5x + y 2 = 0. Ans. x 5 y = 0. 4. Find the equation of the line passing through the point (3, 2) parallel to the line 4x-y-3 = 0. Ans. 4 x - y - 10 = 0. 5. Find the equation of the line passing through the point (3, 0) perpen- dicular to the line 2x + 2/ 5 = 0. 4ns. 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). 4ns. x + 3 y 15 = 0. 7. Find the equation of the line whose intercept on the X-axis is 3 which is parallel to the line x-4y + 2 = 0. 4ns. x-4y-3 = 0. 8. Find the equation of the line passing through the origin and through the intersection of the lines x 2y + 3 = and x + 2y 9 = 0. 4ns. x y = 0. 9. Find the equation of the straight line whose slope is ra which passes through the point PI (xi, y{). Ans. y y\ = m (x Xi). 86 ANALYTIC GEOMETRY 10. Find the equation of the straight line whose intercepts are a and b. Ans. + = l. a o 11. Find the equation of the straight line passing through the points andP 2 (x 2 , 2/2). . (y 2 - ?/i) x - (x 2 - xi) y + z 2 i/i - Xiy 2 = 0. 12. Show that the result of the last problem may be put in the form x-xi _ y - 3/1 OJ2 ~Xi 2/2 - Vl Hint. Add and subtract x t y lt factor, transpose, and express as a proportion. 42. 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 + b 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 P l (x^ y^) and has the slope m is /"TT\ / \ (\ ) y f/i = w(a? g^). Proof. First step. Let the equation of the given line be (1) Ax + By + C = 0. Second step. Then, "by hypothesis, (2) Ax l + %! + C = and (3) -| = m. Third step. Solving (2) and (3) for A and C in terms of B, we obtain A = mB and C = B (mx l y-^). THE STRAIGHT LINE 87 Fourth step. Substituting in (1), we have - mBx + By + B(mx l yi) = 0. Dividing by B and transposing, y 2/i = m(x i). Q.E.D. If P 1 lies on the F-axis, x l and y l = b, so that this equa- tion becomes y = mx -f b. 43. 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 respectively, then the equation of the line is Proof. First step. Let the equation of the given line be (1) Ax + By + C = 0. Second step. By definition of the intercepts (p. 66), the points (a, 0) and (0, b) lie on the line; hence (2) Aa + C = 0, (3) Bb + C = 0. Third step. Solving (2) and (3) for A and B in terms of C, we obtain . A = C and B = - C. x of OD and DP on ON. Then the con- B dition that P lies on AB is (1) proj. of OD on ON + proj. of DP on ON = p. By the first theorem of projection (p. 23) we have (2) proj. of OD on ON = OD cos = 'A 2 + B 2 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. , ^irst step. Find the numerical value of V A 2 + B 2 . Second step. Give the result of the first step the sign opposite to that of C, or. if C = ; the same sicjn as that of B^ Third 'step. Divide the given equation by the result of the second The result is the required equation. THE STRAIGHT LIXE 95 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. 92) 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. Ans. x = 5. (b) = , p = 3. Ans. y + 3 = 0. (c)w = -, p = 3. Ans. V2x + V2y-6 = 0. (d) a, = , p = 2. - Ans. z - V3 y + 4 = 0. 3 (e) u = , p = 4. Ans. V2 z - >/2 y - 8 = 0. 4 : . Reduce the following equations to the normal form and find p and (a) 3 x + 4 y 2 = 0. Ans. p = f , w = cos- 1 f = sin- 1 f . (b) 3 x 4 y 2 = 0. Ans. p = f , = COS~ 1 ( - , ti3 COS- * 1 ' 4 ^ \_V41/ 4. Find the perpendicular distance from the origin to each of the folio - ing lines. (a) 12 x + 5 y - 26 = 0. -4ns. 2. (b) x + y + 1 = 0. Ans. (c) 3x -2^-1 = 0. -4ns. 96 ANALYTIC GEOMETRY 5. Derive (VIII) when (a) - 4- y sin o> p = fo the point Pi(x ly y-^) is (IX) d = #! cos co + y sin (> p. Proof. Let ^45 be the given line and let ON be perpendicular to AB. By the second theorem of projection (p. 41) we have proj. of OP l on ON = proj. of OD on ON + proj. of DP l on ON. From the figure, proj. of OPi on 07V = OE p + c?. By the first theorem of projection (p. 23), proj. of OD on ON = OD cos to = a?! cos OD, proj. of DPi on = DPi cos ( - . Hence and therefore From this theorem we have at once the Rule to find the perpendicular distance from a p -f d = a^ cos o> 4- 2/i sin a>, " , d = x l cos -, = -4. What does the negative sign mean ? Ex. 2. Prove that the sum of the distances from the legs of an isosceles triangle to any point in the base is constant. Solution. Take the middle point of the base for origin and the base itself for the X-axis. Then the values of p for the two legs are equal and the values of w are supplementary. Hence, if the equation of one leg in normal form is x cos u -f y sin <* p = 0, then the equation of the*other leg is . x cos (it w) + y sin (it w) p = 0, or x cos w -f y sin w p = 0. JT' / "O / Y' Let (a, 0) be any point in the base. Then the distances from the legs to (a, 0) are respectively a cos a> p and a cos w j), so that the sum of these distances is 2 p, that is, a constant. PROBLEMS 1. Find the distance from the line (a) x cos 45 + y sin 45 - V2 = to (5, 7). (b) fx-$y-l = to (2,1). (c) 3 + 4y + 15 = to (-2, 3). (d) 2 x - 7 y + 8 = to (3, - 5). (e) x - 3 y = to <0, 4). . -2V2. Ans. -f. . 49 J.ns. V53 -4ns. 12 VTo 2. Do the origin and the point (3, 2) lie on the same side of the line . Yes. THE STRAIGHT LINE 3. Does the line 2x + 3y + 2 = pass between the origin and the point (-2, 3)? Ans. No. 4. Find the lengths of the altitudes of the triangle formed by the lines 2 z + 3 y = 0, x + 3y + 3 = 0, and x + y + 1 = 0. Ans. -4=5i - 1 and V. V13 V10 ' 5. Find the distance from the line Ax + By + C = to the point 6. Prove Theorem IX when . . 7. Find the locus of all points which are equally distant from 3x-4y + l = 0and4x + 3y -1 = 0.. Ans. 7 x y = and x + ly 2 = 0. 8. Find the locus of all points which are twjpe as far from the line 12x-f6y . 1 = as from the Y-axis. Ana. 14 x 5y + l = 0. 9. Find the locus of points which are k times as far from 4x 3 y + 1 = /as from 5 x - 12 y = 0. Ans. (52 25 k) x (39 - 60 k) y + 13 = 0. 10. Find the bisectors of the angles formed by the lines in problem 9. Ans. 77x-99y + 13 = and 27x + 21 y + 13 = 0. 11. Find the distance between the parallel lines, 8 ; 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 JT-axis parallel to one aide. 100 ANALYTIC GEOMETRY \ 16. Find the areas of the triangles formed by the following lines. (a) 2 x - 3 y + 30 = 0, x = 0, x + V = 0. Ans. 30. + y = 2, 3x + 4y - 12 = 0, x - y + 6 = 0. ^4ns. li. (c) 3x-4y + 12 = 0, x-3?/4- 6 = 0, 2x-y = 0. ^4ns. 3?. (d) z + 3y-3 = 0, 5 x - y - 15 = 0, x - y + 1 = 0. ^ins. 8. 17. Plot the following lines and find the area of the quadrilaterals of which they are the sides. (a) x = y, y = 6, x + y = 0, 3 x + 2 y - 6 = 0. Ans. 16f . (b) x + 2 y - 5 = 0, y = 0, x + 4y + 5 = 0, 2x + y-4 = 0. ^.ns. 18. (c) 2 - 4 y + 8 = 0, x + y = 0, 2x-y-4 = 0, 2x + y-3 = 0. Ans. 47. The angle which a line makes with a second line. The angle between two directed lines has been defined (p. 21) 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. 11) from the second line to the first line. Thus the angle which L makes with L 2 is the angle 0. We speak always of the " angle which one line makes with a second ______ J^r~ 2 t 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. 11 and p. 21. Theorem X. The angle 9 which the line L, : A& + B iy + d = makes with the line L 2 : A 2 x -f B 2 y + C 2 = is given by (X) tan *=4?f THE STRAIGHT LINE- icr Proof. Let a^ and a z be the inclinations of Z x and 2 respec- tively. Then, since the exterior angle of a triangle equals the sum of the two opposite interior angles, we have In Fig. 1, (*! = -f a 2 , or = i <**, In Fig. 2, a a = TT + a l9 or = TT -f (a x a a ). ' (2) And since (5, p. 13) tan (TT we have, in either case, ) = tan <, tan 6 = tan (o^ <* 2 ) tan 01 tan a tan a-L tan (by 13, p. 13) But tan ! is the slope of L 1 and tan a z is the slope of Z 2 ; hence (Corollary I, p. 77) _^l , ^2 14- Beducing, we get tan = Q.E.D. Corollary. T/* T/I! ane? w 2 are #Ae slopes of two lines, then the angle 6 which the first line makes with the second is given by tong= 102 ANALYTIC GEOMETRY Ex. 1. Find the angles of the triangle formed by the lines whose equations are M:Qx y - 6 = 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 i, so that M takes the place of LI in Theorem X and L of L 2 . Hence At = 6, J5i=-l; Then and hence A* = 2, B 2 = - 3. A 2 B l - tan^l = -2 + 18_16 12 + 3 ~ 15 B is the angle which L makes with N, and by Corollary III, p. 78, B = . C is the angle which N makes with M, so that if tanO = ^2 = 6, Bo = - 1. 24 + 6 30 15 tanC = 36n = 32 = 16' we must set Hence and We may verify these results. For if B = , then A = (7; and hence (6, p. 13, and 1, p. 12) tan A = true for the values found. tanC 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. 86) (1) y-BrriMx-S). THE STRAIGHT LINE 103 The slope of the given line is ra 2 = 1, and since the angle which (1) makes with the given line is , we have (by the Corollary), 3 it mi 1' tan- = - , 3 1 + ?ni 1+mi whence mi = - = (2 + V3).. 1-V3 Substituting in (1), we obtain y _ 5 = - (2 + V3) (x - 3) L or (2 + V3)x + y - (11 + 3V3) = 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=0 makes with 2x+y 2=0; also the angle which the second line makes with the first, and show that these angles are supplementary. 3 it TC ns ' ~7~' 7' 4 4 2. Find the angle which the line (a) 2 x 5 ?/ + 1 = makes with the line x 2 y + 3 = 0. (b) x -4- y 4- 1 = makes with the line x y + 1 = 0. (c) 3x 4y + 2 = makes with the line x + 3y 7 = 0. (d) 6x-3y + 3 = makes with the line x = 6. (e) x 7y + l = makes with the line x + 2y 4 = 0. In each case plot the lines and mark the angle found by a small arc. Ans. (a)tan-i(- T V); (b); (c)tan-i(V); (d) tan-i(- |) ; (e) tan- *(&). A 3. Find the angles of the triangle whose sides are x + 3y 4 = 0, 3x - 2 y + 1 = 0, and x - y + 3 = 0. Ans. tan-^- V), tan~i(i), tan~i(2). Hint. Plot the triangle to see which 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+3=0, y = 2, x-4y + 3 = 0. Ans. tan-!(5), tan~i(- ), tan~!(- - 1 /-). 5. Find one exterior angle and the two opposite interior angles of the triangle formed by the lines 2x-3y-6=0, 3x+4y-12 = 0, x-3y+6=0. Verify the results by formula 12, p. 13. 104 ANALYTIC GEOMETRY 6 . Find the angles of the triangle formed by3x+2y-4=0, a-3y+6=0, and 4 x 3y 10=0. Verify the results by the formula tan A + tan B + tan C = tan A tan B 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 - 3y + 2 = 0. Ans. 5x - y - 9 = 0. 4 (b) (1, - 3), -, x + 2y + 4 = 0. 4iw. 3x + y = 0. (c) (2, - 5), -, x + 3y - 8 = 0. .4ns. z - 2y - 12 = 0. (d) (xi, ^i), ,y = mx + b. Ans. y - y l = -^ ^- (x - Xi). L m tan (e) (xi, 2/1), 0, 4x + By + C = 0. Ans. y-y l = " (x - 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 ii:3x-4y-3 = and i 2 :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 LZ makes with it. Ans. 7x-7y + 9 = 0. 48. Systems of straight lines. An equation of the first degree in x and y 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 6 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). THE STRAIGHT LINE 105 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 tJie 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. 84. It has already been applied in solving Ex. 2, p. 102, 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 x 2 -f y 2 = 4 in but one point. Solution. First step. The equation represents the system of lines whose slopes are f (Theorem I, p. 51). Second step. The coordinates of the inter- section of the line and circle are found by solv- ing their equations simultaneously (Rule, p. 69). Substituting the value of y in the line in the equation of the circle, we have or 25x 2 + 24 bx + (16 ft 2 - 64) = 0. 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& 2 -64) = 0, whence Third step. Substitute these values of 6 in the equation of the first step. We thus obtain the two solutions and 106 ANALYTIC GEOMETRY PROBLEMS 1. Write the equations of the systems of lines defined by the following conditions. (a) Passing through (- 2, 3). fa) Having the slope |. . (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 X-axis = 6. (g) Having the slope ^. (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 x - 3 y + 4 k = 0. (b) kx-3y -1 = 0. I (c) x + y - k = 0. (d) x + k = 0. (e) x + 2 ky 3 = 0. (f ) 2fcc-3y + 2 = 0. (g) x cos a + y sin a + 5 = 0. Hint. Beduce 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 -3y + k = Q passes through (- 2, 1). Ans. k = 7. (b) the line 2 kx 5 y + 3 = has the slope 3. Ans. k - 1 /. (c) the line x -f y k = passes through (3, 4). Ans. k = 7. (d) the line 3z-4?/ + A; = Ohas intercept on X-axis 2. -4ns. k = 6. (e) the line x 3 ky + 4 = has intercept on F-axis = - 3. Ans. k = |. (f ) the line 4x 3y-f6fc = 0is distant three units from the origin. -4ns. k = f . 4. Find the equations of the straight lines with the slope T 5 ^ which cut the circle ic 2 -f y 2 = 1 in but one point. Ans. 5x + 12 y = 13. 5. Find the equations of the lines passing through the point (1, 2) which cut the circle x 2 + y 2 = 4 in but one point. -4ns. y = 2 and 4 z -f 3 y = 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 -f y 3 = 0. THE STRAIGHT LINE 107 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 3 x 4 y = 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. 49. 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) Ax + By + k = O, 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. 78). 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 P l (XD y^ through which it passes. If PI lies on (XI), then Ax -f- By^ + k 0; and hence k = Ax l BI/I. That is, the value of k may be chosen so that the locus of (XI) passes through any point" Pj. 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-3?/-4 = 0. Solution. Apply the Rule, p. 105. First step. The system of lines parallel to the given line is 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 108 ANALYTIC GEOMETRY 50. The system of lines perpendicular to a given line. Theorem XII. The system of lines perpendicular to the given Ax + By + C = Q is represented by (XII) Bx - Ay + k = O, where k 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. 78) A R BA =0. It remains to be shown that all lines perpen- dicular to the given line are represented by (XII). Any line perpendicular to the given line is determined by some point -Pi ( x i, ?/i) through which it passes. If P l lies on (XII), then whence k = A That is, the value of k may be chosen so that the locus of (XII) passes through any point P x . 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 Zi : 5 2 -f y sin 3 p 2 ) = is constant and equal to k. 112 ANALYTIC GEOMETRY Proof. Let PI (x l} y-^ be any point on L. Then x l cos 2 jt? 2 ) = 0, x-i cos a)! 4- y\ sin o>i #1 and hence k = - . a?! cos 2 ^> 2 . The numerator of this fraction is the distance from L^ to P lt and the denominator is the distance from Z 2 to PI (Theorem IX, p. 97). Hence k is the ratio of the distances from Z t and L 2 to any point on L. Q.E.D. Corollary. If k = 1, then L is the bisector of one of the angles formed by LI and L 2 . That is, the equations of the bisectors of the angles between two lines are found by reducing their equations to the normal form and adding and subtracting them. For when k = 1 the numerical values of the distances from L\ and Lz to any point of L are equal. The angle formed by L l and Z 2 i n which the origin lies, or its vertical angle, is called an internal angle of L l and L 2 5 and either of the other angles formed by L l and L 2 is called an external angle of those lines. From the rule giving the sign of the distance from a line to a point (p. 96) it follows that L lies in the internal angles of LI and L 2 when k is nega- f tive, and in the external angles when k is posi- /, tive. If the origin lies on L l or Z 2 , 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 y + 2 = and 3x 4y 2 = 0, without finding the point of intersec- tion, which (a) passes through the origin. (b) is parallel to5-2y + 3 = 0. (c) is perpendicular to3x 2y + 4 = 0. Ans. (a) 5x- 7y = 0; (b) 5x-2y -50 = 0; (c) 2z + 3y - 58 = 0. THE STRAIGHT LINE 113 2. Find the equations of the lirfes which pass flirough the vertwtes of the triangle formed by the lines 2x-3y + l = 0, x - y = 0, and 3x + 4y-2 = which are (a) parallel to the opposite sides. (b) perpendicular to the opposite sides. Ans. (a) 3x + 4y- 7 = 0, Ux-2ly+ 2 = 0, 17 x - 17 y + 5 = ; (b) 4x'-3y- 1 =.0, 21 x + 14 y- 10 = 0, 17x + 17y-9 = 0. 3. Find the bisectors of the angles formed by the lines 4x 3y 1 = and 3x 4y + 2 = 0, and show that they are perpendicular. Ans: 7x 1y + 1 = and x + y 3 = 0. 4. Find the equations of the bisectors of the angles formed by the lines 5x 12y + 10 = and 12 x 5y + 15 = 0. Verify the results by Theorem X. 5. Find the locus of a point the ratio of whose distances from the lines 4x -3y + 4 = 0and5x-f 12y-8 = 0isl3to5. Ans. 9x + 9y-4 = 0. ^ 6. Find the bisectors of the interior angles of the triangle formed by the lines 4 x 3 y = 12, 5x 12 y 4 = 0, and 12 x - 5 y 13 = 0. Show that they meet in a point. Ans. 7 x - 9 y - 16 = 0, 7 x + 7 y - 9 = 0, 112 x - 64 y - 221 = 0. 7. Find the bisectors of the interior angles of the triangle formed by the lines 5 x 12 y = 0, 5 x + 12 y + 60 = 0, and 12 x 5 y 60 = 0, and show that they meet in a point. Ans. 2 y + 5 = 0, 17 x + 7 y = 0, 17 x - 17 y - 60 = 0. 8. The sides of a triangle are 3x + 4y-12 = 0, 3x - 4y = 0, and 4x + 3y + 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 + 6 = and through the intersection of2x y + 3 = and x - 3y + 2 = 0. Ans. 19x + 3y + 26 = 0. Hint. The systems of lines passing through the points of intersection of the two pairs of lines are and 2ar y + 3 + k'(x 3y + 2)=0. These lines will coincide if (Theorem III, p. 79) 1-fc 2 + 6fc Letting p be the common value of these ratios, we obtain l + fc=2p + p*', l-k=- P -3pk', and From these equations we can eliminate the terms in pic* and p, and thus find the value of k which gives that line of the flrst system which also belongs to the second system. 114 ANALYTIC GEOMETRY 10. Find the equation of the line passing through the intersection of 2x + 5y 3 = and 3x 2y - 1 = and through the intersection of 6 = 0. Ans. 43z - 35y - 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. 11. Find the equations of the three diagonals of the complete quadrilateral formed by the lines x + 2y = 0, 3z-4y + 2 = 0, x y + 3 = 0, and 3x-2y + 4 = 0. Ans. 2x - y + 1 = 0, x + 2 = 0, 5x - Gy + 8 = 0. 12. Show that the bisectors of the angles of any two lines are perpen- dicular. 13. Find a geometrical interpretation of k in (XI) and (XII). 14. Find the geometrical interpretation of k in (XIII) when LI and L z 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. CHAPTER V THE CIRCLE AND THE EQUATION 2 + y z + I>3c +Ey + F= O 52. The general equation of the circle. If (a, ft) is the center of a circle whose radius is r, then the equation of the circle is (Theorem II, p. 51) (1) x 2 + y 2 - 2 ax - 2/3y + a? + ft 2 - r 2 = 0, or (2) (v-af + d, -?) = ,*. In particular, if the center is the origin, a = 0, ft = 0, and (2) reduces to (3) x z + y* = r\ Equation (1) is of the form (4) x 2 + if + Dx + Ey + F = 0, where (5) D = - 2 a, E = - 2 ft, and F = a 2 + f? - r 2 . Can we infer, conversely, that the locus of every equation of. the form (4) is a circle? By adding J D 2 + \ -E 2 to both members, (4) becomes (6) (x + %Dy+(y+iEy = i(D 2 + E 2 -F). In (6) we distinguish three cases: If D 2 -f- E' 2 4 F is positive, (6) is in the form (2), and hence the locus of (4) is a circle whose center is ( J Z), J E) and whose radius is r = Vz> 2 + E 2 4 F. If D 2 + E 2 &F = 0, the only real values satisfying (6) are x = ^D, y \E (footnote, p. 52). The locus, therefore, is the single point ( Z>, J E). In this case the locus of (4) is often called a point-circle, or a circle whose radius is zero. If D 2 + E 2 4 F is negative, no real values .satisfy (6), and hence (4) has no locus. 115 116 ANALYTIC GEOMETRY The expression D 2 + E 2 4 F is called the discriminant of (4), and is denoted by . The result is given by Theorem I. The locus of the equation (I) v? + y 2 + Vx + Ey + F = 0, whose discriminant is = D 2 + E z ~ 4 F, is determined as follows : (a) When is positive the locus is the circle whose center is (%D, %E) and whose radius is r = Vl> 2 + E' 2 4 F = % V. (b) When 1*5 zero the locus is the point-circle ( Z), ^ ). (c) When ts negative there is no locus. Corollary. When E = /i or 5 x 2 + 5 y 2 - 14 x + 7 y 24 = 0. The center is the point (|, T ^), and the radius is \ V29. PROBLEMS 1 . Find the equation of the circle whose center is (a) (0, 1) .a-nd whose radius is 3. (b) (- 2, 0) and whose radius is 2. (c) (- 3, 4) and whose radius is 6. (e) ( J 8 ) and whose radius is /3. * The radius is easily obtained, since V2 is the length of the diagonal of a square whose side is one unit. We may construct a line whose length is Vw by describing a semicircle on a line whose length is n + 1 and erecting a perpendicular to the diameter one unit from the end. The length of that perpendicular will be V. Ans. x 2 + y z 2 y 8 = 0. -4ns. x 2 + y' 2 + 4 x = 0. Ans. x 2 + y' 2 + 6 x - 8 y = 0. Ans. x 2 + y 2 2 ) passes through the points (0, 0), (8, 0), (0, - 6)., Ana. x 2 + / 2 _8x + 6y = 0. (c) passes through the points (4, 0), ( 2, 5), (0, - 3). Ans. 19x 2 + 192/ 2 + 2x- 47 y- 312 = 0. (d) passes through the points (3, 5) and (3, 7) and has its center on the JT-axis. Ans. x 2 + y 2 + 4 x - 46 = 0. (e) passes through the points (4, 2) and (6, 2) and has its center on the T-axis. Ans, x 2 + 2/ 2 + 5 y - 30 = 0. (f) passes through the points (5, 3) and (0, 6) and has its center on the line 2x-3y-6 = 0. Ans. 3 x 2 + 3 y 2 - 114 x - 64 y + 276 = 0. (g) has the center (1, 5) and is tangent to the X-axis. Ans. x 2 + 2/2 _|_ 2 x + 10 y + 1 = 0. (h) passes through (1, 0) and (5, 0) and is tangent to the F-axis. Ana. x 2 + ?/ 2 -6x;t2V5y + 5 = 0. (i) passes through (0, 1), (5, 1), (2, - 3). Ans. 2x 2 + 2?/ 2 -10x + y-3 = 0. (j) has the line joining (3, 2) and (7, 4) as a diameter. Ans. x 2 + y 2 + 4 x - 6 y 13 = 0. (k) has the line joining (3, - 4) and (2, 5) as a diameter. Ans. x 2 + y 2 _ 5 x + 9 y + 26 = 0. (1) which circumscribes the triangle formed by x 6 = 0, x + 2?/ = 0, and x - 2 y = 8. Ans. 2 x 2 + 2 y 2 _ 21 x + 8 y + 60 = 0. (m) passes through the points (1, 2), (- 2, 4), (3, - 6). Interpret the result by the Corollary, p. 89. (n) is inscribed in the triangle formed by4x + 3y 12 = 0, y 2 = 0, -10 = 0. Ana. 36x 2 + 36y 2 - 516x + QQy + 1585 = 0. 4. Plot the locus of x 2 + y 2 - 2x + 4 y + k = for k = 0, 2, 4, 5 - 2, - 4, 8. What values of k must be excluded ? Ans. k > 5. 120 ANALYTIC GEOMETRY 5. What is the locus of x 2 + y 2 + Dx + Ey + F if D and E are fixed and F varies ? 6. For what values of k does the equation x 2 + y 2 4 jc + 2 ky + 10 = have a locus ? Ans. k > -f Vo and A: < - V6. 7. For what values of fc does the equation x 2 + y 2 + kx + F=Q have a locus when (a) F is positive ; (b) F is zero ; (cj F is negative ? ^Ins. (a) fc > 2 \/F and fc < - 2 V* 1 ; (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 + (y - /3) 2 + 2 (x - a) (y ~ /3) cos <* = r 2 . 10. Write an equation representing all circles with ^ the radius 5 whose centers lie on the X-axis ; on the F-axis. 11. Find the number of values of k for which the locus of (a) x 2 + y 2 + 4 kx - 2 y + 5 k = 0, (b) x* + y* + 4'fce - 2y - k = 0, (c) x 2 + y 2 + 4 kx - 2 y + 4 fc = is a point-circle. ^4ns. (a) two; (b) none ; (c) one. 12. Plot the circles z 2 -f y 2 + 4 x - 9 = 0, z 2 + y 2 - 4 x - 9 = 0, and x z + y + 4x -9 + k(x* + y*-4x- 9) = for fc=l, 3, 1, -5, . Must any values of k be excluded ? 13. Plot the circles x 2 + y 2 + 4 x = 0, x 2 + y 2 - 4 x = 0, and x 2 + y 2 + 4 + fc(z 2 + y 2 4x) = for the values of k in problem 12. Must any values of k be excluded ? 14. Plot the circles x 2 + y 2 + 4 x + 9 = 0, x 2 + y 2 - 4 x + 9 = 0, and x 2 + ?/ 2 + 4x + 9 + fc(z 2 + y 2 - 4x + 9) = for k = - 3, - |, - 5, - |, ~~ 1 ~~ f ' ~ 1< What values of A; must be excluded? 54. Systems of circles. An equation of the form x 2 + if + Zta + y + F = will define a system of circles if one or more of the coefficients contain an arbitrary constant. Thus the equation ic 2 + ^ 2 _ r 2 = o 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. 110. THE CIRCLE 121 theorem II. Given two circles, C l :x* + y* + D^x + E 1 y + F l = Q Cz :x* + if + D 2 x + EJJ + F 2 = Oj \ the locus of the equation (II) x 2 + y* + D& + EM + J\ is a circle except when k = l. In this case the locus is a straight line. Proof. Clearing the parenthesis in (II) and collecting like terms in x and y, we obtain Dividing by 1 -4- k we have The locus of this equation is a circle (Theorem I, p. 116). If, however, k = 1, we cannot divide by 1 + k. But in this case equation (II) becomes (A - D 2 )aj + (E l -E 2 )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 C\ and C 2 . Q.E.D. Corollary I. The center of the circle (II) lies upon the line joining the centers of C x and C 2 and divides that line into seg- ments whose ratio is equal to k. For by Theorem I (p. 116) the center of C\ is PI ( - ~t - ^ ) and of Cz ia - \ *J ' -j / Pz(- -- - ) The point dividing P\P>2 into segments whose ratio equals Tc r-t + *(-^) -f +K- is (Theorem VII, p. 32) the point - , - - ^ 1 ~r" ft^ 1 "T~ A/ simplifying, ( - ff ~ f 2 )' wllteh iS th CeDter f (H) ' 122 ANALYTIC GEOMETRY Corollary II. The equation of the radical axis of C and C 2 is (D, -D 2 )x + (E, -Ez)y + (F, - F 2 ) = 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 t and C 2 (Theorem VII, p. 88) and show that it is perpendicular to the radical axis hy Corollary III, p. 78. The system (II) may have three distinct forms, as illustrated in the following examples. These three forms correspond to the relative positions of Ci and C 2 , 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 x 2 + y 2 + 8x - 9 + k(x* + y 2 - 4x - 9) = 0. Solution. The figure shows the circles z 2 + y 2 -f8z-9=:0 and z 2 + 2/ 2 -4z-9 = plotted in heavy lines and the circles corresponding to k = 2, 5, 1, i, - 4, - f, and - | ; 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 Y-axis. THE CIRCLE 123 Ex. 2. Plot the system of circles represented by x 2 + y 2 + 8x + k(x 2 + y 2 - 4x) = 0. X Solution. The figure shows the circles X 2 _j_ y i + 8 x = and x 2 -f y 2 - 4 x = plotted in heavy lines and the circles corresponding to k = 2, 3, |, 5, 1, i, - 7, i, - 4, - 3, and - \. These circles are all tangent to the given circles at their point of tangency. The locus for k = 2 is the origin. Ex. 3. Plot the system of circles represented by x 2 + y 2 - lOx + 9 + k (x 2 + y 2 -f 8x + 9) = 0. Solution. The figure shows the circles x 2 -f y 2 10 x + 9 = and x 2 + y' 2 + 8 x + 9 = plotted in heavy lines and the circles corresponding to * = it 17, i, - 10, - T V, and - -^. 124 ANALYTIC GEOMETRY These circles all cut the dotted circle at right angles (problem 7). For k= |- the locus is the point-circle (3, 0), and for k = 8 it is the point-circle ( 3, 0). In all three examples the radical axis, for which k = 1, is the F-axis. PROBLEMS 1. If Ci and C 2 intersect in PI and P 2 , the system (II) consists of all circles passing through PI and P 2 . 2. If Ci and C 2 are tangent, the system (II) consists of all circles tan- gent to Ci and <7 2 at their point of tangency. 3. The radical axis of two intersecting circles is their common chord, and of two tangent circles is the common tangent at their point of tangency. 4. Find the equation of the circle passing through the intersections of the circles x 2 + y 2 1 = and x 2 + y 2 + 2 x = which passes through the point (3, 2). Ans. 7 x 2 + 7 y 2 - 24 x - 19 = 0. 5. Two circles x 2 + y 2 + DIX + E^y + FI = and x 2 + y 2 + D 2 z + E z y 4>F 2 = intersect at right angles when and only when DiD 2 + EiE 2 - 2 FI - 2 F 2 = 0. Hint. Construct a triangle by drawing the line of centers and the radii to a point of intersection I\. 6. The equation of the system (II) may be written in the form x 2 + y 2 + k'x + F = 0, where Fis constant and k' arbitrary, if the axes of x arid y be respectively chosen as the line of centers and the radical axis of Ci and C 2 . 7. The system in problem C consists of all circles whose intercepts on the F-axis are V F if F<0, which are tangent to the F-axis at the origin if F = 0, and which intersect the circle x 2 + y 2 = F at right angles if P>0. 8. The square of the length of the tangent from PI (1,2/1) to the circle X 2 + y 2 + D X + Ey + F = is Xi 2 + y^ + Dx l + Ey + F. Hint. Construct a right triangle by joining P t and the point of tangency to the center. 9. The locus of points from which tangents to two circles are equal is their radical axis. 10. Find the radical axes of the circles x- + y 2 - 4x = 0, x 2 + y 2 + 6x 8 y ~ 0, and x 2 + y' 2 + G x - 8 = 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. Show that the radical axis of any pair of circles in the system (II) is the same as the radical axis of Oi and C 2 . 14. How may problem 11 be stated if the three circles are point-circles ? CHAPTER VI POLAR COORDINATES 55. 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 the vectorial angle. The yectorial angle is positive or negative as in Trigonometry (p. 11). The radius vector is positive if P lies on the terminal line of 0, and negative if P lies on that line produced through the pole O. 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 0; if negative, produce the 125 126 ANALYTIC GEOMETRY terminal line through the pole and lay off OP eqi,al to the numer- ical value of p. Then P is the required point. In the figure on p. 125 are plotted the points whose polar coordinates are Every point P determines an infinite number of pairs of numbers (p, ff). ___ The values of will differ by some mul- tiple of X, so that if is one value_of 6 the others will be of the form + k 7t, 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, it + <), (p, Unless the contrary is stated, we shall always suppose that is positive, or zero, and less than 2 TT ; that is, < 6 < 2 TT. PROBLEMS '"points (4, |),(0, ^),(_,,'),(4 I !),(-<, IE)'. (o, it). 2. Plot the points (G, ), (-2, f), (3, a), (-4, TT), (6, 0), (- 6, 0). 3. Show that the points (/, ff) and (p, 6) are symmetrical with respect to the polar axis. 4. Show that the points (p, 6), (- p, 6) are symmetrical with respect to the pole. 5. Show that the points ( p, it ff) and (/>, -0) are symmetrical with respect to the polar axis. 56. Locus of an equation. If we are given an equation in the variables p and 6, 'then the locus of the equation (p. 52) is a curve such that : 1. Every point whose coordinates (p, 0) satisfy the equation lies on the curve. 2. The coordinates of every point on the curve satisfy the equation. POLAR COORDINATES 127 The curve may be plotted by solving the equation for p and finding the values of p for particular values of 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 ws 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. 14 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 6 and = TT and solving for p. But other values of 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 axiVif, 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 6. Solution. The discussion enables us to simplify the plotting and is there- fore put first. 1. For = p = 10, and for = it p = 10. Hence the curve crosses the polar axis 10 units to the right of the pole. 2. The curve is symmetrical with cos(-0) = cos0 (4, p. 12). 9 p P 10 Tt It I 12 9.7 lit - 2.0 12 * - 6 8.Z 2* 3 - 5 It Sit _ 7 I 7 4 Tt 5 bit - 8.7 llTT - 9.7 bit 12 12 2.6 Tt -10 respect to the polar axis, for 128 ANALYTIC GEOMETRY 3. As cos is never infinite, the curve does not recede to infinity. Hence the curve is a closed curve. 4. No values of make p imaginary. Computing a table of values we obtain the table on p. 127. 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 2 = a 2 cos 2 0. Solution. The discussion gives us the following properties. 1. For = 0or7rp = 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 (20) = cos 2 (4, p. 12). 4. p does not become infinite. 5. p is imaginary when cos 2 is negative. cos 2 is negative when 2 is in e p 9 p ifc Tt +.7a 6 Tt 12 dh -93 a Tt 4 the second or third quadrant; that is, when STC 7t 7 it bit >20>- or >20> 222 2 Hence we must exclude values of such that 3x it . 7 it 5rt _>0>- a nd >0>-. 444 4 The accompanying table of values is all that need be computed when we take account of 2, 3, and 5. POLAR COORDINATES 129 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 P ~ 1 + cos ^ Solution. 1. For 6 = p = 1, and for 6 '= it p = oo ; so the curve crosses the polar axis one unit to the right of the pole. 2. The curve is not symmetrical with respect to the pole. How may this be inferred from 1 ? 3. The curve is symmetrical with respect to the polar axis, since cos(-0) = cos0 (4, p. 12). 4. p becomes infinite when 1 + cos 6 or cos 6 = - 1 and hence 6 = 7t. The curve recedes to infinity in but one direction. 5."^ is never imaginary. On account of 3 the table of values is computed only to 6 = 7t, and the rest of the curve is obtained from the symmetry with respect to the polar axis. The locus is a parabola. e p e p 1 lit 2.7 it 1.02 12 12 it 1.07 '~3~ 4 "6 it I 1.2 ^4~ 6.7 it 3 1.3 57T 6 14 5tf 12 1.6 \\it 50 12 it 2 2 it CO 5jr PROBLEMS Discuss and plot the loci of the following equations. 1. p = 10. = tan~ 1 l. 5. p sin = 4. 2p " *- * . p ^^ o. v * 3. p = 16 cos 0. 4. p cos = 6. 130 ANALYTIC GEOMETRY 8 1 - 2 cos 9. p = a sin 0. 10. p = a (I cose). 11. /3 a sin 20 = 16. 12. p 2 = 16sin20. 13. p 2 cos 2 20 = a 2 . 14. p = a sin 2 0. p = a cos 2 0. 8 16. p cos e = a sin 2 0. 17. />cos0 = a cos 20. 18. /> = a (4 -f 6 cos 0) for 6 = 3, 4, 6. 15. p = 1 e cos for e = 1, 2, 1 + tan 20. p = a sec & for a > 6, a = 6, a < 6. 21. p = ad. 22. p = a sin 30. /j = a cos 30. 23. Prove that the locus of an equation is symmetrical with respect to 6 = if the results of substituting \- and give equations which 2 22 differ only in form. 24. Apply the test for symmetry in problem 23 to the loci of 4, 5, 10, 11, and 12. 57. Transformation from rectangular to polar coordinates. Let OX and OF be the axes of a rectangular system of coordi- nates, and let O be the pole and OX the polar axis of a sys- tem of polar coor- dinates. Let (x, y) and (p, ff) be respec- ~* tively the rectan- gular and polar coor- dinates of any point P. It is necessary to distinguish two cases according as p is positive or negative. When p is positive (Fig. 1) we have, by definition, (2) cos = j sin = P P whatever quadrant P is in. Hence x = p cos 0, y p sin 9. POLAR COORDINATES 131 When p is negative (Fig. 2) we consider the point P' symmet- rical to P with respect to 0, whose rectangular and polar coordi- nates are respectively ( x, y) and ( p, 0). The radius vector of P', /o, is positive since p is negative, and we can therefore use equations (1). Hence for P' x = p cos 0, y = p sin 0; and hence for P x = p cos } y p sin 0, as before. Hence we have Theorem I. If the pole coincides with the origin and the polar axis with the positive X-axis, then where (x, ?/) are the rectangular coordinates and (p, &) the polar coordinates of any point. Equations I are called the equations of transformation from rec- tangular to polar coordinates. They express the rectangular coordinates of any point in terms of the polar coordinates of that point and enable us to find the equation of a curve in polar coordinates when its equation in rectangular coordinates is known, and vice versa. From the figures we also have * * 2 - 1 - + v* These equations express the polar coordinates of any point in terms of the rectangular coordinates. They are not as convenient for use as (I), although the first one is at times very convenient. Ex. 1. Find the equation of the circle x 2 + y 2 = 25 in polar coordinates. Solution. Substitute the values of x and y given by (I). This gives p 2 cos 2 e + p 2 sin 2 = 25, or (by 3, p. 12) p 2 = 25 ; and hence p = 5, which is the required equation. It expresses the fact that the point (/>, 6) is five units from the origin. 132 ANALYTIC GEOMETRY Ex. 2. Find the equation of the lemniscate (Ex. 2, p. 128) p 2 = a 2 cos 2 in rectangular coordinates. Solution. By 14, p. 13, we have /> 2 = a 2 (cos 2 0-sin 2 0). Multiplying by p 2 , p 4 = a 2 (p 2 cos 2 6 - p 2 sin 2 6). From (2) and (I), (z 2 + y 2 ) 2 = a 2 (x 2 - y 2 ). Ans. 58. Applications. Theorem II. The general equation of the straight line in polar coordinates is (II) p (A cos -f B sin ff) + C = 0, where A, J5, awd C are arbitrary constants. Proof. The general equation of the line in rectangular coordi- nates is (Theorem II, p. 77) Ax -f By + C = 0. By substitution from (I) we obtain (II).' Q.E.D. When -4 = the line is parallel to the polar axis, when B =. it is perpendicular to the polar axis, and when C = it passes through the pole. In like manner we obtain Theorem III. The general equation of the circle in polar coordi- nates is (III) p 2 + P (D cos + E sin ff) + F = 0, where D, E, and F are arbitrary constants. Corollary. If the pole is on the circumference and the polar axis passes through the center, the equation is p 2 r cos B = 0, where r is the radius of the circle. For if the center lies on the polar axis, or X-axis, E = (Corollary, p. 116) ; and if the circle passes through the pole, or origin, F = 0. The abscissa of the center equals the radius, and hence (Theorem I, p. 116) | D = r, or D = 2r. Substituting these values of Z>, E, and F in (III) gives p 2 r cos = 0. Theorem IV. The length I of the line joining two points P 1 (p 1} X ) and P 2 (/o 2 , 2 ) is given by (IV) e = p l t + pS-ap l p t (9 l -Oj. POLAR COORDINATES' 133 Proof. Let the rectangular coordinates of P x and P 2 be respec- tively (a?!, 2/x) and (x 2 , ?/ 2 ). Then by Theorem I, p. 131, #! = p! cos 61, x 2 = p 2 cos 2 , y! = pi sin 1} y z = p 2 sin 2 . By Theorem IV, p. 24, Z 2 = (*! - * 2 ) 2 + ( yi - 2/ 2 ) 2 , and hence I' 2 = (p x cos 1 p 2 cos 2 ) 2 + (p l sin ^ p 2 sin 2 ) 2 . Removing parentheses and using 3, p. 12, and 11, p. 13, we obtain (IV). Q.E.D. PROBLEMS 1. Transform the following equations into polar coordinates and plot their loci. (a) x 3 y = 0. Ans. 6 = tan- 1 ^. (b) 2/-f5 = 0. Ans. p = - m sin Q (c) x 2 + y' 2 = 16. Ans. p = 4. (d) x 2 + y 2 ax = 0. -4ns. p = a cos 0. (f) x 2 y* = a 2 . Ans. /o 2 cos 2 = a 2 . (g) x cos w + y sin w p = 0. Ans. p cos (0 - w) p = 0. (h) (1 - e 2 )x 2 + ?y 2 - 2 e 2 px - e 2 p 2 = 0. 1 e cos (i) 2 xy + 4 y 2 - 8z + 9 = 0. Ans. p 2 (sin 20 + 4 sin 2 0) - 8p cos -f 9 = 0. 2. Transform equations 1 to 21, p. 129, into rectangular coordinates. 3. Find the polar coordinates of the points (3, 4), (- 4, 3), (5, - 12), (4, 5). 4. Find the rectangular coordinates of the points ( 5, J, ( 2, -- J> (3, 7t). 5. Transform into rectangular coordinates p = - -- 1 - e cos 59. Equation of a locus. The equation of a locus may often be found with more ease in polar than in rectangular coordinates, especially if the locus is described by the end of a line of variable length revolving about a fixed point. The steps in the process of finding the polar equation of a locus correspond to those in the Rule on p. 46. 134 ANALYTIC GEOMETRY Ex. 1. Find the locus of the middle points of the chords of the circle C : p 2 r cos = which pass through the pole which is on the circle. Solution. Let P(p, 0) be any point on the locus. Then, by hypothesis, OP = $ OQ, where Q is a point on C. But OP = p and OQ = 2 r cos 0. Hence p = r cos 0. From the Corollary (p. 132) it is seen that the locus is a circle described on the radius of C through O as a diameter. Ex. 2. The radius of a circle is prolonged a distance equal to the ordinate of its extremity. Find the locus of the end of this line. Solution. Let r be the radius of the circle, let its center be the pole, and let P (p, 0) be any point on the locus. Then, by hypothesis, OP = OB + CB. But OP = />, OB = r, and CB = r sin 0. Hence the equation of the locus of P is p = r + r sin 6. The locus of this equation is called a cardioid. P(P,6) 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 -f cos 0). 3. Chords of the circle p = 2 a cos which pass through the pole are extended a distance 26. Find the locus of their extremities. Ans. p = 2 (6 + a cos 0). POLAR COORDINATES 135 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 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. " 5. A line is drawn from a fixed point O meeting a fixed line in PI. Find the locus of a point P on this line such that OPj OP = a 2 . Ans. A circle. 6. A line is drawn through a fixed point O meeting a fixed circle in Pj and PZ. Find the locus of a point P on this line such that r\~p o~p OP = 2 l ' 2 . Ans. A straight line. CHAPTER VII TRANSFORMATION OF COORDINATES 60. 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. 61. Translation of the axes. If the axes be moved from a first position OX and OF to a second position O'X' and O'Y' such that O'X' and O'Y' are respectively parallel to OX and OF, 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 (#, y) and (aJ f , y'). Projecting OP and 00'P,on OX, we obtain (Theorem XI, pi 41) x = x' + h. Similarly, y = y' + k. V Hence, Theorem I. If the axes be translated to a new origin (h, 7c), and if (x, y) and (x', y') are respectively the coordinates of any point P before and after the translation, then = K> +h, 136 A 7 MX TRANSFORMATION OF COORDINATES 137 TO Equations (I) are called the equations for translating the axes, 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 (x, y) lies on the given curve, and since equations (I) are true for all values of (x, y), the new equation gives a relation between x' and y 1 which expresses that P(x', y') lies on the curve and is therefore (p. 46) the equa- tion of the curve in the new coordinates. Ex. 1. Transform the equation when the axes are translated to the new origin (3, 2). Solution. Here h = 3 and k = 2, so equations (I) become x z x' + 3, y = y' 2. Substituting in the given equation, we obtain \ X' or, reducing, z /2 + y' 2 = 25. This result could easily be foreseen. For the locus of the given equation is (Theorem I, p. 116) 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. 51). 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. (a) 3z-4y = 6, (2, 0). Ans. 3z / -4y / = 0. (c) ?/ 2 -6z + 9 = 0, (3, 0). Ans. y^ = Qx'. (d) z 2 + 2/2_i = o, (-3, -2). Ans. (e) y 2 2 kx + k 2 = 0, ( - , V -4ns. (f) z2-4y 2 4-8z+24y-20=0, (-4,3). Ans. 138 ANALYTIC GEOMETRY 3. Derive equations (I) if O' is in (a) the second quadrant ; (b) the third quadrant ; (c) the fourth quadrant. 62. Rotation of the axes. Let the axes OX and OF be rotated about through an angle B to the positions OX' and OF'. The equations giving the coordinates of any point referred to OX and OF in terms of its coordinates referred to OX' and OF' are called the equations for rotating the axes. Theorem II. The equations for rotating the axes through an anyle 6 are - p C^ = ac f cos 9 y* sin 0, [y = a?' sin + y' cos 9. Proof. Let P be any point whose old and new coordi- ^ nates are respectively (x, y) x and (x', y 1 ). Draw OP and draw PM ' perpendicular to OX'. Project OP and OM'P on OX. The proj. of OP on OX = x. (Theorem III, p. 24) The proj. of OM' on OX = x' cos 0. (Theorem II, p. 23) / \ The proj. of M'P on OX = y'cos (| + I (Theorem II, p. 23) \ / = -y'sin0. (by 6, p. 13) Hence (Theorem XI, p. 41) x = x' cos 6 y' sin 0. In like manner, projecting OP and OM'P on OF, we obtain j \ y = x' cos ( ) + y' cos \. / = x' sin0 + ?/'cos0. Q.E.D. If the equation of a curve in # and y is given, we substitute from (II) in order to find the equation of the same curve referred to OX' and OF'. TRANSFORMATION OF COORDINATES 139 Ex. 1. Transform the equation x 2 - y 2 = 16 when the axes are rotated $K , through Solution. Since 4 2 7f 1 and cos = - 4 V2 equations (II) become - V5 Substituting in the given equation, we obtain X or, simplifying, x'y' + 8 = 0. PROBLEMS 1. Find the coordinates of the points (3, 1), ( 2, 6), and (4, 1) when the axes are rotated through 2 2. Transform the following equations when the axes are rotated through the indicated angle. Plot both pairs of axes and the curve. ft , 4 ?/' = 0. Ans. x' 2 = 4. Ans. x' 2 = 4 y'. (b) x 2 + (c) y 2 = (d) x 2 + (e) x 2 + y 2 = r~, 6. Ans. ff ) x 2 4- 2 xy + y 2 -f 4 x 4 y = 0, . Ans. 4 3. Derive equations (II) if 6 is obtuse. 63. 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. 140 ANALYTIC GEOMETRY Theorem III. If the axes be translated to a new origin (A, k) and then rotated through an angle 0, the equations of the transforma- tion of coordinates are /j-rjv' f a? = a?' cos t/ f sin + h, \y = x 1 sin -f y' cos -f k. Proof. To translate the axes to O'X" and O'Y" we have, by (I), x = #* -f 7*, y = y* + &> where (#*, ?/') are the coordi- nates of any point P referred to O'X* and O'Y*. To rotate the axes we set, by (ii), x* = je"cos 7/ f/ sin 0, yt = a "sin + y'cos 0. Substituting these values of x* and ?/*, we obtain (III). Q.E.D. 64. Classification of loci. The loci of algebraic equations (p. 10) 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. Proof. Since equations (III) are of the first degree in x' and ?/', 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 to 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 $' and y' the results are of the first degree in x and y. TRANSFORMATION OF COORDINATES 141 65. Simplification of equations by transformation of coordi- nates. The principal use made of transformation of coordinates to discuss the various forms in which the equation of a curve y be put. In particular, they enable us to deduce simple forms 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) J and collect like powers of x' and y'. Second step. Set equal to zero the coefficients of two terms obtained in the first step which contain h and k (or one coeffi- cient containing ff). Third step. Solve the equations obtained in the second step for h and k* (or 0). Fourth step. Substitute these values for h and k (or 0) 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. Simplify the equation y 2 8x + 6 y + .17 = by translating the axes. Solution. First step. Set x = x' + h and y = y' + k. This gives (y f + fc) - 8 (a/ + h) + 6 (y f + k) + 17 = 0, or (1) 6 + A: 2 -Sh + 6* + 17 t = 0. * It may not be possible to solve these equations (Theorem IV, p. 81). t These vortical bars play the part of parentheses. Thus 2 k + 6 is the coefficient of y' and !; 8/* + 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. 142 ANALYTIC GEOMETRY Second step. Setting the coefficient of y' and the constant term, the only coefficients containing h and k, equal to zero, we obtain (2) 2fc+6 = 0, Third step. Solving (2) and (3) for h and Jk, we find k = -3, h = l. Fourth step. Substituting in (1), remember- ing that h and k satisfy (2) and (3), we have The locus is the parabola plotted in the figure which shows the new and old axes. Ex. 2. Simplify z 2 + 4y 2 -2z 16 y + 1 = by translating the axes. Solution. First step. Set z = z' -f A, y = y' + k. This gives Y' ^Y ^f .X / / r / / / x / O (1 -3) x' \ \ \ \ \ s ^ ^ [ (4) -2 -16 A 2 =0. -2h -16k + 1 Second step. Set the coefficients of x' and y' equal to zero. This gives 2/i-2 = 0, 8&-16 = 0. Third step. Solving, we obtain h = 1, A; = 2. Fourth step. Substituting in (4) , we obtain Plotting on the new axes, we obtain the figure. Ex. 3. Remove the zy-term from x 2 + 4 xy + y 2 4 by rotating the axes. Solution. First step. Set x = x' cos Q y' sin 6 and y = x' sin -f y" cos 0, whence COS 2 + 4 sin cos + sin 2 6 2 sin 6 cos + 4 (cos 2 - sin 2 0) + 2 sin cos 4sin0cos0 -f cos 2 7/ 2 = 4, or, by 3, p. 12, and 14, p. 13, (5) (1 + 2sin20)a/ 2 + 4 C os20- x'y' +'(1 - 2 sin 2 0) y' 2 = 4. TRANSFORMATION OF COORDINATES 143 Second step. Setting the coefficient of xfy' equal to zero, we have cos20 = 0. -* ' Y Third step. Hence Fourth step. Substituting in (5), we obtain, since sin = 1 (P- 14), The locus of this equation is the hyperbola plotted on the new axes in the figure. / 5 From cos 2 = we get, in general, 2 6 = h mt, where n is any positive 2 or negative integer, or zero, and hence 6 = h n Then the xy-term may be removed by giving 6 any one of these values. For most purposes we choose the smallest positive value of as in this example. Ex. 4. Simplif /" ic^ "I" 6 iK^ 1 i o /j* ., 4- ?/ &oZuion. + 4 = by First step. translating the axes. Set ~h x = x' + h, y = y / + Jc. We obtain x' 2 + 3 A 2 + 12 h + 12 sp. Set equ > constant t( 3^ + ( . Solving, x' 4 ?/' + h s + 6 A 2 + 12 h + 4 al to zero the coefl rm. This gives , = 0, - 4 k + 4 = 0. ^0, (6) z'3 + 3/i 1 + 6 5cient I / / r~ > o / i)' HJrij i t P h 8 4 / - Third step / ff p. Substituting in (6), we obtain Fourth ste whose locus is the cubical parabola in the figure. 144 ANALYTIC GEOMETRY PROBLEMS 1. Simplify the following equations by translating the axes. Plot both pairs of axes and the curve. (a) x 2 + 6x + 8 = 0. Ans. x' 2 = 1. (b) x 2 - 4 y + 8 = 0. Ans. x'* = 4 y'. (c) x 2 + ?/ 2 + 4 x - 6 y - 3 = 0. ^ins. x' 2 -f y' 2 = 16. (d) ?/ 2 -6x-10y + 19 = 0. Ans. y' 2 = 6x'. (e) x 2 - y 2 + 8x - 14 y - 33 = 0. Ana. x' 2 - y /2 = 0. (f) x 2 + 4 y 2 - 16 x + 24 y + 84 = 0. 4na. x' 2 + 4 y' 2 = 16. (g) y 3 + 8 x 40 = 0. Ans. 8 x' + ?/' 3 = 0. (h) x 3 - y 2 + 14 y - 49 = 0. ^Ins. ?/ /2 = x /3 . (i) 4x 2 -4xy4- y 2 -40x+ 20y+99 = 0. Ana. (2x' - y')* - 1 = 0. 2. Remove the xy-term from the following equations by rotating the axes. Plot both pairs of axes and the curve. (a) x 2 - 2 xy + y 2 = 12. Ans. y /2 = 6. (b) x* - 2 xy + y 2 + 8x -f 8 y = 0. Ans. V2 y' 2 + 8x' = 0. (c) xy = 18. .4ns. x' 2 - y' 2 = 36. (d) 25 x 2 + 14 xy -f 25 y 2 = 288. Ans. 16 x /2 -f 9 y /2 = 144. (e) 3x 2 - lOxy -f 3 y 2 = 0. Ana. x /2 - 4 y /2 = 0. (f ) 6 x 2 + 20 V3 xy + 26 y 2 = 324. ^Ins. 9 x /2 - y /2 = 81. 66. Application to equations of the first and second degrees. In this section we shall apply the Eule of the preceding section to the proof of some general theorems. Theorem V. By moving the axes the general equation of the first degree, Ax + By + C = 0, may be transformed into x' = 0. Proof. Apply the Eule on p. 141, using equations (HI). Set x = x' cos 6 y 1 sin + h, y x' sin 6 + y 1 cos + k. This gives (1) A cos e + B sin x' A sin + B cos + Bk + C = 0. TRANSFORMATION OF COORDINATES 145 Setting the coefficient of y' and the constant term equal to zero gives (2) A sin 6 + B cos = 0, (3) Ah + Bk + C = 0. From (2), tan0 = -> or = tan-M- From (3) we can determine many pairs of values of h and k. One pair is h= -> k = 0. Substituting in (1) the last two terms drop out, and dividing by the coefficient of x' we have left x' = 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 A'-axis. This theorem is evident geometric- ally. For x 1 = 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 x' = is a straight line. Theorem VI. The term in xij may always be removed from an equation of the second degree, Ax* + Bxy + Cy a + Dx + Ey + F = 0, by rotating the axes through an angle such that (VI) tan20 = A-C 146 ANALYTIC GEOMETRY Proof. Set and This gives sin cos -f C sin 2 x = x' cos y f sin y = x' sin -f y' cos 0. 2J. sin cos -}-.B(cos 2 sin 2 0) + 2 C sin cos B sin cos + C cos 2 4- D cos -f E sin x 1 D sin + cos y' + .F = 0. Setting the coefficient of x'y' equal to zero, we have (C A) 2 sin cos + B (cos 2 sin 2 0) = 0, or (14, p. 13), (C - A) sin 2 + B cos 2 = 0. 7? Hence tan 2 = 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 by rotating the axes the constant term is unchanged unless the new equation is multiplied or divided by some constant. For the constant term in (4) is the same as that of the given equation. Theorem VII. The terms of the first degree may be removed from an equation of the second degree, Ax* + Bxy + C?/ 2 -f- Dx + Ey -f F = 0, . by translating the axeSj provided that the discriminant of the terms of the second degree, A = B 2 4 A C, is not zero. Proof. Set x = x' + h, y = y' + k. This gives (5) Ax' 2 -f Bxy -f + 2.4A + & x' + 5/i, + 2C& -f # y + Bhk + Dh + Ek = 0. TRANSFORMATION OF COORDINATES 147 Setting equal to zero the coefficients of x' and y 1 , we obtain '(6) 2Ah + Bk + Z> = 0, (7) Bh + 2 Ck + E = 0. These equations can- be solved for h and k unless (Theorem IV, p. 81) 2A _ B B '"2C* or B* - 4 A C = 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 be transformed by translating the axes, the coefficients of the terms of the second degree are imchanged 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 &ero the locus of an equation of the second degree has a center of symmetry. For if the terms of the first degree he removed the locus will be symmetrical with respect to the new origin (Theorem V, p. 66). If A= B 2 4.4(7=0, equations (6) and (7) may still be solved for h and k 2A B D if (Theorem IV, p. 81) - = ^ = > when the new origin (h, k) may be any x> 2 C lit 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 2 + 4a;y + 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 h = 2, k = 0. For these values the given equation reduces to The locus consists of two parallel lines and evidently is symmetrical with respect to any point on the line midway between those lines. 148 ANALYTIC GEOMETRY MISCELLANEOUS PROBLEMS 1. Simplify and plot. (a) y 2 - 5 y + 6 = 0. .;(e) x 2 + xy + y 2 = 8. -' (b) x 2 + 2xy + y 2 - 6x - Qy + 5 = 0. (f) x 2 - 9y 2 - 2x - 36y + 4 = 0. (c) 7/ 2 + 6x-10y + 2 = 0. (g) 25t/ 2 -16x 2 + 50y-119 = 0. _J(d) x 2 + 4y 2 - 8x-16y = 0. (h) x 2 -f 2xy + y 2 - 8x = 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 (^, k) must we translate the axes to transform (1 - e 2 ) x 2 + y 2 2px + p 2 = into (1 - e 2 ) x 2 + y 2 2 e^px - e*p~ = ? 4. Simplify the second equation in problem 3. 5. Derive from a figure the equations for rotating the axes through -f - Tt ~ and -- 5 and verify by substitution in (II), p. 138. 2 6. Prove thnt every equation of the first degree may be transformed into y' = by moving the axes. In how many ways is this possible ? 7. The equation for rotating the polar axis through an angle is = 6'+<}>. 8. The equations of transformation from rectangular to polar coordi- nates, when the pole is the point (A, k) and the polar axis makes an angle of with the X-axis, are x = h + pcos(0 + 0), y k + p sin (0 + 0). 9. The equations of transformation from rectangular coordinates to oblique coordinates are x = x' + y' cos w, y = y' sin o>, if the JT-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 \f/) x = x J -- h y - - > sin a) BID. w sin (a sm w where w is the angle between OX and OF, is the angle from OX to OX', and ^ is the angle from OX to OY'. CHAPTER VIII CONIC SECTIONS AND EQUATIONS OF THE SECOND DEGREE 67. 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) ep 1 ecosO where e is the eccentricity and p is the distance from the directrix to the focus. Proof. Let P be any point on the conic. Then, by definition, FP _ ~ EP From the figure, FP = p and EP = HM = p -f p cos 0. Substituting these values of FP and n EP, we have or, solving for p, p cos 6 1 6 COS Q.E.D. * Because these curves may be regarded as the intersections of a cone of revolution with a plane. 149 150 ANALYTIC GEOMETRY From (I) we see that 1. A conic is symmetrical with respect to the principal axis. For substituting e f or 6 changes only the form of the equation, since cos ( 6) = cos 0. 2. In plotting, no values of need be excluded. The other properties to be discussed (p. 127) show that three cases must be considered according as e = 1. The parabola e = 1. When e = 1, (I) becomes p ~ 1 - cos 0' and the locus is called a parabola. 1. For = p = oo, and for = ?r p="|- The parabola therefore crosses the principal axis but once at the point O, called the vertex, which is ^ to the left of the focus F } or mid- way between F and DD. 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 it for which p is infinite. TT 3. When increases from to ? D then cos decreases from 1 to 0, 1 cos increases from to 1, p decreases from oo to p, and the point P (p, 0) describes the parabola from infinity to B. When increases from to TT, then cos decreases from to 1, 1 cos increases from 1 to 2, p decreases from p to ^? & and the point P (p, 0) describes the parabola from B to the vertex 0. CONIC SECTIONS 151 On account of the symmetry with respect to the axis, when 3 TT increases from TT to -^- > P (p, 0) describes the parabola from ,, to B*\ and when 6 increases from - - to 2 TT, from 13' to infinity. a 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. The ellipse, e<\. ep 1. For e = p The hyperbola, e > 1. ep e 1 - l-e 1. For = p = l-e l-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 point A on the hyperbola to the left ofF. A on the ellipse to the right of F. As -^ = I when e 1 l ~ e< jfe - , then FA may be greater, equal to, or less than FH. As >1 (numerically) when e>l, l-e then ; so A lies to the left of H. J 1-fe p. pis For0 = 7f/> = ep p. pis 1+e 1 + e positive, and hence we obtain a point positive, and hence we obtain a sec- ond point A' to the left of F. A' to the left of F. As - <1, then p1; and hence there are no values of for which p becomes infinite. 3. When increases from to , then cos decreases from 1 to 0, 1 e cos increases from 1 e to 1 ; hence p decreases from to ep, L & and P (p, 0) describes the ellipse from A to C. "When increases from to it, 2 then cos decreases from to 1, 1 e cos increases from 1 to 1 + e ; hence p decreases from ep to , 1 -(- 6 and P (p, 0) 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 dosed curve. The hyperbola, e>l. 2. p becomes infinite if 1 e cos = 0, or cos = - - e As e>l, then -<1; and hence e there are two values of for which p becomes infinite. 3. When l increases from to cos -1 ( - ) then cos decreases from 1 to - e l ecos0 increases from 1 - e to ; hence ep p decreases from to oo, and P(p, 0) describes the lower half of the left-hand branch from A to infinity. When (1 \ 7f - I tO > e/ 2 then cos decreases from - to 0, e 1 e cos increases from to 1 ; hence p decreases from oo to ep, and P (p, 0) describes the upper part of the right-hand branch from infinity to a When increases from to TT, then cos decreases from to 1, 1 e cos increases from 1 to 1 4- e ; hence p decreases from ep to 1+e and P (p, 0) 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 153 PROBLEMS 1. Plot and discuss the following conies. Find e and p, and draw the focus and directrix of each. W, = -^ - Substituting in (III) and dropping primes, we obtain x* = 2py. Q.E.D. Y' 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 2 -f Ey = and Cy 2 -f- Dx = 0, After rotating the axes the whole figure is 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 2 + 4 y = and find the focus and directrix. Solution. The given equation may be written 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 I. Ex. 2. Find the equation of the parabola whose vertex is the point (X (3, - 2) and whose directrix is parallel to the F-axis, if p = 3. Y 1 D a A" X ^""~ N / % i-2J \ / \ / 5 i / \ CONIC SECTIONS 157 Solution. Referred to O'X' and O'F' as axes, the equation of the parabola (Theorem III) k -,,'2 _ A />.' ' -O O' x The equation for translating the axes from to (y are (Theorem I, p. 136) x = x' + 3, y = y' -2, whence (5) x' = x - 3, y' = y,+ 2. Substituting in (4), we iHkin as the re- quired equation (y + 2)* = 0(a&-8 or y 2 - 6 x + 4 y -f 22 = Referred to O'JT' and O'F', the coordinates of F are (Theorem III) (|, 0) and the equa- tion of DD is x' = .f . By (5) we see that, referred to OX and OF, the coordinates of .Fare (f, 2) and the equation of DD is x = |. PROBLEMS 1. Plot the locus of the following equations. Draw the focus and direc- trix in each case. (a) y2 = 4x. (d ) 3,2 _ 63 = 0. (b) 2/2 + 4x = o. (e) z 2 + 10 y = 0. (c) x 2 -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 (h, 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 2 = 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 (a, /S) ? -4ns. (x a) 2 = 2p (y /3). 5. Transform to polar coordinates and discuss the resulting equations (a) y 2 = 2 px, (b) x 2 = 2py. 6. Prove that the abscissas of two points on the parabola (III) are propor- tjonal to the squares of the ordinates of those points, Ans. (y - 4) 2 = 12 (x - 3). Ans. (2/ + 3) 2 = -8(x-2) Ans. (y-7) 2 = 16(x + 5). Ans. y-k* = Sx-h. 158 ANALYTIC GEOMETRY 70. Simplification and discussion of the equation in rectan- gular coordinates. Central cpnics, e J 1. When e ^ 1, equation (II), p. 154, is (1 - e 2 )x 2 + y 2 -2 e 2 px - e 2 p 2 = 0. To simplify (Kule, p. 141), set which gives - 2 e' 2 p - 2 e 2 pk -e 2 p 2 Setting the coefficients of x' and y' equal to zero gives 2 h (1 - e 2 ) - 2 e 2 p = 0, 2 k = 0, whence (3) h = k = 0. Substituting in (2) and dropping primes, we obtain or ^ 2 g 2 ^ 2 = 1. This is obtained by transposing the constant term, dividing by it, and then dividing numerator and denominator of the first fraction by 1 e 2 . The ellipse, e < 1. From (3) it is seen that h is posi- tive when e l. From (3) it is seen that h is nega- tive when e > 1. Hence the new ori- gin O lies to the left of the focus F. e 2 Further, > 1 numerically, so 1 - e 2 h > p numerically ; and hence the new origin lies to the left of the directrix DD. CONIC SECTIONS 159 The locus of (4) is symmetrical with respect to YY' (Theorem V, p. 66). Hence is the middle point of A A'. Construct in either figure F' and D'D' symmetrical respectively to F and DD with respect to YY'. Then F' and D'D' are a new focus and directrix. For let P and P f be two points on the curve, symmetrical with respect to FF 7 . 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 IT Hi P Jit and D'D' as 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. 66), 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 co6rdiriates of the focus F in either figure are 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 The coordinates of F' are therefore rM- 1 e 2 I t) The new equation of the directrix DD is x = ^ ^ JL e 160 ANALYTIC GEOMETRY For from (1) and (3), x = 1-e 2 (Theorem II) and dropping primes, we obtain x = = y'. Substituting in x = p P P Hence the equation of D'D' is x = _ 2 - 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 T 2 W r^ + - ^ ' ^**/v-k- n = 1. (1 _ 1-e Its foci are the points ( - 2 > ) \ 1 e I and its directrices are the lines x = The ellipse, e<\. For convenience set ep _e*~ (5) a = 1-e 2 I-. c = 2 and 6 2 are the denominators in (4) and c is the abscissa of one focus. Since e2-n2 a 2 - 6 2 = (1 - e 2 ) 2 1.- 1-e 2 TAe hyperbola, e>l. For convenience set (6) o= ep 1-e 2 1-e 2 a 2 and - 6 2 are the denominators in (4) and c is the abscissa of one focus. Siuce e > 1, 1- e* is negative; and hence a, 6 2 , rnd c arepositfiue. We have at once ,,9^9 ^5^2 a 2 + ft 2 = _ e 2) and a 2 (1 - e 2 ) and a 2 _ 7 ~ (1 - 1-e 2 c (1-e 2 ) 2 1-e 2 1-e 2 Hence the directrices (Lemma) are Hence the directrices (Lemma) are a 2 the lines x = c the lines x = By substitution from (5) in (4) we By substitution from (6) in (4) we obtain obtain A* CONIC SECTIONS 161 I The ellipse, e < 1. The intercepts are x = a and 2/ = 6. -4.4' = 2 a is called the major axis and BB' = 2b the minor axis. Since a 2 b 2 = c 2 is positive, then a > 6, and the major axis is greater than the minor axis. The hyperbola, e>l. The intercepts are x = a, but the hyperbola does not cut the Y-axis. AA' = 2 a is called the transverse axis and BB' = 26 the conjugate axis. 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 :/' 2 v/ 2 where 2 a is the major axis and 2 b the minor axis. If c 2 = a z ft 2 , then the foci are ( c, 0) and the directrices a 2 are x = c Equations (5) also enabje us to express e and p, the constants of (I), p. 149, in terms of a, 6, and c, the constants of (V). For Hence we may restate the Lemma as follows. Theorem VI. The equation of an hyperbola whose center is the origin and whose foci are on the X-axis is < vi > S-S =1 - where 2 a is the transverse axis and 2b the conjugate axis. If c 2 = a 2 + ft 2 , then the foci are ( c, 0) and the a 2 directrices are x = c Equations (6) also enable us to express e and p, the constants of (I), p. 149, in terms of a, 6, and c, the constants of (VI). For = a and M = v ' c = and 162 ANALYTIC GEOMETRY The ellipse, e\. In the figure OB = b, OA' = a ; and since c 2 = a 2 + 6 2 , then BA' = c. Hence to draw the foci, with O as a center and radius BA', describe arcs cutting XX' at F and F'. Then F and F' are the foci. If a = b, then (VI) becomes x 2 - y 2 = a 2 , whose locus is called an equilateral hyperbola. Transform (VI) by rotating the axes through an angle of (Theo- rem II, p. 138). We obtain Theorem VET The equation of an hyperbola whose center is the origin and whose foci are on the Y-axis is < vm > -+ =l ' X' D' D where 2 a is the transverse axis and 2b is the conjugate axis. Ifc 2 = a 2 + b 2 , the foci are (0, c) and the directrices a 2 are the lines y = . c CONIC SECTIONS 163 The ellipse, e<\. The essential difference between (V) and (VII) is that in (V) the de- nominator of x 2 is larger than that of y 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>l. The essential difference between (VI) and (VIII) is that the coeffi- cient of y 2 is negative in (VI), while in (VIII) the coefficient of x 2 is nega- tive. (VI) and (VIII) are called the typical forms of the equation of an hyperbola. An equation of the form Ax 2 + Cif + F = 0, where A, C, and F are all different from zero, may always be written in the form 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 ft are both positive, a 2 will be equal to the larger denominator and b' 2 to the smaller. 2. An hyperbola if a. and ft have opposite signs, a 2 will be equal to the positive denominator and b 2 to the negative denomi- nator. 3. If a and ft are both negative, (11) will have no locus. Ex. 1. Find the axes, foci, directrices, and eccentricity of the ellipse 4 x 2 + y 2 = 16. Solution. Dividing by 16, we obtain *+* = !. 4 16 The second denominator is the larger. By comparison with (VII), W- = 4, a 2 = 16, c 2 = 16 - 4 = 12. Hence 6 = 2, a = 4, c = Vl2. The positive sign only is used when we extract the square root, because a, b, and c are essentially positive. 164 ANALYTIC GEOMETRY Hence the major ?aaa_AA' = 8, the minor axis BB' = 4, the foci F and F' are the points (0, "^12), and the equations of the directrices DD and D'l/ a 2 16 4 _ are y = = - -' Vl2. c Vl2 Vl2 4 1 From (7) and (9), e = - - and p = = - Vl2. 4 V12 3 PROBLEMS 1 . Plot the loci, directrices, and foci of the following equations and find = 81. (e) 9y 2 - 4x 2 = 36. (b) 9 x 2 - 16 y 2 = 144. (f ) x 2 - y 2 = 25. (c) 16 x 2 + y 2 = 25. (g) 4 x 2 + 7 y 2 = 13. (d) 4x 2 + 9y 2 = 36. (h) 5x 2 - 3y 2 = 14. 2. Find the equation of the ellipse whose center is the origin and whose foci are on the .X-axis if (a) a = 5, b = 3. Ans. 9x 2 + 25 y 2 = 225. (b) a = 6, e = $. Ans. 32 x 2 + 36 y 2 = 1152. (c) 6 = 4, c = 3. Ans. 16 x 2 + 25 y 2 = 400. (d) c = 8, e = f. 4ns. 5x 2 + 9y 2 = 720. 3. Find the equation of the hyperbola whose center is the origin and whose foci are on the JT-axis if (a) a = 3, 6 = 5. Ans. 25 x 2 - 9y 2 = 225. (b)a = 4, c = 5. Ans. 9x 2 16 y 2 = 144. (c) e = f , a = 5. Ans. 5 x 2 - 4 y 2 = 125. (d) c = 8, e = 4. 4ns. 15x 2 - y 2 = 60. 4. Show that the latus rectum (chord through the focus perpendicular to 252 the principal axis) of the ellipse and hyperbola is . a 5. What is the eccentricity of an equilateral hyperbola ? Ans. 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? * -aO 2 , (if ~ W _ -. . (* ~ <*)* (V - W , ~~ + ~~- l -~ ~ CONIC SECTIONS 165 71. 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. 149) will be perpendicular. If the equation of an hyperbola is given in typical form, then the equation of the conjugate hyperbola is found by changing the signs of the coefficients of x 2 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) 16z 2 - ?/2 = 16 and -I6x* + y*= 16 are conjugate hyperbolas. They may be written a;2 y 2 X 2 y 2 ___ = land __ + ^ = 1 . The foci of the first are on the Jf-axis, those of the second on the F-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 c 2 (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 2 = 1 + 16, while in the second 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. 59) which are called the asymptotes of the hyperbola. Thus the asymptotes of the hyperbola (2) bV - aY = <*?& 'are the lines (3) bW - aY = 0, or (4) bx -f ay = and bx ay = 0. 166 ANALYTIC GEOMETRY Both of these lines pass through the origin, and their slopes are respectively (5) --and-- 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 ly y-^) be a point on either branch of (2) near the first of the asymptotes (4). The distance from this line to P^Fig., p. 167) is (Rule, p. 97) x/>>. T bx l -\- ay i + V& 2 -f a 2 Since P x lies on (2), b^x^ ahj^ = Factoring, bx l + ay = Substituting in (6), d = + a 2 bx As PI recedes to infinity, x and y become infinite and d approaches zero. For bxi and ay^ cannot cancel, since Xi and y 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 16a;2 _ y z - o and -!Qx* + y 2 = 0, which are the same. An hyperbola may be drawn with fair accuracy by the fol- lowing Construction. Lay off OA = OA ' = a on the axis on which the foci lie, and OB = OB' = b on the other axis. Draw lines through A, A 1 , 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 167 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 ( i a, 6) will lie on each asymptote (4) . Half the diagonal will equal c, the distance from the origin to the' foci, because c 2 = a 2 + 6 2 . 72 . The equilateral hyperbola referred to its asymptotes. The equation of the equilateral hyperbola (p. 162) is (1) x 2 - y 2 = a 2 . Its asymptotes are the lines x y = and x -f y = 0. These lines are perpendicular (Corollary III, p. 78), and hence they may be used as coordinate axes. Theorem X. The equation of an equilateral hyperbola referred to its asymp- totes is (X) 2jcy = a z . Proof. The axes must be rotated through to coincide with the asymptotes. Hence we substitute (Theorem II, p. 138) X' -}- y / X = V2 in (1). This gives (z' + y') a (- 2 y = Or, reducing and dropping primes, 2xy = Q.E.D. 168 ANALYTIC GEOMETRY 73. 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. Proof. Let P be any point on the ellipse. By definition (p. 149), r = e-PE, r' = e- PE'. Hence r + r 7 = e (PE + PE') = e-HH'. From (7), p. 161, e = - , a and from the equations of the direc- trices (Theorem V), a 2 c Hence r + r / = --2 = 2o. Q.E.D. Theorem Xn. The difference of the focal radii from any point on an hyperbola is equal to the transverse axis 2 a. Proof. Let P be any point on the hyperbola. By definition (p. 149), r = e-PE, r' = e PE'. Hence r' - r e (PW PE) = e- HH'. From (8), p. 161, e= -, and from the equations of the direc- trices (Theorem VI), Q.E.D. Hence r'-r = --2- = a c 74. 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 f 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 169 If the pencil be tied to the string at P, and both strings be pulled hi 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 DZ>. 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 DD 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) 4z 2 - ?/ 2 = 36. (c) 16z 2 - y 2 + 64 = 0. ^Jb) 9 a 2 - 25 y 2 = 100. (d) 8 x 2 - 16 y* + 25 = 0. 2. Prove Theorem IX for the asymptote which passes through the first and third quadrants. 3. If e and ef are the eccentricities of two conjugate hyperbolas, then 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, yi) on the parabola y* = 2px is f-, 170 T ANALYTIC GEOMETRY 8. The focal radii of a point PI(ZI, y { ) on the ellipse 6 2 x 2 + a 2 ?/ 2 = a 2 6 2 are r = a ex\ and r' = a + exi. 9. The focal radii of a point on the hyperbola 6 2 x 2 a 2 ?/ 2 = a 2 6 2 are r = exi a and r' e&i + a when PI is on the right-hand branch, or r = - exi a and r' = - exi + a when Pj. 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. 75. 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. 140), 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. 145, the term in xy may be removed by rotating the axes. Hence we only need to consider an equation of the form (1) Ax* + Cy* + Dx + Ey + F=0. It is necessary to distinguish two cases. CASE I. Neither A nor C is zero. CASE II. Either A 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 A = B 2 4 A C is not zero, and hence (Theorem VII, p. 146) we can remove the terms in x and y by translating the axes. Then (1) becomes (Corollary I, p. 147) (2) Ax 12 + cy 2 + F 1 = 0. We distinguish two types of loci according as A and C have the same or different signs. CONIC SECTIONS 171 Elliptic type, A and C have the same sign. 1. F / ^ 0.* Then (2) may be written 1 = 1, where 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 x 2 v 2 written + = 1, a ' ' where Hence the locus is an hyperbola whose foci are on the Y-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 A = and C = 0, so that (1) becomes (3) Cif + Dx + Ey + F = 0. For if A T and (7=0, (1) becomes ^z 2 + Dx + Ey + F=0. Rotate the axes (Theorem II, p. 138) through by setting x=y',y= x'. This equation becomes Ay 2 + Ex' -Dy' + F=0, which is of the form (3). By translating the axes (3) may be reduced to one of the forms (4) Cif + Dx = or (5) cy + F r =o. For substitute in (3), x = x" -f A, y = y' + &. This gives (6) C?/ /2 + Da/ + 2 C'fc I ?/' + CW =0. If we determine h and k from + Dh + Ek + F then (6) reduces to (4). But if D = 0, we cannot solve the last equation for h, so that we cannot always remove the constant term. In this case (6) reduces to (5). * Read " F' not equal to zero " or "F' different from zero." 172 ANALYTIC GEOMETRY Comparing (4) with (III), p. 155, the locus is seen to be ¶b- I ~F' ola. The locus of (5) is the pair of parallel lines y = ^ when F 1 and C have different signs, or the single line y when F' = 0. If F 1 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, which may be coincident. By moving the axes its equation may be reduced to one of the three forms Ax 2 + Cy* + F* = 0, Cy 2 + Dx = 0, Cif + F* = 0, where A , C, and D are different from zero. Corollary. The locus of an equation in which the term in xy is lacking, Ax * + C f + Dx + E?/ + F = 0, will belong 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) ? / D E \ Ans - -'- 2. To what point is the origin moved to transform (3) into (4) ? into (5) ? 3. Simplify Ax* + Dx + Ey + F=Ql)y translating the axes (a) if E -^ 0, (b) if E = 0, and find the point to which the origin is moved. A, (a) A* + Ey = 0, (- A , * 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 173 4. To what types do the loci of the following equations belong ? (a) 4z 2 +?/ 2 - 13x + 7y - 1 = 0. (e) x 2 + 7 y 2 - Sx + 1 = 0. (b) y 2 + 3x - 4 y + 9 = 0. _^C (f) x 2 + y 2 - 6x + 8 ?/ = 0. (c) 121 x 2 - 44 y 2 + 68 x - 4 = 0. ij y(g) 3x 2 -4?/ 2 -6y + 9 = 0. (d) x 2 +4y-3 = 0. ! '(h) x 2 -8x + 9?/- 11 =0. (i) The equations in problem 1, p. 148, wfiich do not contain the xy-term. 76. Construction of the locus of an equation of the second degree. To remove the ajy-term from (1 ) Ax 2 it is necessary to rotate the axes through an angle B such that (Theorem VI, p. 145) (2) tan 2 = B A -C while in the formulas for rotating the axes [(II), p. 138] we need sin and cos 0. By 1 and 3, p. 12, we have (3) cos 2 6 = + taii 2 20 From (2) we can choose 2 in the first or second quadrant so the sign in (3) must be the same as in (2). 6 will then be acute ; and from 15, p. 13, we have 1 - cos 2 I + cos 2 B (4) In simplifying a numerical equation of the form (1) the com- putation is simplified, if A = J5 2 4 A C = 0, by first removing the terms in x and y (Theorem VII, p. 146) and then the xy-term. Hence we have the Rule to construct the locus of a numerical equation of the second degree. First step. Compute A = B 2 4 A C. Second step. Simplify the equation by (a) translating and then rotating the axes if A = 0; (b) rotating and then translating the axes if A = 0. 174 ANALYTIC GEOMETRY Third step. Determine the nature of the locus by inspection of the equation ( 75, p. 170). Fourth step. Plot all of the axes used and the locus. In the second step the equations for rotating the axes are found from equations (2), (3), (4), and (II), p. 138. But if the 2-y-term is lacking, it is not necessary to rotate the axes. The equations for translating the axes are found by the Rule on p. 141. Ex. 1. Construct and discuss the locus of 4y 2 + 12x-6y = 0. Solution. First step. Here A = 4 2 -4-1-4 = 0. Second step. Hence we rotate the axes through an angle such that, by (2), _fiU\ tan2 ^T^H-U-<: Then by (3), cos 20 = -!, 2 1 and by (4), sin0 = and cos0 = -- V5 V5 The equations for rotating the axes [(II), p. 138] become Substituting in the given equation,* we obtain Z'2-Ay'^O. V5 It is not necessary to translate the axes. Third step. This equation may be written q Hence the locus is a parabola for which p = -, and whose focus is on the F'-axis. Vs *When A=0 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 It can be shown that when A = the locus is always of the parabolic type. CONIC SECTIONS 175 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- V 2V5 y A^s trix is the line y' = (Theorem 2V5 IV, p. 155). The old coordinates of 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' ' D and substituting in the equation given above. Ex. 2. Construct the locus of 5 x 2 + 6 xy + 5 y 2 + 22 x 6 y + 21 = 0. Solution. First step. A- 6 2 -4-5- 5^0. Second step. Hence we translate the axes first. It is found that the equa- tions for translating the axes are and that the transformed equation is From (2) it is seen that the axes must be rotated through Hence we set y i\ " + and the final equation is Third step. The simplified equa- tion may be written Hence the locus is an ellipse whose major axis is 8, whose minor axis is 4, and whose foci are on the Y"-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 9, may be obtained from (4). In is example tan0= = -- ' cos0 V5 > method given in the footnote, p. 28. this example tan0= = =-. 7^ = 2, and the Jf'-axis may be constructed by the cosfl V5 V5 176 ANALYTIC GEOMETRY PROBLEMS 1. Simplify the following equations and construct their loci, foci, and directrices. (a) 3z 2 - 4zy + 8z - 1 = 0. Ana. z" 2 - 4y" 2 j- 1 = 0. (b) 4z 2 + 4zy + y 2 + 8z-16y = 0. 4ns. 5z' 2 - 8 V5y' = 0. (c) 41 z 2 - 24 zy + 34 y 2 + 25 = 0. Ans. x /2 + 2 y" 2 + 1 = 0. (d) (e) y 2 + 6z-6y + 21 = 0. -4na. y' 2 -f 6z' = 0. (f) z 2 -6zy + 9y 2 + 4z- 12y + 4 = 0. Ana. y" 2 = 0. (g) 12 zy - 5y 2 + 48 y - 36 = 0. Ans. 4 z" 2 - 9 y" = 36. (h) 4z 2 -I2xy + 9y 2 + 2x- 3y-12 = 0. ^4ns. 52 y //2 - 49 = 0. (i) 14z 2 -4zy + ll?/ 2 -88z + 34 y + 149 = 0. -4ns. 2 z //2 + 3 y //2 = 0. (j) 12z 2 + 8zy + 18?/ 2 + 48z + 16y + 43 = 0. Ans. 4z 2 + 2y 2 = 1. (k) 9z 2 + 24zy + 16y 2 -36z-48y + 61 = 0. Ana. z' /2 + l = 0. (1) 7 z 2 + 50 xy + 7 y 2 = 50. Ana. 16 z /2 - 9 y /2 = 25. (m) z 2 + 3 xy - 3 y 2 + 6 z = 0. ulns. 21 z //2 - 49 y" 2 = 72. (n) 16z 2 - 24 xy + 9y 2 - 60z - 80y + 400 = 0. Ana. 2/" 2 -4z" = 0. (o) 95 z 2 + 56 xy - 10z/ 2 - 56z + 20y + 194 = 0. Ana. 6z" 2 -y //2 + 12=:0. (p) 5 z 2 - 5 xy - 7 ?/ 2 - 165 z + 1320 = 0. -4ns. 15 z //2 - 11 y" 2 - 330 = 0. 77. 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 z 2 + 4 y 2 = k. Solution. Since the coefficients of z 2 and y 2 have the same sign, the locus belongs to the elliptic type (Corollary, p. 172). When k is positive the locus is an ellipse ; when k = the locus is the origin, a degenerate ellipse ; and when k is negative there is no locus. CONIC SECTIONS 177 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. J Ex. 2. Discuss the system of conies represented by 4 x 2 16 y 2 = k. Solution. Since the coefficients of x 2 and y 2 have opposite signs, the locus 3P belongs to the hyperbolic type. The hyperbolas will all have the same asymptotes (p. 165), namely, the lines x 2 y = 0. The given equation may be written 4 16 The locus is an hyperbola whose foci are on the X-axis when k is positive and 178 ANALYTIC GEOMETRY on the F-axis when k is negative. For k = the given equation shows 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 F-axis. Ex. 3. Discuss the system of conies represented by y 2 = 2 kx + 16. Solution. As only one term of the second degree is present, the locus belongs to the parabolic type (Corollary, p. 172). The given equation may be simplified (Rule, p. 141) by translating the axes to the new origin ( , Y We thus obtain \ k * g 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 k = 0, the locus is the degenerate parabola y = 4. In the figure the locus is plotted for k = 4, 2, 1, f , 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 y = 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 179 PROBLEMS 1. Plot the following systems of conies. (a) ~ + | - *. (b) y* = 2kx. (c) H - = *. (d) x 2 = 2ky - 6. x 2 w 2 2. Plot the system -- (- - t 1 for positive values of k. What is the locus if k = 16? Show how the foci and directrices behave as k increases or decreases and approaches 16. 3. Plot the following systems of conies and show that all of the conies of each system have the same foci. 4. Plot and discuss the system fcx 2 + 2 y 2 - 8x = 0. 5. Show that all of the conies of the f ^ O vdi-s -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) (y 2 -4x) + A;(y 2 + 4x) = 0, k = + 1, -1. (b) (x 2 + y 2 - 16) + fc(x 2 - ?/ 2 - 4) = 0, k = + 1, - 1, - 4. (c) (x 2 + y 2 - 16) + fc(x 2 - y 2 - 16) = 0, k = + 1, - 1. 6. 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. 7. 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. 8. Show that a conic or degenerate conic may be found which satisfies five conditions, and formulate a rule by which to find its equation. Find the equation of the conies (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). The circle whose radius is a and whose center is the center of a central conic is called the auxiliary circle. 9. The ordinates of points on an ellipse and the auxiliary circle which have the same abscissas are in the ratio of 6 : a. 10. Show that the locus of xy + Dx -f Ey + F = is either an equilateral hyperbola whose asymptotes are parallel to the coordinate axes or a pair of perpendicular lines. CHAPTER IX TANGENTS AND NORMALS 78. The slope of the tangent. Let P^ be a fixed point on a curve C and let P 2 be a second point on C near Pj. Let P 2 approach P x by moving along C. Then the limiting position 1-iT of the secant through P! and P 2 is called the tangent to C at P!. It is evHent that the slope of P^ is the limit of the slope of PiP 2 . The coordinates of P 2 may be written (x l + h, y -f &), T where h and k will be positive or negative numbers according to the relative positions of P : and P 2 . The slope of the secant through P! and P 2 is therefore (Theorem V, p. 28) (i) As P 2 approaches P! both h and k approach zero, and hence - approaches > which may be any number whatever. The actual value of the limit of may be found in any case from the condi- tions that P! and P 2 lie on C (Corollary, p. 46), as in the example following. 180 TANGENTS AND NORMALS 181 Ex. 1. Find the slope of the tangent to the curve C : 8 y = z 8 at any point Pi (xi, ?/i) on C. Solution. Let PI (xi, y\) and P 2 (x\ -f A, yi -f fc) be two points on C. Then (Corollary, p. 46) 8 A: = Xi 8 + 3xi*ft + 3z Subtracting (2) from (3), we obtain Y 1 1 . /9\ i i V z / 7tj and 8 A ' / or IL ^'/ (3) 8 f _J - W /I Subtrac f 'o / 7 X / / / / / I Factori y * / / Factoring, 8 k = h (3 x x 2 + 3 xji + A 2 ) ; k _ 3 x^ + 3 x-Ji + A 2 A~ 8 Then, as P 2 approaches P l5 A and k approach zero and the 7- -, s i- * . i limit of-= limit of - h 8 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 remarkable 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 1\ on C. First step. Let PI (x lt y-^) and P 2 (x l -f- ^> y\ + &) b& t w 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 2 . 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 solution will contain h and k separately, so that the equation is not solved in the ordinary sense. 182 ANALYTIC GEOMETRY PROBLEMS 1. Find the slopes of the tangents to the following curves at the points indicated. (a) 2/2 = 8 x, PI (2, 4). Ana. 1. (b) x2 + j/2 = 25, Pi (3, - 4). Ans. f . (c) 4x2 + ?/2 = 16? Pl (o, 4). Ana. 0. (d) x2 - 9 2/2 = 81, P! (15, - 4). Ana. - f. 2. Find the slopes of the tangents to the following curves at the point (a) 2/ 2 = (b) 162/ = x 4 . -^ A- ^4ns. O) 4 3*1 /rt\ /~2 i /.2 1A /A >i^ 2/1 2/1 79. 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 (a>i, 2/i) on C. First step. Find the slope m of the tangent to C at P l (Rule, p. 181). Second step. Substitute %i, y^ and m in the point-slope form of the equation of a straight line [(V), p. 86]. Third step. Simplify that equation by means of the condition that P l lies on C (Corollary, p. 46). Ex. 1. Find the equation of the tangent to C : 8 y = x* at PI (xi, ?/i). 3xi 2 Solution. First step. From Ex. 1, p. 481, the slope is m = - Second step. Hence the equation of the tangent is 3X!\ y - yi = - - (* - i), or (1) 3 Xi 2 x - 8 y - 3 xi 3 + 8 yi = 0. Third step. Since PI lies on C, 8 yi = Xi 3 . Substituting in (1), we obtain (2) 3xi 2 x-87/ -2xi 3 = 0. TANGENTS AND NORMALS 183 The normal to a curve C at a point P l on C is the line through P l perpendicular to the tangent to C at P lt Its equation is found from that of the tangent by the Rule on p. 105, using Theorem XII, p. 108. 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. 108) (3) 8x + 3xi 2 y + fc = 0. If PI lies on this line, then (Corollary, p. 46) whence k = 8 Xi 3 x^y^. Substituting in (3), the equation of the normal is 8 x + 3 Xi 2 y 8 Xi 3 Xi' 2 yi = 0. PROBLEMS 1 . Find the equations of the tangents and normals at PI (xi, y\) to the curves in problem 2, p. 182. Ans. (a) yiy = 3(x +i), y\x + 3 y = x$j\ + 3 yi. (b) Xi 3 x - 4 y = 12 y lf 4 x + Xi 3 y = 4 x x + (c) Xix + yiy = 16, y x x - iy = 0. (d) XiX - yiy = 4, y& + iy = (e) 2. Find the equations of the tangents and normals to the following curves at the points indicated. (a) y 2 - 8 x + 4 y = 0, (0, 0). <4ns. 2x-y = 0, x + 2y = 0. (b) xy = 4, (2, 2). .4ns. x + y = 4, x-y = 0. (c) x 2 4y 2 = 25, PI(XI, 2/1). ^Ins. XiX-4yiy=25, 4y 1 x+Xiy=5x 1 yi. (d) x2 + 2xy = 4, PI(XI, yi). ^Ins. (xi + yi) x + Xiy = 4, x x x - (xi + yjy = xi 2 - Xiyi - y'i 2 . (e) y 2 = (f) 6 2 x 2 -4n. 6 2 xix - a^ii/ = a 2 6 2 , a 2 yix -f ft^iy = (a 2 (g) x 2 - y 2 + x3 = 0, (0, 0). Ans. y = x, x = =F y. 184 ANALYTIC GEOMETRY 80. Equations of tangents and normals to the conic sections. Theorem I. The equation of the tangent to the circle at the point PI (x l} y-^) on C is (I) oc& + yjj r*. Proof. Let PI (xi, yi) and P 2 (xi + A, 2/1 + k) be two points on the circle C. Then (Corollary, p. 46) + ( yi + fc) 2 = r 2 , Subtracting (1) from (2), we have Y' Transposing and factoring, this becomes whence Ji 2 yi + k is the slope of the secant through PI and P 2 . Letting P 2 approach PI, h and k approach zero, so that m, the slope of the tangent at P 1? is ,. .. . 2xi + h Xi m = limit of or The equation of the tangent at PI is then (Theorem V, p. 86) Xi y - 2/1 - -(*. 1), xix + y\y = %i 2 + yi 2 * But by (1), X! 2 + yi 2 = r 2 , so that the required equation is *i* + y\y - r 2 . Q.E.D. In like manner we may prove the following theorems. Theorem II. The equation of the tangent at P l (x l} y^) to the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 is b 2 x^ + a 2 y t y = a z b z ; hyperbola b 2 x 2 a 2 y 2 = a 2 b 2 is I^X^K (tfy^y = a z b z ; parabola y 2 = 2px is yy = p(oc + TANGENTS AND NORMALS 185 Theorem III. The equation of the tangent to the locus of Ax 2 + Bxy + Cif + Dx + Ey + F = the point PI(X\, yi) on the locus is Theorem III may be stated in the form of the Rule to write the equation of the tangent at P\(x^ y-^ to the locus of an equation of the second degree. First step. Substitute x^x and y^y for x 2 and y*, - for xy, and - and for x and y in the given equation. 2 Z Second step. Substitute the numerical values ofx l and y lt if given, in the result of the first step. The result is the required equation. From Theorem II, by the method on p. 183, we obtain Theorem IV. The equation of the normal at PI (x^ y^) to the ellipse b z x 2 + a 2 if = aW is a?y& - b^y = (a 2 - ft 1 ) hyperbola bV a 2 ?/ 2 = a*b 2 is a*y& + b z x y = (a 2 + 6 2 ) parabola y 2 = 2px is yjc + py = oc^y^ PROBLEMS 1. Find the equations of the tangents and normals to the following conies at the points indicated. (a) 3x2 _ 10 y2 = 17, ( 3 , i). ( C ) 2x2 - y* = 14, (3, - 2). (b) 7/ 2 = 4x, (9, - 6). (d) x2 + 5y2 = 14, (3, 1). (e) x 2 -xy + 2z-7 = 0, (3, 2). (f) xy-y 2 + 6x + Sy -6 = 0, (-1,4). The directed lengths on the tangent and normal from the point of contact to the ^Y-axis are called the length of the tangent and the length of the normal respectively. Their projections on the X-axis are known as the subtangent and subnormal. 2. Find the subtangents and subnormals in (a), (b), (c), and (d), prob- lem 1. Ana. (a) - ^ ft ; (b) - 18, 2 ; (o) - f , 6 ; (d) f , - f . 3. Find the lengths of the tangents and normals in (a), (b), (c), and (d), problem 1. Ans. (a) \ VT81, T ^ VTsT ; (b) 6 VlO, 2 VlO ; (c) VlO, 2 VlO ; (d) i V34, V4. 186 ANALYTIC GEOMETRY 4. Find the subtangents and subnormals of (a) the ellipse, (b) the hyper- bola, (c) the parabola. 62 (c) -2 -4ns. a 2 - 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 P! 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. -4ns. x\. 13. In an equilateral hyperbola the length of the normal at P! is equal to the distance from the origin to PI. 81. Tangents to a curve from a point not on the curve. Ex. 1. Find the equations of the tangents to the parabola y 2 = 4 x which pass through P 2 ( 3, - 2). . Solution. Let the point of contact of a line drawn through P 2 tangent to the parabola be PI. Then by Theorem III the equation of that line is Since P 2 lies on this line (Corollary, p. 46), (2) - 2 T/! = - 6 + 2 Xi ; and since PI lies on the parabola, (3) ?/i 2 = 4 xi. 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). TANGENTS AND NORMALS 187 If (1, 2) be the point of contact, the tangent line is, from (1), or x y + 1 = 0. If (9, - 6) be the point of contact, the tangent line is -6y = 2x + 18, or x-3y + 9 = 0. The method employed may be stated thus : Rule to determine the equations of the tangents to -a curve C passing through P 2 (# 2 , y^) not on C. first step. Let PI (xi, y-^ be the point of tangency of one of the tangents, and find the equation of the tangent to C at P l (Rule, p. 182). Second step. Write the conditions that (x z , y z ) satisfy the result of the first step and (x^ y^) the equation of C, and solve these equa- tions for x l and y. 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 2 + y 2 = 25, (7, - 1). Ans. 3x - 4y = 25, 4x + 3y = 25. (b) y' 2 = 4x, (- 1, 0). Ans. y = x + l, y + x + l = 0. (c) 16x 2 + 25 y 2 = 400, (3, - 4). Ans. y + 4 = 0, 3x - 2y = 17. (d) 8 y = x 3 , (2, 0). Ans. y = 0, 27 x - 8 y - 54 = 0. (e) x 2 + 16?/ 2 -100 = 0, (1, 2). 'Ana. None. (f) 2xy + z/ 2 = 8, (- 8, 8). Ans. 2x + 3y-8 = 0, 4z + 3y-f8 = 0. (g)- 2/ 2 + 4z-6y = 0, (- *, -1). Ans. 2x-3y = 0, 2x-y + 2 = 0. (h) x 2 + 4 y = 0, (0, - G). Ans. None. (i) x 2 -3y 2 + 2x + 19 = 0, (-1, 2). Ans. x-f-3?/-5 = 0, x-3y + 7 = 0. (j) y2 = x ^ (.^ o). Ana. 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. Ana. (a) 7x-y = 25; (b) x = 1 ; (c) 12x-25y = 100; (f) x = l; (g) x-2y = 0; (i) y = 6. 188 ANALYTIC GEOMETRY 82. Properties of tangents and normals to conies. Theorem V. If a point moves off to infinity on the parabola y 2 = 2px, the tangent at that point approaches parallelism with the X-axis. ' r Proof. The equation of the tangent at the point PI (xi, yi) is (Theorem II, p. 184) y\y = px Its slope is (Corollary I, p. 77) Vi As PI recedes to infinity y becomes infinite, and hence m approaches zero, that is, the tangent approaches parallelism with the -3T-axis. Q. E. u. Theorem VI. If a point moves off to infinity on the hyperbola 62x 2 _ a 2 ^2 _ a 2^ the tangent at that point approaches coincidence with an asymptote. Proof. The equation of the tangent at the point PI (xi, y\} is (Theorem II, p. 184) (1) b 2 XiX a 2 yiy = a 2 6 2 . Its slope is (Corollary I, p. 77) m = - d 2 yi As PI recedes to infinity x\ and y\ become infinite and m has the inde- terminate form 00 But since PI lies on the hyperbola, Dividing by a 2 ?/] 2 , transposing, and extracting the square root, ^=>RTi : ay \ ^ 2/i 2 Multiplying by - , a TANGENTS AND NORMALS 189 From this form of ra we see that as y\ becomes infinite m approaches -i the slopes of the asymptotes [(5), p. 166], as a limit. The intercepts of a a 2 6 2 (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 VII. 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 (zi, y\) on the ellipse & 2 z 2 + a 2 ?/ 2 = a?W to the focus F' (c, 0) (Theorem V, p. 161) is (Theorem VII, p. 88) yix + (c xi) y cyi = 0, and the equation of P\F is yix - (c + zi) y + cyi = 0. The equation of the tangent AB is (Theorem II, p. 184), .A" We shall show that the angle which AB makes with PF' equals the angle which P\F makes with AB. By Theorem ., p. 100, tan - ~ But since PI lies on the ellipse, 2 cyi - (a 2 - 6 2 ) and (Theorem V, p. 161) 22 _ Hence a 2 - & 2 = c 2 . 52 a' 2 cy l - cyi (a 2 - In like manner tan ; and since and are both less than it, = . That is, AB bisects the external angle of FPi and F'P, and hence, also, CD bisects the internal angle. Q.E.D. In like manner we may prove the following theorems. Theorem VIII. The tangent and normal to an hyperbola bisect respec- tively the internal and external angles formed by the focal radii of the point of contact Theorem IX. The tangent and normal to a parabola bisect respectively the internal and external angles formed by the focal radius of the point of contact and the line through that point parallel to the axis.* These theorems give rules for constructing the tangent and normal to a conic by means of ruler and compasses. Construction. To construct the tangent and normal to an ellipse or hyper- bola at any point, join that point to the foci and bisect the angles formed by these lines. To construct the tangent and normal to a parabola at any point, draw lines through it to the focus and parallel to the axis, and bisect the angles formed by these lines. The angle which one curve makes with a second is the angle which the tangent to the first makes with the tangent to the second if the tangents are drawn at a point of intersection. Theorem X. Confocal ellipses and hyperbolas intersect at right angles. Proof. X,et be an ellipse and hyperbola with the same foci. Then (3) o 2 - 6 2 = a' 2 + 6' 2 . For if the foci are ( c, 0), then in the ellipse c 2 = a 2 - fc 2 and in the hyperbola c 2 a' 2 4- i' a (Theorems V and VI, p. 161). This theorem finds application in reflectors for lights. TANGENTS AND NORMALS 191 The equations of the tangents to (2) at a point of intersection PI (xi, y\) are (Rule, p. 185) It is to be proved that the lines (4) are perpendicular, that is (Corollary III, p. 78), that X! 2 tti 2 (5) Since PI lies on both curves (2), we have ^ + ^ = land^ a 2 6 2 a' 2 Subtracting these equations, we obtain But from (3), a 2 - a' 2 = 6 2 + 6' 2 , and hence (6) reduces to (5) and the lines (4) are perpendicular. Q.E.D. In like manner we prove Theorem XI. Two parabolas with the same focus and axis which are turned in opposite directions intersect at right angles. Hence the confocal systems in problem 3, p. 179, are such that the two curves of the system through any point intersect at right angles. PROBLEMS 1. Tangents to an ellipse and its auxiliary circle (p. 179) at points with the same abscissa intersect on the JT-axis. 2. The point of contact of a tangent to an hyperbola is midway between the points in which the tangent meets the asymptotes. 3. The foot of the perpendicular from the focus of a parabola to a tan- gent lies on the tangent at the vertex. 4. The foot of the perpendicular from a focus of a central conic to a tan- gent lies on the auxiliary circle (p. 179). 5. Tangents to a parabola from a point on the directrix are perpendicular to each other. 6. Tangents to a parabola at the extremities of a chord which pass through the focus are perpendicular to each other. 7. The ordinate of the point of intersection of the directrix of a parabola and the line through the focus perpendicular to a tangent is the same as that of the point of contact. 192 ANALYTIC GEOMETRY 8. How may problem 7 be used to draw a tangent to a parabola ? 9. The line drawn perpendicular to a tangent to a central conic from a focus and the line passing through the center and the point of contact inter- sect on the corresponding directrix. 10. The angle which one tangent to a parabola makes with a second is half the angle which the focal radius drawn to the point of contact of the first makes with that drawn to the point of contact of the second. 11. The product of the distances from a tangent to a central conic to the foci is constant. 12. Tangents to any conic at the ends of the latus rectum (double chord through the focus perpendicular to the principal axis) pass through the intersection of the directrix and principal axis. 13. Tangents to a parabola at the extremities of the latus rectum are perpendicular. 14. The equation of the parabola referred to the tangents in problem 13 is x 2 - 2 xy + y 2 - 2V%p (x + y) + 2p 2 = 0, or (compare p. 10) x^ + y^ = v_p V2. 15. The area of the triangle formed by a tangent to an hyperbola and the asymptotes is constant. 16. The area of the parallelogram formed by the asymptotes of an hyperbola and lines drawn through a point on the hyperbola parallel to the asymptotes is constant. 17. Find the length of the tangent to a parabola at an extremity of the latus rectum and restate the equation of the parabola in problem 14 in terms of this length. 83. Tangent in terms of its slope. The coordinates of the points of intersection of a line and conic are found by solving their equations (Rule, p. 69), which are of the first and second degrees respectively. To solve their equations we eliminate x or y* as may be more convenient, and thus obtain an equation of one of the forms (1) Ay* + Bij+C = 0, Ax 2 + Bx+C = 0. If the discriminant A = B 2 4 A C is zero, the roots of (1) are real and equal (Theorem II, p. 3), and hence the points of intersection of the line and conic will coincide, that is, the line is * If one variable is lacking in either equation, we usually solve that equation for the other variable. But for our purposes we always eliminate the variable which occurs in both equations. TANGENTS AND NORMALS 193 tangent to the conic. The equation obtained by setting the dis- criminant 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 resulting quadratic equal to zero. Ex. 1. Find the condition for tangency of the line - + - = 1 and the parab- ola y 2 = 2 px. Solution. Eliminating x by solving the first equation for x and substituting in the second, we get by 2 + 2 apy - 2 abp = 0. The discriminant of this quadratic is A = (2 ap) 2 - 4 b ( - 2 abp) = 4 op (op + 2 6 2 ). Hence the condition for tangency is 4 ap (ap + 2 & 2 ) = or ap (ap + 2 6 2 ) = 0. Ex. 2. Find the equations of the lines with the slope $ which are tangent to the hyperbola x 2 6 y 2 + 12 y 18 = and find the points of tangency. Solution. The lines of the system (1) V = have the slope $ (Theorem I, p. 51). Y Solving (1) for x and substituting in the given equation, (2) y 2 + (4fc - 6)y + 9 - 2 A: 2 = 0. Hence the condition for tangency is (4 k - G) 2 - 4 (9 - 2 fc 2 ) = 0. Solving this equation, k = or 2. Substituting in (1), we get the required equations, namely, (3) x-2y = 0, x-2y + 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 k = 0, (2) becomes (y - 3) 2 = ; .-. y = 3 ; if k = 2, (2) becomes (y + I) 2 = ; .:y = - 1. 194 ANALYTIC GEOMETRY Hence 3 and 1 are the ordinates of the points of contact. Then, from (1), if k = and y = 3, we have x = 6 ; if k = 2 and y = 1, we have x = 6. Hence, if k = 0, the point of contact is (6, 3) ; if k = 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. The method of obtaining equations (3) may be summed up in the Rule to find the equation of a tangent in terms of its slope m. First step. Find the condition for tangency of the line y = mx -f k and the given conic. Second step. Solve the equation found in the second step for k and substitute the values found in y = mx -f- k. The equations obtained are those required. By means of this Rule we may prove Theorem XII. The equation of a tangent in terms of its slope m to the circle x 2 + y 2 = r 2 is y mx r Vl + m 2 ; ellipse b 2 x 2 + a' 2 y 2 = a 2 b 2 is y = mx hyperbola b 2 x 2 a 2 y 2 = a 2 b 2 is y mx 2 w 2 b z ; D parabola y 2 = 2px is y = mx + -- PROBLEMS 1. Determine the condition for tangency of the loci of the following equations. (a) 4x 2 + y 2 - 4 x - 8 = 0, y = 2 x + k. Ans. k 2 + 2 k - 17 = 0. (b) z 2 + ?/ 2 = r 2 , 4y-3z = 4A;. Ans. . a? 6 2 a ft .^_?! = i, +?=i. . of 6* a ft of p *In these problems it is assumed that the constants involved are not zero. TANGENTS AND NORMALS 195 2. Find the equations of the 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) y 2 = 4 x, slope = \. Ans. x-2y + 4 = 0. (b) x 2 + y 2 = 16, slope = f . Ans. 5x + 3y20 = 0. (c) 9x 2 + 16 y 2 = 144, slope = - \. Ans. x + 4 y 4 VlO = 0. (d) x 2 - 4 y 2 = 36, perpendicular to6x-4y + 9 = 0. Ans. 2x + 3y (e) x 2 + 2 ?y 2 - x + y = 0, slope = 1. (f) xy + y 2 4 x + 8 y = 0, parallel to 2 x - 4 y = 7. ^Ircs. x ' = 2 y, x - 2 y + 48 = 0. (g) x 2 + 2xy + y 2 + 8x - 6y = 0, slope = f. ulns. 4 x - 3 y = 0. (h) x 2 + 2 xy 4 x + 2 y = 0, slope = 2. . y = 2 x, 2 x - y + 10 = 0. 3. Find the equations of the common tangents to the following pairs of conies. Construct the figure in each case. (a) 7/2 = 5x, 9x 2 + 9y 2 =16. Ans. 9 (b) 9x2 + 162/2 = 144, 7x2-32y2 = 224. Ans. x - y 5 = 0. (c) x 2 + y 2 = 49, x 2 + 2/2 _ 20y + 99 = 0. Ans. 4x-3y + 35 = 0, 3x 4y + 35 = 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. 79). 4. Two tangents, one tangent, or no tangent can be drawn from a point Pi ( x i yi) to the locus of (a) y 2 = 2px according as ?/i 2 1px\ is positive, zero, or negative. (b) 6 2 x 2 -+ a 2 y* = a 2 6 2 according as fc^ 2 + a 2 yi 2 - a 2 6 2 is positive, zero, or negative. (c) 6 2 x 2 - a 2 y 2 = a 2 6 2 according as 6 2 X! 2 - afyi 2 - a?V* is negative, zero, or positive. 5. Two perpendicular tangents to (a) a parabola intersect on the directrix. (b) an ellipse intersect on the circle x 2 + y 2 = a 2 + fc 2 . (c) an hyperbola intersect on the circle x 2 + y 2 = a 2 ft 2 . CHAPTER X CARTESIAN COORDINATES IN SPACE 84. 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 (a, y} (p. 18). 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 2. This deter- mination is accomplished as follows : Let there be given three mutually perpendicular planes inter- secting in the lines XX' 9 YY', and ZZ 1 which will also be mutually perpendicular. These three planes are called the coordinate planes and may be distin- guished as the JCF-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 co6rdinate planes and cutting the axes at 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 f being to the right, that on ZZ' being upward. Y Y' is supposed to be perpendicular to the plane of the paper, the positive direction being 4n front of the paper, that is, from the plane of the paper toward the reader. 196 CARTESIAN COORDINATES IN SPACE 197 Any point P in space determines three numbers, the coordinates of P. Conversely, given any three real numbers x y y, and 2, 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 (aj, y, z), and the symbol P(x, y> z) is to be read, "The point P whose coordinates are x, y, and ." The coordinate planes divide all space into eight parts called octants, designated by 0-XYZ, O-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 coord i- x 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 off OA = 7 on OX, draw AQ parallel to OF and equal to 6 units on OY, and QP parallel to OZ and equal to 10 units on OZ. 198 ANALYTIC GEOMETRY 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, - 1 4,C), (9, 7, -2). 3. Where can a point move if x = ? if y = ? if z = ? 4. Where can a point move if x = and y = 0? if y = and z = 0? if z = 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 (x, y, z) with respect to the -XT-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, 2) 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 -XT-plane ? of x if P is in the YZ-plane ? of y if P is in the Z JT-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 Y-axis ? of x and y if P is on the Z-axis ? 9. A rectangular parallelepiped lies in the octant O-XYZ with three faces in the coordinate planes. If its dimensions are a, 6, and c, what are the coordinates of its vertices ? 85. Orthogonal projections. Lengths. The definitions of the orthogonal projection (p. 22) 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 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. From the preceding definitions follows at once as on p. 24, CARTESIAN COORDINATES IN SPACE 199 Theorem I. Given any two points x z &! = projection of 7/2 i/ 1 = projection of z z z = projection of y lt z^ and P 2 (# 2 , 2/2> upon JOT', upon YY*, upon Theorem II. The length I of the line joining tivo points (x\, V\, i) and P 2 (x 2 , y z , z z ) is given by (H) / A '} ^t*^^L. j* B D / / / x The proof is similar to that for the plane, p. 24. If we construct a rectangular parallelepiped by passing planes through PI and P 2 parallel to the coordinate planes, its edges will be paral- lel to the axes and equal numerically to the projections of PiP 2 upon the axes. PiP 2 will be a diagonal of this parallelepiped, and hence t 2 will equal the sum of the squares of its three dimensions, that is, of the numerical values of x\ ic 2 , y\ y 2 , and zi z 2 . PROBLEMS 1. Find the length of the line joining (a) Px(4, 3, - 2) to P 2 (- 2, 1, - 5). Ans. 7. (b) P! (4, 7, - 2) to P 2 (3, 5, - 4). Ans. 3. (c) P!(3, - 8, 6) to P 2 (6, - 4, 6). Ans. 5. 2. Show that the points (- 3, 2, - 7), (2, 2, - 3), and (- 3, 6, - 2) are the vertices of an isosceles triangle. 3. Show that the points (4, 3, - 4), (-2, 9, - 4), and (-2, 3, 2) are the vertices of an equilateral triangle. 4. Show that the points (-4, 0, 2), (-1, 3 Vjj, 2), (2, .0, 2), and (-1, V3, 2+2 V6) are the vertices of a regular tetraedron. 5. What does formula (II) become if PI and P 2 lie in the XF-plane ? in a plane parallel to the XF-plane ? 6. The coordinates (x, y, z) of the point of division P on the line joining -Pi (i, yi 21) and P 2 (z 2 , y 2 , z 2 ) such that the ratio of the segments is p. p := X are given by the formulas PP 2 x = 1 + X Hint. This is proved as on p. 32. V = y\ 200 ANALYTIC GEOMETRY 7. The coordinates (x, y, z) of the middle point P of the line joining PI (a*i, 2/i, zj) and P 2 (x 2 , y 2 , 22) are 8. 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), X = \. Ans. (tf, |, |). (b) (- 1, 4, - 6), (2, 3, - 7), X = - 3. Ans. (f, f, - y). (c) (8, 4, 2), (3, 9, 6), X = - I. Ans. (V, f , 0). (d) (7, 3, 9), (2, 1, 2), X = 4. Ans. (3, |, y). 9. Show that the points (7, 3, 4), (1, 0, 6), and (4, 5, - 2) are the vertices of a right triangle. 10. 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, J. (c) (- 3, 4, 2), (7, - 2, 6), and (2, 1, 4). Ans. 1. 11. Find the lengths of the medians of the triangle whose vertices are the points (3, 4, - 2), (7, 0, 8), and (- 5, 4, 6). Ans. Vll3, V89, 2 V29. 12. Show that the lines joining the middle points of the opposite sides of the quadrilaterals whose vertices are the following points bisect each other. (b) (8, 4, 2), (0, 2, 5), (- 3, 2, 4), and (8, 0, - 6). (a) (0, 0, 9), (2, 6, 8), (- 8, 0, 4), and (0, - 8, 6). (c) PI(XI, 2/1, Zi), P 2 (x 2 , 2/2, z 2 ), P 3 (3, 2/3, s), P4(4, 2/4, z 4 ). 13. 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). 14. Find the coordinates of the point of intersection of the medians of the triangle whose vertices are any three points Pj, P 2 , and P 3 . Ans. [i (xi + x 2 + X 3 ), | (2/1 + 2/2 + 2/s), | (Zi + z 2 + z 8 )']. 15. 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. 16. 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). 17. The points (xi, 2/1, Zi), (xi + a, 3/i + , Zi), and (xi, 2/1 + a, Zi + a) are the vertices of an equilateral triangle. CHAPTER XI SURFACES, CURVES, AND EQUATIONS 86. 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 o'f 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 PS is the circle in which the sphere whose center is PI and whose radius is r intersects the plane which is perpendicular to PzPs at its middle point. These two kinds of loci must be carefully distinguished. 87. 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 . 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 2 , and A,, it satisfies two conditions, namely, of being equidistant from /', and P 2 and from P 3 and 7' 3 . 201 202 ANALYTIC GEOMETRY 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. 46. 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 (II), p. 199, P X P = V(z - 3)2 + y* + ( Z + 2)2. .-. V(x - 3)2 + y2 + (z + 2)2 = 4. Simplifying, we obtain as the required equation z 2 + y 2 + z 2 - 6z + 4 z - 3 = 0. That ^js is the equation of the locus should be verified as In Ex. 1, p. 45. 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 = constant. PROBLEMS 1 . Find the equation of the locus of a point which is (a) 3 units above the JTF-plane ; (b) 4 units to the right of the FZ-plane ; (c) 10 units back of the ZX-plane. , e * y\, 2. Find the equation of the plane which is parallel to (a) the JTF-plane and 4 units above it ; (b) the ZJT-plane and 3 units in front of it ; (c) the rZ-plane and 7 units to the left of it. 3. Find the equation of the sphere whose center is ( 0, but there is no locus if k < 0. Hence the surface lies entirely to the right of the FZ-plarie. If k increases from zero to infinity, the radius of the circle increases from zero to infinity while the plane z = 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. 207) to be a parabola whose equation is (compare Ex. 1, p. 206) These parabolas are found to have the same value of p, namely, p = 2, and their vertices recede from the YZ- or ZJT-plane as k or k' increases numerically. PROBLEMS 1 . Discuss the loci of the following equations. (a) x 2 + z 2 = 4 x. (f ) x 2 + 2/2 - z 2 = 0. (b) z 2 + y 2 + 4z 2 = 16. (g) z 2 - 2/2 _ Z 2 = 9. (c) z 2 + 2/2 _ 4z 2 = 16. (h) z 2 + ?/ 2 - z 2 + 2xy = 0. (d) 6o; + 42/4- 3z = 12. (i) x + y - 6z = 6. (e) 3x -f 2y + z = 12. (j) y 2 + z 2 = 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 + z 2 -4z + 4 = 0. 4. 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. x 2 + y 2 - 3 z 2 - 6 z + 9 = 0. 5. Find the equation of the locus of a point whose distance from the point (0, 4, 0) is three fifths its distance from the ZJT-plane and discuss the locus. Ans. 25x 2 + 162/2 + 25 s 2 - 200 y + 400 = 0. 210 AXALYTIC GEOMETRY Consider now in turn loci defined by equations of the first and second degree in x } y, and z. EQUATIONS OF THE FIRST DEGREE 93. Plane and straight line. We confine ourselves to the two theorems. Theorem V. Plane. The locus of the general equation of the first degree in x, y, and z, (V) Ax +- By + Cz + D = O, isM plane. ^ The proof follows the method explained in problem 2, p. 209. Theorem VI. Straight line. The locus of two equations of the first degree, - C^z + D 1 = 0, is a straight line unless the coefficients ofx, y, and z are proportional. The proof follows at once from Theorem V and Elementary Geometry. If the coefficients are proportional, it is readily seen that the traces (p. 208) of the planes are parallel (p. 78), and hence the planes are parallel. 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. PROBLEMS 1. Find the intercepts on the axes and the traces on the coordinate planes of each of the following planes and construct the figures. (a) 2 + 3y + 4z-24 = 0. (e) 5z-7y-35 = 0. (b) 7z-3y + z-21 = 0. (f) 4z + 3z + 36 = 0. (c) %'-7y-93 + 63 = 0. (g) 6y - 83 - 40 = 0. (d) 6x + 4y - z + 12 = 0. (h) 3z + 5z + 45 = 0. 2. Find the points in which the following lines pierce the coordinate planes and construct the lines. (a) 2z + y-z = 2, x-y + 2z = 4. (c) x + 2y = 8, 2x - 4y = 7. (b) 4z + 3?/-6z =12, 4z-3?/ = 2. (d) y + z = 4, x-y + 2z=10. SURFACES, CURVES, AND EQUATIONS 211 3. Find the equation of the plane which passes through the points (1, 1, -1), (-2, -2, 2), and(l, -1, 2). Ans. x-3y-2z = Q. 4. Find the equation of the plane whose intercepts are a, 6, c. ' 5. The system of planes passing through the line AIX + Biy + Ci + DI = 0, A 2 x + B^y + C 2 z + A = is represented by AIX + #iy + Ci + A + A: (^I 2 x + B 2 y + C 2 z + J> 2 ) = 0, where k is an arbitrary constant. 6. Find the equation of the plane determined by the line 2x + y 4 = 0, y + 2 z 0, and the point (2, - 1 , 1). EQUATIONS OF THE SECOND DEGREE The locus of an equation of the second degree, of which the most general form is (1) Axt + Bif + Cz* -f Dyz + Ezx + Fxy + Gx + Hy + Iz+K = 0, is called a^uadric surface or conicoid. By methods similar to those employed in Chapter VIII, p. 172, it may be shown that the locus of (1) is a pair of planes, one plane, a straight line,* or one of the loci about to be discussed. 94. The sphere, f Theorem VII. The equation of the sphere whose center is the point (a, /?, y) and whose radius is r is (a?-a) 2 + (2/-^) 2 + (^-y) 2 = r 2 , or (VII) 2 + 2/ 2 + * 2 - 2 ax - 2ptj -Zyz + a* +^ + y 2 - r 2 = O. Proof. 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 (II), p. 199, and squar- ing, we obtain (VII). Q.E.D. *For example, the locus of x 9 + 7/ 2 = is the Z-axis. It is to be regarded as a special case of a cylinder (Theorem IX, p. 213). t 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. 212 ANALYTIC GEOMETRY Theorem VIII. The locus of an equation of the form (VIII) v2 + y* + z* + a x + ify + j[z + K = o is determined as follows : (a) When G* + H* + 7 2 4 K > 0, the locus is a sphere whose center is ( J G, J H, J /) 0, the locus of equations (1) is a pair of lines ; if k = 0, it is a single line (the Z-axis) ; and if k < 0, equations (1) have no locus. f*^~\ Similarly, the intersection with a plane parallel to the ZJT-plane, y = k, is a straight line whose equations are (Rule, p. 207) and which is therefore parallel to the Z-axis. The intersection with a plane parallel to the -ZT"-plane is the parabola 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 JTF-plane y 2 = 4 x, 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 IX. 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. 214 ANALYTIC GEOMETRY 96. Cones. Ex. 1. Determine the nature of the locus of the equation 16 x 2 + y' 2 - z 2 = 0. Solution. Let PI (xi, y^ Zi) ne a point on a curve C in which the locus ffltersect- plane, for example, z = k. Then (1) 16xi + y 1 a-2! 1 a = 0, z l= k. The origin OTies on the surface (Theorem III, p. 207). It may be shown* that the line OPi lies entirely on the surface, and therefore that the surface is a cone whose vertexis tl> origin. In the same way the locus may be shown to be a cone whenever the equa- tion of the surface is homogeneous f in the yariables x, y, anjjs. Hence Theorem X. The locus of an equation which is homogeneous in the variables x, y, and z is a cone whose vertex is the origin. PROBLEMS 1 . Determine the nature of the following loci ; discuss and construct them. (a) x 2 + ?/ = 36. (e) x 2 - y 2 + 36 z 2 = 0. (b) x 2 + 2/ 2 = z 2 . - (f ) y 2 - 16 x 2 + 4 z* = 0. (e) y' 2 + 4 z 2 = 0. (d) x 2 - z 2 = 16. (g) x 2 + 16y 2 - 4x = 0. (h)' x 2 + yz = 0. 2. Find the equations of the cylinders whose directrices are the following curves and whose elements are parallel to one of the axes. (a) y 2 + z 2 4 y = 0, x = 0. (c) & 2 x 2 - a 2 ^ 2 = a 2 6 2 , z = 0. (b) z 2 + 2 x = 8, y = 0. (d) y 2 + 2pz = 0, x = 0. 3. Discuss the following loci. (a) x 2 + y' 2 = z 2 tan 2 7. (c) z 2 + x 2 = y 2 tan 2 /3. (e) y 2 + z 2 = r 2 . ? + P-?- which contains only the squares of the tive coefficient., is called an hyperboloid of one sheet. The sections of the sur-. face formed by planes parallel to the AT-plane (Rule, p. 207) are ellipses, while those formed biplanes parallel to either of the other coordinate planes are conies of the hyperbolic type. T\v > systems of straight lines lie .oJL_the hyperboloid and it is therefore called a ruled surface. The loci of the equations are also hyperboloids of one sheet. 216 ANALYTIC GEOMETRY The locus of the equation (4) y with two negative coefficients, is called an hyperboloid of two sheets.* Sections formed by planes paral- lel to the rz-plane (Rule, p. 207) are conies of the elliptic type, while those formed by planes parallel to either of the other coordinate planes are hyperbolas. The^loci of the equations (5) X ,2 ~2 ~2 ~.2 ^2 J_ ^L _ 1. 1 ^ l^_ . ^_ _ 1 .> I 7 o o *J . 70 I *rc General Library SSSSJKS 1 University of California (B139s22)4/6 Berkeley I /' 66 UNIVERSITY OF CALIFORNIA LIBRARY