' /vi. W/r /"/' '"^- ANALYTIC GEOMETRY BY W. A. WILSON, Ph.D. ASSISTANT PROFESSOR OF MATHEMATICS YALE UNIVERSITY AND J. I. TRACEY, Ph.D. INSTRUCTOR IN MATHEMATICS YALE UNIVERSITY 3i*;< D. C. HEATH & CO., PUBLISHERS BOSTON NEW YORK CHICAGO ■ . ' ' ' / Copyright, 191 5, By D. C. Heath & Co. 2E 2 ■Engineering. PREFACE The purpose of this book is to present in a short course those parts of Analytic Geometry which are essential for the study of Calculus. The material has been so arranged that topics which are less important may be omitted with- out a loss of continuity, and the text is therefore adapted for use in classes which aim to cover in one year the funda- mental principles and applications of both Analytic Geom- etry and Calculus. For such classes Chapters I-VI, VIII, and the earlier sections of Chapters VII and IX form a course which can be completed in about thirty-five lessons. In order to provide material for study if more time is avail- able, a discussion of the general equation of the second degree and short chapters on Tangents and Normals and Solid Analytic Geometry have been included. In the preparation of the book the authors have used as a foundation a pamphlet written by Professor William Beebe for use in his classes. The authors wish to express their thanks to Dr. David D. Leib and to Dr. Levi L. Conant for many valuable criticisms and suggestions. 6S4544 m Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/analyticgeometryOOwilsrich CONTENTS Introduction CHAPTER I — COORDINATES AND EQUATIONS SECTION 1. Introduction 7 2. Position ; Cartesian Coordinates 7 3. Notation 8 4. Plotting of Points 9 5. Directed Lines 11 6. The Distance Formula 12 7. The Point of Division Formulas 13 8. Geometric Applications 14 9. Angle between Two Lines 16 10. Inclination and Slope 17 11. The Slope Formula 17 12. Parallelism and Perpendicularity 18 13. The Angle Formula 18 14. Equations and Graphs 20 15. Plotting of Graphs 21 16. Derivation of Equations 24 17. Functional Variables 27 18. Functional Notation 28 19. Graphs of Functions 29 20. Discussion of Equations 30 21. Intercepts 31 22. Symmetry 31 23. Extent _ . . 33 24. Symmetrical Transformations 35 25. Transformations of Equations not Altering their Loci . 36 26. Equations whose Graphs cannot be Plotted ... 37 27. Intersection of Curves 38 V VI CONTENTS CHAPTER II — THE STEAIGHT LINE SECTION 28. Equations of the Straight Line 29. The Point Slope and Two Point Forms 30. The Slope Intercept Form 31. The Intercept Form .... 32. Lines Parallel to the Axes 33. The Linear Equation 34. Relations between Linear Equations 35. Application to the Solution of Linear Equations 36. Geometric Conditions Determining a Line 37. The Normal Form 38. Reduction of a Linear Equation to the Normal Form 39. Distance from a Line to a Point .... 40. Systems of Straight Lines 41. Systems of Lines through the Intersection of Two Given Lines ...... 42. Plotting by Factoring . 40 40 41 41 42 44 45 47 50 50 51 54 57 59 61 CHAPTER III — THE CIRCLE 43. The Equation of the Circle 44. The General Equation of the Circle . . . . 45. Identification of the Center and Radius . . . 46. Special Forms 47. The Equation of the Circle Derived from Three Con- ditions ......... 48. Length of a Tangent 49. Common Chord of Two Circles 50. Radical Axis 51. Systems of Circles 64 66 66 67 69 70 72 73 75 CHAPTER IV — THE PARABOLA 52. Conies .... 53. The Equation of the Parabola 54. Discussion of the Equation 55. Latus Rectum . 56. To Draw a Parabola . 57. Mechanical Construction of the Parabola 77 77 78 79 79 80 CONTENTS Vll SECTION 58. The Parabola as a Conic Section 59. The Path of a Projectile . PAGE 82 83 CHAPTER V — THE ELLIPSE 60. Fundamental Constants . 61. The Equation of the Ellipse 62. Discussion of the Equation 63. Second Focus and Directrix 64. Focal Radii 65. Latus Rectum . 66. Ellipse with Foci on the ?/-axis 67. The Circle as a Limiting Form 68. Construction of an Ellipse by Points 69. Mechanical Construction of the Ellipse 70. The Ellipse as a Section of a Cone . 85 86 86 87 89 90 90 91 91 93 94 CHAPTER VI — THE HYPERBOLA 71. Constants 96 72. The Equation of the Hyperbola 97 73. Discussion of the Equation 98 74. Second Focus and Directrix 98 75. Latus Rectum 99 76. Hyperbola with Foci on the y-axis 100 77. Focal Radii 101 78. Asymptotes 101 79. Conjugate Hyperbolas ....... 102 80. Equilateral Hyperbolas 104 81. Construction of a Hyperbola by Points .... 104 82. Mechanical Construction of the Hyperbola . . . 105 83. The Hyperbola as a Conic Section 106 CHAPTER VII — TRANSFOR:\LiTION OF COORDINATES AND SIMPLIFICATION OF EQUATIONS 84. Change of Axes 85. Formulas of Translation 107 107 Vlll CONTENTS SECTION PAGE 86. Formulas of Rotation 108 87. Application to the Conies 110 88. Test for Axes of Symmetry Ill 89. Simplification of Equations by Translation . . . 112 90. Discussion of the Equation Ax"^ -\- Cy"^ + Dx -\- Ey + F = 114 91. Limiting Forms 116 92. General Statement 116 93. Generalized Standard Equations of the Conies . . 117 94. Simplification of Equations by Rotation . . .119 95. The General Equation of the Second Degree . .121 96. The Characteristic 121 97. Test for Distinguishing the Conies .... 122 98. Suggestions for Simplifying the Equation of a Conic . 128 99. The Conic through Five Points 124 CHAPTER VIII — POLAR COORDINATES 100. Definition 101. Relations between Rectangular and Polar Coordinates 102. Polar Curves 103. The Equation of the Straight Line 104. The Equation of the Circle 105. Discussion of Polar Curves 106. Rotation of Axes 107. Symmetrical Transformations 108. The Equation of the Conic 109. Equations Derived in Polar Coordinates 127 128 129 131 131 132 136 137 138 140 CHAPTER IX — HIGHER PLANE CURVES 110. Algebraic and Transcendental Equations 111. The Exponential Curve 112. The Logarithmic Curve 113. Applications . 114. Periodic Functions 115. The Sine Curve 116. Circular Measure 117. The Cosine Curve 118. The Tangent Curve 143 143 144 145 147 147 149 149 150 CONTEXTS IX 119. Multiple Angles .... 120. Sum of Functions .... 121. Inverse Trigonometric Functions . 122. Parametric Equations 123. Parametric Equations of the Circle 124. Parametric Equations of the Ellipse 125. The Path of a Projectile 126. The Cycloid .... 127. The Epicycloid and Hypocycloid 128. The Involute of the Circle 129. Algebraic Curves 130. The Witch 131. The Strophoid 132. The Ovals of Cassini PAG» 150 151 152 153 154 154 155 157 159 161 162 162 163 164 CHAPTER X — TAXGEXTS AX^D XORMALS 133. Definitions 134. Slope of Tangent and Xormal to the Circle at a Given Point 135. Slope of Tangent and Xormal to the Ellipse at a Given Point 136. Slope of Tangent and Xormal to the Parabola and Hy- perbola at a Given Point . . . 137. Equations of Tangent and Xormal at a Given Point . 138. Tangent and Xormal. Subtangent and Subnormal 139. Tangent having a Given Slope 140. Tangent from a Given External Point .... 141. The Parabolic Reflector 142. Angle between the Focal Radii and the Tangent to an Ellipse at Any Point 143. Diameters of a Conic 144. Conjugate Diameters of the Ellipse .... 145. Conjugate Diameters of the Hyperbola .... 166 166 167 168 169 171 171 172 174 175 176 178 179 CHAPTER XI — SOLID ANALYTIC GEOMETRY 146. Introduction 181 147. Coordinates ......••• 1^1 148. Radius Vector and Direction Cosines .... 182 CONTENTS 150. 151. 152. 153. 154. SECTION 149. Distance between Two Points Direction of a Line .... Angle between Two Lines The Locus in Solid Geometry The Normal Equation of the Plane Planes Parallel to One or More Coordinate A 155. The General Equation of the First Degree 156. Angle between Two Planes 157. The Intercept Equation of the Plane 158. The Equations of the Line 159. The Symmetrical Equations . 160. The Two Point Equations 161. The Projection Forms 162. Reduction of the Equations of a Line to the Forms 163. Cylindrical Surfaces 164. Surfaces of Revolution . 165. Discussion of Surfaces 166. Quadric Surfaces .... 167. The Ellipsoid 168. The Hyperboloid of One Sheet 169. The Hyperboloid of Two Sheets . 170. The Elliptic Paraboloid . 171. The Hyperbolic Paraboloid Formulas and Equations xes Standard PAGE 183 184 185 187 187 188 189 190 190 192 192 193 193 194 196 197 199 200 200 202 203 204 204 206 ANALYTIC GEOMETRY 3 ^ \5 ' AI^ALYTIC GEOMETRY INTRODUCTION DEFINITIONS AND FORMULAS FROM ALGEBRA AND TRIGONOMETRY 1. The Quadratic. — An expression reducible to the form Ax"^ -{- Bx+ C, where A, B, and C are any constants, is called a quadratic in X. 2. Solution of the Quadratic. — The roots of any quadratic equation may be found by completing the square, or by using the quadratic formula. (a) To solve by completing the square, transpose the constant term to the right-hand member and divide both sides of the equation by the coefficient of x"^. In the re- sulting equation add to both members the square of half the coefficient of x and extract the square root. (b) If the above rule is applied to the general quadratic Ax"^ + Bx -{- C = 0, we get ^ -B±s/&-4.AC 2A which is known as the quadratic formula. To solve an equation by means of this formula simply substitute for A, B, and C in the formula their values in the given equation. 3. The Discriminant. — The expression under the radical, B^ ^ 4: AC, is called the discriminant oi the quadratic. It 1 2 ANALYTIC GEOMETRY is denoted by the G)eek letter A. The quadratic formula shows that the roots of a quadratic equation are imaginary, if A < ; real and equal, if A = ; real and unequal, if A > 0. 4. Logarithms. — The logarithm of a number to a given base is that exponent by which the base must be affected in order that the result may be equal to the number. Thus if 6^ = N, we say x is the logarithm of N to the base b and write X = logj, JSf. The base b may be any positive number different from and 1. Two systems of logarithms are in common use: the natural, or Naperian, where the base is e = 2.71828 . . . ; and the common, or Briggs, where the base is 10. Important properties of logarithms to any base are the following : log,l = 0; log,6 = l; M log, MN= log, 31 + log, N; log6^= ^^g, M- log, AT; log, N^=k log, N; log,^i^= ^ log, N) log,A^ = ^^g^=log,A^.log,6. 5. Angles. — In trigonometry an angle is supposed to be generated by revolving a line, called the generating Ime, about the vertex from one side of the angle, called the initial line, to the other side, called the terminal line. If the rotation is counter-clockwise, the angle is called posi- tive ; if clockwise, it is called negative. Degree Measure. — The ordinary unit of angle measure- ment is one ninetieth part of a right angle, which is called a degree. One sixtieth part of a degree is called a minute and one sixtieth part of a minute is called a second. Thus INTRODUCTION 3 60" = one minute ; 60' = one degree ; 360° = one complete revolution = four right angles. Circular Measure. — The unit of circular measure is a central angle subtended by an arc equal to the radius of the circle. It is called a radian. We find at once that IT radians = 180° ; 1 radian = : 1° = -— - radians. 180 From this definition we have the important result that in any circle the length of an arc equals the product of the measure of its subtended central angle in radians and the length of the radius. 6. Trigonometric Functions. — From any point P of the terminal line VP of the angle = A VP drop a perpendic- ular on the initial Ime of the angle (produced if necessary). The ratios of the sides of the triangle thus formed are called the trigonometric ratios or functions of the given angle. ^/^ ~v c A M k. ^sT A Vv V^ i < ANALYTIC GEOMETRY They are . . MP . VM . ^ MP sm 6 = ; cos 6 = ; tan 6 = ; VP VP VM' G^(iO = ; sec^=-— -: cot » = -——. MP VM' MP In these ratios the side of the triangle opposite the angle is considered positive when it extends up from the initial line, negative when down; the adjacent side is con- sidered positive if it extends to the right of the vertex, negative if to the left ; the hypothenuse is always positive. 7. Relations between the Functions. — Eor any angle, /, 1 . /J cos ^ sec 6 = ; cot ^ = — cos ^' sin^' 1 sin 6 1 tan^ sin_^, cos d' CSC 6 = -^ ; sin'' 6 + cos^ ^ = 1 ; cot e = 7-^ ; 1 + tan2 $ = sec^ 6 •, tan^ = ^^: 1 + cot^ $ = csc^ 0. 8. Reduction of Angles. — sin (— ^) = — sin.^; sin C|±^J=cos^; cos (- &)= cos e ; cos f'^±o\=T sin $ ; tan (-^) = -tan^; tan f^±e\=Tcote; sin (it ±0)=T sinO] cos (tt ± ^) = — cos ^ ; tan (tt ± ^) = ± tan 6. INTRODUCTION 9. Special Angles. — Angle TT TT TT T TT 37r 2ir 6 4 3 2 2 Sine 1 V2 V3 1 — 1 2 2 2 Cosine 1 V3 2 V2 2 1 2 - 1 1 Tangent V3 3 1 V3 00 00 10. Formulas for the Sum and Difference of Two Angles. sin (6 ± )= sill cos ± cos ^ sin <;^ ; cos (^ ± <^) = cos cos <^ qp sin ^ sin <^ ; tan(^±<^)^ tan^±tanc^^ ^ 1 T tan ^ tan <^ 11. Double and Half Angle Formulas. — • sin 2 = 2 sin 6 cos 6', cos 2 ^ = cos2 - sin2 6* = 1 - 2 sin^ = 2 cos^ ^ - 1 ?^ - 2 tan ^ tan 2 ^ = sui 1 - tan2 ^ ' — cos tan - 2 /I - cos /I -h c =\/ ; cos-=\/ — - — \ 2 2^2 -4 cos — cos 1 — cos sin ^ + cos sin 1 + cos ^ 12. Triangle Formulas. — In an}^ triangle ABC, — = — = (Law of Sines) ; sm A SYD.B sm C a^ = b- -\- c"^ — 2 be cos A (Law of Cosines). 6 ANALYTIC GEOMETRY 13. The Greek Alphabet — Letters Names Letters Names A, a Alpha N, V Nu B, p Beta H, 1 Xi r, y Gamma O, o Omicron A, 8 Delta n, TT Pi E, c Epsilon P, 9 Eho z, C Zeta 2, ^ Sigma H, rj Eta T, r Tau @, e • Theta Y, V Upsilon I, t Iota $, (^ Phi K, K Kappa X, X Chi A, X Lambda ^, «A Psi M, /a Mu n, w Omega CHAPTER I COORDINATES AND EQUATIONS 1. Introduction. — The chief feature of Analytic Geom- etry, which distinguishes it from the geometry which the student has hitherto studied, is its extensive use of algebraic methods in the solution of geometric problems. Just as the use of symbols in algebra makes possible the ready solution of many problems which would be difficult if not impos- sible by the j^rocesses of arithmetic, so the use of algebraic reasoning simplifies much of geometry and widens its scope. The student is already familiar with some of these applica- tions ; for example, the theorems stating the numerical properties of a triangle are derived in whole or part by algebraic reasoning. In trigonometry also the greater part of the reasoning is of the same character, and further simplification is gained by the introduction of negative numbers. For example, the law of cosines, a^ = 6^ -^ c- — 2 6c cos A, states in compact language the contents of three theorems of plane geometry, since cosine A is positive or negative according as angle A is acute or obtuse. Other examples of a similar character may occur to the student. 2. Position ; Cartesian Coordinates. — The most important help in the extension of geometric analysis has been found in the location of points in a plane or in space by Cartesian coordinates, an adaptation to mathematics of the means of expressing position used in ordinary life. AYe ordinarily locate a point by stating its direction and distance from a fixed reference point, as — Philadelphia is 7 8 ANALYTIC GEOMETRY --5 X' I I I I I -5 A r, X B P{4,-3) ninety miles southwest of Xew York. Less commonly, its position is fixed by its distances from a pair of fixed per- pendicular lines. An illustration of this is found in the familiar direction given in a city to go, for example, eight blocks east and three north to reach a desired destination. Both of these methods are used analytically ; the second is simpler and T^ill be discussed first. In the figure the reference lines are the perpendiculars ^'Xand yy. These are called the axes of coordinates or coordinate axes, and their intersection is called the origin. The two axes di- vide the plane into four quadrants numljered as in trigonometry. The position of a point P is determined by measuring its distance from Y' Y along a parallel to X'X, and its distance from X'X along a parallel to Y'Y. Distances meas- ured to the right of I"' l"or up from X'X are called positive^ those to the left or down negative. The measures of these distances, with the proper signs prefixed, are called the coordinates of the point : the one measured along a line parallel to X'X is the j:-co6rdinate or abscissa ; the one parallel to Y'Y is the z/-coordinate or ordinate. Thus in the figure the abscissa of P is BP= -|- 4, its ordinate is AP = — 3. The coordinate axis X'X is called the axis of abscissas or A:-axis, and the axis Y'Y the axis of ordinate s or z/-axis. 3. Notation. — In naming a point by its coordinates, write them in a parenthesis, putting the abscissa first. Thus in the figure Pis the point (4, — 3). In case the coordinates -5 COORDINATES AND EQUATIONS 9 are variable or unknown, the abscissa is denoted by the letter x, and the ordinate by y. Fixed points of which the coordinates are not known or are arbitrary, will be dis- tinguished by means of subscripts, being lettered Pi, Pg, etc., and represented by the coordinates (xi, y^, {x2, 2/2), etc. 4. Plotting of Points. — It will be seen that any point in a plane is fixed by means of its coordinates ; for the abscissa locates it on a parallel to the ^/-axis and the ordinate on a parallel to the a;-axis, and these meet in one point. To locate a point whose coordinates are given (or as commonly stated, to plot the point), first choose a convenient unit of measure, then measure off the abscissa on the a;-axis —(4 4) C-'? '^^j- - X31-2-)- A' Y' and the ordinate from the end of the abscissa. Thus to plot (—3, 2) count 3 unit's to the left on the a>axis and 2 units up. This is especially convenient when using coordinate paper, i.e. paper ruled with sets of parallel lines. The system of coordinates which has been described is the rectangular system and is a particular case of Cartesian coordinates. In the general Cartesian system the axes are not necessarily perpendicular, but may be oblique. All of the above definitions hold for. the oblique system, which is illustrated above. 10 ANALYTIC GEOMETRY PROBLEMS 1. Plot accurately the points : (a) (2,5),(3,0), (-V2,5)|^ (6) (-3,0),(5, -8),(1,V12); (c) (2, -6), (-3,-3), (0, -Vt); (d) (-1.2), (-2, -3), (-V3,0). 2. Draw the triangle whose vertices are as follows : {a) (3,-1), (-2,5), (-8, -4); (6) (2,4), (4, V8)^(3, 11); _ (c) (5, 3), (5 + 4V2, 3 + 4V2), (5 + 8V2, 3 - 8 V2). 3. Draw the quadrilateral whose vertices are as follows : (a) (-2,0), (2,4), (6,0), (2, -4); (5) (0, -6), (8, 2), (0. 6), (-8, 2). 4. Construct with compasses on coordinate paper V^. V3, Vo, and estimate the value of each. Hint. — Each radical will be the measure of the hypothenuse or leg of a right triangle. 5. Draw a circle of radius 10 with its center at the origin. Draw radii at 20^ intervals up to 180^. Estimate from the figure the coordi- nates of the ends of these radii. 6. From the figure of problem 5 make a table of sines and cosines of the angles drawn. 7. What is the abscissa of all points on the y-axis ? the ordinate of all points on the a;-axis ? What are the coordinates of the origin ? 8. To what quadrants Is a point limited if its abscissa and ordinate have like signs ? unlike signs ? 9. If a point moves on a parallel to the x-axis, which of its coordi- nates remains constant ? which if it moves on a perpendicular to the a:-axis ? 10. (a) What is the locus of a point whose abscissa is 6 ? whose ordinate is — 6 ? (b) What is the locus of all points having the same abscissa? having the same ordinate ? 11. What is the locus of points whose abscissas and ordinates are (a) equal ? (6) numerically equal, but of unlike sign ? Why ? '• 12. A square whose side is 2 a has its center at the origin and its sides parallel to the axes. Eind the coordinates of its vertices. COORDINATES AND EQUATIONS 11 V 13. An equilateral triangle of side a has one vertex at the origin and the opposite side parallel to the ?/-axis. Find the coordinates of the other vertices. ^^^^ ^^ V3, ± ^) , or (- | V3, ± f) ' -^ 14. A rhombus has one angle of 60^ and two vertices at (0, 0) and (a, 0). Find the coordinates of the other vertices if (a) both are in the first quadrant; (&) one is in the second quadrant. . , . (a aV3\ /3a aV3\ •si 15. A regular hexagon of side a is placed so that one diagonal lies along the x-axis and the center is at the origin. Find the coordinates of the vertices. 5. Directed Lines. — We have referred to tlie advantage of using positive and negative lines in trigonometry and have defined the signs of coordinates. In analytic geometry we constantly use directed lines, that is, lines whose lengths are reckoned as positive or negative according to the direc- tion in which they are read. For example, if the positive direction is from left to right, and AB is 8 units long, then A B AB = 8, while BA = — 8. Thus, changing the direction of reading a line changes its sign, i.e. BA = — AB. In adding or subtracting line segments great care should be taken to avoid errors of sign. It is advisable for begin- ners to read all segments in one direction in performing such operations, later reversing segments as required. For example, let us find the relation between AB, BC, and AC in the figure. A C B We have AB=AC-{-CB = AC- BC. In the Cartesian coordinate system the positive direction on all lines parallel to the x-axis is to the right; on lines not parallel to the x-axis the positive direction is upward. All lines should be read m the positive direction unless it is 12 ANALYTIC GEOMETRY intended that their lengths are to be considered negative, Since coordinates are the measures of directed lines, they must be read up or to the right if positive, and doAvn or to the left if negative. Exercise 1. Prove that for any position of the point O on a directed line passing through the points A and B, AB = AC+ CB. (Three cases.) Exercise 2. Prove that the distance between any two points hav- ing the same ordinate is the difference of their abscissas, or d = xi — X2. (Three cases. ) M 6. Distance Formula. — The distance between two points, Pi (a^i, y\) <^f^nd Pj (^2j yi), li'liose coordinates are known is given by the formula. For But and P^M=^ P^H^- HM= OK- OB = x^ - x^, MP, = MK^ KP, = KP, - BP, = y,- y,. Substituting, we have the formula. Note. — In this demonstration and those which follow, the fact that the formulas are true for all positions of the points and lines involved is of fundamental importance. The student should draw the figure for several possible positions and satisfy himself that the demonstration covers all cases. COORDINATES AND EQUATIONS 13 7. Point of Division Formulas. — Tlie coordinates {xq, 2/0) of a poi7it Pq dividing a line joining Pi(x^, y^) and Pgfe ^2) in a given ratio r^ : r^ are given by the formidas _ ^2X1 4- r,x. Xq ^ ) ri4-r Hq ; ''1+12 (2) In either figure, we have, by similar triangles, P,P,:P,P, = P,E:P,K But PiPo : AA = n:ro by hypothesis. Also P^E = OB - 0A = .To - x„ and PoF= OC-OB^x^- Xq. Substituting, Solving for Xq, ' 1 _ '<^0 Xn -^l JCf) ~~~' »//n ^'1 + ^2 We derive the formula for ?/o from the triangles PqFPj and PiEPq in the same manner. When Pq is the mid-point of the line, r^ = r^ and these formulas become : which are called the mid-point formidas. (2 a) Exercise 3. Derive the distance formula when ,' ^- M^ }■ W(ci') Pi lies in the fourth quadrant and P2 in the second ; "^'.^ lb) Pi lies in the second quadrant and Po in the third ; (c) Pi lies in the third quadrant and P2 in the fourth. V^' 14 ANALYTIC GEOMETRY Exercise 4. State the rule expressed by (a) the distance formula, (6) the mid-point formula. Exercise 5. Derive the mid-point formulas directly from a figure. ^v Exercise 6. Derive the point of division formulas for Pq on P1P2 produced. Hint. — Note the sign of ri : r2. -; ■' ^ y\\ ' ■ 8. Geometric Applications. — By means of these formulas and others which will follow, many of the theorems of plane geometry can he proved in a very simple manner. '^ --■' Example. — Show by analytic means that the mid-point of the hypothenuse of any right triangle is equidistant from the vertices. Solution. — Let the lengths of the legs be a and b. Place the right angle at the intersection of the coordinate axes. Then the vertices are (0, 0), (a, 0), (0, b). Let (7 be the mid-point of the hypothenuse. By formula 2 a its coordinartes are ( -,-l. The dis- tance formula gives O ' '> The same result is obtained for CB and AC, which proves the theorem. In this problem the lengths of the sides were expressed by letters so as to insure that OB A should represent any right triangle. The justi- fication for placing the right angle at the origin lies in the geometric axiom that figures may be moved about in space without altermg their size and shape. PROBLEMS 1. Find the length of the line joining (1, 3) and (- 2, 6). Solution. — CM (1, 3) Pi and (-2,6) Po. Thenxi=l, ?/i = 3. and Xg = — 2, 2/2 = 6. Substituting in the distance formula, d = V(l + 2)2 -t- (3 - 6)2 = V18 = 3v^. COORDINATES AND EQUATIONS 15 2. Find the lengths of the sides of the triangles whose vertices are (a) (4,3), (2, 2), (-3, 5); (6) (-3,1), (7, -2), (-6,5); (c) (5,4), (-3,2), (-3, -6); (d) (4,5), (2,-3), (-6, -3). _ _ _ Ans. (a) V5, V34, V53. y'Z. Show that the following are the vertices of isosceles triangles : •wrJr- («) (2,4), (5,1), (6,5); '^ (&) (2,6), (6,2), (-3, -3). 4. Show that (3, 3), (-3, -3), (3V3, -3^3) are the vertices of an equilateral triangle, and find the length of its medians. 5. Show that (3, 1), (6, 5), ( — 1, 4) are the vertices of a right tri- angle. What is its area and the distance between the mid-points of its legs ? 6. Show that (15, 8), (23, - 7). (8, - 15), (0, 0) are the vertices of a square. Find the lengths of its diagonals. V 7. Show that the points (3, 4), (2, 1 -h Vl2), (4, 3) lie on a circle with its center at (1, 1). ^^ >j 8. Find the coordinates of the point dividing in the ratio 3 : 5 the segment whose extremities are : (a) (5, 6) and (13, -10); (6) (- 1, 5) and (3, -9); (c) (-6, 3) and (8, -7). . Ans. (a) (8,0). 9. In each of the following find the coordinates of the point three fourths of the distance from the first to the second point : (a) (5, 6) and (13, - 8) ; (6) (-1,5) and (3, -4); (c) (- 6, 8) and (3, - 2). Ans. (a) (11, - 4.5). i| Hint^^ Fi^7 determine the ratio of the segments. / ^ UO. Find the point of intersection of the medians of the triangles ^ whose vertices are the points in Problem 2. Ans. (a) (1, -V). ZZinf. — We know from plane geometry that it is two thirds of the dis- tance from a vertex to the middle of the opposite sidei 11. Find a point 10 units distant from the point (—3, 6) and with the abscissa 3. 12. Find a point equidistant from the points (13, 8), (6, 15) and (-4,-9). /r^'- 16 ANALYTIC GEOMETRY 18. Two of the vertices of an equilateral triangle are (1, 4) and (3, — 2). Find the other vertex. ■■^ 14. Prove analytically that : (a) the diagonals of a square are equal ; ;^,. > " -^"n^^iTi^ (b) the diagonals of a rectangle are equal ; ' ^ /" , (c) a line joining the middle points of two sides of a triangle is one half the third side ; (d) the median of a trapezoid is one half the sum of the bases. 15. Prove analytically that in any triangle : (a) the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides decreased by twice the product of one of those sides and the projection of the other upon it ; (b) the sum of the squares of two sides is equal to twice the square of one half the third side, increased by twice the square of the median on that side ; (c) the sum of the squares of the medians is equal to three fourths the sum of the squares of the three sides. 16. Prove analytically that : (a) the sum of the squares of the four sides of a parallelogram is equal to the sum of the squares of the diagonals ; (b) the sum of the squares of the four sides of any quadrilateral is equal to the sum of the squares of the diagonals increased by four times the square of the line joining the mid-points of the diagonals. 9. Angle between two Lines. — Tlie angle of intersection of two directed lines is understood to he the angle between their positive directions ; i.e. the angle of intersection of AB and CD is BED. There- fore in Cartesian geometry the value of any angle lies between and tt. For ex- ample, suppose the line CD to be rotated around E so " y^ ^^'^' ^ that the angle BED in- creases from until ED reaches and passes the position EA. After that the positive direction on (7Z> becomes EC and the angle between the lines is BEC. jj.-^' {C^' d^^. . 4--^ -"f*^ uot^Ti-.-^ COORDINATES AND EQUATIONS 17 10. Inclination and Slope. — Tlie inclination of a line is its angle of intersection with the x-axis. If it is parallel to the a>axis, its inclination is 0. TJie slope of a line is the tangent of its inclination. We denote the inclination by the letter a, the slope by m. Thus ?7i = tan a. When m is positive, a is acute ; when m is negative, a is obtuse, and conversely. The slope fixes the direction of a line. Thus a straight line is determined when its slope and one point on it are known. ^ 5 -11. The Slope Formula. — The slope of a line passing through the points Pi(Xi, y^) and P2(^25 .^2) *^ given by the formula 9 y m — Xi — Xo (3) Take the case where Pi lies in the first quadrant and Po in the third. T y JMvPi) / i' K\ X PiiXo.lJn) H M Then KP, MP, tan « = .-f^ = jTiTr- AR P2M Now MP, = MK + KP, = KPi - BP, = y,- 2/2, and P23/= P^H^ HM= OK- OB = x, - x^. Substituting, we have the formula. Exercise 7. Derive the slope formula when (a) Pi is in the fourth quadrant and P2 in the second ; (6) Pi is in the second quadrant and P2 in the third ; (c) Pi is in the third quadrant and P2 in the fourth ; . {d) both points are in the first quadrant and the inclination is greater than 90°. Exercise 8. State the rule expressed by the slope formula. 18 ANALYTIC GEOMETRY 12. Parallelism and Perpendicularity. — If tivo lines are parallel their slopes are equal, and conversely; if they are perpendicular their slopes are negative reciprocals, and conversely. If Zi and ^2 axe parallel, obviously «, = a^, whence the slopes are equal. If \ is perpendicular to l^, we have a-^ = ac,-\-'^, whence tan «! = — cot ct2 = — tan tto This gives mi = '-, or ??ln mi»?i2 = — 1. Let the student prove the converse theorems. Summarizing, we have : Condition for parallelism, jHi = ttZj ; Condition for perpendicularity, mjiti, = — 1. (4) (5) F yAtto \ \ ^' / / \^ 13. The Angle Formula. — Tlie angle between two lines is given by the formula tanp m^ — m^ 1 -{-m^m^ (6) X m^ denoting the slope of the line of greater inclination. COORDINATES AND EQUATIONS 19 Obviously a^ = 05 + ^ and fi =a^— a^, wheuce tanfl= tan «, - tan «, 1 + tan «! tan Og _ TTli — 7712 1 + mi??i2 Exercise 9. Derive the angle formula for the case when the point of intersection is below the a;-axis. PROBLEMS 1. Find the slope of the line joining (a) (2, 3), (3, 5) ; (6) (3,8), (-6, -Q)-, (c) (V2, V3), (_ V3, V2); (d) (a + &, c), (a, b-\- c). ^' 2. What is the inclination of the line joining (a) (3, 3) and (- 3, - 3) ; (6) (-3, 2) and (-4, 3); (c) (5, 0) and (6, \/3) ; _ (d) (-3,0) and (_2,V3);. (e) (0, -4) and (-3, 1); (/) (0, 0) and (- V3, - 5) ? Ans. (a) 45°. /jA-o 3. Prove by means of slopes that (2, 3), (6, — 3) and (— 2, 9) are on the same straight line. 4. Prove that (4, 7) is on the line joining the points (6, 6) and (2, 8), and is equidistant from them. V^. Prove that the line joining (6, - 3) and (2, 8) is perpendicular !b the one joining (0, 0) and (11, 4). 6. What is the inclination of a line parallel to Y'Y? perpendicu- lar to Y' Y? What are the slopes of these lines ? |/ 7. Find the angle between y^ (i) lines (a) and (d) in Problem 2 ; ' *''(ii) lines (&) and (e) in Problem 2 ; (iii) lines (c) and (/) in Problem 2. Ans, (i) 15°. >i 20 ANALYTIC GEOMETRY JL-^ J^^^' Find the angle between the line from (4, — 2) to (— 3, 6) and e line bisecting the first quadrant. 9. Prove by means of slopes that the following points are the vertices of a right triangle and find its angles : (a) (5, 0), (8, 4), (1, 3) ; (6) (5, 5), (2, 8), (10, 10). 10. Prove by means of slopes that the following points are the vertices of a parallelogram and find its angles : (a) (-1, -2), (0,1), (3,4), (2,1); {h) (3, -3), (4,0), (7,3), (6,0). 11. Prove that (0, - 2), (4, 2), (0, 6), and (~ 4, 2) are the ver- tices of a square. J. 12. Find the slopes of the sides and the angles of the triangle whose vertices are : (a) (- 2, 2), (4, 2), (1, 5) ; : (&) (1, 1), (- 1, - 1), (5, - 5). j^ 13. Prove analytically that (a) the diagonals of any square are perpendicular; (6) the median of a trapezoid is parallel to the bases ; (c) the lines joining the mid-points of the sides of any quadrilateral form a parallelogram ; {d) the lines joining the mid-points of the sides of a rectangle form a rhombus. 14. What is the relation between two lines for w'hich wii = — m-i ? 15. Three vertices of a parallelogram are (8, 6), (2, 5), and (4, —2). Find the fourth vertex. 16. Find a point on the line joining (2, 3) and (—4, 6) and 2 units distant from the latter point. 17. Derive formula (5) from formula (6). 14. Equations and Graphs. — The coordinates x, y, if no restriction is placed on their values, represent any point in the plane. If, however, the values of x and y are subject to certain conditions, points having these coordinates will lie upon certain lines or curves. For examj)le, if y is unre- stricted, but X always equals — 6, all such points will lie upon a line parallel to the ?/-axis and 6 units to the left. Again, the locus of all points whose abscissas and ordinates COORDINATES AND EQUATIONS 21 are equal is the bisector of the first and third quadrants. Such restrictions upon coordinates are expressed by means of equations. Thus, the equation of the first locus is x = — G, of the second x = y. As a general definition, we have: The equation of a locus is. an equation satisfied by the coor- dinates of all points lying on the given locus, and conversely. The curve which contains all points whose coordinates sat- isfy the given equation and no other points is called the locus or graph of the equation. The derivation of equations of loci and the study of graphs by means of their equations form the chief part of elementary Analytic Geometry. 15. Plotting of Graphs. — When a locus is given by its equation, the shape of the curve is sometimes evident from the form of the equation, as in examples mentioned above, and as we shall see in more complicated problems later. Usually the graph is constructed by a process called plot- tnig the graph of the equation, in which we proceed as follows : Solve the equation for y in terms of x. Set X equal to convenient positive and negative values (gen- erally integral) and compute the corresponding values of y. Each pair of values of x and ?/ is a solution of the equation, and hence the point of which these are the coordinates lies on the curve by the definition of the locus. Make a table of values by arranging these pairs in order according to the magnitudes of the values of x. Plot the points thus tabidated and join them by a smooth curve in the order of the table. This gives an approximation to the true curve which becomes more exact when a larger number of points is plotted. Remarks. — It is sometimes more convenient to solve for x in terms of y. In this case the above is applicable on interchanging x and y. 22 ANALYTIC GEOMETRY If the solution for y in terms of x involves a square root, the double sign must be used with the radical and in general two points will be found for each value of x. A similar remark applies to a solution for x in terms oft/. Sometimes the values of a; or y will be so large that it is difl&cult to draw the curve on the plotting paper. In this ease use one scale for x and another for y. "When the points plotted do not show the form of the curve, look for an error in the table of values. If there is no error, assume intermediate fractional values of x or y and plot the corresponding points. The table of values should be extensive enough to show the form of the curve completely. If it extends to infinity, the plot should contain all parts with considerable curvature and should extend far enough to show the direction of the curve beyond the limits of the paper. ^ Example 1. — Plot the graph of 3 jC -{-?/ — 4 = 0. Solving for ?/, Substituting a: = — 2, — 1, 0, 1, 2, 3, we get 2/ = 10, 7,4, 1,-2, -5 respectively. Thus points etc. oil the line are (— 2, 10), (-1,7), (0, 4), etc. The table of values is at the left and the plot at the right. X y -2 10 -1 7 4 1 1 2 - 2 3 — 5 Example 2. — Plot the graph of y- — 2 y = :r — 3. Here it is easier to solve Y ' for X in terms of y : x = y'--2y + S. In the plot each unit on the ?/-axis has been taken equal to two units on the X-axis. This is because the values of x are so much larger than those of y that if the same scale were used on both axes, the curve would be hard to draw. In such cases the scale used should be indicated on the graph. y X -3 18 -2 11 -1 6 3 1 2 2 3 3 6 4 11 5 18 ^- COORDINATES AND EQUATIONS 23 Q L 'Example 3. —Plot the graph of 4 x^ + y"- = 20. X y -3 imag. -2 ±2 -1 ±4 ± 2V5 = ±4.5 1 ±4 2 ±2 3 imag. SoMn.o: of X, for ?/ in terms y =± V20 — 4 x^ _,= ± 2V5-r/:^. Here no part of the curve can lie outside of X = + 3 and x = — 3, "Whereas there are two points corresponding to X = ± 2. To find out where the curve crosses the x-axis, we set ?/ = in the equa- tion and solve, getting x = ± \/5= ± 2.2+. Adding the points (2.2+, 0) and (- 2.2+, 0) to the plot, we are able to draw the entire curve. Example 4. — Plot the graph of xy = — 4, or y — )..^ X y 1 -4 2 -2 3 -f 4 - 1 6 -1 8 _ 1 1. X y - 1 4 -2 2 -3 -4 f 1 -6 -8 1 Since x cannot equal 0*. we must take fractional values of x between ± 1 in order to find the shape of the curve. These give additional points : X y ! X y i - 8 1 -i 8 1 - 16 -i 16 Adding these to the plot, we see that the curve goes to infinity along the x- and 2/-axes in the second and fourth quadrants. * This is due to the fact that the substitution of x = in the equation involves division by 0, an impossible operation. Since y becomes indefi- nitely great as x approaches 0, it is sometimes stated that for x — 0,y = », but this abbreviation should not be allowed to obscure the fact that for a; = there is no point on the curve. 24 ANALYTIC GEOMETRY PROBLEMS Plot the graphs of the following equations : 1. X = 6. J 11. 9 r;2 + 2/2 = 36. 2. y =0. 18. x2 + 2 2/2 = 2. 3. y = -2. 19. 25 x2 + 16 ?/2 = 400. 4. Sy = x. 20. x2 - ?/2 = 9. 5. y=-2x. 21. ^2 = 25 + x2. 6. 3x + 4:y =0. : 22. x2 + 4 X + 4 = ?/2 - 3 2/. 4 X - 3 ?/ = 6. 23. x2 - 9i/2 = 36. 8. x + y = 1. 24. 9x2-2/2 + 9 = 0. 9. ?/2 = X^. 6 25. X2/ = 12. ^10. y = x^ — ox -{- 4. 26. 2xy =- 15. 11. x^-Sx+2y = 6. 27. ?/ = X'^ — 9 x2. 12. X = y^ — 6 y ■}- ^. 28. 2/^ = x2. 13. 2/2 - 2/ = X + 2. 29. xy = y + 2. 14. x^ + y^= 16. 30. xi/ = 3x — 1. 15. j,2 + 2/2 = 25. 31. xy -y^- = 10. (Solve for x.) 16. x2 + 4 2/2 = 4. 32. X + 2/ — 2 x?/ = 0. 33. Find two of the above curves whicl 1 pass through (a) (0,0); (6) (-1, -^ 1) ; (c) (1,1). 16. Derivation of Equations. — The derivation of equations of loci or curves is a process depending largely upon the ingenuity of the solver of the problem. However, the following general suggestions will be helpful. 1. Take the origin and axes in a convenient position. The origin will usually be a fixed point mentioned in the prob- lem, or the point midway between two such points, with one of the axes passing through them. This does not involve any loss of generality as far as the locus is concerned, since the locus is independent of its position in the plane ; while a correct choice of axes will greatly simplify the form of the equation and the work of deriving it. 2. Mark P (x, y) as any point satisfying the given condi- tions and therefore on the curve. Draw its coordinates if necessary. 3. Draic any line suggested by the data of the problem. COORDINATES AND EQUATIONS 25 It should contain all the given 4. Express the conditions of the problem in an equation con- taining X, y, and the given constants. For this purpose it will be necessary to lind some formula or principle of geometry relating to the coordinates and given constants. 5. Simplify the equation. numerical or arbitrary con- stants, but no variables other than X and y. Example 1 . — Find and plot the locus of points distant 6 units from the point (7, — 4). SoIutio7i. — Here the coordinate axes are fixed by the statement of the problem. Take any point P{x, y) as one satisfying the conditions of the problem and draw CP. Since CP has the constant length G, the distance formula is suggested. This gives at once r)=V(x-7)2 4-(?/+4)2, or (x-7)2+(j/+4)2=36. We can either plot the curve by points, or, since we know that it is a circle, draw it with compasses. ^ Example 2. — A rock in the ocean lies 6 miles off a stretch of straight coast and a ship moves so as to be always equidistant from the rock and the coast. Find the locus of the ship. Solution. — Take the coast as the .r-axis and let the ?/-axis pass through the rock, which will have the coordinates (0, G). Take a point P(x, y) and draw its ordinate ^IP. Then OA = x, AP = y. P is to be equidistant Y ^'' P{x,y) R -{pfi) i?^ i X ^'■/^-f^ ; 26 ANALYTIC GEOMETRY from R and OA. This suggests drawing RP, AP being already drawn. By the conditions of the problem RP = AP. To get the length of RP use the distance formula, which gives RP = Vx=^ + (2/ — t))'^- But its equal AP = y .-. Vx2 + (?/ — 6)2=2/, whence a;2_ 12 y + 36 = 0. The derivation of the above equations shows that the coordinates of all points which lie on the locus satisfy the equation obtained. To show that the equation fulfills the second requirement of the definition of the locus, namely, that any point the coordinates of which satisfy the equation, lies on the locus, it is merely necessary to see if the various steps taken in the derivation of the equation can be retraced. To illustrate, in Example 2 any point P{x, y)i whose coordinates satisfy the equation x^ — 12 2/ + 36 = 0, must also satisfy the previous equation since it can be re- duced to that form by reversmg the ste ps by which the latter was obtained. But the relation •\/x^ + (?/ — 6)2 = ?/ simply says that the point P is equidistant from the a;-axis and R, which was the condition stated in the problem. PROBLEMS 1. What is the equation of a straight line (a) parallel to the cc-axis and 6 units above it ; . . f\ (h) parallel to the ?/-axis and 6 units at the left ? ^ 2. Find and plot the equation of the locus of points (a) three times, (6) m times as far from the x-axis as from the y-axis. < 3. Solve Example 2, page 25, when .(a) the origin is at J? ; ^ (6) the origin is midway between and B. /,4. Find and plot the equation of ±he locus of points equidistant f^«°^ K(«) (3.^ ^^d (6, i ^ ; I7 ^ (6) (-4, 3) and (6, 0); (c) (-5,0) and (0,4); -^ ^ [ (.d) (0, 0) and (a, 0). lO I II COORDINATES AND EQUATIONS ' 27 / lb. Find and plot the equation of the circle having V (a) radius 3 and center (— 5, 6) ; (&) radius 6 and center (1, — 3) ; (c) radius 2 and center (0, 0) ; (cZ) radius r and center (^, k). ' ' 6. Find and plot the equation of the locus of points equidistant ftom (^a) the line y -Q and the point (3, 0) ; (6) the line u; = 4 and the point ( - 2, 0) ; (c) the line x = — 8 and the point (0, — 5) ; {d) the line x = and the point (p, 0). ^ ^/ 7. Find the locus in Problem 6 when the distance from the point' f is twice that from the line. 8. Find the locus in Problem 6 when the distance from the line is twice that from the point. 9. Find the equation of the straight line passing through ^'^' {a) (3, 4) and (5, - 2) ; (c) (1, 1) and ( - 2, - 2) ; (6) (4, 3) and (- 1, 6) ; (d) (a, 0) and (0, h). \^ 10. Find the equation of the straight line passing through (— 2, 3) and of inclination («) 120^ (6) ^; (c) 150°; (d) ^. 11. The distance between two fixed points is 2 c. Find the locus of a point moving so that the sum of the squares of its distances from the points is 4 c^. 12. The ends of a straight line of variable length rest on two per- pendicular lines. Find the locus of the middle point if the area of the triangle formed is constant. ^*" 13. The base of a triangle is of length 2 a. Find the locus of the ^j^vertex if the vertical angle is 90°. "N 17. Functional Variables. — The symbols used in mathe- / ^ matics represent two kinds of quantities, variables and ^ constants. A variable is a quantity which takes on an unlimited num- ber of values ; e.g. the velocity of a falling body, the abscissa of a point moving along a curve, etc., are variable quantities. A constant is a quantity having a fixed value. There are two kinds of constants : absolute constants, which have the 28 ANALYTIC GEOMETRY same value in all problems, as 2, — 6, tt ; and arbitrary constants, which may have any value assigned, but keep the same value in a given discussion, as the quantities m^ and 77I2 in the discussion of parallel and perpendicular lines in § 12. When two variables are connected by some law such that the value of one depends upon that of the other, the first variable is said to be a function of the other. The first variable is called the dependent and the second the independent variable. Thus the area of a circle is a function of the radius, the pressure of steam is a function of the temperature, etc. Such relations are commonly expressed by means of equa- tions, as, in the case of the formula expressing the area of a circle, A = irr'^. 18. Functional Notation. — The expression f{x) is used to denote any function of x\ it is read "function of a;," or "/ of «." Similarly f(x, y) stands for a function of both X and y. To denote a different function of x, some other letter is used, as F(x) or g(x). Since functional relations are usually expressed by means of equations, the expression f(x) is ordinarily an abbreviation for some combination of terms containing x and no other variable. If an equation in x and y is solved for y, we regard x as the independent variable and 2/ as a function of x. Thus, if x and y are connected by the relation ic^ -f- 4 ?/2 = 4, y=±^ V4— x~ may be denoted by the abbre viation y=f(x), where f(x) stands for the quantity ±-i-V4 — .'c^. The symbols /(I), /(— 2) represent the values of the func- tion when 1 and —2 respectively are substituted for x. For this function /(l)=:±iV4^^ = ±iV3,/(-2)=±iV4^^ = 0. If it is not desired to solve the equation for either variable, we may transpose all terms to the first member and denote this form of the equation by the expression /(a;, y) = 0. COORDIXATES AND EQUATIONS 29 19. Graphs of Functions. — If we plot the graph of an equation expressing a functional relation, the graph may be regarded as a pictorial representation of the relation between the variables. Con- sider, for example, the rela- tion between the velocity of a ball thrown vertically up with an initial velocity of 64 feet per second and the height which it has attained at any time. By a physical formula ^1 = 642 _ 64 h. Let each unit on the .r-axis represent 1 foot and each unit on the ?/-axis represent a velocity of 1 ft. per second. For each value of h the correspond- ing ordinate represents the velocity of the ball at that height. In presenting statistical information the graph is con- stantly used. Here, as a general thing, the functional rela- tions cannot be expressed by means of equations, and the graphs are plotted from observed data. The following table gives the debt of the United States for the years 1875 to 1910, the unit being $100,000,000, and the adjoining graph presents the same mformation. "o'|64 16 55 32 '45 48 32 64 Debt i>* Year Hundreds of MiLLTOXS 1875 22 1880 21 1885 19 1890 15 1895 17 1900 21 1905 23 1910 27 1875 1885 1895 1905 30 ANALYTIC GEOMETRY PROBLEMS 1. Write the equation 2x^ + y^ = 4:xy in the three forms y =f{x)^ a^=/(2/), and/(x, y) =0. Solution. — To express y in terms of x, solve the equation for y, ■ 2 Hence y =f{x) = 2 a: ± a; V2. lu like manner ^ =/(?/) = 2/ db i2/V2, and f{x,y)=2x-^-^zy + y^ = 0. ^^z. "Write each of the following equations in the three forms y =f{x), x=f{y), and/(.r, y)=0: (a) xy + y = 6; ^c) x^ -{- 2x = y'^ - 6y ; Xb) oi' + 9^2 ^ 9; {d) y={x-\- If. *y^. If f (rr) = a;3 _ 2 :e2 + Sx, find the value of /(I). /(O), /(- 3), 4. If /(x) = 6x* - os;2 + 3, show that/(- a:) =/(x) identically. 5. If /(a:, 2/) = 8x2 + 3?/- ?/2, show that /(- x, 2/) =/(x, y) iden- tically. Does/(x, - ?/) =/(:'•, y) ? 6. Write a function of x and 2/ such that (a) f(x, - y) =f(x, y) identically; (&) /(- X, - y) =f(x, y) identically ; {c) f{-x,y) =/(x, y) identically. 7. Express the radius of a sphere as a function of the surface and plot the graph of the function. 8. The area of a triangle of variable base and altitude is always 16 square feet. Express the base as a function of the altitude and plot the graph of this function. 9. A point moves around the circumference of a circle of radius 5. Express its distance from one end of a fixed diameter as a function of its distance from the other end and plot the graph of this function. : 10. A rectangle of varying size has one side 3 inches longer than the other. Express the area in terms of the shorter side and plot the graph of this function. 11. From an almanac find the length of daylight on the first of each month during the year and draw a graph illustrating this data. 20. Discussion of Equations. — Since it is jDossible to plot but a few points on a curve, the graph is always more or COORDINATES AND EQUATIONS 31 less inaccurate. To discover the properties of a curve it is necessary to study, or discuss, its equation. The purpose of this discussion is threefold : it gives exact information re- garding the curve, it furnishes a check upon the accuracy of the plot, and it usually facilitates the labor of plotting. The properties that can be conveniently studied in most curves are the intercepts, symmetry, and extent of the curve. C ^ 21. Intercepts. — The intercepts of a curve are the dis- y tances from the origin to the points where it meets the axes. To hnd the .T-intereepts, set y = and solve the resulting equation ; to find the ^/-intercepts, set x = and solve. y; ly 22. Symmetry. — Definitions. TJie axis of symmetry of ^\/^tico points is the perpendicular bisector of the line joining them. The center of symmetry of two points is the point mid- way between them. A curve is symmetrical with respect to an axis or to a center ichen each point of the curve has its symmetrical point on the curve. Thus the points (3, 4) and (—3. 4) are symmetrical with respect to the 2/-axis ; (3, 4) and (3, —4) with respect to the a>axis ; and (3, 4) and (—3, — 4) with respect to the origin. From the figure we can readily establish the general principle : Two 2^oints are symmetHcal ivith respect to the x-axis ivhen their abscissas are the same and their ordinates differ only in sign ; to the y-axis ichen their ordinates are the same and their abscissas differ only in sign ; to the origin ichen their resp)eG- five coordinates differ only in sign. The proof is left to the student. Y ^(-^vl/i) P.( ^I'^i) ^ L-— "x- P-i^-'^v-yi) A( ^Wi) 32 ANALYTIC GEOMETRY Applying this theorem to curves, we see that a curve is symmetrical w^ith respect to the 2/-axis, if for each point (x, y) on the curve, the point (— X, y) is also on the curve. The symmetry of a curve may be determined by inspec- tion of the equation accord- ing to the following theorem : If the substitution in an equation of — j: for X ("gives an equation reducible to itsl i/-axis ; — y for y \ original form, the curve is sym- r^-axis ; —xioTX and —yiory i metrical with respect to the J origin. The proof of the first statement is as follows : Let the equation be written in the form f(x, y) = 0. Let P{Xi, yi) be an}' point on the curve and Q(— a:-, 2/1) be its symmetrical point with respect to the ^/-axis. By hypothesis, f(-x, y)=f{x, y) ; hence f(-x„ y,) But P is on the curve, hence /(x^, y{) = 0. Therefore /(— Xi, 2/1) = 0, and hence Q is on the curve. Thus the definition of symmetry is satisfied and the theorem proved. The other cases are treated in exactly the same manner. The above figure is the graph of x'^—2y—l = 0, an equation satisfying the test for symmetry with respect to the ^/-axis. Solving for x, we have x= ± V2y -h 1, and the table of values illus- trates various pairs of symmetrical points. A curve which is symmetrical with respect to both the x and ?/-axes is symmetrical with respect to the origin, but the converse is not true (e.g. y = x^). Moreover, a curve may ±1 _ ± V3 ± V5 i3 COORDINATES AND EQUATIONS 33 have axes of symmetry other than the coordinate axes. In this case, if there are two perpendicular axes of symmetry, their intersection will be a center of symmetry. Exercise 10. Prove that if Pi(xu yi) and Piixz, 2/2) are sym- metrical with respect to the x-axis, then x\ = X2, and yi = — y^. Exercise 11. Write a proof of the third test for symmetry. 2B. Extent. — To investigate the extent of a curve, we solve its equation for y in terms of x, and for x in terms of y. If either operation gives rise to radicals of even degree involving one of the variables, say x, then values of x which make the expression under the radical negative must be ex- cluded, for the corresponding values of y would be imaginary. To illustrate, consider the equation, 4 x^ — 9 ?/2 = 36. Solving, y = ± sVcc^ — 9, x =± fvl/H-^. For a;2 < 9^ or for vahies of x between ± 3, x2_9 js negative, and there- fore y is imaginary. Hence such values of x must be excluded from the table of values and the curve lies wholly without that part of the plane bounded by the lines x—±3. As 2/2 4- 4, > for all values of y, no value of y needs to be excluded. PROBLEMS 1. Discuss the equation 4:x'^ + y^ - IQ x = and plot its graph. Solution. — The x-intercepts are and 4 y ±3.4 ±4 ±3.4 Symmetry : With re- spect to the X-axis. X = 2 ± W16 — ?/2. Hence y^ must be < 16 and the curve lies between the lines-y = ±4. y = ± V4x(4-x). If X is > 4 or < 0, the factors under the radical have unlike signs and their product is negative. Hence such values of x must be excluded and the curve lies between the lines x = and x = 4. the only y-intercept is 0. r 7/ =4 34 ANALYTIC GEOMETRY 2. Discuss the equation y = x^ — a^x and plot its graph. Solution. — This equation contains an arbitrary constant a. It will have a different graph for each value given a, but the properties obtained by the discussion will be common to all such curves. Intercepts : Putting x = 0, the y-intercept is 0. Putting y = 0, x^ — a^x = ; the x-intercepts are 0, ± a. Symmetry : The substitution of — x for x, or — y for y, changes the equation, but if — x is put for x and also — y for y, we get — ?/= (— x)3 — a2( — x), which reduces to y = x^ — a^x. Thus the curve is symmetrical with respect to the origin. Extent : Since this equation is of the third degree, it will have at least one real solution for any value of y. Hence we shall not solve it for X. It is already solved for y and no radicals appear, hence no values of X need to be excluded, for none make y imaginary. Thus x may take on any value. If the equation is put in the form y = x{x^ — a^), it becomes evident that when x > a, y is positive, and when x approaches infinity, y approaches infinity. Thus the cui»ve extends to infinity in the first quadrant. Owing to the symmetry, the path where x < is similar, extending to infinity in the third quadrant. We could now sketch the curve roughly. A table of values for X positive gives a more accurate graph. In making the table we assign a any convenient value, as 1. X y 1 2 6 3 24 etc. Discuss each of the following equations and plot their respective graphs : .' Z. 4x'^ — y = 0. 9. ^■' 4. 2/^-8x+ 16 = 0. 10. x2 - ?/2 + 12 «/ = 0. j^. x2 + ?/2 - 16 a: = 0. 6. a;2 + 4?/2 4-4?/ = 0. 7. x2 + 2y2 ^ 16. >^. 9 ^2 _ a;2 _ 36 = 0. 11. xy-9 = 0. 12. 2 5C2/ + 15 = 0. 13. 2/2 = 8x3. ^4. y = x^ — 9x. COORDINATES AND EQUATIONS 35 15. X^ = 9 2/3. 16. y + {x- 2)3 = 0. 17. xy^ = 36. 18. x(y' + 1) _ 12 = 0. 19. x^y = 1. 20. 4 a:2 + 3 y2 = 0. 21. y^ = ax?. 22. y- = 2px. 23. X2 + 2/2 = (j2 24. a2 62 25. 5c2 y2 _ ^ a^ 62 • 26. x^ = 2py. 27. y = ax'\ 28. xy = 2 «2. 24. Symmetrical Transformations. — In case the tests for symmetry of § 22 are not satislied, the substitutions used will transform the given equation into a new equation. For example, if we substitute — x for x in the equation of a curve symmetrical with respect to the ?/-axis, the equation is unchanged ; but if the curve is not symmetrical with re- spect to the y-Sixis, a new equation is obtained. The ques- tion at once arises : what relation has the locus of the new equation to that of the given equation '? In this case it can be shown as in the proof of § 22, that for any point P(xi, ?/,) on the locus of the given equation /(^'> y) = ^) the symmetrical point Q(— Xi, y^ lies on the locus of the new equation f(— x, y) = 0. For this reason the curves are said to be symmetrical to each other with respect to the y-Sixis, and the transformation is called a symmetrical transformation. Similar reasoning applies to the other sub- stitutions. Summarizing, we have the following definition and theorem : Two curves are symmetrical to each other icith respect to an axis or to a center, ichen each point of one curve has its symmet- rical point on the other. If in an equation substitution is made of — xioxx { the locus of the new equation ) y-axis ; —1/ for y is symmetrical to that of the • .r-axis ; — JT for ^ and —y for y ^ old with respect to the J origin. 36 ANALYTIC GEOMETRY Example — Write the equation of the curve symmetrical to y = x^ — 5x + Q with respect to the ?/-axis, and plot both curves on the same axes. Putting — X for x, we get y = x^ + 5 x -\- 6. The continuous line is the graph of the original equation, the dotted line that of the new one. Note that the table of values for y = x^+5x + 6 may be obtained from that for y =x^ — 6x -\- 6 by merely changing the signs of the values of x. Tables of Values y = x^ — 5x -{- Q y = x^ -\- 5x -\- 6 X y 6 X y 0" 6 1 2 - 1 2 2 -2 3 -3 4 2 -4 2 5 6 - 5 6 25. Transformations of Equations not Altering Loci. — The following transformations of the equation leave the form of the locus unchanged. («) Eearrangement of terms, solving for one variable, multiplying by a constant. These do not alter the relation between x and y. (b) The symmetrical substitutions. (c) Interchange of variables. The effect of this is to interchange the axes. r PROBLEMS 1. Write the equation of the curve symmetrical to the given curve with respect to the aj-axis, and plot both curves on the same axes. (a) 1/ = 3 x2 — 4 ; (c) y = x^ + 4:x; ^b) a;2 + 4 ^2 - 4 ^ = 0; (d) y = x^ - 2 x\ \y^2. Write the equation of the curve symmetrical to the given curve with respect to the 2/-axis and plot both curves on the same axes. i-fti) ?/2 = 3x-4; (c) y = a;3 + 4a;; (6) x2 + 4 1/2 - 4 X = ; (d) 2/ =: x3 - 2 x2. COORDINATES AND EQUATIONS 37 ^/^. Write the equation of the curve symmetrical to the given curve with respect to the origin and plot both curves on the same axes. (a) y = 3a;2-2x + 4; ^ (c) y = a:^ - 4x2; (&) x = 2^2_62/ + 8; (d) y = 5x-3. 4. Plot on the same axes the graphs of (a) 2/"^ = 3 X + 5 and x^ = 3 ?/ + 5 ; (&) 2/ = (x — l)3and x=(y— 1)3; (c) x2 — 4 2/2 = 4 and 'ip- — \x^ —\\ (rZ) y2 = ^3 and x2 = ?/3 ; (e) y = 7? and ?/ = Vx. (-N , ■ 26. Equations whose Graphs cannot be Plotted. — Since real numbers alone can be used as the coordinates of a point, an equation which is satisfied only by imaginary values of the variables does not define a curve and has no locus. Such equations may be most easily distinguished by completing squares. If then every term is positive, the equation is ob- viously satisfied only by imaginary values of x and y. PROBLEMS 1. Show that x2 - 2 X + 2 2/2 + 8 ?/ + 14 = has no locus. Solution. — Completing the squares in z and y, we have (x-l)2 + 2 (2/ + 2)2 = -5. For any real values of x and y, (x — 1)^ and {y + 2)^ are positive or zero-; hence no real values of x and y can satisfy the equation. >- 2. Examine the following equations for the existence of a locus : (a) x2 + 6 X + 2/2 _ 4 ^ _^ 17 _ 0; /^A/) y(^) x2-3x + 2/2 + 42,=_7; (c) (x + 2)2 4- j/2_6y + 8=0; ^^^ y(d) x2 + 16 = 0; VV--J/A: -(g) a;4 + 4?/2 + 4 = 0; *^ ^ , (/) X2 + 4X2/ + 42/2 + 16 = 0. 3. For what values of k has the equation x^ -\- y'^ ■\- k = no locus ? 4. What is the locus of (a) x2 + 2/2 = 0; ih) (X- 3)2 +(2/ + 3)2=0? 38 . ANALYTIC GEOMETRY 5. For what values of k will the locus be imaginary, a point, or a curve in each of the following : (a) x2 + 92/2 = ^•2-4; (6) x'^-^x + y^ + Sy-\-k = 0; (c) x'^ + 4:x + y'^ — Q y - k = ; (d) x'^-9y^ + k = 0. 27. Intersection of Curves. — It follows from the definition of the locus of an equation that a point lies on two curves if and only if its coordinates satisfy the equation of each. Hence we conclude that the coordinates of points of inter- section may be obtained by solving the equations simul- taneously. If there are no real solutions, the curves do not intersect. PROBLEMS 1. Find the points of intersection of the curves x^ + y- = 16 and y'^ = 6 X. Solution. — Solving simultaneously, a; = 2 or -8. When a; = 2, y =±^12; when cc= — 8, ?/=±V— 48. Thus there are but two pairs of real solu- tions, giving the two points (2, Vl2) and (2, — Vi2). The figure shows the curves and their points of intersec- tion. Find the points common to the loci of the following equations and plot the loci. Note. — In many cases the points of intersection and the intercepts, together with considerations of symmetry, will be sufficient for plotting the curves without calculating tables of values. • 6. a;2 + 2/2 _ 25 = 0, ^ 2. 4x-y + 2 = 0, X -\-Sy + 1Q = 0. --/ . 3. x-y = 0, 2/2 _|. 6 X = 0. 4. 2x-\-ny-15 = 0, 7x- by -12 = 0. 5. x + y — S = 0, x^-^y -\-2 = 0. y -2x-\-n = 0. 7. a:2 ^_ 2/2 _ 25 = 0, a;2 _ 3 y _ 21 = 0. S. 8. a:2 + ?/ = 7, y2 _ 81 = 0. 9. 9x^ -1-4 2/2 = 36, 9x^+lQy = 33. COORDINATES AND EQUATIONS 39 10. a;2 _ y2 + 36 = 0, 14. 3a;2 + 4 2/2 := 12, y-2-3x-3'2 = 0. x^-y^ = - 10. 11. y'^ = 2ax, 15. y = x^, 7? = ay. y = x'^ -\-2x. 12. X- + 6 ?/2 z= 20, 16. x'^ + y- -Qx = 0, xy = 4. y- = x^. 13. a:2 + ?/2 - 8 a: = 0, 17. xy = 4, 2/2 _ 8 X - 2 = 0. 2/2 + a;2 = 9. r7-=^ ^ P) ^h-^jJU^^r^, CHAPTER II THE STRAIGHT LINE 28. Equations of the Straight Line. — Several properties of the straight line, such as its determination either by two points or by a point and a direction, lead to relations between the coordinates of its tracing point that can be expressed in the form of an equation. There are several forms of the straight line equation, but for a par- ticular line^ each one may in general be reduced to any of the others. 29. The Point Slope and Two-Point Forms. — Let the line be fixed by a point and a direction. Let the point be P^{x,, y^ and let the direction be given by the slope m = tan a. Let the tracing point be P{x, y). By the slope formula, m = ^ ~ ^S £C -^ X^ or- y-yi = m{x-Xi). (7) This is the point slope form of the straight line equation. It is used in writing the equation of a line when one point and the slope are known. It may also be used when two points are known, for then the slope of the line can be found at once by the slope formula (3). In the latter case it is often convenient to use the two-i)oiyit form, which is de- rived as follows ; Let the line be determined by the points 40 THE STRAIGHT LINE 41 Pi(xi, yi) and P2(^2> 2/2)) and let the tracing point be P{x, y). The slope formula gives us at once m = y ~y^ and m = ^ — ^ • Equating these we have the desired form, (7a) 30. The Slope Intercept Form. — If the slope and the ^/-intercept are given, the line is determined by a point on the 2/-axis (0, 6), and the slope m. Substituting in (7), we have or •^ "}■. y -b = m(x - 0), J ^y y=mx^h, //v^V-^sf which is called the slope intercept form. This form is used in finding the equation of a straight line when the slope and the ^/-intercept are known. Con- versely, if an equation can be reduced to form (8), its locus is a line whose slope is the coefficient of x and whose y-m- tercept is the constant term in the reduced form. 31. The Intercept Form. — Suppose the intercepts of the line are given ; let the ^/-intercept be h and the ic-intercept a. Here the line is determined by two points (a, 0) and (0, h). We have at once, P{x,y) '111 = — Then by (7) a which reduces to ? + «=!. a b (9) This is called the intercej^t form. It is used in writing the equation of a straight line when the intercepts are known. ( MAm-Q \ ^ 42 ANALYTIC GEOMETRY 32. Lines Parallel to the Axes. — Th^ eqaation of a line parallel to the ?/-axis cannot be written in forms (8) or (9), since there is no ^/-intercept. The equation of such a line is obviously ^ X = a. Similarly the equation of a line parallel to the .T-axis has the form /-' , Exercise 1. Derive the intercept form geometrically * when (a) a is positive and h negative ; (6) a is negative and h positive ; (c) both a and 6 are negative. Exercise 2. Derive the slope intercept form geometrically * for the cases given in Exercise 1. PROBLEMS 1. Derive the equation of the line determined by the points (3, 1) and (5, 4). Also reduce the result to the slope intercept form. Solution. — By the slope formula rn = f . Using the point slope form, y-l = i(.T-3), which reduces to 3x — 2y — 7 = 0. To reduce this to the slope inter- cept form, solve for y : y = ^^ — h This form shows that the slope is f and the ?/-intercept is — §. Equations can be readily checked by substituting the coordinates of known points or drawing the line from the intercepts and observing whether or not it satisfies the given conditions. J^ 2. Derive the equation of the line determined by the given points in each of the following, and find its slope and intercepts. (a) (-2,2), (6,5); . (c) (7, o), (_4, -3); (&) (4, 2), (6, - 5) ; ^.(d) (8, - 5), (- 4, 3). * I.e. without using the slope formula. (0 a = -3, a ~6 ' a = -4, a 27r ~ 3 ■ V(k) b = -5, a = tt; (0 b = 6, a = 37r 4 ^■"^ THE STRAIGHT LINE 43 ^Z. Derive the equation of the line («) Passing through (3, 5) and of slope | ; ^^b) Passing through (— 5, 2) and of slope — i ; (c) Passing through (—5, — 1) and of slope 2 ; (d) Passing through (0, 0) and of slope — | ; (e) Passing through (5, 1) and parallel to the y-axis; /(/) Passing through (1, 5) and parallel to the x-axis ; (g) Passing through (8, 2) and of inclination 30° ; (h) Passing through (— 2, 5) and of inclination 135° ; (i) Passing through (1, 5) and of inclination 90°. Alls, (a) 3x-2?/ + 1 =0; (g) x- ^/Sy = 8-2^/S. w 4. Derive the equations of the lines from the given data. . ^ v (a) Intercepts 2 and- '— 6 ; /w^"""^^ lA^) Intercepts - 6 and — | ; (c) Intercepts | and ^ ; (d) Intercepts — | and 2 ; (e) m = I, b=—6; /if) m=-2,b = S; y (g) m = S, a = 4: ; (^) m =— 2, a = — 3; Ans. (a) 6x- 2y. -12=0; (i) x-VSy -\-S = 0. • \^'r^. Find the angle between the 'line* 2x — Zy = b and each of the lines in Problem 4. ^ > Uj^ ^««- («) 37° 52'. .< 6. Keduce each of the following straight line equations to the inter- cept and slope intercept forms if possible, and draw the Unes. (a) 3x-4i/-8 = 0; (e)4x-3y = 0; ..^N ^_l=i. (/)5x + |2/ = 4; • / ' ^^ ^ 3 5 ' {g) 2a;-3?/ + 3 = 0; VW^ <(c) Zx-2y = Q; {h)x-Sy = 0. ^ )^{d)Sy + 2=0; -/s 7. Find the equation of the line passing through the point (1, 3) and forming with the axes a triangle of area 8. 8. Show that all lines for which b = a have the same slope. < 9. The distance between two fixed points is 2 c. Find the equa- tion of the locus of a point moving so that its distances from the fixed points are equal. * More properly we should say " the line whose equation is 2 a; — 3 ?/ = 5," but the close relation between a curve and its equation has given rise to the custom of using the abbreviated phrase. 44 ANALYTIC GEOMETRY 10. The distance between two fixed points is 2 c. Find tiie equa- tion of tlie locus of a point moving so that the difference of the squares of its distances from the fixed points is k. j ^ 11. The base of a triangle is fixed in length and position. Find the M locus of the opposite vertex if the slope of one side is twice the slope of the other and (^d) the base lies on the a;-axis ; (b) the base lies on the ?/-axis. 12. Show analytically that the centers of circles passing through two fixed points lie on a straight line. 13. Derive the straight line equation x = 7iy -\- a, where n = cot a, and a is thq^jintercept. J/a/-^ , xi. Find the equations of the two lines through the origin which tri- A ^ sebt the triangle formed by the line - -f- ^ = 1 and the coordinate axes. '^ 15. Two straight lines of slope — 1 are tangent to a circle whose center is the origin and whose radius is 4. Find the equations of the lines. 16. Find the equation of a straight line through the point (4, 3) and having equal intercepts. 33. The Linear Equation. — Theorem I. Every straight line is defined by an equation of the first degree, in one or two variables. Proof. It has been shown that every straight line cut- ting the ?/-axis may be defined by the equation y = mx -\- b. In case the line is parallel to the 2/-axis, its equation is of the form x = a. Both of these are of the first degree. Theorem II. Conversely, the locus of every first degree equation in one or tico variables is a straight lirie. Proof. The general form of the equation of the first degree is ^x + 5y + C = 0, (10) where A, B, and C may have any values, zero included. When B =^ 0, this may be solved for y, giving ^ B^ b' THE STRAIGHT LINE 45 This is of the form y = mx + h and therefore defines a — 4 — C straight line with slope —^ and t^-intercept -— - • C When B = 0, vre have x = • This is of the form x = a and therefore defines a straight — C line parallel to the ^/-axis and with ic-intercept — — . Thus the theorem is proved for both cases. Note. — For this reason an equation of the first degree is called a linear equation, the term being used by analogy for all equations of the first degree in any number of variables. Exercise 3. Transform the general equation Ax -\- By -\- C = into the intercept form. When is this impossible ? 34. Relations between Linear Equations. — Theorem I. If ill two linear equations in two variables the coefficients of the variables are proportional, the lines defined by those equations are parallel, and conversely. r, W - "' Proof. Let the equations be ^ y ^ / 1, ^ Ax + 57/ + O = 0, and A'x + B'y + C = 0. Solving for y, I V AC. A C 11 = X and y = — -^,x — ^;r,' ^ B B B' B' These are of the form y = mx + b. Hence m = — — and m' = — — • But by hypothesis A: A = B :B' oy A: B = A \ B'. Thus we have ??i = m', and the lines having equal slopes are parallel. A/t - / 3 G) - ^ The converse and the special case where B = B* = are left to the student. Exercise 4. Assume the lines parallel and prove that the coeffi- cients of X and y are proportional. ^ V- <^H -f I V- ^ ^^ - i %^ ^ 46 ANALYTIC GEOMETRY Theorem II. In any tico linear equations, Ax + By+ (7=0, A'x 4- B'y +C' = 0, if AA' = — BB', the lines defined by the equations are x>erpen- dicular, and conyerselv. Proof. Solving the given linear equations for y, as before, we have A , , A' m = ana m = • B B' m. , f A\r A'\ A A' Then ?)i?7i = BJ\ B'J BB' But by hypothesis AA' = — BB', hence mm' = — 1, and the condition of perpendicularity is satisfied. Exercise 5. Assume the hnes perpendicular and prove that AA' = - BB'. Theorem III. If in any tivo linear equations Ax + By + C=0, A'x + B'y + C =0, we have A:A' = B:B'=C:C', the lines defined by the equations are identical, and conversely. Proof. Prom the assumed proportions ^ = 9L ovb = b\ ■ B B'' Also A A' , T f — = — - , or m = ??i as beiore. B B" Hence the lines have the point (0, b) in common and also have the same direction. They are therefore identical. Exercise 6. Assume the lines identical and prove that A:A' = B:B'=C.C'. With the aid of these theorems it is possible to determine by inspection of their equations whether two lines are iden- / THE STRAIGHT LINE 47 tical, parallel, perpendicular, or otherwise. This is of con- siderable aid in checking the solutions of problems involving straight lines. Suppose, for example, that it is desired to write the equation of a line passing through the point (4, — 2) and perpendicular to the line 10 x — 4 ?/ + 11 = 0. From the given equation the slope of the line is f, and hence the slope of the required line is — | (§ 12). Substi- tuting in the point slope equation and reducing, we get 2x + oy-{-2=0. This line is evidently perpendicular to the given line, for AA' = 20 and BB' = — 20. Also it passes through the given point, for 2«4-}-5-— 2 + 2 = 0. The same problem may also be solved by using Theorem II. Any line having the form 4:X-\-10y = k will be per- pendicular to the given line 10 x — 4.y -\- 11 = 0. Since the line passes through (4, — 2), A; = 4 . 4 + 10 ( - 2) = 16 - 20 = - 4. Hence the required equation is 4:X-\-10y = — 4:,ov2x-\-oy-{-2 = 0. (\ 35. Application to the Solution of Linear Equations. — Two equations of the tirst degree in two variables are classified as (a) Simultaneous if they have one solution, w^hich is usually the case ; (6) Incompatible if they have no common solution, as ' 3x -\-7y = 1, \6x + Uy = l', (c) Dependent if they have innumerable solutions, as f 9aj— 6?/ = 3, [Zx-2y = l. It is sometimes convenient to determine the number of solutions without solving. The test by which the class of the equations can be determined is easily deduced from the re- lation between their loci^ and from Theorems I and III, § 34. 48 ANALYTIC GEOMETRY (a) The equations have one and only one solution, if their lines intersect. This is the case when (b) The equations have no common solution, if their lines are parallel. This is the case when A:A' = B:B'. (c) The equations have innumerable solutions, if their lines have more than one point in common and are there- fore identical. This is the case when ■ A:A' = B:B' = C:C\ y-/- 2^- PROBLEMS ji^. Select two pairs each of parallel and perpendicular lines from wie following : (a) y = 2x + ^; (e) x -\- 5y = 0; (b) 2y = -6x-\-9; (/) 6a: + 2^ + l = 0; (c) y = 2« + 9; (g) 5x-y-S = 0] {d) 2y=-\x-\-7 ; ^'{jjA^A V) THE STRAIGHT LINE 49 5. Show that the bisectors of the angles of any rectangle form a square. ^ 6. In the triangle whose vertices gire (0, 0), (6, 0), and (8, 6) find w^^ the equations of its sides ; y^y\^ ^^-'^ 5 ^**K^ the equations of the perpendicular bisectors of the sides ; . (c) the equations of the perpendiculars from the vertices on the opposite sides ; (d) the equations of the medians. ^/V. Show that in the above problem p^a) the perpendicular bisectors meet in a point ; (6) the medians meet in a point; ,.{c) the perpendiculars from the vertices on the opposite sides meet in a point. (In each case show that the coordinates of the intersection of two lines satisfy the equation of the third.) 8. Show that the three points of intersection found in the above problem lie in a straight line. ^^ 9. Show that the medians of any triangle meet in a point. (Take "^he vertices as (0, 0) (a, 0), and (6, c).) --• 10. Show that the perpendicular bisectors of the sides of a triangle meet in a point. -?>-: — 11- Show that the perpendiculars from the vertices of a triangle on the opposite sides meet in a point. 12. Show that the three points of intersection found in Problems 9, 10, and 11 lie in a straight lin6. 13. The vertices of a parallelogram are (0, 0), (6, 0), (4, 8), and (10, 8). Lines are dra^NTi from two opposite vertices to the mid-points of opposite sides. Show that they are parallel and that they trisect one of the diagonals. 14. Find the center and radius of a circle circumscribed about the triangle whose vertices are («) (0,0), (-6,0), (8,4); (6) (4,0), (-4,0), (6,8). Ans. (a) center (—3, 16), radius \/265. !6**-^15. Two opposite vertices of a square are (— 2, 3) and (4, — 5). Find the other vertices and the equations of its sides. ^ 16. Find the fourth vertex of the parallelogram which has three 'of its vertices : (a) (3, 5), (4, 9), (- 2, 6) ; (6) (8, 6), (5, -4), (0, 3). 50 ANALYTIC GEOMETRY 17. Find the equation of a straight line through the origin making an angle of 45° with each of the lines of Problem 1. Hint. — To find the slope of the required line, observe that its inclina- tion is 45° greater or less than that of the given line. Ans. (a) y=— ^xov y = \x. Y z •A B \r/ / b V \ ^ a ^\r 36. Geometric Conditions Determining a Line. — In eacli of the standard equations y = mx + h and - -}- f = 1, we note ah that there are involved two arbitrary constants which rep- resent geometric conditions determining the line. Two of these, the intercepts a and 5, represent distances, while the other, the slope 771, represents a direction. The direction may also he given by the inclination a. There are also two other such constants obtained as follows. Let AB be the given line and OZ a line from the origin perpendicular to it at C. This is called the normal axis of the line. Its inclination is denoted by w, and the perpendicular distance OC of the line from the origin is de- noted by p and called the normal intercept. Thus we associate with any straight line six constants ; three, a, 6, and p, representing distances, and three, m, a, and w, representing directions. Any pair of them (one being a distance) will determine the line (and also the other con- stants), and combined with its current coordinates they will furnish an equation of the line. Of the many possible equations involving these constants only three are commonly used, two of which have been discussed in §§ 30, 31,. 37. The Normal Form. — In this form of the equation the line is determined by x> and w. ^ m y-f^r^^ THE STRAIGHT LINE 51 To derive it, take any point P on the line and draw FD from the foot of the ordinate of P perpendicular to the normal axis at D ; also draw PE perpendicular to FD. It follows at once that Z EFP = w, OD ,^ l!}-p and cos (i) sm cu = X EP y But OD+EP=x>. hy^- .-j'-'^j Substituting, a: cos w + i/ sin (o — /> = 0. This is the normal equation. Here w, the inclination of the normal axis, like a, the inclination of the line, is reckoned from 0° to 180°. The upward direction of p, reckoned from the origin, is taken as positive, and the downward direction, reckoned from the origin, is taken as negative. 38. Reduction of a Linear Equation to the Normal Form. — Let the equation of the given line I be Ax^-By-^C= 0. It is required to find the equation of I in the normal form. A Since the slope of the given line is , the slope of the B B normal axis is the negative reciprocal of this, — A tan o) = — • Elementary trigonometry gives us at once A Hence sina; = B cos CO = A ±V^2 + ^' ±^A'-\-B' Since w < 180°, sin w is positive. Therefore the sign of the radical must be taken to agree with that of B. Thus we have determined sin w and cos co in terms of the coefficients of the given equations. 52 ANALYTIC GEOMETRY Dividing the given equation by V^^ + ^'^ we liave A . B . C -X :2/ + = 0, which is of the form X cos (0 + 2/ sin w — j> = 0. Comparing these equations, we see that p must be -O ■VA' + B' The special cases where A = and ^ = are left to the student. Summing up, we have the following working rule : To reduce the linear equat ion Ax -\- By -{- C = to the normal form, divide it by ± VA^ + B^, giving the radical the sign of B. Example. —Reduce 3x— 4 «/=20 to the normal form. Here VA^ + B^ = 5, and B is negative. Therefore the required equation is o 5 This gives cos w = — f , sin w = f , and p = — 4. Plotting the equa- tion affords an excellent check on the work. The equation Ax-\-By+C=0 can always be reduced to the normal form. For the division by the radical is always possible, since A and B cannot both be zero. This consti- tutes one of the two most important advantages of the normal form. The other consists in its convenience for use in finding the distance from a line to a point. Exercise 7. Derive the normal equation for a line crossing the axes (a) in the second quadrant ; (c) in the fourth quadrant. (6) in the third quadrant; THE STRAIGHT LINE 53 PROBLEMS )( 1. Draw the lines satisfying the following conditions : ^£^ (a) a = -5, a= 30'^; (/) i> = 3, a, = 45°; ^^n(&) 6= 4,«= 60°; ig) a=-6,p = S; \r^ \c) p = - 3, « = 120° ; (/i) 6 = - 5, p = - 4 ; (cZ) a = 5, w = 150° ; (i) « = - 3, ;> = - 3. (e) 6 = - 4, w = 120° ; 3. With ri^Zer and compasses construct the required line, given (a) a and « ; {d) a and w (&) h and « ; (e) 6 and w (c) j9 and «; (/) p and w (^) a and p ; (Ji) h and p. ^3. Write the equations of the lines of Problem 1. (^US^ T Ans./{a) x-V3^ + 5 = 0; {g) 3cc-4?/ + 15 = 0. '>^4. Find the distance of a line from the origin when the intercepts are : \ ^(a) 3 and 4 ; (6) 3 and - 4 ; (c) — 6 and 8 ; y.5. Find the equations of the lines from the following data : ^, >((a) p = - 6, w = 45° or 135° ; , ' . ^ /I (&) J9= 5, w = 150°; ^ ' ' (d) 6 and 9 ; (g) 7 and 2 ; ■X(e) 5 and — 12 ; (h) 7 and Vl5. (/) -5 and 12; A)is. (a) V (c) ;9 = 4, « = 60° or 120° ; {d) p= 3, w = 0°, 90°, or 180°; (e) p= 6, a> = 75°. (sin 75° = cos 15 3o^J l + cos30° N 1^6. Write the equation of the line passing through the point (5, — 10) and having ^ ^a) 0,^135°; (c) p= -2; ^/)oih> (&) 0, = 120° ; {d) p=-6. Ans. (a) x-y -15=0; (c) 4 x +3 ?/ + 10 = 0. y^7. Reduce to the normal form and verify the work by constructing the line from its intercepts in each of the following cases : r • •' {a)y = ?>x + b, *^)3x + 2y = 2, ' «', /l*1- ^^s3 Ms. (a) VTO.-^'-'),' / VA 56 ANALYTIC GEOMETRY y f^^^h^-^^^ Find the distance from the intersection of the lines (a) x-4y- 12 = 3 X — 4 ?/ - 2 = to the line a- + ?/ + 10 = ; ^(6) Sx- y- 2=0 5x+ y—12 = to the line 12 x — 5?/ + 23 = 0. S^4. Find the areas of the triangles formed by the three lines in (a) and (6) in Problem 3. Ans. (a) jf^. 5. Find the areas of the triangles whose vertices are (a) (0, 0), (1, 2), (5, 7) ; (c) (- 7, 8), (- 7, - 6), (1, - 4) ; (6) (-1,3), (2,-4), (0,6); (d) (5,4), (8,3), (6,5). Ans. (a) I ; (c) 56. 6. Find correct to one decimal place the areas of the polygons having the following vertices : (a) (6,0), (7,10), (-3,5), (-2, -6); (6) (2, 0), (1, 4), (- 2, 2), (- 1, - 3), (1, - 2). Ans. (a) 94.5; (6) 17.5. 7. Find the equations of the bisectors of the angles between the lines Sx — 4y = 12 and 12x + 5y = S0. Solution. — Let the lines meet at E ; there are two bisectors EF and EG. Let Piixi, ?/i) lie upon EG, the bisector of the angle BEC. Then by plane geometry the perpendiculars HP^ and KPi are of equal length. To find the length of HPi put 3 X — 4 ?/ = 12 into the nor- mal form. Then HP,= Similarly OXi 5 ^Ml 13 13 30 13' But HPi is negative and KPi positive. Therefore HPi^ = - KPi, or 12 Xi 5yi . 30 5 13 13 13* Simplifying and dropping primes, 21 a; + 77 ?/ + 6 = is the equation of EG. In like manner the equation of EE is 99x — 27?/ -306 = 0. A convenient check on the accuracy of the work is the application of the test for perpendicularity to the two solutions. THE STRAIGHT LINE 57 ^^^"^y^ 8. Find the equations of the bisectors of the angles between , ^ ■ (a) 3a: + 42/ = 5, (6) 3x - 4?/ == 7, ^ .. "^ ^ 4a; + 32/ = 6; 4x-32/ = 8. * x/^. Find the locus of points twice as far numerically from the first line as from the second in each of the groups in Problem 8. Ansrjsia) 5x-\-27j^7 = 0, llx -f lOy - 17 = 0. /^^^ y 10. Find the locus of points the sum of whose distances from each of the pairs of lines in Problem 8 is 12. Ans.'{a) 7x + 7y — 71 =0. 11. Find the locus of points 6 units distant from the line (a) 3x + 4?/ = 5; (c) 15x - Sy = 12. (6) 12x- oy = n ; Ans. (a) Sx-\-iy = S6, Sx + 4:y = -25. 12. A triangle has vertices (3, 0) and (— 1, 3). Find the locus of the third vertex if the area is to be 16. 13. Find the equation of the line halfway l)etween the parallels {a) 2x + y = 7, (b) Sx-2y = 9, 2x+y = 10; Sx-2y = l6. ^ 14. A triangle has the vertices (0, 0), (12^, 0), and (8, 6). y (a) Find the equations of the bisectors of the angles. (6) Show that these bisectors meet in a point. (c) Find the coordinates of the center and the radius of the inscribed circle. 40. Systems of Straight Lines. — When one of the con- stants of a straight line equation is arbitrary and the other absolute, as in y = 3x + k, the equation defines a line for each value assigned to the arbitrary constant, but each line so defined has the property given by the numerical con- stant. Thus in the above equation k is the ^/-intercept, and it may have any value ; but the slope of each line is 3, and by assigning all possible values to k we have all possible lines of slope 3. Such a group of lines is called a system, and the arbitrary constant is called the parameter of the system. We have already seen that any of the standard forms of the straight line equation involves two arbitrary constants 58 ANALYTIC GEOMETRY which correspond to the geometric conditions defining the line. Hence if one of these constants is given a numerical value and the other left arbitrary, vre get a system of straight lilies characterized by the geometrical property defined by the numerical constant, and with one geometrical property left arbitrary or variable. This correspondence is general, as will be seen later in the case of the circle and the conies. PROBLEMS In each of the following problems draw several lines of the system. 1. Write the equations of the systems of lines defined as follows: (a) passing through (5, — 1) ; Ans. ?/ -f 1 = m{x — 5). (&) passing through (— 2, 2) ; (/) having the slope — |; (c) passing through (6, — 3) ; {g) having the slope I ; (c?) with the ^/-intercept — 2 ; {h) distance from origin 13 ; (e) with the a;-intercept — 3 ; (i) distance from origin 5. 2. What geometric conditions define the systems of lines repre- sented by the following equations • Hint. — Put each equation in a standard form. ^y {a) x + k = 0; {e) 2x-ky + 2 = 0; (6) 3 :k + 4 y + ^• = ; lf)kx + ky -Q = ; > (c) x + ?jky-b = 0; ^f gr) kx + !/ VH^ = 6 ; {d) kx-\-2y -Q = 0; (^h) y - 1 = mx + 2 m. Ans. (c) a:-intercept 5 ; (g) normal 6. « 3. Determine k so that 1 (a) the line Sx + iy =k passes through (—2, 1) ; (6) the line Sx — 4y = k passes through (— 3, 4) ; ' (c) the line kx — 4y =— S has the slope 2 ; (d) the line Sx-4y-k=0 has its :c-interce"pt = 3 ; "^ (e) the line 3x — 4^•?/+5 = has its ^/-intercept = — 3 ; (/) the line 4:X— Sy + k = is distant 5 units from the origin. A?is. (a) k=-2; (c) k = S ; (e) k=-j%. V 4. Write the equation of the system of lines perpendicular to ij y = mx + b. 5. Find the equation of the system of lines whose normal axis has the slope — f"^. 6. Find the equation of the system of lines whose normal axis has the inclination — - . 6 THE STRAIGHT LINE 59 7. "What is the equation of the system of lines tangent to a circle having the center at the origin and the radius 10 ? 41. Systems of Lines through the Intersection of Two Given Lines. — Suppose it is desired to find the equation of the system of lines through the intersection of two given lines. One way of doing this would be to solve the two given equations for the point of intersection and to use the point slope equation. A method more expeditious and of great theoretical importance follows. Theorem. Let /j = and 1^ = be the equations* of two intersecting lines I^ and L, and let k be an arbitrary constant. TJien J^-{- k ' L = defines a system of lines passing through the intersection of the given lines. Proof. We first observe that l^-]- k • L = defines a system of straight lines with k as a parameter, since it is obtained by adding two first degree equations. Let the point of intersection P have coordinates (a, b). By the locus definition the equations 1^ = and 1^ = are both satisfied by x = a and y = b, since P{a, b) lies on both curves. That is, the quantities li(a, b) and ^(a, b) are both zero. Therefore ?i(o, b)-\- k • /^(ft, b)= for all values of k. Hence every line of the system Z^ -f ^' • ?, = passes through P, since the coordinates of P satisfy the equation.f * The equation h = stands for an expression of the form Ax-{-B7j + C = 0, and would more properly be written h{x, y)=0, but the variables are omitted for convenience. t We can also prove that every line passing through P belongs to the system. Let /g = be another line through P, and let Pi(xi, y{) be some other point on l^. If the coordinates of P^ be substituted in li + k -12 = 0, we obtain an equation h{^\, 2/i)+A--/2(a^i, 2/i)=0 in which k, being the only unknown, «. an be found. For this value of k, the locus of the equation l^ + k • 1-2 = passes through Pj, since its coordinates make the left hand member equal to zero. As it already passes through P, it must be identical with l^, having two points in common with it. 60 ANALYTIC GEOMETRY Example. — Find the line containing the point (2, 3) and the inter- section of?i: Sx + y — 5 = and Zo: x + 2 y — S — 0. The system Zi + A* ■ Zo = contains the intersection, and k may be found for tlie assumed point by substituting the coordinates of that point in the equation of the system, thus : Sx + y -6 + k{x + 2y-3)=0 becomes 6 + 3 - 5 -f A:(2 + 6 - 3) = 0, and A: = — 4 ; from which we have the result 11 x — 3 ?/ — 13 = 0. Verification. Solving the given equations simultaneously, the inter- section is (I, f). These coordinates and the coordinates (2, 3) both satisfy the new equation. The above proof will apply to any two equations of the form f(x, y) = 0, as well as to two linear equations. The general principle may be stated as follows : Tlie equation f(x, y)-\-k - g(x, y) = defines a system of curves passing through the intersections of the curves defined by the two equations fix, y) = and g(x, y) = 0. PROBLEMS 1. Find by the method of the example given above the equation of the system of lines passing through the intersection of Sx — 2y = 16 and S x + 6 y = I. Find the particular line of the system passing through (— 5, 4). i- 2. Find the line of the above system which has the slope |. Hint. — The equation in k is 3 a; — 2 t/ — 16 + A--(3 a; + 6 ?/ — 1) = 0, or (3^• + 3)x + (6A;-2)^/-fc-16 = 0. Solve this equation for y and the coefficient of x is the slope. V 3. Find the line of the above system which is parallel to the a;-axis ; also the line which is parallel to the ?/-axis. . 4. Find by the method of the illustrative example the equations of the lines determined by V (a) P(4, 3) and the intersection ol 'ix— 2 y = 7 and x — 2y = S ; (6) P(3, 0) and the intersection ofSx—2y = — 12 and Sx + 2y=0; (c) P(— 5, 0) and the intersection oi Sx + 2y = o and x = 6 ; (d) P(— 4, 1) and the intersection of x — y-\-6 = and x+y— 5=0. 5. Find by the method of Ex. 2 two lines of the system containing the intersection of the first pair of lines, one parallel, the other perpen- dicular to the third line : (a) 3 x — 5 y + 5 = 0, 5 X + 4 2/ = 10, and x + y = 6; THE STRAIGHT LINE 61 (b) X + y = 0, X - y = 0, and 2 X + S y + 6 = ; (c) 6x — 5y + i = 0,x + y = l, and x = S; (d) 6 X - 3 y + 9 = 0, 4: X + 3 y = 10 and bx-9y + 10 = 0. 6. Solve Problem 4 (a) by finding the point of intersection and using (7). 7. Find and plot the equations of two curves which pass through the intersections of (a) ?/2 = 8r/:+16, (6) x^ + y^- = 13, 2/2 = 4 X + 16 ; x2 - 2/2 - 5. 42. Plotting by Factoring. — Theorem. If 1^ = and Zg = define the lines Ii and l^, the locus o/ ?i • /^ = consists of these lines. Proof. This follows directly from the locus definition. For the coordinates of a point (a, b) on either line, say l^, satisfy its equation l^ = 0. But if li(x, y) vanishes for x = a and y =b, the product of li(x, y) and /oC^j y) must vanish for the same values, i.e. (a, h) lies on the locus of Z^ • /, = 0. Conversely, if (a, h) is on this locus, one of the factors l^(x, 2/), ?2(^j y) must vanish for x = a, y—h. Hence (a, h) lies on the line defined by the equation formed by setting this factor equal to zero. This proof is made clearer by application to a particular example. Consider the equation x^ — ^ y"^ — X — 2 y = 0. Factoring, we have {x^2y)(x-2y-l) = 0. Now the coordinates of any point on the line x-\-2 y = 0, as (2, — 1) make the first factor zero and hence the product is zero. Therefore all points on .^' + 2 ?/ = lie on the locus of the given equation. The same is true of x — 2 y — 1 — 0. Conversely, if a point lies on the locus oi x^ — 1 y- — x — 2 y =0, its coordinates satisfy the equation and must make at least one of the factors of the left-hand member zero. Therefore 62 ANALYTIC GEOMETRY all points of the given locus belong to one of the lines x-\-2y = 0,x-2y-l = 0. This reasoning a23plies to all equations of the form f(x, y) = 0, whether they deline straight lines or not, although it is useful chiefly for linear factors. It leads to the follow- ing rule. Rule for Plotting by Factoring. — Transpose all terms to the first member. Factor as far as possible ; set each factor equal to zero, ayid plot the resulting equations on the same axes. PROBLEMS 1. Plot by factoring the pairs of lines defined by the following equations : (a) a:2_ 9?/2 = 0; tyj tC(6) center at (— 8, — 6) and radius 10 ; (c) center at (4, — 3) and radius 5 ; (d) center at (— 8, 6) and radius 8; (e) center at (6, 0) 3.nd radius 6. Ans. (a) a;2 + y2 _ 6a- - 10 ?/ + 9 = 0. yC 3. Write the equations of the circles satisfying the following condi- tions and draw the figure in each case : (a) tangent to x axis, radius 3, abscissa of center 6 ; (b) tangent to y axis, radius 4, ordinate of center — 5; (c) tangent to both axes, radius 16 ; (d) passing through origin, radius 5, abscissa of center —4 ; (e) passing through origin, radius 13, abscissa of center — 5; Rw,-i)0 passing through(2, 2), radius 5, ordinate of center — 1 ; <^ ' '^ Ans. (a) x2 + 2/2 _ 12 X ± 6 2/ + 36 = ; (d) x'- + i/+ 8x±6y=0. 4, Prove that an angle inscribed in a semicircle is a right angle. . Solution. — Let the radius of the circle be r; take the diameter as the X-axis and the center as the origin. Let P {x, y) be any point on the cir- cumference. We have to prove that BPC is a right angle. Now the coordinates of B are (— r, 0) and of C, (r, 0). Hence by the slope for- mula, the slope of BP is -^ ; that of CP X— r have mm = , -^ r.2 x-\- r Calling these slopes m and m', we /2_ a;^ — r'^ But P(x, y) is on the circle. Therefore, x2 _|_ y-i — p2^ or ?/2 = /-"^ — :c2. This give s , r2 — x2 -, mm' = — ■ = — 1, 3.2 — f>% which proves that BP and CP are perpendicular, or that BPC is a right angle. v^ 5. Prove without the use of slopes that an angle inscribed in a emicircle is a right angle. 6. Prove that a perpendicular from any point on the circumference of a circle to a fixed diameter is a mean proportional between the seg- ments into which it divides the diameter. 66 , } ANALYTIC GEOMETRY ,■(/>' 7. Prove that angles inscribed in the same segment of a circle are equal. Hint. — Take the chord whose ends are (c, d) and (— c, d) and use the angle formula. 8. Prove that a line from the center of a circle bisecting a chord is perpendicular to it. Hint. — Let the ends of the chord be (— r, 0) and (b, c). 44. The General Equation of the Circle. — If we expand the standard form (13), we get ^\^ Y~Kf/"^ This is of the form i^ ^^...^A..--^ x'^ + y' + Dx+Ey^F = 0, (14) which may be called the general form of the circle equation. Conversely, this form represents a circle, since by complet- ing squares it is possible to reduce it back to form (13), which we know is the equation of a circle of center (Ji, h) and radius r. (Xote exception, § 45.) The general equation of the second degree has the form Ax" + Bxy + Cy'' + Dx + Ey + F=0. Comparison with (14) shows that this can be reduced to form (14) if B — and A = C, by dividing by A. Hence we have the theorem : An equation of the second degree in two variables defines a circle when, and only when, the coefficients of x^ and y"^ are equal and the term in xy is ynissing. 45. Identification of Center and Radius. — To find the cen- ter and radius of a circle when the equation is given, reduce the equation to form (14). Example. — Find the center and radius of the circle 2 ic2 - 8 x + 2 ^2 4- 6 ?/ - 21 = 0. Dividing by 2 and completing the squares, we have a;-2 _ 4 x -h 4 + 2/2 + 3 ?/ + I = -y- + 4 -h f , or {x-2y~+{y + iy=3-^-. THE CIRCLE 67 This is of form (13) and therefore represents a circle whose center is (2, — I) and whose radius is ^V67 = 4.1 approximately. Degenerate Forms. — Sometimes on completing the squares the term corresponding to r"^ is negative or zero. In the former case, it may be shown by the methods of § 26 that the locus is imaginary and we say that the graph is an imaginary circle. In the latter case, the radius of the circle is zero, and the graph is called a point circle. Such forms are sometimes known as degenerate forms. 46. Special Forms. — Comparing the general form (14) and the expansion of the standard form (13) in § 44, we see that D = -2 7i,E = -2k, and F=h^+ A:^ _ ^,2, The first two of these relations enable us to determine the center by inspection. They also enable us to determine the following special forms : ii D = 0, h = and the center is on the ?/-axis ; ii E = 0, k = and the center is on the a;-axis ; if Z) = and E = 0, the center is at the origin ; if i^" = 0, /i^ + k^ = r-, and the origin is on the circumfer- ence. Exercise 1 . Prove by considerations of symmetry that if Z) = 0, the center is on the ?/-axis. PROBLEMS iX^l. Prove that each equation defines a circle. Also determine the center and radius and draw each circle. f,\ (a) (a; + 3)2+(y_4)2 = 25; v'i:^" \^) Xb) (a; + 4)2+(y-5)2 = 0; '/^y-^ j^yt^) 15 x^ + lb y'^ = o X ; ^ y Sy 4 \/2. 7. Find the values of k for which each of the equations of Problem 6 defines a point. /8. Find the points of intersection and the equation of the common chord of the circles : V{a) cc2 + y-2 + 8 2/ = 20, (6) x^ + ?/2 + 4 a; - 60 = 0, :c2 + j/2 _ 8 .c = 4 ; x-2 + ?/2 _ 4 ?/ - 36 = 0. 9. (a) The point (|, \) bisects a chord of the circle x- + y- = 25. Find the equation of the chord and its length. (6) The same for (4, 7) and a:2 + ?/2 _ 4 :>: - 6 ?/ - 12 =0. Ans. (a) 7 s: + ?/ - 25 = 0, 5\/2 ; (6) a; + 2?/-18 =0, 2V5. 10. Find the equation of a circle tangent to the line 4x + 3?/ — 7 =0 and having its center at the origin. 11. Find the locus of the vertex of a right triangle which has the ends of its hypothenuse at (a) (2, 3) and (- 2, 6) ; (6) (- 4, 3) and (0, 5). General Hint. — In each of the following locus problems, the directions of § 16 as to choice of axes should be followed. After deriving the equation, the locus should be identified and drawn, and its relation to tlie given data stated. 12. {a) The distance between two fixed points is 2 c. Find the locus of a point moving so that the sum of the squares of its distances from the points is 4 d^. (6) Solve the same problem when the sum of the squares is any constant, as k. Ans. The locus is a circle, whose center is midway THE CIRCLE 69 between the given points and whose radias is -^ — - • If ^• < 2 c^, there is no locus. ^3. The base of a triangle is fixed in length and position. Find the locus of the opposite vertex if —7 1 (a) the vertical angle is 90° ; y^6) the vertical angle is 45° ; (c) the vertical angle is any constant ; (d) the median to one of the variable sides is constant. 14. The ends of a straight line of constant length rest on two per- pendicular lines. Find the locus of the middle point. 15. Find the locus of a point whose distance from one fixed point is k times its distance from another. 16. Find the locus of a point if the sum of the squares of its dis- tances from (a) the sides, (6) the vertices, of a given square is constant. 17. From one end of a diameter of a circle whose radius 'S r chords are drawn and produced their own length. Find the locus oi the ends of these lines. /is. From the point (- r, 0) chords are drawn in the circle a;2 + y-2 = r^. Find the locus of their mid-points. 47. The Equation of the Circle Derived from Three Condi- tions. — The correspondence between the geometrical condi- tions necessary to determine a circle and the algebraic conditions necessary to determine its equation are analogous to those discussed for the straight line. The number of in- dependent constants in the general linear equation' is two, for if we divide each term by A, we have the form x+B'y-hC' = 0. Geometrically, a straight line is determined by two points, or in general by two geometric conditions. Both forms (13) and (14) of the circle • equation involve three arbitrary con- stants. Geometrically a circle is determined by three points not on a straight line. A circle may be determined in other 70 ANALYTIC GEOMETRY ways than by the condition of passing through three points, but the determining condition is in any case threefold. To derive the equation when the geometric conditions are given, it is necessary to express these conditions in three equations involving h, k, and ?- (or D, E, and F) and solve them simultaneously. The method is illustrated by the following : Example. — Find the equation of the circle determined by (1, 7), (8, 6), and (7, -1). ^ , I ;. , , Each pair of these coordinates tnust satisfy the circle equation (13). Hence (1 - h)'^ + (7 - k^ = r^ \ i*>^-^^ v ^. (7 _ A)2 + (_ 1 _ A,-)^ ^ ^2,^ V^' Solving these equations simultaneously, /i = 4, A: = 3, and r = 5. Therefore the equation is {x — 4)^ +(?/ - 3)- = 5'^, which reduces at once to x^ + ?/2 — 8 x — ^y = 0. We may also solve the problem by using form (14), which gives l + 49 + Z) + 7£' + i^=0, 64 + 36 + 8Z) + 6E'+i^ = 0, Solving, i> = - 8, E=-6, and F = 0. A convenient graphical check on either solution is to draw the circle with compasses and observe whether or not it passes through the three given points. 48. Length of a Tangent. — Let t be" the length of the tangent P^T, let (h, k) be the coordinates of the center O, rj^x'y^) and let r be the radius. Then Using the distance formula, this becomes 1^ = (^x,-hy + {y,-ky-r\ (15) THE CIRCLE 71 It is observed that the expression for t is the same in form as the equation of the circle, with the coordinates of Pi substituted for the variables. Hence, to find the length of a tangent from a given point to a given circle, transpose all terms of the given equation to the first member, substitute in the expression thus formed the coor- dinates of the given point, and extract the square root. ISTote the similarity of this to the formula for the distance from a line to a point. Since the right-hand member of (15) is the same in form as the standard equation of the circle, it may also be written f = X,' + j/r -f Dx, + Ey, -h F. (loo) Example. — Find the length of the tangent from (5, 6) to the circle x" +y- — ix -\- 6y — S =0. Using formula (loa), we have f2 = 52 _[. 6-2 - 4 • 5 + 6 . 6 - 3 = 74. t = VtT PROBLEMS /T Find the equation of the circle determined by the points : <^) (-1,2), (4, 2), (-3,4); (d) (- 5, 0), (0, 4), (2, 4) ; (b) (0, 0), (6, 0), (0, - 8) ; (e) (2, 0), (0, 0) ,(- 2, - 2) ; (c) (1, 1), (1, 3), (9, 2) ; (/) (6, 0), (- 2, 4), (0, - 6). ^ Find the equation of the circle which (a) has the center (3, 4) and passes through (4, — 3) 5 (b) has the center (r, 0) and is tangent to the ^/-axis ; ^ --^^. (c) passes through (2, 0) and (6, 0) and is tangent to the ?/-axis ; ^d) has the line joining (6, 2) and (— 8, 6) as a diameter; (p) passes through (3, 2) and ( — 2, — 1) with center on the x-axis ; (/) circumscribes the triangle whose sides are the lines x = 5, y =-S, x-2y=l ; (g) passes through (9, 8) and is tangent to both axes. Ans. (6) y'^ =2rx — x^ ; (e) 5x2+ 5?/2-8x = 41; ig) x^ + 1/2 _ lOx - 10 ?/ + 25 = or x2 + ?/2 - 58 X - 58 ?/ + 841 = 0. 72 ANALYTIC GEOMETRY ♦^^3. Find the equation of the circle which 1^-/ " (^) is tangent to the line x -\- 2y — 10 = and passes through the ^^^^oints (3, 3) and (0, 0) ; (6) has the center (1, 5) and is tangent to the line Sx — 4y-\-7 = 0', (c) is tangent to the lines 2 x — Sy = and 3 a: + 2 z/ — 13 = and passes through the point (— 4, 6) ; (d) is tangent to the line 3a: — 4?/ = 2at the point (2, 1) and passes through the origin. Ans. (a) (x - 1)2 + (y - 2)^ = 5, or (X--2^^)2+ (y +1)2=8/^5. (c) (X + 2)2 +(y- 3)2 = 13, or (x + 22)2+ (?/-7)2 = 325; ^^-^ (d) 2x^ + 2y^ + 7x-24y = 0. 4. Find the length of the tangent from the point (6, 8) to each circle of Problem 1, page 67. *^5. Find the least distance from the point (6, 8) to each circle of Problem 1, page 67. Ans. (a) v 97 — 5. 6. Find the coordinates of the points of contact of the tangents from the given point to the given circle : (a) x^-\-y^-6x-6y-l = 0, (-4,4), (6) x2 + 2/2=17, (5,3). Hint. — The coustruction of the tangents by the usual geometric process will suggest a method of solution. 49. Common Chord of Two Circles. — To find the common chord of two circles we may solve the equations simul- taneously, to get the points of intersection, and then use the two-point form of the line equation. A better method will now be illustrated. Let the two circles have the equations x'+f—Sx-4.y-U=0, (a) and x''-{-7f—2o = 0. (b) Subtracting, Sx-\-4.y-{-19 = 0. (c) By the theorem of § 41 the locus of (c) contains the points common to (a) and (b). Since it is of the first degree, it defines a straight line. Hence this line is the common chord. e..^i9^^ PC A-- THE CIRCLE 73 The equation of the common chord of two circles can always be obtained by subtracting the equation of one of them from that of the .other, as above. For, since noth- ing above the first degree appears in the equation of a circle except the characteristic terms x^ + y"^, which are always present, these will cancel by subtraction, leaving a linear equation, whose locus contains the points of inter- section of the two circles. Exercise 2. Derive the general equation of the common chord of two circles. Ans. (D— D'yx +(E — E')y + (F— F')= 0. Exercise 3. Prove that the common chord of two circles is perpen- dicular to their line of centers. Hint. — The line of centers of two circles is determined by the points (h, k) and (h', k'). Hence its slope is k-k' Compare this with the y h—h' slope of the common chord, found by solving tlie answer to Exercise 2 for y, and expressing the result in terms of h, k, h', and k'. -■y 50. Radical Axis. — Con- sider the following two ex- amples. Example 1. — Find the equation of the common chord of the circles (x + 3)2 + 2/2 = 25 and (x — 5)2 + y^ = 4, and the intersections of the circles. Applying § 49, the equation is found to be 16 x = 37 ; but the points of intersection are imagi- nary, as is seen by inspection of the equations, since the sum of the radii is 7, while the distance between the centers is 3 + 5 = 8. Example 2. — Find the equation of the common chord of the circles (X + 3)2 + 2/2 = 25 and (x - 5)2 + y^ = 9. Here the distance between the centers is the same as the sum of the radii and the circles are tan- gent to each other at the point (2, 0). Also we have by subtrac- 74 ANALYTIC GEOMETRY tion X = 2. whose locus, passing through the point (2, 0) and being perpendicular to the line of centers, is the common tangent. Thus vre see that the application of the rule for finding a common chord gives a common chord when the two circles intersect, a common tangent when they are tangent, and linally, when tliev do not meet, a line which has no apparent significance, ^^'e shall hnd, however, that the line in all three cases has a common property ; viz. it is the locus of the point from which tan- gents to the two circles are equal. To prove this, let t and f be equal tangents from P to the circles C and C respectively, of which the equations are x'- + f- + D:c + Ey-^F=0 X'- -f ?/- + D'x + E'y ^F' = 0. From § 48, (15a), t- = x'- + ?/- + Dx + Ey -^ F. Also r- = x"- + 1/ + D'x + E'y -f F'. Hence f - t'- = (D - D')x +{E- E')y j^F-F'. By hypothesis, t = f, or t- — t'- = 0. Therefore, (D- iy)x -\- (E - E')y -\- F - F' = (16) is the equation of the locus of P. But this is the equation obtained by eliminating the terms in a;2 and y-. Hence we have the theorem : TJie locus of the point from which the tangents to two circles are equal is the locus of the equation obtained by eliminating the terms in a-^ and y- between the equations of the circles. This locus is called the radical axis. When the circles intersect, it is their common chord ; and when they touch, it is their common tangent. and ^ ^ THE CIRCLE 75 PROBLEMS lA. Find the equations of the radical axes and the points of inter- section of the following pairs of circles : (a) x2 + ?/2_^4x + 6 2/=7, id) y?^y^\^x^-Zy-^2=^, a;24.2/2_2a:-6?/ + 5 = 0; a;2 + ?/2 + 3x + 5?/-28 = 0; (6) a;2+2/2_5x-5?/=-ll, (e) a:2_^y2_2x+4?/-4=0, a:2 + y2_9^0; 2a:2 + 2i/2_4a^ + 6y_7^0; ^c) a;2 + ?/2 + 8a: + 18?/ = l, (/) a:2+7/H7x-5 2/ = 8, 3a;2+32/2 + 20a:+44yzzll; x2 + ?/2+8a:-4?/=3 5. 2. Find a point from which equal tangents may be drawn to each of thertollowing circles : (a) a;2_^y2 + 3^^4.2?/ = 6, i^) x2+y2^4^_32/^5^0, a;2 + 2/^ _ 3 X - 2 ?/ = - 2, 2a:2-f-2?/24.5a:_6y+l =0, a:2 + 2/2_4a;_8y =_12; 332+^2^5 a._4y^5^0. 3. Prove that the radical axes of any three circles taken in pairs meet in a point or are parallel. ^ •^. In each pair of circles in Problem 1 find the line of centers, and /V prove that it is perpendicular to the radical axis. 51. Systems of Circles. — The geometric constants of the circle designated by U^ k, and r determine the center and radius. If two of these constants are restricted by assign- ing conditions to them, numerical or otherwise, the equation • will represent a system of circles defined more or less com- pletely according to the nature of the conditions. Thus, if 7i = 2 and k = 4, we have a system of concentric circles of varying radius. Again, it h = k and r = 10, we have a system of equal circles whose centers lie on the line X — y = 0. Systems having the Same Radical Axis. — If we use the theo- rem of § 41, we find that C+k- C = 0, where k is any constant and C = and C" = are the equations of two circles, will define a system of circles passing through the points of intersection of G and C. But the radical axis of any two circles passes through the points of intersection, real or imaginary ; hence this equation defines a system of circles having the same radical axis. 76 ANALYTIC GEOMETRY The figure below illustrates the system of circles having the same radical axis as 2 1 ,,2 1 4 o' _ qa — and 0. x" + ?/2 - 21 X + 54 These meet at the points (6, 6) and (6, — 6) ; the equation of the radical axis is x= Q. The equation of the system is ^2 + ?/- + 4 a; - 96 + kif + /- - 21 a; + 54) = 0. /fc=-"xr :\i^ I I I I 4\ -is 1 IN I I I I In' the figure the circles drawn in heavy lines are the given circles and the others are the circles corresponding to A: — 11 4 — J- — 1 — 1 — A_3 _6 Note that when k approaches — 1, the circle becomes of infinite size and approaches as a limit the radical axis. CHAPTER IV THE PARABOLA 52. Conies. — The locus of a point ichich moves so that its distance from a fixed point is in a constant ratio to its distance from a fixed line is called a conic. The fixed point is called the focus and the fixed line the directrix. The line through the focus perpendicular to the directrix is called the principcd axis. The constant ratio is called the eccentricity and is represented by the letter e. There are three kinds of conies, classified as follows : e = 1, the parabola ; e < 1, the ellipse ; e > 1, the hyperbola. 53. The Equation of the Parabola. — Let AB be the direc- trix, F the focus, and DF the principal, axis of the parabola. By the definition of the parabola, there is a point 0, halfway between the focus and the directrix, which lies upon the locus. This is called the vertex. To derive the equation take the prin- cipal axis as the a:-axis and the vertex as the origin. Call the distance between the directrix and the focus p. Then F has the coordi- nates [-^, |, and the directrix has the equation x = D Y y"^ JS D F(P/2,0) X 1 V 1 II \v A \^ 2 .\ c/ 77 78 ANALYTIC GEOMETRY . Let P{x, y) be any point on the locus. Draw NP perpen- dicular to AB, and join FP. By definition FP = XP, that is, yj(x-ijj+f = x+^. Squaring and reducing, y' = 2px. (17) This is the standard equation of the parabola when the vertex is taken as the origin and the principal axis as the a>-axis. If the principal axis is taken as the ?/-axis and the vertex as the origin, the variables are interchanged and the equation is x' = 2py. (17a) If the focus is in a negative direction from the vertex, the right-hand members of these equations have minus signs. The simplicity of these equations is due entirely to the choice of the coordinate axes. If the axes are chosen in any other way, a different and more complicated equation is obtained, but the curve itself is unaltered. Therefore we shall study -the properties of the parabola and other conies by means of the simple forms and reserve the general equa- tions for a later chapter. Exercise 1. Derive equation (17a) from a figure. 54. Discussion of the Equation. — The form of equation (17) shows that the parabola is symmetrical with respect to the principal axis, and that the parabola crosses this axis at the vertex only. Since negative values of x make y imaginary, the curve lies wholly to the right of the vertex and has no point nearer the directrix. It recedes indefinitely from both axes, since y = ± V2 px increases indefinitely with x. A different parabola is obtained for each value assigned to 2^- THE PARABOLA 79 f Y L J .^_- B X F [1 L' V D X A Exercise 2. Derive the equation of the parabola, taking the focus at the left of the directrix, and discuss as in this section. Ans. y- =— 2 px. 55. Latus Rectum. — The chord L'L through the focus parallel to the directrix is called the latus rectum. By definition of the parabola, the distance of L from the focus is the same as that from the directrix. Hence FL = LN=FI) = i) and the length of the whole latus rectum is 2 p. 56. To Draw a Parabola. — Consider the equationa.'^ = — 8y. This is of the form x"^ = — 2 py, and hence is the equation of a parabola. To draw the curve we can make a table of values as usual, or proceed according to one of the two following methods. Method 1. From the form of the equation, 2^j = 8, or J) = 4. Since the sign before 2 p is minus, the curve lies wholly below the ic-axis. Hence the focus is f 0, — -^ V 2 or (0, — 2), and the equation of the directrix is 2/ = 2. Meas- uring off p = 4 to the right and left of the focus gives the ends of the latus rectum. These together with the vertex make three points on the curve, which suffice for a sketch, as the general shape of a parabola is known. 80 ANALYTIC GEOMETRY Method 2. If a more accurate graph is desired, find the focus and directrix as above. Then with the focus as a center and ^ v-arlina -,. \-? T A 5 -Y . \ F "^^ r \ V a radius r > ^- 9 draw an arc. This will cut the line parallel to the di- rectrix and r units from it in two points which lie on the parabola by definition. By using coordinate paper and varying r, as many points as are desired mav be readilv constructed. 57. Mechanical Construction of the Parabola. — Place a right triangle with one leg CE on the directrix. Fasten one end of a string whose length is CD at the focus F and the other end to the triangle at D. With a pencil at P keep the string taut. Then FP = CP ; and as the triangle is moved along the directrix the point P will describe a parabola. PROBLEMS 1. "With coordinate paper and compasses plot a parabola having the latus rectum : , , (a) 12; {h) 16. 2. Find the coordinates of the focus and the equation of the direc- trix of each of the following parabolas and draw the figure to scale : (a)2/2^8a;; (c) 2 x^ + 5?/ = ; (e)?/2 = 6aa;; 3. What is the latus rectum of each parabola in Problem 2 ? THE PARABOLA 81 4. Write the equations of parabolas satisfying the following data : (a) directrix ?/ =— 8, focus (0, 8) ; ^/(b) directrix x = 6, focus ( — 6, 0) ; ^(c) directrix y = 8, vertex (0, 0) ; {f^{d) latus rectum 12, vertex (0, 0) ; ^^ie) vertex (0, 0), one point (3, — 4). j^y^ Ans. (a) x^ = 32y. Derive the equation of the parabola when the principal axis is (a) as the :/>axis, with the origin at the focus ; (6) as the y-axis, with the origin at the focus ; .l/^) b^ the a"-axis, with the origin on the directrix ; A^) as the ?/-axis, with the origin on the directrix. Ans. (a) 2/2 = 2px +p^; (c) y^ = 2px — p^. 6. Find the locus of the center of a circle passing through a given point and tangent to a given line. 7. Find the equation of the circle circumscribing the portion of the parabola x~ = 2py cut off by the latus rectum. Ans. 2x^ + 2y^ — bpy = 0. ^8. Find the points of intersection of the parabolas x^ = 2py and y^ = 2px. 9. In the parabola y^ = 2px an equilateral triangle is inscribed with one vertex at the origin. Find the length of a side. Ans. 4p Vs. -XIO. Show that the distance of any point of the parabola y"^ = 2px from the focus is a: + -. (This is called the focal radius of the point.) 11. Show that the latus rectum of any parabola is a third propor- tional to the abscissa and ordinate of any point upon it. 12. Find the area of the right triangle inscribed in the parabola y^ = 2px which has a double ordinate as a hypothenuse and a vertex at the origin. Ans. ip'^. flZ. One end of a chord through the focus of a parabola which has the a;-axis as the principal axis and the origin as the vertex is the point (—8, 8). Find the coordinates of the other end. 82 ANALYTIC GEOMETRY 58. The Parabola as a Conic Section. — In tlie right circular cone 0-AB pass a plane through the yertex and the diameter of the base AB. It will be perpendicular to the base and will contain the elements OA and OB, and also the axis and the center of every section parallel to the base. Draw a plane perpen- dicular to the plane OAB intersecting it in tlie line MX parallel to OA, and cutting the surface of the cone in CXD. Through P, any point of CXD, and X pass planes par- allel to the base, inter- secting the plane OAB in HX and FG, which are the diameters of the circular sections thus formed. Now QP, the intersection of the }»lanes FPG and CXD, is perpendicular to OAB and hence to the lines FG and XM. Taking X as the origin and XM as the ^axis, the coordinates of P are x = XQ and y = QP. Then i/=qP'=Fq By elementary geometry, QG. FQ = HX, and Sg = ^^, or QG = NQ OH' HX OH X. Substituting, if = HX' OH •T. As HX and OH are constants, being independent of the position of P, ^^ may be taken as 2 p. Thus the section CXD is a parabola. THE PARABOLA 83 59. The Path of a Projectile. — A ball is thrown horizon- tally from a point above the ground. Find its path, neglect- ing air resistance. Let V be the velocity of projection. Take the point of projection as the origin and a horizontal line through it as the a^axis. Let P{x, y) be the position of the ball after t seconds. If V is measured in feet per sec- ond, the ball will have moved horizontally a distance vt feet, that is, x = OQ = vt. But during the same time, by a principle of physics, the ball will have fallen a distance i gt"^ where g = S2 nearly. Hence Eliminating t between these equations. fi — 2^2 -■ — y- 9 Therefore the locus is a parabola having its vertex at and its latus rectum of length Z V' 9 The figure is drawn for -y = 64 feet per second. Later we shall find that the path of a projectile hurled in any direction is a parabola. PROBLEMS 1. The altitude of a right circular cone is 10 inches and the diameter of the base is 8 inches. If the vertex of a parabola constructed as in § 58 bisects an element of the cone, find its latus rectum, Ans. — \/29. 29 2. If two parabolic sections are made as in § 58, prove that their latus rectums are proportional to the distances of their vertices from the vertex of the cone. 84 ANALYTIC GEOMETRY 3. If the vertex of a parabolic section is made to approach the vertex of its cone, what is the limiting form of the parabola? of its equation ? 4. A bomb is dropped from an airship traveling east at 30 miles per hour at a height of 1 mile. Write the equation of its path, re- ferred to the starting point as an origin. How far east will it be when it strikes the ground ? Ans. 799 feet. 5. A ball was thrown horizontally on a level field at a height of 5 feet and struck the ground 100 feet from the starting point. What was its initial velocity ? " Q. An aviator flying horizontally at 45 miles per hour wishes to lit a target on the ground. He estimates his height above the ground at 1000 feet. How far away from the target should he be when he drops his projectile ? Ans. 522 feet. 7. An airship which flies at 30 miles per hour has a device for re- leasing a projectile when it is directly over the center of the target. How large must the target be to insure hitting it when the height -of the airship is (a) ^ mile ; (6) 2 miles ; (c) li miles '? Ans. (a) 1130 feet wide. '~ 8. Sketch the parabolas satisfying the following data by finding the axis, vertex, and ends of the latus rectum : ^ J (a) directrix y = 5, focus (4, — 3) ; X / Ans. Axis x = 4, vertex (4, 1), ends of latus rectum / (12, -3) and (-4, -3). y^^/^b) directrix x = -S, focus (— 1, 4) ; ^■j^^^^^ji^ (c) directrix y =Q, focus (4, — 2) ; '' (d) directrix x = 6, focus ( — 6, — 6) . '^-'9. Derive the equations of the parabolas of Problem 8. ,, ' . / Ans. (a) x^—Sx+16y = 0. 10. Show that the circle x- — 2 ax + y'^ = and the parabola y'^ = 2px meet in but one point unless a>p. n.1. Find the locus of the mid-points of the ordinates of the parab- ola y^ = 2 px. 12. Find the locus of the ends of the latus rectum of the parabola y'^ = 2px when p is allowed %o vary. CHAPTER V THE ELLIPSE 60. Fundamental Constants. — In § 52 the ellipse was de- fined as a conic of eccentricity less than 1. In the figure let AB be the directrix, F the focus, and DF the principal axis. . .p — r**- — -« — >k a — Jy' Since e < 1, there are two ^ points V and V on the principal axis, one on either side of the focus, satisfying d the definition of the ellipse. These are called vertices, the point C midway between f^ them is called the center, and the distances between Fand C, and F and C designated by a and c respectively. Be- tween the four constants a, c, p, and e exist certain im- portant relations which we now proceed to derive. Applying the definition of the ellipse to the points Fand V, we have FV' = eDV', or a-{-c = e(c-\-p + a) and VF=eDV, ov a — c = e{c+p — a). ^ Adding and subtracting these equations, a = e(c-j-p), or c + /> = -, (18) e and c = ae. (19) The first relation gives the distance between the center and the directrix in terms of a and e, and the second that between the center and the focus. Exercise 1. Show how to find Fand F' with ruler and compasses, when the directrix, focus, and eccentricity are given. 85 86 ANALYTIC GEOMETRY 61. The Equation of the Ellipse. — To derive tlie equation of the ellipse take the principal axis for the a:-axis and the center for the origin. Then F has the coordinates (— ae, 0) and the directrix has the equation it' = e Let P be any point of the locus. Then FP=e'NP. By the distance formula z> Y ^^-r,?/) N (..^■' v\ iF{-ae,o)0 Jd jv'X > J II _»-..^^ b ' d' FP = V(x + 0£f H- 2/^ and Substituting these values e V(a.' + ae)- -{- y"- = e[ x -\--\= ex -\- Squaring and collecting terms, (l-e'-)x^ + f- = a\l-e''). a. Dividing by a-(l e2), ^ + y =1. Since e < 1, the quantity a-{l — e-) is positive and < «-. Calling this 6-, we have the standard form of the equation of the ellipse when the center is taken as the origin and the principal axis as the x-axis, namely : 1, (20) or 62x'2 + a-if = ci-h- 62. Discussion of the Equation. — Intercepts and Extent. — The intercepts of equation (^10) are obviously ± a on the a>axis, and ± 6 on the 2/-axis. Solving for ?/, we have y=±- Va- — .i--, a form which a shows that all values of x numerically greater than a must THE ELLIPSE 87 be excluded. Similarly all values of y numerically greater than h must be excluded. Hence the ellipse lies wholly within the rectangle whose sides are x = ± a, y = ± b. Axes of Symmetry. — The form of equation (20) shows that the ellipse is symmetrical with respect to both the O/'-axis and the y-axis, and hence is symmetrical with respect to the origin 0. The segment of the prin- cipal axis intercepted by the ellipse, namely W, is called the major or trans- verse axis. The segment BB' of the 2/-axis is called the minor or conjugate axis. The major axis is of length 2 a and the minor axis of length 2 6. is called the center of the ellipse and is the mid-point of every chord of the ellipse whicli passes through it, since it is the center of symmetry. From the definition of b (§ 61), we have at once E Y E' X B --.^^ n' D V\ F M F'JV D'X whence (21) a relation showing that a is always greater than either b ore. As equations (18), (19), and (21) are mutually independent, all of the five constants of the ellipse can be found if any two of them are given. Exercise 2. minor axis. Find the distance from the focus to the ends of the 63. Second Focus and Directrix. — On OV^ take OF' = 0F and OB' = OD. Draw D'E' parallel to DE. Then F' is also "a focus and D'E' the corresponding directrix of the ellipse. This is shown as follows : 88 ANALYTIC GEOMETRY Let P be any point on the ellipse. Draw F'P, and draw PN' perpendicular to D'E'. Calling FP = e • PN\ and sub- stituting the corresponding values, the result is V( ae xy + ?/' = e' — ^ Squaring both sides and col- lecting terms, this reduces to equation (20). Hence the ellipse of eccentricity e and having D'E' for directrix and F' for focus is the same as that referred to DE and F. Therefore the ellipse has two foci, (± c, 0), and two direc- trices, X = ± a PROBLEMS l/'l. Find the semi-major axis, tlie semi-minor axis, the eccentricity, the coordinates of the foci, and the equations of the directrices of the following ellipses : (a) 9 x^ + 2b y'^ = 225. (SoZw^io/i. — Dividing by 225, we have 25^9" ' which is of the standard form. Hence a = 5, and b = 3. By (21) c = 4 and by (19) e = |. The foci are (± 4, 0) and the directrices x =±6 J. (&) 9^2 + .36 ?/2 = 324 ; V(e) x^ + 2y^ =2; '-'■y k &, « = 5 and 6 = 3, and the equation is of the form a2 62 Hence the vertices are (0, ±5), the foci (0, ±4), e = |, the directrices are ?/ =± 6|, and the latus rectum ^i-. -l 2. Investigate the following equations as in Problem 1. /• (a) 36a:2 + 4?/2 = 144; (d) 9 x^ + ^y'^ = 4:-, YaA- ^h) 3x2 + 21/2 = 1; (e) 16 a;2 + 9 ?/2 = 144 ; V (c) 5 x2 + 3 ?/2 = 1 ; (/) 25 x^ + 4y^= 100. '{^ fi ' 3. Write the equation of the ellipse with the center at the origin, given : >{a) focus (0, 6), directrix y = S; .^(&) vertex (0, 4), focus (0, 3) ; (c) directrix x = 6, distance between foci 7 ; (d) latus rectum —, vertex (0, — 13) ; lo (e) one end of minor axis ( V2, 0), latus rectum 2. Ans. (a) 4 x2 + 2/2 ^ 48 ; (^d) 169 x2 + 25 2/2 = 4225. My^. Write the equations of the following ellipses with the center 1, there are two points of the hyperbola, V and F', on the principal axis, one on each side of the directrix. As in the ellipse these are called vertices; the point C midway between them is called the center ; and the distances between V and C, and F and C are des- ignated by a and c respectively. Applying the definition of the hyperbola to the points V and V\ we have and VF= e • DV or c— a = e[a —(c—p)'], y'F= e . V'D or c -\- a = e^a -\- (c — jj)]. Subtracting and adding these equations, a = e(c—2J) or c - p a and e c= ae. (22) (23) 96 THE HYPERBOLA 97 The first relation gives the distance between the center and directrix in terms of a and e, and the second that be- tween the center and focus. Exercise 1. From the definition of the hyperbola find D V smdV'D in terms of p and e. . p e±l Exercise 2. Using the results of Exercise 1, show that C is to the left of the directrix. 72. The Equation of the Hyperbola. — To derive the equa- tion of the hyperbola take the principal axis as the a>axis and the center as the origin. Then F has the coordinates (ae, 0), and the directrix has the equation x = -- e Let P be any point on the locus. Draw PiV per- pendicular to AE and join FF. Then FF=e . NF. But FF = V{x - aef -f 2/^ and NF = x a Hence ^(x — aey-\-y^ = elx — -\ Squaring and collecting terms, (1 - e2)a;2 -f 2/2 = a\l - e"), or t a2(l - e2) = 1. In this form the equation of the hyperbola appears identi- cal with that of the ellipse (§ 61). However, e > 1 and a2(l — e^) is negative. Calling IP- = (^{^^ — 1), we have the hyperbola equation in its usual form, when the center is 98 ANALYTIC GEOMETRY taken as the origin and the principal axis as the x-axis, namely, ---=1, (24) or 61^2 — ahf = ci^W. 73. Discussion of the Equation. — Intercepts and Extent. — Solving equation (24) for x, we have x= ±- ^if + 5^, which h shows that the .T-intercepts are ± a, and that no value of y needs to be excluded. Solving for y, y =z ±- -y/x^ — a"^, showing that there are a uo y-intercepts and also that y is imaginary for all values of X between ± a, but if x is numerically greater than a, y is real and increases indefinitely as | x | increases. Thus the hyperbola consists of two infinite branches, lying without the lines x = ± a. Symmetry. — The form of (24) shows that the hyperbola is symmetrical to the x- and ?/-axes and hence to the origin. The segment V'Voi the principal axis is called the transverse axis and is of length 2 a. The segment B'B of the ?/-axis, where B'O = OB = b, is called the conjugate axis. From the definition of 6 in § 72 we have at once c- = a2 + &2^ (25) a form showing that for the hy2:>erlx)la c is greater than either a or 6, while there is no restriction on the relative sizes of a and h. It is for this reason that the terms major and minor axis are not used. 74. Second Focus and Directrix. — As in the case of the ellipse there is a second focus and a second directrix. For on the principal axis take OF' = OF and 0Z>' = OD, and THE HYPERBOLA 99 erect the perpendicular D'E'. Then the same curve will be generated by usmg the focus F', the directrix D'E', and the same value for e. For we have F'P = ey'P- but F'P = V(ae + if + y\ and .Y'P = .r + ^^ Therefore and this reduces to (24). Hence the hyperbola has two foci, (±c, 0), and two directrices, x = ±-' 75. Latus Rectum. — The chord L'L through either focus perpendicular to the principal axis is called the latus rectum and it is seen (as in the case of the ellipse) that its length- 2b'^ IS a Exercise 3. Show that the length of the latus rectum is also 2 ep. PROBLEMS 1. Find the semi-transverse au{ semi-conjugate axes, the coordinates ol the foci, the eccentricity, and the equa- tions of the directrices of the hyperbola having the equation 25 x- — 9 y- = 225. Solution. — Dividing by 225, we have the standard form 9 25 Hence « = 3, 6 = 5, c= VsJ, and e = iV^. The foci are (± V34,0), and 9 the directrices are a: = ± Vsi As the we lay points. vertices (zb 3, 0) give only two points, off the semi-latus rectum — = ^ from the foci, getting four more 100 ANALYTIC GEOMETRY / 2. Find the properties of the following hyperbolas as in Problem 1 : (a) 9x--^-16y2= 144; (d) x^ - 2y' = l; VV\> ^(&) 25x2 _ 9^-2 = 225 ; "(e) 3x2 - 5 2/2 ^ 15 . /^'X ^'^ ""'" 9^2 ^9. (/) 25x2 -15 2/2 = 144. ^ ^-:.3- i'i"d the equation of the hyperbola having its center at the k\ rajaigin and its foci on the x-axis, if y(a) a = 4, 6 =-; (d) a = 5, (e) c = 5, e = 3' a=2\/G; ^ (c) 6 = 4, e = 2 ; )((/) c = V5, e = V5 ^, ^?iS. ep e2-l y^4. Express a in terms of e and p. 5. Express c in terms of e and p. 6. Find the locus of a point moving so that the difference of its distances from the points (i 13, 0) is 24. A71S. 25 x2 - 144 ?/2 = 3600. 7. Solve Problem 6 when the points are ( ± c, 0) and the difference of the distances is 2 a. 8. Define the hyperbola from the conditions of Problem 7. ^ 9. Find the equation of a hyperbola having its center at the origin \iE«id its foci on the x-axis and passing through : (a) (3, 2) and (- \/3, 1) ; (ft) (3, 0) and (5, - 4). 10. Find the equation of a hyperbola whose vertex is midway be- tween the center and the focus. Ans. Sx^ — y- = S a?-. 76. Hyperbola with Foci on the F-Axis. — If we describe a hyperbola with the foci on the i/-axis, with its transverse axis 2 a, distance between the foci 2 c, and eccentricity e, the effect on the original equation will be to inter- change the variables x and 2/, and the equation will be- come a2 IP- or hhf — ci?x^ = (J^IP-. The vertices of this hy- F V r^ P D 1 Q X U' ,' E' THE HYPERBOLA 101 perbola are (0, ± a), the foci (0, ± c), and the direc- trices , a e Exercise 4. Derive equation (24a) as in § 72. 77. Focal Radii. — As in the ellipse the distances of any point on the hyperbola from the foci are called the focal radii and designated by p and p. From the ligure of § 74 we have / p = FP= e'NP=elx--]=ex- a, and p' = F'P = e.N'P = e(x-^-]=ex-\-a. Subtracting these, we have p — p = 2 a. Hence the dif- ference of the distances of ciny point of the hyperbola from the foci is constant and equcd to the transverse axis. This agrees with the results of Problems 7 and 8, page 100, which show that the hyperbola may be defined as the locus of a point moving so that the difference of its distances from two fixed points is constant. 78. Asymptotes. — If a line is drawn through the center of a hyperbola intersecting it at P, and P is made to move off to infinity along the curve, the line will turn about and approach one of two limiting lines. We call these asymptotes, defin- ing them as lines through the center of the hyperbola along which the curve re- cedes to infinity. The equation of any line PP^ through the center or origin is ?/ = 7nx. The coordinates of its intersections with the hyperbola are given by solvmg this simultaneously with \ \ Y y^^p / J' ;?o X ^ V 102 ANALYTIC GEOMETRY -^ = 1. Substituting, we have = 1, which gives ^^ x = ± Now if P is to move off to infinity, x must become in- finite, i.e. the denominator must become zero. This gives 6^ — ahn~ =0 or m = ± - . a Hence the slopes of the asymptotes are ± - and their a equations y = ±lx, (26) a or bx ± ay = 0. These equations may be combined by § 42 in the form b^x^-aY=0, or ^-^1 = 0. (26a) a^ b~ In case the vertices are on the ?/-axis, x and y are inter- changed, giving -- — — = 0. In either case, to find the equa- o} b'^ tions of the asymptotes, ivrite the equation of the hyperbola in standard form, replace the 1 by 0, and factor. 79. Conjugate Hyperbolas. — Since the relative sizes of a and b are immaterial, a hyperbola may be described having the segment 2 b for a transverse axis and the segment 2 a for a conjugate axis. The hyperbola with transverse axis A' A (= 2 a) along the a^axis and conjugate axis B'B (= 2 b) /v»2 nil along the ?/-axis has the equation ^= 1. The hyper- bola with transverse axis B'B (=2 6) along the ^/-axis and THE HYPERBOLA 103 conjugate axis A' A (= 2 a) along the ic-axis has the equation y~ X- =1. The two hyperbolas («) ^-!^=i =»"'^ (*) ?!-^!=i are called conjugate hyperbolas. The transverse axis of each is the conjugate axis of the other. Since c^ = a- + 6-, the foci of both hyperbolas are at the same distance from the center. Those of (a) are ( ± c, 0) ; those of (h) are (0, ± c). If we let e^ and e, be the respec- tive eccentricities, we have ei= -, eo= -, whence e^ : 62= & : a. a ' h The directrices of (a) are x = ± -, oi (b) y = ±—. From the equations it is evident that conjugate hyperbolas have the same asymptotes. This property is useful in sketching a pair of conjugate hyperbolas. Draw the rectangle of the given axes, that is, a rectangle having its sides of length 2 a and 2 b and sym- 104 ANALYTIC GEOMETRY metrical with respect to the axes, and draw the diagonals. As these have slopes ± - , they are the asymptotes. The circle a circumscribing the rectangle has the radius V'a^ -\-b~ = c, and therefore intersects the axes at the foci of the required hyperbolas. AVith the rectangle and asymptotes as guiding lines the curves can be sketched accurately. Exercise 5. Prove that the directrices of a hyperbola cut the con- jugate hyperbola at points on the circle drawn through the four foci. When a = b, the equation i/ = a- 80. Equilateral Hyperbolas. of the hyperbola reduces to and tlmt of the conjugate to 7/2 _ 3^2 _ ^2^ The asymptotes are inclined to the axes at 45° and are per- pendicular to each other. Such hyperbolas are called equi- lateral or rectangular hyperbolas. 81. Construction of a Hyperbola by Points. — A hyperbola may be plotted accurately by forming a table of values from the equation or by proceeding according to methods similar to those used for the ellipse (§ 68). PROBLEMS 1. Find the semi-transverse and semi- conjugate axes, the coordinates of the foci, the eccentricity, and the equations of the directrices of the hyperbola whose equation is 25 x- - ^ y^ -{- 225 = 0. )SoZ?J (c) focus (1, 2), directrix a; = — 4, e = 2; (d) focus (3, —4), directrix 1/ = 0, e = _ 4 3 > (e) focus (—2, 3), directrix 2/ =—1, e = f . ^ 7. Derive the equations of the hyperbolas of Problem 6 y^ / z^.. ^^*- («) 3 ic2 - 84 X - 2/2 + 4 ?/ + 536 = 0. '8. Show that the distance of any point on an equilateral hyperbola from the center is a mean proportional between the focal radii. 9. Show that for a pair of conjugate hyperbolas e^ + ei^ — e^e'?. 82. Mechanical Construction of the Hyperbola. — Place thumb tacks in a drawing board at F and F' ^ the foci. Let a string be tied to a j)encil at P and looped about the tacks as in the figure. If the ends are drawn in to- gether, P will describe a hyperbola since {F'F+FP)-F'P ^' \ remains constant. For F'F is constant ; hence F'P — FP is a constant. 106 ANALYTIC GEOMETRY 83. The Hyperbola as a Conic Section. — Let the cutting plane intersect the given cone so as to meet some of the elements produced. Follow the construction of § 70 and choose the axes in the same man- ner, viz., origin at C, and a; -axis 3IX. Then Also FQ^MQ ^^^^ QG^m, KC MC HC CN Substituting, ^ MC'CN ^ or if MC = CN=a and KC-IIC = h\ ¥ Vf Z)^ f = ^aiQ'NQ). But whence 3IQ = X -{- a, NQ = x — a, ^ a^- ^ (£- If- Hence the plane cuts the surface of the cone in one branch of a hyperbola and the conical surface formed by prolonging the elements through the vertex in the other branch. CHAPTER VII TRANSFORMATION OF COORDINATES AND SIMPLIFICATION OF EQUATIONS 84. Change of Axes. — The position of the axes of coordi- nates to which a given locus is referred is arbitrary. It may be changed at will and the equation of the locus altered to correspond by substituting for the former coordinates their values in terms of coordinates measured from the new axes. The advantage to be secured by a change of axes is usually a simplification of the equation, and this is best secured by choosing a position of symmetry when possible. When the new axes are drawn parallel to the old, tli6 transformation of the equation is called a transformation by translation. When they are drawn through the same origin oblique to the axes but still perpendicular to each other, the transformation is called a transformation by rotation. 85. Formulas of Transla- tion. — In the figure let the original set of axes be OX and 0Y\ the new axes O'X' and 0' Y' with the origin at 0', which has the coordi- nates (li, Jc) with reference to OX and OY. Let P be any point in the plane which has coordinates (x, y) with reference to the old axes, and (x^, y') with reference to the new ones. Then 107 1 Y Y P(x,y) D E 1 {^\u) O' {h,k) B\ X' C A X 108 ANALYTIC GEOMETRY (27) x = EP, a/ = DP, h=--OC = ED, y = AP, y' = BP, k = CO' = AB. From the figure EP = DP - DE = DP -\- ED and AP= BP + AB. Hence x = x' + h, and y = y' + k. These are the formulas of translation. By using these substitutions a given equation is trans- formed into a new equation in x' and y' , which is the equa- tion of the locus of the given equation with respect to coordinate axes drawn through the point (Ji, k) parallel to the old axes. 86. Formulas of Rotation. — Let the original axes be OX and OY, the new axes OX' and OY', and the angle of rota- tion 6. P is any point in the plane with coordinates (x, y) and (xf, y') with refer- ence to the old and new axes respectively. Then X =0A, xf = OD, y = AP, y' = DP. Since the sides of angle CPD are perpendicular to those of ^, OPi) = ^. Hence we have CD = y sin (9, CP = y' cos 0, OB = x' cos 0, BD = x'smO. But x=OA=OB- CD and y = AP=BD+ CP. Substituting, we get X = x' cos — y' sin 6, (28) 1/ = x' sin 6 + y' cos 6, which are the formulas of rotation. TRANSFORMATION OF COORDINATES 109 These substitutions transform a given equation into a new equation with variables x' and ?/', in which the coordi- nate axes are drawn through the old origin but inclined at an angle Q to the old axes. PROBLEMS General Directions. — Usually the equation obtained by transforma- tion is a known form and its locus can be identified. When this is not the case, the table of values should be computed from the new equation and the curve plotted on the new axes, since they are axes of symmetry. Fre- quently the intercepts with respect to both old and new axes, together with considerations of symmetry, make an extended table of values un- necessary. V >4ua Translate the origin to the point indicated and transform the ,^ quation to correspond. Draw the curve and both sets of axes. y^ (a) 4x2-?/-^ -8 X + 4?/ + 4 = 0, (1, 2). A)i8. 4x'"2— ?/'2 -f-4 = 0. ^(6) 2x2 + 2/- 5 = 0, (0, .5); (c) 2/2 _ 8^^ ^_ 24, (.3,0); n r ^(d) x2 4- 2/- = 6x - 4 2/, (3, - 2) ; r :^.t^ \ . (e) 2/ = ^' + 6x2+8x + 3, (-2, 3); ^v(/) y = b + (x-a)% (a,b); y^ig) 10x2 - 64x + 9y2 4- 18 2/ - 71 = 0, (2, - 1). K 2. Rotate the axes through the angle indicated and transform the equation to correspond. Draw the curve and both sets of axes. Identify known loci. (a) xy = 8,d=^; ^b) x2 - 2/2 = 16, = ^ ; (c) xy=-S,d = ~; ^ (d) x2 + 2/2 - 4x2/ + 9 = 0, ^ = ^; /(e) x2 + 4x2/ + 2/2= 16, ^ = ^; ^'(/) 29 x2 - 24 X2/ + 36 2/2 = 180, ^ = sin -i | ; (g) 17 x2 - 312 xy + 108 y^ = 900, d = sin -i f . Ans. («) x'2 - 2/'2 = 16 ; (/) 4 x'2 + 9 2/' 2 = 36. ;^3. Prove that x^ -\- y^ = r^ is unchanged by rotating the axes through any angle. 110 / ,^(1 ANALYTIC GEOMETRY t 4. Find the coordinates of P(4, 3) and P(2, —1) when the origin is translated to («) (6, 5) ; {h) (4, - 2) ; (c) (- 2, - 1) , {d) (- 4, - 5). ^^^ JL. ^'is- («) (-2,-2), (-4, -6). ^ 5. Find the coordinates of (a) (3, 5) ; (6) (- 2, 2) ; (c) (4, - 2) ; (cZ) (- 6, 6) when the axes are rotated 45°; — 46°; 90°. Ans. (a) (4V2, V2), (- \/2, 4V2), (5, - 3). 87. Application to the Conies. — Let us derive the equation of the conic, taking the directrix as the y-Sixis and the prin- cipal axis as the a;-axis. AVe have by definition FP=eNP, ,P{x,y) whence ^(x— py-^y- = ex, or {x-py-\-y' = e''x\ (29) np,o) This is a general form of the conic equation, true for all values of e. From it by translation of axes we can derive the forms previously obtained. For the parabola e = 1. Then (29) reduces to 2/^ = 2 2^x — p^. As the vertex is halfway between the directrix and the focus, its coordinates are (^, Oj. Thus to move the origin to the vertex we substitute cc = x' -}-^-, y = y' . Simplifying and dropping primes, we get 2/2 = 2 px, the standard equation of the parabola. For the ellipse we have seen in Chapter V that the dis- tance from the directrix to the center is j) + c. Therefore TRANSFORMATIOX OF COORDINATES HI the coordinates of the new origin are {^p + c, 0). Substi- tuting x = a;' + j> + c, ?/ = ?/' in (29), we get (a;'4-c)2 + 2/'2 = eV+i> + c)2. But c^ae^p -\- c=-. Using these relations and dropping primes, we obtain the equation, ^1 (x + aey + 2/2 = e'^fx + ^Y, " ^J, ^ which can easily be reduced to (l-e2)^.2^2/2 = a2(l-e2). Dividing by the right-hand member, and putting a2(l — e^) = b^, we have the standard form The central equation of the hyperbola can be derived in the same way. The student should work out the reduction for himself, bearing in mind that for the hyperbola the dis- tance from the directrix to the center is j:> — c = — - , and that 52 = a2(e2-l). ^ 88. Test for Axes of Symmetry. — To find axes of sym- metry parallel to the coordinate axes, proceed as follows: Solve the equation for y in terms of x. If the solution is of the form y = Z: ±f(x), the line y = k is an axis of sym- metry. This is due to the fact that each value of x, as X = a, gives two values of y, and thus two points, one f(a) above the line y = Jc, the other f(a) below. Similarly, if the solution for x is of the form x = h ± y(y), the line x=h is an axis of symmetry. Since the lines x = h and y = Jc are perpendicular to each other, a curve symuietrical with respect to both of these lines has their intersection (h, k) as a center of symmetry. (See § 22.) 112 ANALYTIC GEOMETRY 89. Simplification of Equations by Translation. — In case the curve has two axes of symmetry, they can be found by the method just given and the translation made by formula. When this is not true, the second method illustrated in the following examples must be used. Example 1. — Simplify x'^ + 4 y^ - 2 x - 24: y -\- 21 = 0. Solution. — First Method. Solving for X aud y, we have x = l±ny)* and y = ?>±fj{x). Hence the axes of symmetry are x= \ and ?/ = 3 and the center of symmetry is'(l, 3). Substitute X = x' + 1 and y = y' -\-Z in the original equation, and we have (x' + 1)2+4 {y' + 3)2 - 2 {X' + 1) - 24 (^/' + 3) + 21 = 0. x'-^ + \y'-^ = 16. 16 4 Thus the curve is ap ellipse with its center at (1, 3) and semi-axes 4 and 2. Second Method. In the given equation substitute x — - x' + h and y = y' -\- k. It becomes, after collecting coefficients, x'2 + (2 ;i - 2) x' + 4 ?/'2 + (8 yfc — 24) y' + (/i2 + 4 ^-2 - 2 A - 24 A: + 21) = 0. If the curve is symmetrical with respect to the new axes, there can be no terms of the first degree in the new equation. This is the case if 27i - 2 = and 8 A; - 24 = 0, or A = 1, k = 3. Hence the center of symmetry is (1, 3) and the transformed equation becomes x'2 + 4 ?/'2 — 16 = as before, on substituting h = 1, ^ = 3. * In solving for x, the terms not involving x may be disregarded, as we are interested only in showing that the solution is of the form x = h±/{y). TRANSFORMATION OF COORDINATES 113 Example 2. — Simplify y^ -{- Sx + 'iy - 20 = 0. Solution. — If the equation is solved for x no axis of symmetry is re- vealed, and so we must use the second method. Put z = x' -\- h, and y = y'-\-k. Collecting coefficients, the equation becomes ?/'2 + 8x'+ (2;t + 4)?/' + (A-2 + 8/?. + 4^--20) =0. Evidently the terms in y'~ and x' cannot be eliminated. But if 2 A: + 4 = and A;2 + 8 A + 4 ^ — 20 = 0, the terms in y' and the constant term vanish. These equations give k =—2, h = 3. The new equation then is y'2=-Hx', which is of the form y^ =—2px. The curve is a parabola which has its vertex at (3, — 2) and its axis y =—2. PROBLEMS ^ 1. Find the center of symmetry by solving for x and y, and move the origin to this point. Draw the curve and both sets of axes. (a) 9 x2 - 36 X + 4 2/2 + 16 ^ + 16 = ; (b) 16 x2 - 64 a: - 25 2/2 - 336 = ; (c) 4 x2 - ?/2 + 32 a: + 4 ?/ + 60 = ; /(d) a;2 ^_ y2 _ 18 X + 6 y + 54 = ; (e) x^-\-x + 3y'--9y + ^ = 0; (/) 6 ic2 - 4 / 6. Find the locus of the middle points of chords of the ellipse rrf^(}x2 Jf- 9 y^ = 144 drawn from one end of : ^(a) the major axis ; (b) the minor axis. 7. Find the locus of the centers of all circles tangent to the circle x2 + ?/2 — q2 and the line y=b. 8. From one focus of the ellipse bh:^ + a-y- = a-b"^^ focal radii are drawn and bisected. Show that the locus of the points of bisection is an ellipse, and find its center and foci. / " 90.- Discussion of the Equation Ax"" + Cy^ -{- Dx -\- Ey -\- I F — 0. This represents the general form of a quadratie in X and y with the xy term lacking. The only restriction on TRANSFORIVIATION OF COORDINATES 115 the coefficients is that A and C cannot both be zero. We desire to find under what conditions the locus is a parabola, an ellipse, or a hyperbola. Case 1. When either A or C is zero. Suppose ^ = 0. Here the equation has the form yvv '* C. If D =^ 0, by a proper translation we can remove ''^ the y term and the constant, and obtain an equation of the form Cy"^ -\- Dx' = 0, of which the locus is a parabola. If i> = 0, the y term can be removed by translation, and we get Cy^ -\- F' = 0, of which the locus is imaginary if F' and C have the same sign ; the new a"-axis if i^' = ; and a pair of lines parallel to the cc-axis if F' and C have unlike signs. A. similar discussion holds when C = 0. Case 2. Wien A and C are of like sign. Removing the first degree terms by translation, we have Ax'^ + Cy'^ = F'. This is evidently an ellipse if F' has the same sign as A and C ; a point if i^"' = ; and an imaginary locus if F' has the opposite sign to that of A and C. ^ Case 3. Wlien A and C are of urdike sign. Removing the first-degree terms, we have A.x'^ + Cy'^ = F'. This is evidently a hyperbola unless F^ = 0, in which case the locus is a pair of lines intersecting at the new origin. * By actual substitution of x = x' + ^ and tj = ij' + k,\t can be shown that (o) translation does not affect the coefficients of the highest powers of the variables ; (b) that, if neither A nor C is zero, the first-degree terras can be removed ; (c) that, if one variable appears only to the first degree, the first power of the other variable and the constant terra can be removed ; (d) that, if one variable is missing, the first power of the other can be removed. 116 ANALYTIC GEOMETRY 91. Limiting Forms. — We have seen in § § 58, 70, 83 how the various conies may be formed by passing a plane through a cone. In the previous section we have found that in gen- eral the locus of the equation Ax^ -\- Cy"^ -\- Dx -\- Ey + F = is a conic, but that for particular values of the coefficients the locus is imaginary, or is a point, a line, or a pair of lines. These degenerate cases may be regarded as limiting forms of sections of a cone. The discussion of the different cases is as follows : (a) If the vertex of the cone is moved away from the base indefinitely, the cone approaches the cylinder as a limiting form. Here the elements are parallel and the para- bolic section approaches a pair of straight lines. If the cutting plane becomes a tangent, the limit is a double straight line. If the plane is moved farther, there is no in- tersection, or an imaginary one. (h) If the plane forming a circular or elliptical section is moved towards the vertex, the section grows smaller, until at the vertex it becomes a point. If it is moved still far- ther, there is no intersection, or an imaginary one. (c) Any section of a cone formed by a plane passing through an element is a triangle, hence, if the plane forming a hyperbolic section is moved to the vertex, the section be- comes two intersecting straight lines. 92. General Statement. — The preceding analysis may be summed up as follows : A quadratic equation of the form Ax" + Cy + Dx -\-Ey-bF=0 always defines a conic or one of its limiting "forms, the coeffi- cients being any real-numbers, zero included. If either A or C is zero, the conic is a parabola or in special cases two parallel straight lines distinct, coincident, or imag- inary. TRANSFORMATION OF COORDINATES 117 If A and C have the same sign, the conic is an ellipse, or in special cases a circle, point, or imaginary ellipse. If A and C have unlike signs, the conic is a hyperbola, or in special cases two intersecting straight lines. Ji^J^ 93. Generalized Standard Equations of the Conies. — The equations derived in Chapters IV, V, and VI we have seen to be of a special character, 'since they are applicable only to the case that the vertex or center is at the origin and the principal axis is the x- or ^/-axis. We are now in a position to generalize these equations and obtain standard forms where the only restriction is that the principal axis is paral- lel to one of the coordinate axes. Consider an ellipse with the center at (h, k), semi-axes a and b, and transverse axis y = k. It is required to find its equation. If the origin is translated to (h, k), we know from Chap- ter V that the transformed equation will be /y.'2 ,,'2 — 4-^ = 1. a" 62 The problem then is the reverse of that considered in § 89, viz., the transformed equation is given and the equation with reference to the old axes is to be found. Hence the substitutions are those of § 85 reversed, that is, x' = x — h, and y' = y — k. These give ' as the required equation. \ V ' '- r Vi In the same way we discuss the other cases and^bta^i the following equations : ^ Parabola — 'Vertex (/i, k), axis y = k\ (y-ky~ = 2 p(x - h) ; (30) " " " x=:h: (x-hy = 2p(y-k); (30a) lla ANALYTIC GEOMETRY Ellipse — - Center (h, fc), axis ?/ = A; : _,. (x-hf , (y-kf + (31) (( a X = A: (y^Z^ + iA^JH^l; (31a) Hyperbola— Tz/ - *)^ Center (7^% axis ?/ = A- : i:^^^-iL^ = l; (32) ^ = n: (K^:^^(AIZ31^1. (32a) 6» These forms are easy to remember if one bears in mind that they are merely generalizations of the simple forms y'^ = 2px, x- = 2py, etc. To reduce an equation to orie of these forms, it is necessary only to complete the squares. Example 1. — Write the equation of the ellipse which has the center (0, 1), one focus (—4, 1), and minor axis 6. Solution. —Here the principal axis is y = l. From the given data & = 3 and c = i, whence a = 5. Therefore the equation is 25 9 Example 2. — Simplify the equation 2x^' - Sy'^ -\-8 x - G y -h 11 = 0. Find the center, semi-axes, vertices, and foci, and draw the curve. Solutio7i. — Completing squares, 2(a: + 2)2-3(2/ + l)2=-6 3 2 W[ The center is (— 2, — 1) and since the ^ term in {y + 1)''^ is positive, the trans- verse axis is X— — 2. Hence a=\2, b = V3, and c = Vs, The vertices are and the foci (-2, — 1±V5)> TRANSFORMATION OF COORDINATES 119 PROBLEMS ^ 1. Write the equations of the following ellipses in the general form, the major axis being first parallel to the x-axis, and second parallel to the y-Rxis, unless the given data fix its direction, (a) a = 5, ?> = 3, center (2, 3) ; (6) a =i,b =2, center (-1,2); y (c) a = f, 6 = I, center (- 2, 3) ; (d) vertices (±5, 3), focus (3, 3) ; -7Cj(e) vertices (— 3, 1), (7, 1), focus (0, 1) ; (/) foci (±3, - 2), directrix x = — 4 ; (gr) minor axis 12, e = f^g, focus (0, 0) ; s^ (A) major axis 6, e = ^, focus (1, 1) ; Ans. id) 5; + (^^=l; 2o lb ^^^ 169 36 4 2. Write the equations of the hyperbolas satisfying the data of Problem 1, if possible. If not possible, explain why and revise data to make a similar problem for the hyperbola and solve. Ans. (d) not possible, for c 0, A' and C are of unlike sign, and the conic is a hyperbola* * Or one of the limiting forms discussed in §§ 90, 01. TRANSFORMATION OF COORDINATES 123 98. Suggestions for simplifying the Equation of a Conic. — If the conic is an ellipse or hyperbola (that is, if A ^ 0) determine the coordinates (h, k) of the center of symmetry and remove the tirst-degree terms by translation. Then rotate the axes as in the example of § 94. If the conic is a parabola (A =0), the substitutions for rotation should be made first, and then equations of condi- tion involving h and k can be formed. A convenient check on the accuracy of the rotation sub- stitutions is the fact that A -\-C is imchanged until the new equation is simplified. (See Exercise, § 96.) PROBLEMS Simphfy the following equations. In each case draw the conic and the three sets of coordinate axes. 1. 73x2 + 72x?/ + 52?/2-218x- 116 y -f 07 = 0. i Ans. 4x"2 + y"2 = 4. . 2. -^2 _ 24 a-?/ + 11 1/2 + 72 a:- 116?/+ 204 =0. Ans. a:"2 - 4 ?/"2 = _ 4. 3. 18 x2 - 48 a;?/ + 32 y^ _ 120 x + 35 ?/ + 200 = 0. /Ty .^ - /,' ^ - ■ Alls. 2y"'^ = Zx". 4. 6 x2 + 13 x?/ + 6 2/2 - 7 X - 8 ?/ + 2 = 0. Ans. Two Hnes. 5. 2 x2 + 4 xy + 4 ?/2 + 2 X + 3 = 0. Ans. Imaginary. 6. 41x2 + 24x?/ + 34?/2_ 90x+ 120I/ + 225 = 0. Ans. Point. 7. 16 x2 + 24 x?/ + 9 ?/2 - 60 X - 45 ?/ + 50 = 0. Ans. 4 x"2 = 1. 8. 66 x2 + 24 xy + 59 ?/2 + 396 x + 72 ?/ + 444 = 0. Ans. 3 x"2 + 2 ?/"2 = 6. 9. x2 + 24 x?/ - 6 ?/2 + 4 X + 48 ?/ + 34 = 0. Ans. 3 2/"2 - 2 x"2 = 6. 10. 16 x2 + 24 x?/ + 2/2 _ 60 X + 80 ?/ + 200 = 0. Ans. x"2 + 4?/'^ = 0. ,.•11. 7x2-48x?/-7?/2=0. ^^ Ans. x"'^ - y"-^ = 0. 12. 2 x2 + 6x?/ + 10 ?/2 - 2 X - 6 ?/ + 19 = 0. Ans. Imaginary. 13. 9a;2_24x2/ + 16^2_3^_1_ 4y _6 =0. Ans. Two lines. 124 ANALYTIC GEOMETRY 14. 13x-^ + 10 a;?/ + 13 2/2 - 42 X + 6 ?/ - 27 = 0. Ans. 9x"-^ + 4:y"-^ = S6. 15. 13x2 + 10x2/ 4-13 y2 + 16 X- 16?/ + 16 = 0. Ans. Point. 16. x2 + 4x?/ + ?/2- 2x- 10?/ - 11 = 0. Ans. S x"-^ - y"-^ =9 . 17. 3 x2 + 8 x?/ - 3 ?/2 _ 10 X - 30 ?/ + 20 = 0. Ans. ?/"2 _ .7.//2 _ 4 18. 6 x2 — 4 X2/ + 9 ?/2 — 40 X + 30 ?/ + 55 = 0. Ans. x"2 + 2 ?/"2 = 2. ^ 19. 3x2-2V3x?/ + ?/2_ (8+ 12V3)x+(12-8V3) ?/- 12 = 0. ^ns. ?/"2 = 4x. 20. The ends of the base of a triangle are (0, ) and (4, 0), Fina the locus of the vertex if the sum of the slopes of the sides is {a) j\, 2 (6)1, (c) — ~"- Simplify the equation found and draw the locus. 21. Solve the previous problem when the given points are (0, 0) and (a, 0) and the sum of the slopes is any constant k. Then show that the locus is a hyperbola passing through the given points and 2 having its axes of symmetry inclined at an angle I arc tan 22. The difference of the base angles of the triangle whose base joins (0, 0) and (4, 0) is 45°. Find the locus of the vertex. Ans. x"2 -y''-^ = 2V2 or x"2 - y"-^= - 2 V2. 99. The Conic through Five Points. — The general equation of the second degree involves six arbitrary constants. As in the case of the circle, however, one of these can be divided out ; hence, if we can express five of the coefficients in terras of the sixth, the conic is completely determined. Geomet- rically, this means that the conic is completely determined by five points, or by five other geometric conditions. To find the equation of a conic so defined we proceed to form equations between the coefficients of the general equa- tion and solve them. For example, suppose the conic is to pass through the points (4, 2), (2, 4), (—3, 1), (1, — 3), and (0, 0). Substituting these in the general equation, Ax" 4- Bxy + CY +- Dx -\- Ey + F= 0, TRANSFORMATION OF COORDINATES 125 wehave: 16^ + 8^+ 4. C -{- 4. D + 2 E + F = 0, 4.A + SB + 16C+2D + AE + F=0, 9A-3B+ C-SD-j- E + F = 0, A-SB+ 90+ D-3E-\-F=0, F=0. Solving these, we find ^ = C= ^, D = E = ~ ^. This oO o makes the general equation, on dividing out the B and clear- ing of fractions : x^ -\- m xy + y- — 70 X - IQ y = 0. The value of A shows that the conic is a hyperbola. Its center is found by the method of § 89 to be (ff , ff)- The inclination of its axes is 45°, since tan 2 ^ = oo . This in- formation, together with the live points given, is sufficient ' for drawing the curve. The graph follows. A shorter method of solv- ing the above problem is given by the theorem of §41. Call the first four points P, Q, E, and S. The equations of the lines PQ and BS are x -\- y — 6 = and a; + 2/ + 2 = 0. ^lul- tiplying these together, the equation of the pair of lines is x-^ -\- 2 xy + y^ — 4:X — 4: y - 12 = 0, by § 42. Similarly the equation of the pair of lines RQ and iSP is 15 0.-2 - 34 xy 4- 15 /- + 2Sx-\- 2Sy- 196 = 0, 126 ANALYTIC GEOMETRY Now the intersections of these two equations are the points P, Q, R, and S. Hence by § 41 15 x" -34:xy -{- 15 if -\- 2S x -\- 28 ij - 196 + k(x''-i-2o^!j + 7f-4.x-4:y-12) = (a) is the equation of a system of curves passing through these four points. As this is of the second degree for all values of k, they are all conies. To find the conic of this system passing through the fifth point (0, 0), substitute its coordi- nates for the variables. This gives - 196 -12k = 0, whence k = — ^^-. Putting this value for k in (a) and col- lecting terms, we have the same result as before. PROBLEMS 1. Find the equation of the conic passing through the following points : (a) (3, 2), (1, 2), (3, 0), (2, |), (Y, 1) ; (6) (1, 6), (- 3, - 2), (- 5, 0), (3, 4), (0, 10) ; (c) (3, 1), (3, 5), (6, 2), (6, 10), (11, 5) ; (cO (1, 1), (- 1, 2), (0, _ 2), (- 2, - 1), (3, -3). 2. Find the equation of the parabola passing through the following points : (a) (0,0), (4,0), (0,3), (-8,8); (6) (0, 1), (4, 7), (4, 1), (12, 7) ; (c) (1, 1), (4,0), (0,4), (9, 1); Hint. — One of the conditions is B^ — ^AC = 0. 3. Find the equation of a conic with axes of symmetry parallel to the coordinate axes and passing through : (a) (-4,5), (6,0), (-4, -5), (4,3); (&) (7,1), (6, -2), (7, -o), (-2, -2); (c) (-3, -4), (-1, -3), (1,0), (.3,5). 4. Find the equation of a conic symmetrical with resi^ect to the origin and passing through : (a) (2,4), (3, 1), (-2,0); (ft) (1,2), (-1, 1), (3,2); (c) (0,0), (0,4), (4,0). CHAPTER VIII POLAR COORDINATES 100. Definition. — Some topics in analytic geometry can be better investigated by the use of polar coordinates than by rectangular coordinates. In the polar system the posi- tion of a jjoint is fixed by measuring a distance and a direc- tion instead of by the measures of two distances. This is essentially the same s^^stem as that of bearing and distance used in surveying, contrasted with that of latitude and longitude used in geography. Choose a fixed point as the origin, called the pole, and a fixed line OA through it, called the polar axis. Then any " \>^ point P is determined if we know its distance from and the angle that OP makes with OA. The measures of the distance OP and the angle AOP are called the polar coordinates of P and are designated b}^ p and 6. The distance p is called the radius vector of P, and is called the vectorial angle. Polar coordinates do not obey the conventions of the rectangular system as to their direction and magnitude. As in trigonometry the radius vector may be rotated indefi- nitely in a counter-clockwise or clockwise direction, mak- ing 6 take on any positive or negative value. Distances measured on the terminal line of 6 away from the pole are 12.1 128 ANALYTIC GEOMETRY positive; those measured in the opposite direction, on the terminal produced, are negative. (See the right-hand dia- gram on page 127.) Hence every pair of real numbers (p, 6) determines one point "vrhich may be located according to the following rule. EuLE FOR Plotting. — Taking the ptolar axis as an initial line, lay off the vectorial angle 6, counter-clockwise if positive, clockwise if negative. Then measure off the radius vector p, on the terminal of 6 if positive, on the terminal of produced through the pole if negative. Since 6 and ^ -|- 2 tt have the same terminal line, a point may he represented by an indefinite number of pairs of coordinates. Thus in the ad- ^^"'^ joining figure ws may take for the coordinates of P, p = 8, ± 2 mr. Since $ = — - or — 6 6 the terminal line of pro- duced is the terminal of ^ ± tt, a second set of coordinates is P = '=1 TT or - ± 2 nir. Ordinarily we keep 6 within 6 the limits ± tt, or and 2 tt. 101. Relations between Rectangular and Polar Coordinates. — Take the pole at the origin of rectangular coordinates and the polar axis as the positive half of the ic-axis. Erom POLAR COORDINATES 129 the figures and the definitions of the trigonometric func- tions it is evident that for P(x, y) = P(p, 0) in any quadrant the following formulas are true : X = p cos 6, y = p sin 9, (34) ^^ p2=:x'^ + y\ e = arctan^. (35) These equations enable us to transform rectangular equa- tions and coordinates into polar forms, and conversely. NoTE/ — In transforming the rectangular coordinates of a point into polar coordinates, care should be taken to group together the correspond- ing values of p and d. 102. Polar Curves. — The definitions of equation and locus in polar are the same as those in rectangular coordinates, if (p, 6) is substituted for (x, ?/) (§ 14). The equation in polar coordinates is derived as in the rectangular system (§ 16). In a few cases the polar equation may be obtained best by deriving it in the rectangular form and then substituting for X and y their values in terms of p and 6 and vice versa. Plotting in polar coordinates resembles that in rectangular coordinates. The equation should usually be solved for p and a table of values formed, taking values of 6 at intervals of 30° (or sometimes 15°). When the curve has symmetry, it is usually unnecessary to carry the table through more than two quadrants. Example. — Plot the ellipse 2 p — p cos = 6. Solving, p = - — ^ — -. 2 — cos 6 Table of Values e° p e° p 6.0 210 2.1 30 5.3 240 2.4 60 4.0 270 3.0 90 3.0 300 4.0 120 2.4 330 5.3 150 2.1 360 6.0 180 2.0 300 130 "I. ANALYTIC GEOMETRY ,'S6(!^^^ PROBLEMS V 1. Plot the points: (a) (0, ± 30°) ; (b) (±10,420^); (c) ^8, M (8, ~Y What symmetry has each pair of points ? 2. Plot ( 8, — ^ j . Plot points symmetrical to this with respect to the pole, polar axis, and the 90° axis, and find their coordinates. 3. As in Problem 2 find the points symmetrical to : («)(ie,i^');W (-6,-2-');(e)(-10,-?^). 4. Fix P(p, 6) in any quadrant and find its symmetry to : (a) P (/), - d) ■ {b) P(- p, 6) ; (c) P(p, TT - 6). 5. Find the rectangular coordinates of the following points : (.) (10,^); (.)(4,^f); (c)(6,|); (d) (4,.); (e) (-2,^); (/) (-8,^). Ans. (a) (8\/8, 8). 6. Find two pairs of polar coordinates for each of the following points and plot the point in each case : (a) (-1, Vr); _ (&) (2, -2); (c) (5, -12); (d) (2, -2V3); (e) (-V2, - V2). ylns. («) (2, 120^), (-2, 300"). •j^ 7. Plot each of the following equations and identify each curve by . W\ , transforming its equation to rectangular coordinates : \\/ / (a) p'= 4 ; ' (d) pcos0=^ -2; (g) p = 2 a cos 9 ; (b) ^ = arc tan I ; (e) p = ±Q] {h) p sin = 2 a. (c) p =2 cos ; (/) tan = §. < 8. Draw each of the following curves and transform its equation into polar coordinates : (a) y^ox; (g) x^ + ?/ + 4 2/ = ; (h) Sx + 2y = 0; (h) x^ - y'^ = IQ ; (c) x-a = 0; (i) y^ - x^ = 16; \W V ?,£. (a) tan ^ = 3 ; (c) p cos ^ = a ; ■(/) p = 4cos0; (h) p^ cos 2 ^ = 16 ; {k) /)2 sin 2 ^ = a^. POLAR COORDINATES 131 p{p,e) 103. The Equation of the Straight Line. — The general equation of the straight line in polar coordinates is not as convenient as the equation in rectangular coordinates and will not be discussed. The special cases where the line is parallel or perpendicular to the polar axis or passes through the pole lead to very simple equa- tions. In the figure let the line / be perpendicular to the polar axis OA and have the polar intercept a. For any point F{p, 6) on the line it is evident that cos 6 =, whence the equation of the line is ^ p cos 6 = a. (36) Similarly the equation of a line parallel to the polar axis, of 90° intercept a, is p sin 6 = a. (36a) For a line passing through the pole the equation is evidently e = c. (37) Exercise 1. By transforming the normal form of the straight-line equation into polar coordmates show that the equation of any line in polar coordinates is p cos (6 — u) = p. 104. The Equation of the Circle. — As in the case of the straight line the general form of the circle equation is not often used. Several special forms, which are common, are : Circle icith center at pole, radius r : p = '' ; (38) Circle imtli center at (/•, 0), radius r : p = 2r cos 9 ; (39) Circle ivith center at TT '' o ;? 2 radius r: p = 2 r sin 6 ; (39a) n 132 ANALYTIC GEOMETRY Circle through pole, with polar intercept a and 90° inter cept b : p = a cos + & sin 8. (40) p(p.e) The first equation is ob- vious. The second may be readily derived frora the ad- joining figure, and the third from a similar figure. The simplest way of getting the fourth equation is to ob- serve that its rectangular equa- tion is X- -\- y~ — ax — by = 0. Transforming this into polar coordinates, we have the de- sired result. Exercise 2. Derive equation (40) directly from a figure. Hint. — Show that the radius vector is in length the sum of the projec- tions upon it of the x- aud y- intercepts of the circle. 105. Discussion of Polar Curves. — As in the case of rec- tangular equations, the study of polar curves is facilitated by discussion of the equation in conjunction with plotting. The topics discussed are of much the same character. (a) Intercepts. — The intercepts on the polar and 90° axes are the values of p for ^ = 0°, 90°, 180°, and 270°. (6) Symmetry. — It is easy to establish that in accord- ance with the definitions of § 22, (p, 6) is symmetrical to (— p, 6) with respect to the pole ; to (p, — 9) with respect to the polar axis ; and to (p, tt — 6) with respect to the 90° axis. Hence we have the following tests for symmetry : If the substitution of — p for p ( does not change the form of the 1 the p)ole; — 6 for 6 equation, there is symmetry icith \ the polar axis; IT — 6 for 6 [ respect to J the 90° axis. POLAR COORDINATES 133 Exercise 3. Prove that (p, ^)"is symmetrical to : (^(i) ( — p, ^) with respect to the pole ; (5) (p^ _ ^) with respect to the polar axis ; (c) (p, TT — ^) with respect to the 90'^ axis ; (d) (p, TT + ^) with respect to the pole * ; (e) (_ p, TT _ ^) with respect to the polar axis* ; (j^ (_ p,_ ^) with respect to the 90° axis.* (c) Limiting Values of O. — li the solution for p gives p = a± Vj\0j, the equation /((9) = usually determines limit- ing values of 6, since values of making f(0) < make p imaginary and must be excluded. (d) Extent of the Curve. — If any values of make p in- finite, these values determine the direction in which the curve extends to infinity. If p is never infinite, the values of 6 which make p take on its greatest numerical value should be found. Likewise the values of for which p = 0, or for which p takes on its least numerical value, should be found. Example 1. — Discuss and plot the locus of the equation p = 2 e 0° 90° 180° 270° p 1 2 CO 2 1 + cos 6' (a) Intercepts (&) Symmetry : With respect to the polar axis, since cos (— 0) = cos e. Hence the table of values is unnecessary for 6 > 180°. (c) Limiting values of 6 : None, since no radical is involved. (d) Extent : p is infinite when 1 + cos ^ = 0, i.e. when cos = - 1, or ^ = TT. p is evidently never 0. Its least value occurs when 1 -h cos ^ is greatest, i.e. when cos ^ = 1, or ^ = 0. Here p = 1. * Exercise 3, d, e, and /, gives three alternative tests for symmetry. Their existence is due to the double representation of a point in polar coordinates. 134 ANALYTIC GEOMETRY Table of Values e (radians) (degrees) 1 + cos fl p ±0 2.00 1.0 ~ 6 ± 30 1.87 1.1 ^3 ± 60 1.50 1.3 TT "^2 ± 90 1.00 2.0 ' 3 ± 120 0.50 4.0 ^ 6 ± 150 0.13 15.0 — TT ± 180 0.00 X Ti'ansformation to rectangular coijrdinate.s shows that this curve is a parabola. Example 2. — p- = a- cos 2 6, or p = ± aVcos 2 d. (a) The intercepts on the polar axis are ± a ; on the 90^ axis they are imaginary. (6) The three tests for .symmetry are satisfied, for cos2(-P) = cos2 ^, etc. Hence vakies of d up to 90" only are needed. (c) p is imaginary' when cos 2 ^ < 0, i.e. when 90^ < 2 ^ < 270=, or 45= < ^ < 135=. (d) The greatest numerical vahie of p occurs when cos2^=l, or e = or 180=. This value is a. p = when cos 2 ^ = 0, or when 2 ^ = 90=, 270=, etc., or 6 = 45=, 135=, 225=, and 315=. Table of Values, (a — 10) e •2 cos 2 e p Vcos -.i e 1.00 ±1.00 ±10 15= 30 .87 ± .93 ± 9.3 30= 60 .50 ± .71 ±7.1 45^ 90 .00 .00 0.0 POLAR COORDINATES 135 Example 3. — p = 1 + 2 sin ^. (a) 0° 1 90° 180^ 1 270° - 1 (h) Since sin (tt-^) = sin d, the curve is symmetrical with respect to the 90° axis. (c) No values of 6 are excluded. {d) p is never infinite; its greatest value occurs when sin ^ = 1, or e = 90°. p ^0 when 1 + 2 sin ^ = 0, or sin ^ = - ^. Hence the curve passes through the pole when ^ = 210^ or 330°. Table of Values 9 sin 9 p .00 1.00 30 .50 2.00 60 .87 2.73 90 1.00 3.00 120 .87 2.73 150 .50 2.00 180 .00 1.00 sin 9 p 210 -.50 .00 240 - .87 - .73 270 - 1.00 - 1.00 300 -.87 - .73 330 -.50 .00 360 .00 1.00 PROBLEMS 1. Identify and draw the locus of each of the following : (a) p cos e — 4 = ; (g) P^ + 3p-4 = 0; (&) p cos ^ + 4 = ; (h) sec2d = 2; (c) P-4; (0 ^ = ^; id) tan ^ = V3 ; D (e) p - 8 sin ^ = ; (J) p = -Q sine; (/) /) + 7 cos ^ = ; (k) p = 3 cos^; (?) p- - 2 p (cos — sin ^) — 7 = 2. Transform into polar coordinates : (a) Ax + By + C = 0; (&) x2 + y^ -\- Dx + Ey -\- F= 0. Discuss and plot the locus of each of the following : X 3. (a) p = a (1 — cos d); (c) p = a (1 — sin 6) ; u lb) p = all + cos d); (d) p = a (1 -i- sind). (Cardioids.) 136 ANALYTIC GEOMETRY 4. (a) p = a -{- b cos 6 ; p = a — b cos 6 ; 5.'\a) p = a tan- ^ sec ^ ; (c) p = a + ti sin^ ; (rZ) p — a — b sin ^, (Lima^ons : three forms according as a <, =, or > b.) (?)) p = a cot-^ csc^. (Semi-cubical parabolas.) 6. (a) p = asece ±b; (b) p = acscd ± b. (Conchoids of Nicomedes ; three forms according as a <, =, or > 6.) p = 2a tan^sin^; (?>) p = 2 a cot ^ cos ^. (Cissoids.) p = a sin 2 6 p = a sin 3 6 p = a sin 5 ^ (6) p = « cos 2 ^. (Four-leafed roses.) (5) p = a cos 3 ^. (Three-leafed roses.) (?)) p = a cos 5 0. (Five-leafed roses.) a2 sin 2 ^. (Lemniscate.) ad. (Spiral of Archimedes.) 6«^. (Equiangular spiral.) = a. (Lituus.) 106. Rotation of Axes. — Rotation of the polar axis does not affect the vahie of p. The figure shows that if the co- ordinates of P with refer- P(P,0) {Pid') ence to OA are (p, 6), and to 0^1' are (p, 0'), while angle ^4' 0-4 is «, we have as a formula of rotation e = 6' + a. (41) By rotating the polar axis the standard equations of curves may be Avritten in a variety of forms. Translation of the pole is seldom necessary and is best performed by turning the equation into rectangular coordi- nates. , ♦ Express e in radians, POLAR COORDINATES 137 107. Symmetrical Transformations. — Equations may also be transformed by the substitutions for symmetry. As in the case of rectangular coordinates it can be shown that the effect of such a substitution is to transform the locus into a curve symmetrical to the given locus with respect to the center or axis involved. Thus we have the following : If in an equation substitution is made of — p for p - 6 for 6 TT-e for the locus of the new equation " is symmetrical to that of the old with respect to the PROBLEMS pole ; polar axis ; 90° axis. 1. In the following find the new equation on rotating the axes through the indicated angle, and draw curve and both axes : (a) p = a tan-^ d sec 6, u = 90= ; (c) p = a sin 3 6, a = 30° ; • C&) p = asece ±b, a = 90' ; {d) p = acosd, a = 90° ; (e) p = a -\-b cose, a=: 90^. Ans. (rt) p — — acot:^ dcscd. 2. Show that the first of each of the following pairs of equations may be transformed into the second by a proper rotation of the polar axis; (a) p = a {I + cosd), p = a (1 - sin ^) ; (6) p =asin2^, p =acos2^; (c) p = a sin S d, p = a cos 3 6 ; (rf) p- = a'^ sin 2 6, p^ = a- cos 2 d. 3. Find the new equation when the substitution for symmetry with respect to the 90° axis is made, and draw both curves on the same axes : (a) p = a{l - cos d) ; (c) p = a tan^ ^ sec ^ ; (e) p = a cos 3 ^ ; (6) p = a + 6 cos ^ ; (rf) p = 2 a tan ^ sec ; (f)p = a cos 6. Ans. (a) p = «(1 + cos^). 4. To each equation apply the substitution for symmetry with re- spect to the polar axis and draw both curves on the same axes. (a) p = a(l - sin ^) ; (c) p = a sin 3 ^ ; (6) p = a — 6 sin ^ ; (cl) p = a sin 6. Ans. (a) p = a(l + sin^). 138 ANALYTIC GEOMETRY 108. The Polar Equation of the Conic. — If the pole is taken at the focus and the directrix is perpendicular to the polar axis as in the figure, we have at once from the defini- tion of a conic, OP = eMP, whence p = e(2^ + p cos 6), which gives P = ep (42) 1 — e cos 6 If the directrix is taken par- allel to the polar axis, the equation becomes ep 9 = (42«) 1 — € sin 6 The transformations of rotation and symmetry give eight forms of the conic equation in which the focus is at the pole and the directrix perpendicular or parallel to the polar axis. They are : _^ gp . g.-, n=- i ; p = =*= — i . l±ecos^ 1 ± e sin In the first set the principal axis is the polar axis ; in the second it is the 90° axis. To sketch a conic it is sufficient to put the equation in standard form and find the intercepts. The standard form determines the principal axis and eccen- tricity, and the intercepts give the vertices and the ends of the latus rectum. By using the polar equation many of the properties of the conic may be derived more easily than by the use of the rectangular equation. Some of these are indicated in the following problems. PROBLEMS 1. Transform the equation of the conic, x^ -\- y^ = e- {x + p)-, into the polar form. 2. Show that the conic is symmetrical with respect to the principal axis. Does the polar form show any other axis of symmetry ? POLAR COORDINATES 139 3. Find the latus rectum of the conic. Ans. 2 ep. 4 Reduce to the standard polar form each of the following conies, find e and p, and draw the curve and its directrix : 16 , - 5 («) ^=2-cos^' ^'^ ^ = l + cos^' 8 12 1 -2 sin 0' ''^ '^ 4 + cos^' (^^^ = 2 + 3'cos^' (^) ^ = i_2cos^' Q Alls, (a) p = e = 1 n = 16. ^ -' ^ 1 — ^cos^' 2'^ ep 5. Show that if p =- ;,is the equation of a hyperbola, the ^ 1 — e cos d .11 inclinations of the asymptotes are given by cos d = ±-. e 6. In the hyperbolas of Problem 4, draw the asymptotes. 7. Find the polar intercepts of the conic. From these show that (a) the major axis of the ellipse is — ^^ ; (h) the transverse axis of the hyperbola is — ^ , ^^ e--l 8. Show that the conic never crosses the directrix. Hint. — The equation of the directrix is p cosO = — p. Points of inter- section are found by solving simultaneously. 9. Derive the equation of the conic when the equation of the directrix is : (a) pcosd=p; (b) p?,'m.d=p. 10. Draw the conic with it-s directrix for each form : ^P . n,\ .- ^P {^) p-^^ — .; i^) p 1 4- e cos ^ ' 1 + e sin ^ ' ep ep 1 + e cos d 1 + e sm d ep ep P = -. -'y P~ P = 1 — e cos d 1 — e sin 6 ^_P . «__ ep 1 — e cos 6 1 — e sin 6' 11. Show that the locus ot p =. a sec- -is a parabola. 140 ANALYTIC GEOMETRY 109. Equations derived in Polar Coordinates. — The equa- tion of a locus may sometimes be derived more readily in polar than in rectangular coordinates. This is especially true when the locus is described by the end of a line of varying length revolving around a fixed point. The process is substantially the same as when rectangular coordinates are used. The pole should be chosen at a convenient place, usually a fixed point about which some line in the problem revolves. Then a point (p, 0) satisfying the conditions is chosen, and relations between p, and the given arbitrary constants are obtained by means (usually) of trigonometric formulas. Example. — A chord through the end of a diameter of a circle is extended the length of the diameter. Find the locus of the end of this line. Choose one end of the given di- ameter as the pole, and the diame- ter as the polar axis. Let 2 « be the diameter of the given circle. Then from the figure : OB = 2 a cos e, p = OB + 2 a. Therefore the equation is P = 2 a(l + cos^), and the required locus is a cardioid. p(p.e) PROBLEMS 1. Find the locus under the conditions of the above problem when the chord is extended a distance 2 b. Ans. The limagon. 2. The radius of a circle whose center is the origin is prolonged a distance equal to the ordinate of its extremity. Find the locus of the end of this line. Ans. The cardioid, p = a(l + sin ^). 3. Derive a formula for the distance between two points in terms of their polar coordinates. Hint.— Use the law of cosines. POLAR COORDINATES 141 4. Derive the polar equation of the circle having the center (pi. ^i). Ans. /)2 + pi2 _ 2 pi p cos (d - ^i) = r^. 5. Derive the polar equation of the ellipse when the pole is at the center. A^is. p^(l — e^cos^d) = b'^. Hint. — Use the law of cosines and the fact that the sum of the focal radii is 2 a. 6. Find the locus of the vertex C of a triangle whose base AB is fixed and which has the angle A = 2 C. Ans. p = a (2 cos ^ + 1), the trisectrix.* Hiiit. — Use the law of sines. 7. A fixed point A is at a distance a from a fixed line BC. From ^ a line is drawn cutting BC in B and points P and P' are chosen on this line b units from D on either side. Find the locus of P and P' as D slides along BC. Ans. p = a sec 6 ± b. 8. A tangent is drawn to a circle whose center is the origin and terminated by the x- and ?/-axis. Find the locus of its mid-point. A71S. p sin 26 = a. 9. Find the locus of the mid-points of chords drawn from the end of a fixed diameter of a circle. 10. A line of length 2 a has its extremities on two fixed perpen- dicular lines. Find the locus of the foot of a perpendicular from the intersection of the fixed lines to the line of constant length. Ans. p — a sin 2 6. 11. Given the circle p = 2 a cos 6 and the line p cos 6 = 2 a. From the pole a chord OB is drawn, meeting the line in C. Find the locus of a point P on the line OC, if OP always equals BC. Ans. p =2 a tan 6 sin 6, the cissoid. 12. The base of a triangle is in length 2 a. Find the locus of the vertex if the product of the sides is equal to a-. Ans. p^ = 2o2 cos 2 6., the lemniscate. ^m^ — Take the pole at the center of the base and use the law of cosines. * Since the vertical angle C is one third of the exterior angle at B, this form of the lima^on can be used to trisect any angle, and so is called the trisectrix. 142 ANALYTIC GEOMETRY REVIEW PROBLEMS Identify and sketch the following curves : 1. p = 6sin6. 14. p cos ^ = — 5. 2. p= ^ ^. • 15. p= ^ 4. p = Sd. 6. p=10 sin 2 d. 6. p = 6 tan sin ^. 7. p = 4 -\- 6 sec ^. 8. p 11. p sin 6* =10. 2 - sin 1 + cos^ 3. pr=6 + 4sin0. ig. pzr5-10cos^. 17. p^ = 8 TT. 18. p2 = 4 cos 2 d. 19. p = 8 cos 3 e. 3 20. p = 1 + sec ^o 1 +2COS0 21. p2 = 4sin2 2^. 9. p = 8cos2^. 22. p= -2cot6'csc9. 10. p = 8- 10 sin 0. 23. 2p — pcos<9 = 8. 24. tan ^=1. 12. p_4sec2-. gg p = 10 sin 3 0. 13. p = 6 sin 5 0. 26. p = 8 tan- 6 sec 0, ^Wou^ Y^ ^ I.-/ -^ f--^ Jrl CHAPTER IX HIGHER PLANE CURVES 110. Algebraic and Transcendental Equations. — The equations in Cartesian coordinates which we have hitherto treated have been algebraic equations, i.e. have involved only integral and fractional powers of x and y. Any equation which is not algebraic (e.g. y + sin x = 0) is called transcendental, and functions defined by such equa- tions are called transcendental functions. The elementary equations of this class are those in which the exponential, logarithmic, trigonometric, and inverse trigonometric func- tions are used. In this chapter we shall discuss a number of curves de- fined by transcendental equations and algebraic equations of degree higher than the second. Such curves are called Idgher plane curves. 111. The Exponential Curve. — This is defined by the ex- ponential equation where h is any positive constant. The quantity h is called the base. If the exponent is fractional and involves even roots of b, only the positive value of the root is used. Discussion. — For simplicity consider the case that 6 > 1. When X = 0, y = b^ = \. If ?/ = 0, we would have = 6^, which is impossible for any value of x. Therefore the curve crosses the ^/-axis one unit above the origin and does not cross the a;-axis. For all values of x, positive and negative, y is positive, since any power of a positive number is positive. When 143 144 ANALYTIC GEOMETRY X > 0, b"" increases with x, since 6 > 1. Therefore in the first quadrant the curve recedes to infinity away from both the X- and y-Sixes. Since b'"" = — , when x approaches — co, b"" y approaches as a limit. Thus in the second quad- rant the curve approaches the a>axis as an asymptote. In the following figure the curve is plotted for two values of b. The Base e. — The most important case of the exponential is for the base e(= 2.718+), the base of the natural system of logarithms. In the figure the dotted line is the curve oi y = e"" and the heavy line that of y = 2'. X 2/=«^ y = '2' -2 0.135 0.250 1 0.368 0.500 1. 1. 1 2.718 2. 2 7.389 4. 3 20.086 8. Exercise 1. Discuss the effect upon the graph of increasing b. Exercise 2. Discuss the exponential equation for : (a) 6 < 1 ; (6) b = 1. 112. The Logarithmic Curve equation This is the graph of the y = log^ X. The base b is always positive and different from 1, and usually greater than 1. Since in Algebra the logarithm of a number to a given base is defined as that power of the base which makes the result equal to the given number, the equation y = loQf^x may be written in the form x = b". This is the exponential equa tion with the variables interchanged. Thus the logarithmic HIGHER PLANE CURVES 145 equation is obtained from the exponential y = 5^ by solving for X and interchanging variables. Logarithmic and expo- nential functions are said to be the inverse of each other. The discussion of the logarithmic equation y = log^ x will therefore follow from that of the exponential equation y = h'^ on interchanging variables. Thus the logarithmic curve crosses the avaxis at (1, 0) and does not cross the ?/-axis. For .r > 1, 2/ > and as x =*co, y = cc ; for x < 1, y < and as x = 0, 2/ = — GO. Since x = b", the abscissa x is positive for all real values of y, whether positive or negative. This is the reason for the statement that " logarithms of negative numbers do not exist." The following graph is the logarithmic curve for b = 10. The table of values may be computed by using a table of logarithms or by writing the equation in the form x = 10^ and computing values of x corresponding to various values of y. Note that the logarithmic curve is symmetrical to the. exponential curve with respect to the line y = x. y = logio X X y X y 1.0 .1 -1 3.1 .5 .01 -2 10 1 .001 -3 100 2 .0001 -4 etc. etc. etc. etc. 113. Applications. — The exponential and logarithmic curves show clearly the method of variation of these func- tions. Thus for & > 1, the exponential function increases with great rapidity as x approaches go, while the logarithmic function increases very slowly. This symbol is used for the words " approaches as a limit." 146 ANALYTIC GEOMETRY These functions not only have a theoretical importance, but constantly appear in the statements of the laws of Phys- ics and other sciences. For example, the adjoining figure represents a weight W suspended by a /^ "'N^ rope wrapped several times about a wooden beam and kept from falling by a tension T. The relation between W and T is given by the equation where x represents the number of times the rope is wound around the beam and /x is a constant de- pending on the friction between the rope and the beam. For a hemp rope on smooth oak /x = 3.34, nearly. PROBLEMS 1. Plot the system of curves defined by ?/ = &^ for 6 = 1, 2, d, and 10 ; all on the same axes. 2. Plot the system defined hj y — h' ior b = 1, ^ and \ ; all on the same axes. 3. Plot the logarithmic curve for the bases 10, 3, e, and 2 ; all on the same axes. 4. Plot the graphs of : («) y = 2-^; (c) y = x^^; (e) y = e-^ ; (b) y=4-'; (d) y = Q-; (/) y = e^-. 5. Plot the graphs of : (a) y = Logo X ; (c) 2x = logio y ; (6) y = S\og2X; (d) x = logio2^ 6. Write each of the equations of Problem 4 in logarithmic form. Ans. (a) X = — logo y. 7. Write each of the equations of Problem 5 in exponential form. Ans. (a) X = 2y. 8. Discuss the following equations for symmetry and piot their graphs : pX _|_ p~X {(^) y = — . This is called the hyperbolic cosine and written y — Qosh X. HIGHER PLANE CURVES 147 oX __ P~X (b) y = , or y = sink x. (c) y=- (e" + e""), the equation of the catenary.* 9. Show that the graphs of y = b"" and y = (-\ are symmetrical with respect to the ?/-axis. 10. Plot the curve i/ = 3^ and estimate the logarithms to the base 3 of the integers from 2 to 10. 11. Plot the graph of the function W = Te^^, where T = 100 lbs. and fj. = l.l. Estimate from the graph how many turns of the rope would be required for this tension to support 1 ton. 2 tons. 10 tons. Hi7it. — This makes ei^ = 3, nearly. 114. Periodic Functions. — A i^eriodic function of a variable is a function ichose values are repeated at definite intervcds as the variable increases. Cosine x is an example. It passes, successively through all values from +1 to —1 and back to + 1 as its angle increases from to 2 tt, and repeats these values in the same order as often as the angle x increases by 2 tt ; or n times for a change of 2 mr in the angle. The interval of repetition is called the period of the function. If k is the period of f(x), then f{x-\-'k) = f(x). The maximum numerical value of the function is called the amplitude. The table illustrates these definitions for several func- tions. /(^•) Period Ampli- tude Sina: 27r 1 Cosx 27r 1 Tana: TT CO Sin 2 X TT 1 Cos- 2 IITT 1 n 115. The Sine Curve. — This is plotted from the equation y = sin x, where x is to be reckoned in circular, or radian measure. *This is the form assumed by a perfectly flexible cord or chain sus- pended between two points. 148 ANALYTIC GEOMETRY We may assume values for x at any convenient interval, as ^ , and take the corresponding values of sin x from a table of natural sines. In the accompanying table the values of x in degrees are found in column 2 and their cir- cular measures to be used in plotting are in colunm 1. The circu- lar measure of x is the abscissa and the value of its sine is the or- dinate. In plotting it is con- venient to lay off on the .T-axis — = .52 units 6 and, using this as a unit, to mark off the various abscissas. After the values from the table up to 2 tt have been plotted the curve is constructed through one period, and the X (radians) X (degrees) sin x cos X .00 1.00 ^= .52 6 30 .50 .87 1.05 60 .87 .50 1 = 1-^^ 90 1.00 .00 2.09 120 .87 -.50 2.62 150 .50 -.87 7r = 3.14 180 .00 -1.00 3.66 210 - .50 -.87 4.19 240 -.87 - .50 4.71 270 - 1.00 .00 5.24 300 -.87 .50 5.76 330 - .50 .87 2 TT = 6.28 360 .00 1.00 6.81 390 .50 .87 7.33 420 .87 .50 etc. etc. etc. etc. same values of y may be used to construct it through suc- cessive periods. Since the angle x may be taken negative as well as positive, the curve may be extended indefinitely in the negative direc- tion. If we substitute {— x, — y) for {x, y) in the equation, we nave — y = sin (— a;) = — sin x, or y = sin x^ HIGHER PLANE CURVES 149 and the form of the equation is unchanged. Hence the curve is symmetrical with respect to the origin. 116. Circular Measure. — For plotting the sine curve x can be measured in degrees, and any convenient unit taken as a degree, but the scale of measurement in x will not corre- spond to that in y unless circular measure is used. Circular measure gives the same scale for both coordinates. For in radians, arc angle = or X radius ' AB But y = sin x = r AC T So as the angle x is generated by rotating the side OA from the initial position OB, the lengths of AB and AC in terms of the radius as a unit are the values of x and sin x given in columns 1 and 3 of the table of values for plotting. This is only true where x is expressed in circular measure. 117. The Cosine Curve. — The curve of cosines is plotted from the equation y = cos x. This curve has the same period and amj)litude as the sine curve, and in fact has the same shape ; but it passes through (0, 1) instead of the origin. This is true because the cosine of any angle is equal to the sine of an angle 90° greater, or cos x = sin (90° + x). Hence the table of sines can be used for plotting the cosine curve by moving the whole column of sine values up 90°, so that the sine curve becomes the cosine curve when the origin is advanced to the point [ ^, OV 150 ANALYTIC GEOMETRY 118. The Tangent Curve. — The curve of tangents is plotted from the equation y = tan x. It passes through the origin and its period is tt. The value of y is infinite when TT X is an odd multiple of -, and the curve passes through all values of v between a: = — - and a; = - , and so on at inter- 2 2 vals of TT, the period. The table of values follows. X tan X IT — oo 2 TT ~3 -1.73 ■K ~ 6 -.58 0. TT .58 TT 3 1.73 W 2 x 119. Multiple Angles. — Consider y=?>mnx. The mul- tiple n cUcides the period of the function by n, but does not alter the anqjlitude. To prove this, substitute x -\- "^ f or x. n We have • / , 2 ttX • ^ , o \ sm n X -\ = sm (nx + J tt) = sm nx. Hence -^ is the period of sin nx. That the amplitude i? n unaltered is obvious. HIGHER PLANE CURVES 151 Compare this with y = n sin x, where the period is un- altered, but the amplitude is multiplied by n. All of this appears clearly in the plots. Thus we have for y = sin x, the period is 2 tt and y = 1, when x = 90° ; for ?/ = sin 2 x, the period is w and y = 1, when x = 45° ; for y = 2 sin x, the period is 2 tt and y = 2, when x = 90°. 120. Sum of Functions. — Sometimes it is necessary to plot functions which are the sum of two or more trigono- metric functions. Time is saved in making the table of values by writing the values of the various terms in ]3arallel columns and summing corresponding values to get the various ordinates. Thus to plot y = sin 2 a; + 2 cos x, make a table of values for sin 2 x and 2 cos x in parallel columns and write the sum of values corresponding to the same value of x in a third column for the ordinate to be plotted. An alternative method is to plot the curves y' = sin 2 x and y" = 2 cos X on the same axes and obtain the correspond- ing ordinates by measurement, method. The figure illustrates this 152 ANALYTIC GEOMETRY 121. Inverse Trigonometric Functions. — The relation ex- pressed in the inverse notation by 2/ = sin~^ x reads in direct notation x = sin y. y = cos~i X = arc cos x. y = sin~i X = arc sin x. Y The curve of inverse sines is therefore the same as the curve of sines in shape, period, and amplitude, but the axes are interchanged. The notation arc sin x is preferable to sin~^ x. PROBLEMS 1. Discuss and plot the graph of : (a) y = cos X ; (c) y = sec x ; (6) y = cot x; (d) y = esc x. 2. Find the period and amplitude and draw the graph of (a) y = s'mSx; (d) y = SsecSx; lb) y = cosbx; (e) ?/ = |cos|x. (c) y = tan -. 3. Plot the graph of (a) y = arc tan x ; (b) y = arc sec x ; (c) y = 2 arc cos 2x: 3 arc sin - . id) y o 4. Draw on the same axes the graphs of : raW^^^'^^' (b) \y = ^^^^^ f(.^ f^ = tanx, ^ ^ \^y = CSC x ; ^ ^ \^y = sec x; ^^ ]y = cot x. What relation holds between corresponding ordinates of each pair of curves ? V^j (yjJjO^A > /^yjb^ 3 /- - HIGHER PLANE CURVES 153 5. Draw each of the following curves. For any value of ar, draw ordinates at ± ^, tt ± a*, 2 tt ± a:, and then write the trigonometric for- mulas suggested by the figure. (a) 2/ = sinx; (c) ?/ = tan x ; (e-) «/ = sec x ; • (6) y = cos %\ ((Z) y = cot x ; (/) 2/ = esc x. Ans. (a) sin (— .r) = — sinx ; sin i^TT ± x) = ^ sin x ; sin (2 TT ± ar) = ± sin X. 6. Plot the graphs of the following equations : (rt) y = cos 2 a: — sin -; (c) ?/ = a: + 2 sin a; ; (b) y =zsecx — x; (d) y = cos| + sin 2 x. 7. Determine which of the functions of Problem 6 are periodic and find the periods. Alls, (a) and (cZ) have the period 4 tt ; (6) and (c) are not periodic. 122. Parametric Equations. — In some curves the coor- dinates are functions of a third variable such that the condi- tions determining the curve may conveniently be expressed by two equations between these three variables instead of by a single equation in x and y. This third variable is called a parameter, and the two equations jjarametric equations of a curve. If t is the parameter, the parametric equations of the curve are usually of the form x =f(t), y = g{t). The parameter usually represents some geometric mag- nitude, or the time during which the point tracing the curve has been in motion ; but it may be chosen in any manner that is convenient. Thus for the semi-cubical parabola y"^ = a^^ we may assume a parameter t and write x = at^, y = aH^, since by eliminating t between these equations we obtain the original form. To plot the graph of such equations we take various values of the parameter and compute the corresponding values of x and y. Then plot the points {x, y) as usual. The values of the parameter appear only in the table of values, and not in locating the point. 154 ANALYTIC GEOMETRY 123. Parametric Equations of the Circle. — As usual wo take the origin at the center of the circle. Then the circle is generated by rotating OP. Taking 6 as the parameter, we have at once from the figure and the definitions of the sine and cosine : X = r cos B, y = r sin 6. These are the parametric equations of the circle. Erom the table of values below points on the curve are (10, 0), (8.7, 5), (5, 8.7), (0, 10), (-5, 8.7), (- 8.7, 5), (- 10, 0), (- 8.7, - 5), etc. Table of Values for r = 10 P{x,y) 9 X y 10 30 8.7 5. 60 5. 8.7 90 10 120 -5. 8.7 150 -8.7 5. 180 - 10. 0. 210 -8.7 - 5 Note that if we eliminate between the parametric equa- tions, we get the standard form x"^ -\- if = r^. 124. Parametric Equations of the Ellipse. — Suppose we have a circle of radius a and center 0, in which OA is any radius and AMsl perpendicular from A upon a fixed diameter. 5 is a point on the radius at a distance b from and P is the projection of B on AM. It is required to find the locus of P Take the fixed diameter as the X-axis and the center as the HIGHER PLANE CURVES loo origin. Let (x, y) be the coordinates of P and 6 = /.MO A be the parameter. Then we have at once from the figure, X = OM = OA cos ^ — a cos 0, y = MP= OB sin (9= 6 sin 0. These are the parametric equations of the locus. It remains to show that the curve is an ellipse. Solving the equations for cos 6 and sin 6 respectively, squaring and adding we get «2 ^ b^ cos2 ^ + sin2 ^ = 1, the equation of the ellipse. The circles described by B and A are called the auxiliary circles of the ellipse. By drawing them on coordinate paper, the various positions of P can be readily found and any ellipse of given semi-axes plotted with accuracy. 125. Path of a Projectile. — If an object moves with a velocity of V feet per second along a line inclined to the x- axis at an angle a, it is clear from the adjoining figure that at the end of each second its distance from the ?/-axis has been increased b}' V^= V cos a feet, and that from the .r-axis by Vy = V sin a feet. These quantities, V^ and V^,, are called the X- and y-components, respec- tively, of the velocity, and when multiplied by the time give the total displacement parallel to the X- and y-axes. In the case when the motion is in a vertical plane, the vertical velocity does not remain constant, but is affected by the force of gravity. In this case we find in Mechanics that the height is given by the formula h = vt— 16 f^, where r ^ Vy^ yy y y. X 156 ANALYTIC GEOMETRY li is the height in feet, t the time in seconds, and v the ver- tical component of the initial velocity in feet x>er second* Using these principles, let ns solve the problem : Find the. path of a projectile discharged at an angle «= arc cos -| with an initial velocity of 80 feet per second. Taking the point of projection as the origin, the ?/-axis vertical to the earth's surface and the x-axis horizontal in the direction of projec- tion, let F(x, y) be the position of the projectile after t seconds. The horizontal component of the initial velocity is Vx = 80 cos a = 80 • f = 48 ; the vertical component is F^ = 80 sin a = 80 • f = 64. As the horizontal velocity remains con- stant, X = Vxt = 48t. (a) Using the above formula, y = Vyt - \6 f^ z= Q4: t - 16 f^ (b) Equations (a) and (h) are the parametric equations of the locus. To identify the curve, eliminate t and we have ?/ = gX — xfi^ ? evidently a parabola. To get the range, set y = 0, and we have x = 192 feet. If we com- plete the square, we have (X- 96)2= -144(2/ -64). Hence the vertex is (96, 64), i.e. the greatest height reached is 64 feet. PROBLEMS 1. Plot the graphs of the following parametric equations : 1 t^ (a) x = St,y = —; ,(e) x = ti-l,y=-; (b) X = t"^, y = 2 t ; (f) x = acosd, y = b sinS . (c) X = 5 cos 6, ?/ = 3 sin ^ ; (g) x cos d = a, y = a tan d ; (d) x = a cos'^^, y = bsin^d] (h) x = 2 + 3cos ^, ?/ = 3sin^ - 1. 2. Eliminate the parameter in each pair of equations in Problem 1. 3. Form parametric equations for the folium of Descartes x^ + y^ = 3 axy, make a table of values and plot the curve. Hint. — Set y = tx. , 3 «« 3 af^ *In this and all that follows, we disregard air resistance. The results are therefore first approximations to the true ones. HIGHER PLANE CURVES 157 4. A straight line has the a;-intercept 5 and the inclination 30^. Find its parametric equations when the parameter t is the distance of the tracing point from the intersection with the a:-axis. Ans. x = b+ «— , y =Z' 6. Find the locus of the projectile, the range, and highest point reached from the following data : («) initial velocity 576 feet per second, angle of projection 45°; (6) initial velocity 400 feet per second, angle of projection 30° ; (c) initial velocity 800 feet per second, angle of projection 45°. 6. In Problem 5 (a), the projectile is fired up a hill of slope ^. Find how far up the hill it will strike the ground. Ans. 5796 feet. 7. Find the equation of the locus of a projectile of initial velocity 16.7:2 Fand angle of projection a. Ans. y = a: tan a — F^cos^a 8. In Problem 7 find the range and greatest height. Ans. Range — — ■ ; greatest height — . o2 o4 9. A fly crawls on a spoke of a wheel from the hub toward the rim at a rate of 4 feet per minute. The wheel makes 2 revolutions per minute. Find the equation of the path described by the fly. A}is. (in polar coordinates) p = - • IT 10. A point moves around a circle of radius 10 feet 50 times per minute. Write the parametric equations of the circle, using the time during which the point has been in motion as a parameter. 11. The parametric equations of the path of a point are X = a cos kt, y = o, where x is measured in feet, t is measured in seconds and a and k are constants. Plot the locus. How often does the point pass through the origin ? 126. The Cycloid. — If a circle rolls along a straight line the curve traced by a point on the circumference is called a cycloid. The straight line is called the base. 158 ANALYTIC GEOMETRY As the circle rolls it describes the length of the circum- ference on the base at each revolution, and a proportional distance for any part of a revolution. Hence the point of the circle touching the base is always at a distance from the starting point equal to the length of the arc sub- tending 6. In the figure the tracing point P starts from the point of contact 0, which is taken as the origin, and the base is taken as the a>-axis. Call the angle which the radius to the tracing point makes with the vertical, 6. Then OA = PA = rO, since arc = angle x radius. This gives x=OA-BA = rO - r sin 0, y = AC- DC= r-r cos 6, which are the parametric equations of the cycloid. llie Prolate and Curtate Cycloids. If the tracing point is taken anywhere on the radius of the rolling circle or the radius produced, the curve is called the x>rolate cycloid when P is without the circle and the curtate cycloid when P is within. The term trochoid is also used for both prolate and curtate cycloids. Let a be the distance from the center to the tracing point. The parametic equations of both prolate and curtate cycloids are X = rO — a sin 0, y = r— a cos 0. HIGHER PLANE CURVES 159 They are derived in precisely the same manner as those of the ordinary cycloid. PROBLEMS 1. Sketch the path of a point on a 28-inch bicycle wheel for two revolutions. Find the base of the cycloid and the coordinates of the tracing point when the wheel has made one third, one half and two thirds of a revolution. 2. Plot the cycloid for r = 10, forming a table of values for ^, X, and y. 3. Eliminate 6 between the parametric equations of the cycloid. Hi7it. — 6= arc cos ( ^ ^ ] and vers ^ = 1 — cos 6, Ans. x = r arc vers " — V2 ry — y'^. r 4. Write out in full the derivation of the parametric equations of (a) the curtate cycloid ; (6) the prolate cycloid. 5. Trace the curtate cycloid though two periods when r = 6 and, a = 5, and find the coordinates of the tracing point when 6 = 120^. 6. Plot the prolate cycloid for r — lO and a = 12. 7. Show from the equations of the cycloid that y is a, periodic function of x, of period 2 ttv. 8. Construct the figure and derive the equations of the cycloid when d > 90°. 9. Show that the first arch of the cycloid is symmetrical with respect to the line x = irr. Hint. — Show that for any value of ?/, as 2/1, 6 = di, ov 2'ir — di. From this get two values of x, Xi and x.2, and show that x.i~ tt r = tt r — Xp Then apply the definition of symmetry. 127. The Epicycloid and Hypocycloid. — When the base is a circle instead of a straight line, the curve is called an epicycloid if the generating circle rolls on the outside of the base, and a hypocycloid if it rolls inside. Note. — The epicycloid is the basis of the " Theory of Epicycles " which was used in the Ptolemaic system of astronomy to explain the motions of the planets. 160 ANALYTIC GEOMETRY Let the tracing point start from A and the generating circle move on the outside until its contact with the base is at D. Thus the arc AP of an epicycloid is described and the arc PD has rolled over the arc AD of the base. Let the angle which the generating radius CP makes with the line of centers be <^, and the angle of this line CO with the ic-axis be 6. Using circular measure, since arc AD = arc PD, BO = r. Now CPN is the supplement oi + cfi, and But and sin CPN = sin (0 + cf>), cos CPX= - cos {0 + cf>). x= OB=OE + XP = (R + r) cos $ — r cos (6 + <^) y=BP=EC-NC = {R-\- r) sin O — r sin (^ + <^). (a) These equations contain two parameters, 6 and (f>. cf> may be eliminated by substituting its value from (a), which gives x = (i? + ?') cos O—r cos — ^ Of r y =(^R -\-r) sin O — r sin — ^t_ $^ HIGHER PLANE CURVES 161 If r is taken negative we have the equations of the hy- pocycloid : a; = (i2 - r) cos ^ + r cos ?^^^ 0, ^ r y ==(E - r) sine -r sin ^^^ 6. r The Astroid. — In the hypo- cycloid B = 4:1', the rolling circle makes four revolutions in pass- ing around the base, forming the astroid, or hypocycloid of four cusps. 128. The Involute of the Circle. — A string is wound about the circumference of a circle. One end is fastened to the circumference and the string is unwound. If the string is kept stretched the curve traced by the free end is called the involute of the circle. If it begins to unwind at A, the arc AP of the involute is traced when the string has unwound as far as B, and the part unwound is BP. BP is tangent to the circle at B, and The equations are BP=BA = Be. x = R cos + Bd sin 9, y = EsiiiO-Pe cos 0. Note. — The teeth of large-geared wheels are cut in the shape of epi- or hypocycloids as the best forms to avoid the binding and friction that occur between straight teeth. 162 ANALYTIC GEOMETRY PROBLEMS 1. Show that for the astroid the equations reduce to X = B cos-^ 6, y = B sin3 e, 2 2_ 2. whence :r^ + y^ = B^. 2. Simplify the equations of the epicycloid and hypocycloid, and draw the curves when {a) B = r: (h) B = 2 r. 3. Show that the epicycloid in which i? = ?- is a cardioid. 4. Draw the epicycloid and hypocycloid when (a) 4i? = 3r: (6) SB = or] (c) r = 2B. 5. Derive the equations of the hypocycloid from a figure. 129. Algebraic Curves. — Algebraic curves of degree higher than the second have a great variety of forms and are best studied with the aid of Calculus. Two simple examples of these curves are the cubical x^arabola y = a:i^ and the semi- cubical parabola y- = ax^. The majority of the curves in Chapter YIII are found to be algebraic curves if their equa- tions are transformed into rectangular form. Many of these curves are of historical importance, as the limacon, the con- choid, the cissoid, and the lemniscate, given in Chapter YIII, and the witch, the strophoid, and the ovals of Cassini, which will be discussed in the following sections. 130. The Witch . — A circle of radius a is drawn tangent to the :/>axis through the origin, meeting the y-axis in A. Y HIGHER PLANE CURVES 163 Through A a tangent is drawn to the circle, and from a secant is drawn meeting the circle in B and the tangent in C. The locus of the projection of B upon the ordinate of C is called the witch. To find its equation take the angle = ^OC as a param- eter. Then we have at once x= 0D = AC =2 a tsiiie, and y = BP =EB = OB cos = 2 a cos^ $. Eliminating the parameter 0, the equation is The form of the equation shows that the curve is symmet- rical with respect to the ?/-axis and has the a>axis as an asymptote. 131. The Strophoid . — The distance of a fixed point A from a fixed line BC is a. From A a perpendicular ^0 is drawn to BC, also another line meeting BC at E. On this last line points P and P' are taken such that P'E=EP=OE. The locus of the points P and P' is a curve called the strophoid. To find its locus take the fixed line as the 2/-axis and the coaxis through the fixed point. Let the coordinates of P and P' be de- noted by (x, y). Taking the angle OAE as a parameter 6, we have at once that P'E = EP= 0E = at2ine. Hence and X = ± EP cos 6 = ± a sin ; 2/ = 0^ ± -EPsin ^ = a tan ^ (1 ± sin $). 164 ANALYTIC GEOMETRY By eliminating the parameter, we have the equation, y =±x-—^^, Va — X or y^ = x^ — — ■ a— X The curve is evidently symmetrical with respect to the a;-axis and has the line x = a as an asymptote. 132. The Ovals of Cassini. — The locus of the vertex of a triangle which has a constant base and has the product of its other two sides equal to a given constant is called the ovals of Cassini. Call the length of the base 2 a, take the origin at its mid-point, and let the ic-axis lie along the base. Then the conditions of the problem make F'P - FP equal to a constant, which we call c^ since it is essentially positive. Using the distance formula, we have Y ___^,«i^ (f^^fc^ 5^^^ %^ X ■\/{x H- ay -f y^ . -V{x - ay + ?/" = c^- Simplifying, this reduces to (0)2 + 2/^ + a^y _ 4 a'^x'^ = c\ Three types are possible, according to the relative values of c and a. If c = a, this last equation may easily be re- duced to the form, {x''^y''y=2a\x^-y% which becomes on transformation to ^polar coordinates, p2 = 2 a2 cos 2 e, showing that in this case the curve is a lemniscafe. (See page 134.) HIGHER PLANE CURVES 165 PROBLEMS 1. Transform into rectangular form the polar equations of the fol- lowing curves : (a) the lima9on ; (d) the cissoid ; (6) the cardioid ; (e) the four-leafed rose. (c) the conchoid ; 2. Show that the following curves are not algebraic : (a) the spiral of Archimedes ; (&) the lituus. 3. Plot the witch from its parametric equations. 4. Plot the strophoid. 5. Find the polar equation of the strophoid. Ans. p = — a cos 2 d cos 6 6. Transform the rectangular equation of the ovals of Cassini into polar form . 7. Derive from the figure the polar equation of the ovals of CassinL ^1 ^^'(Y-") \y^y[y_: -jrfr^ ' ^^- ' c "- ■^ r ^ 1 K^'^ i!\r .-f- t A? ( CHAPTER X TANGENTS AND NORMALS 133. Definitions. — A line cutting a curve is called a secant. In the figure FQ is a secant. If Q is made to move along the curve towards P, it is clear that PQ will turn about P and approach a limiting position PA, which is called the position of tangency. Therefore The tangent to a curve at a given point is the limiting posi- tion of a secant ichen two of its intersections approach coinci- dence at that point. TJie normal to a curve at a given point is the line perjoen- dicular to the tangent at that point. 134. Slope of Tangent and Normal to a Circle at a Given Point. — To find the slope of a tangent to a curve at a given point Pi(i\, 2/1), we first find the slope of the secant through Pi and a neighbor- ing point P.^(xi + h, 2/1 + k) on the curve. This will be k h proach F^ as a limit, the secant approaches the tan- gent and the slope of the tangent is the limiting As P2 is made to ap- TANGENTS AND NORMALS 167 value of -, which we write lim -, Since both k and h ap- h' h proach zero, the limiting value of - has the form -, which h is called an indeterminate form. To find the value of such a limit special methods must be employed. In the case of the circle we proceed as follows. Since P^ and P2 are both on the circle, their coordinates satisfy the equation of the circle. Hence (x^ + hy+{yy-\-ky = r\ Expanding and subtracting, 2 hx^ + /i2 + 2 ky^ + k"" = 0, h(2x,+ h)=-k(2y, + k), , . k 2 x^ + h and - = — -— ^ • h 2 ?/i + k From this last relation, we have lim^ = -lim?^L±i=_f^. h ^Vi-hk y^ Therefore for the tangent to the circle x^ -\- y"^ = r^, m=-^. (43) As the normal is perpendicular to the tangent, its slope is the negative reciprocal of this expression, or ^ • 135. Slope of Tangent and Normal to an Ellipse at a Given Point. — The method of procedure is the same as that for the circle. Let the equation of the ellipse be 168 ANALYTIC GEOMETRY and the point of tangency be Pi{Xi, 2/1) • The slope of a se- cant through Pi and a neighboring point P2{xi-\-h, 2/1 + k) k on the curve is as before h Substituting the coordinates of Pi and P2 in the given equation, bW + ciV = «'&^ b\x, + hf + a\y^ + ky = aW. Expanding and subtracting, 2 b^x^h + bW + 2 a^^/iA: + a'k^ = 0, 7i(2 t^xi + bVi) = - k(2 ahj^ + o?k), 2b%^±¥h 2 d-yx + a?k Hence ,. k y 2b% + bVi 2b^x^ lim - = — lim ^— ! = — - — -—' h 2 ahj, + a% 2 ahj^ Thus for the tangent to the ellipse, the slope is m = a'!/i (44) The slope of the normal is 136. Slope of Tangent and Normal to the Parabola and Hyperbola at a Given Point. — By applying the method of §§ 134, 135 to the equations of the parabola and hyperbola, y"^ = 2px and 6V _ g^yi _ ^2^2^ ^g obtain the following results : Tangent to parabola : m = — , Tangent to hyperbola : m = -^. (45) (46) TANGENTS AND NORMALS 169 The slopes of the normals are : for the parabola, — ^ ; P for the hyperbola, -^. Exercise 1. Prove formula (45). Exercise 2. Prove formula (46). 137. Equations of the Tangent and Normal at a Given Point. — To find the equation of the tangent or normal to a curve at a given point, it is necessary only to find the slope and then use the point-slope equation of the straight line, y -y^ = m(x-Xi). Exercise 3. Show that the equation of the tangent to the circle x2 + y- = r- at any point Pi(xi, yi) reduces to XiX + yiy = f^. Hint. — Since Pi is on the circle, simplify by using x-^-\-yi^= r^. Exercise 4. Show that the equation of the tangent at any point Pi(^i, y\) is : (a) for the ellipse - a (6) for the hyperbola - a ' , r _ 1 a;ix ^ly _ , ^ _yl-l xix_yiy _ , . (c) for the parabola y- = 2px, y\y =p{x + Xi). PROBLEMS 1. Find the slope of the tangent to each of the following curves foi any point (a:i, y\) on the curve : {a)xy = '2a^; cf^^_yl__-i. (6) y = ax^; ^'^ ^ a^ &2 - -^ ' (c) y^ = ax^; (g) (^x-hy + (y-ky = r^; (d) x^ = 2py; (^h) (y-ky2=2pix-h); (0 x^-y' = a^; ^^ (X - hy^ (y - k^ ^ ^'' a^ + b-^ -^' Ans. (a) -^; (6) Sax^^; (c) ^^^'"- (s)-y;zrr W ^^T^^' W- safety 170 ANALYTIC GEOMETRY 2. In each of the following curves find the slope of the tangent and of the normal at the point indicated : (a) 9 x-' + 2/2 = 18, (1, - 3) ; (e) x' + y'' = 169, (5, - 12) ; (&) x2 + 4 2/2 = 4, (2, 0); _ (/) x' + y'' = 25, (3, 4) ; (c) x2 - 2/2 = 25, (10, 5y3); (g) 2y^ = 5 a:, (10, 5); (d) 5x2-12 2/2=13, (V5,-l); (h) ^2 = 36 2/, (12, 4). 3. Write the equation of the tangent and normal to each of the curves in Problem 2 at the point indicated. Ans. (a) Tangent, Sx — y — 6 = 0; normal, x + Sy + 8 = 0. 4. Find the points at which the tangent to the corresponding curve in Problem 2 has the indicated slope : («) -3; (b) -iV3; (c) -fV3; (d) ^. (e) 1; (/) i; (g) - 1 1 W |. Ans. (a) (1, 3), (- 1,-3). 5. Find the angle between each of the following pairs of curves. (By the angle between two curves is meant the angle between their tangents at the points of intersection. ) (a) a:2 + y2 = 25, (c) x^ - y^ = a\ x-\-y =1 ; xy = V2a^; (b) a:2 + 2/- = 12, (d) ^2 + 4 2/2 = 4, 2/2 = X ; X + 2 2/ = 2. Ans. (a) arc tan |. 6. Prove that the tangents at the ends of the latus rectum of a parabola are perpendicular to each other. 7. Find the angle between the tangents at the ends of a latus rectum of (a) the ellipse ; (6) the hyperbola. 2e Ans. (a) arc tan ^ _ .^. 8. Prove that the tangents at the ends of any chord through the focus of a parabola are perpendicular to each other. 9. An ellipse and hyperbola have the same foci. Prove that the tangents at their points of intersection are perpendicular to each other. Hint. — The value of c is common to the curves. 10. Through what point of the ellipse b-x^ + a'^y- = a^b^ must a tangent and a normal be drawn so as to form with the principal axis an isosceles triangle ? , / -i «" ± b'^ Ans. Vrt-i-f b-y Va2 4-&2 11. Prove that a circle described on a focal chord of a parabola as a diameter is tangent to the directrix. TANGENTS AND NORMALS 171 138. Tangent and Normal. Subtangent and Subnormal. — Let P^The the tangent and P^iVthe normal to the curve in the figure at P^. By the length of the tangent is meant the distance from the point of tangencj to the point where the tangent meets the a/'-axis, i.e. the length of P^ T. Sim- ilarly the length of the nor- mal is the length of P^N. The subtangent is the pro- jection of the tangent on the ic-axis, i.e. TS. The subnor- mal is the projection of the normal on the x-axis, i.e. JSN. The direction of reading the subtangent and subnormal is away from the intersec- tion of the tangent with the aj-axis ; hence they are positive if they lie at the right of this point ; negative if at the left. To find the length of the subtangent and subnormal, let m be the slope and r the inclination of the tangent. Then , SPi y, m = tan t = — — ; = -^' TS TS TS=^ m Hence For the subnormal, the slope of NPi is ??i -i=tanPi.Y/S' = ^= ^^' m NS (47) Then Therefore SN = my,. (48) To find the lengths of the tangent and normal apply the right triangle theorem to the triangles NP^S and SPiT. 139. Tangent having a Given Slope. — Suppose it is re- quired to find the tangent to the curve a;^ + 4 ?/2 = 8 having the slope ^. 172 ANALYTIC GEOMETRY Here tlie slope is given and it is necessary to find the point of contact, which vre call Pi{x^, y^). Since P^ is on the curve, a^i' + 4?/i2 = 8. («) Now a;2 + 4 2/- = 8 is an ellipse, for which a^ = 8 and ¥ = 2. Hence by (44) the slope of the tangent at F^ is -^ But by the con- ditions of the problem the slope is to be i. Hence Sy~2' iP) Solving (a) and (6) simultaneoush', we get X, = t2, 2/x = ± 1. Thus there are two tangents with points of contact (2, — 1) and (— 2, 1) and equations y±l = \{xT 2), which reduce to .T - 2 ?/ T 4 = 0. The method illustrated in the above problem is typical and may be applied to any problem of the same character. 140. Tangent from a Given External Point. — To find the equation of a tangent to a curve from a given external point we proceed as in the following problem. Let the curve be ic^ _|_ 4 ^2 _ g g^j-^j^^ ]^\^q point (1, f ). As before, let the point of contact be P^ix^, y^. Then we have a;,2 + 4 2/i^=8. (a) 2 0^1 The slope of the tangent is, by (44), - equation of the tangent is Zi X-i / \ y-yi = -^{^-^i)' 82/i 82/1 Hence the (p) TANGENTS AND NORMALS 173 Since tliis is to pass through (1, |), we have 82/1 Solving equations (a) and (b) simultaneously, we get a^i = 2 or - I, yi = loT I. Thus there are two tan- gents with points of contact (2,1), and (-1,1). For the first of these the slope is — 4- and the equation 2/-l=-^(x-2), or x-\-2y— 4^ = 0. Eor the second the slope is Jj and the equation or a; - 14 ^ + 20 = 0. PROBLEMS 1. Write the equations of the tangent and normal to each of the following curves at the point indicated : (a) x^ + y^ = lS, (2, -3); (e) y = 2x^, (2, 16); (6) x2 + 2y2=18, (4, -1); (/) y^=-2x^ (-2, -4); (c) x2 - 2y2 = 18, (_ 6, 3) ; (g) V = 9x^ (1, 9) ; (d) 2 2/2 + 5 X = 0, (- 10, 5) ; (h) xy = 24, (4, 6). 2. Find the length of the subtangent and subnormal for each curve in Problem 1. 3. Find the equation of the tangent to each of the curves in Prob- lem 1 parallel to the line 4 .r — 3 ?/ = 6. 4. Find the equation of the tangent from the point (5, — 1) to each of the curves (a), (6), (c), and (d) in Problem 1. 174 ANALYTIC GEOMETRY 6. Find the lengths of the subtangent, subnormal, tangent, and normal at any point (x, y) of (a) 2/2 = 2px ; (c) b'^x^ - aY~ = «^&^ 5 (6) b-^x'^ + a^y^ = a'^b^ ; (d) x^ + y^ = r^. Ans. Subtangent: (a) 2x, (b) ~^, (c) ^, (d) -^'; sub- b-x b'-x X normal: (a) p, (b) --^, (c) "Y' ^^^ ~ ^• 6. Write and simplify the equation of the normal to each of the curves of Problem 5 at any point (xi, yi). 7. Show that a line from the focus of the parabola y^ = 2px to the point where any tangent cuts the ?/-axis is perpendicular to the tangent. 8. Prove that the tangents at the ends of a latus rectum intersect on the directrix in the case of : (a) the parabola ; (&) the ellipse ; (c) the hyperbola. 9. Prove that the tangents to the ellipse 6-x'- + cfiy"^ = a^ft'-^ and the circle r? + y^ = a^ at points having the same abscissa meet on the cc-axis. 10. Prove that the triangle formed by a tangent to the curve 2 xy = a^, the a:-axis, and a line joining the point of contact with the origin is isosceles. 11. Prove that the triangle formed by the tangent to the curve 2 xy = a'^ at any point and the coordinate axes has the area a^. 12. A tangent to the parabola y- = 2px has the intercepts on the axes numerically equal. Find its equation. 13. If a tangent to a parabola meets the latus rectum produced at A and the directrix at B, prove that FA = FB, F being the focus. 14. At what points on the ellipse ^ + ^ = 1 are the tangents equally inclined to both axes ? ^ 15. If d and d' are the distances of the foci of an ellipse from a tangent, prove that dd' is the square of the semi-minor axes. 141. The Parabolic Reflector. — Let TP be tangent to the parabola i/ = 2 px at Pix, y), and let TM be the subtangent TANGENTS AND NORMALS 175 Applying tlie subtangent formula, TM = 2 x; hence, TO = 03/ = X. Xow if F is the focus, this gives TF=x+P. ^^' But FP= XP=x+^^. Hence, TPF is isosceles, i.e. the tangent to a ■parabola makes equal angles icith the focal radius and the prin- cipal axis. This i3rinciple is used in the construction of parabolic reflectors. These are reflectors whose surfaces are gen- erated by revolving a parabola about its principal axis. The lamp is placed at the focus. Then by the laws of optics a ray of light from F to any point on the surface P is re- flected along PE so that angle TPi^= angle HPE. But this makes angle HPE equal to angle PTF by the above proof. Therefore PE is parallel to OX. Thus all the rays of light from a lamp placed at F are reflected along lines parallel to the axis. Exercise 5. V&e the above property of the tangent to devise a method of constructing a tangent to a given parabola at any point by ruler and compasses. 142. Angle between the Focal Radii and the Tangent to an Ellipse at any Point. — Let PT be a tangent and PX sl normal to the ellipse at any point P(x, y) on the ellipse, whose foci are F and F\ We desire to show that these lines make equal angles with the focal radii to P. 176 ANALYTIC GEOMETRY The slope of FP is -1—, of FP, ■' , and of NP. ^^, X' + c X — c ' b-x by § 135. Hence a^y y tan F'P\= ^'^ ^ + ^ = g^ x?/ + g^cy - b'xy ^ ay 62^2 + b'^cx + a2?/2 62a;(x + c) ~ a^b^ -f 62cx ~ 62 ' and tan FPN = y «^.y X — C 62^ 62^?/ — a2a;j^ _p ci2c2/ 62X(X — C) 62cx + a22,2 a2c?/ — c2x?/ cy aW - b'^cx 62- Therefore FPN= FPN\ and the angles made with the tangent, being complementary to these, are also equal. Exercise 6. Show how to draw a tangent to an ellipse at any point. 143. Diameters of a Conic. — The locus of the middle points of a system of parallel chords of any conic is called a diameter of the conic. Let a chord with slope tn meet the ellipse ^' 4- 2/' = 1 a2 62 TANGENTS AND NORMALS 177 in the points Pi(x^, 2/1) and P^^x^^ + h, y^ + A-). Then by § 135 the slope of the chord is a\2y, + k) Now let P(x, y) be the mid-point of the chord ; then 2 .X = 2 X, H- h and 2y = 2y, + lc by the mid-point formulas. Substituting, we have, m = — P„{x+h,xj^-^k) b'^x a-y or (49) which is the equation of the locus. From this equation we see that the diameter is a straight line through the cen- ter of the ellipse of slope m' = . Transferring the m a^m to the other side, we have an important relation between the slopes of a diameter and its chords, &2 mm = — (SO) In similar manner, we may show that the equation of the diameter of the parabola 2/^ = 2 pa;, is m (51) where m is the slope of the system of chords. For the hyperbola we obtain as the equation of the diameter of a set of chords of slope m, &2 y-^ a^m X. (52) 178 ANALYTIC GEOMETRY The relation between the slopes of the chords and their diameter is (53) mm = — • Exercise 7. Prove (51). Exercise 8. Prove (52). 144. Conjugate Diameters of an Ellipse. — The relation mm = o? states the condition satisfied by m and m' if the line y = m'x is the diameter of a set of chords of slope m. But as m and m' can be interchanged in (50) without altering the relation, we see that y = mx is the diameter of the set of chords of slope m\ parallel to the first diameter. Two diameters of slopes m and ??i', such that mm^ = are called conjugate diameters. by § 135. But ^ = 2/1 a' Each diameter is one of the chords bisected by its con- jugate. It is readily seen that tangents at the ends of a diameter are parallel to the conjugate diameter. For, let PifcTi, ?/i) be the end of the diameter y = mx. The slope of the tangent at Pi is -, since Pi lies on y = mx. 62 cr/>i , which is also the Hence the slope of the tangent is slope of the conjugate diameter. This property, together with Exercise 6, supplies a method of constructing a diameter conjugate to a given diameter with ruler and compasses. TANGENTS AND NORMALS 179 145. Conjugate Diameters of the Hyperbola. — The relation between the slope of a diameter of the hyperbola W-:i? — ahp- = a262 and the slope of the corresponding chords has been found to be mm =—• a2 As the equation of the conjugate hyperbola, — b^x"^ + aV = aW, is obtained by substituting — 5^ for b^ and cv^ for — a2, it is easy to see that the corresponding relation for the diameter and chords of the conjugate hyperbola is - = — . That is, it is the same in both the hy- mni' = perbola and its conjugate. As in the case of the ellipse, two diameters of slopes m and m' such that mm' = — are called conjugate diameters. Similarly, the chords bisected by the diameter y = mx are parallel to the conjugate diameter y = vi'x. Also the tan- 180 ANALYTIC GEOMETRY gents to a hyperbola at the ends of a diameter are parallel to the conjugate diameter. Exercise 9. Prove that the tangents at the ends of a diameter of a hyperbola are parallel to the conjugate diameter. PROBLEMS 1. Find the diameter bisecting the system of chords of slope 2 in each of the following conies : (a) a;2 + 4 1/2 = 16 ; (d) 2y^ - 6x = 0; (6) x^ -92/2 = 9; (e) 9 x^ -\-Wy^ = 144 ; (c) a:2 + y^ = 18 ; (/) 16 x^ -9y^= 144. 2. Find the diameters conjugate to those obtained in Problem 1. 3. What is the relation holding between conjugate diameters of the ellipse U^y^ + a^x- = a-b'^ ? 4. Show that conjugate diameters of an ellipse lie in different quadrants. 5. Show that conjugate diameters of a hyperbola lie in the same quadrant. 6. Prove that the major axis of an ellipse is greater than any other diameter. Hint. — Show that 4 a^ is greater than the square of any diameter. 7. Lines are drawn joining the ends of the major and minor axes of an ellipse. Show that the diameters parallel to these are conjugate. 8. Prove that the tangents to an ellipse at the ends of a diameter make equal angles with lines joining these points to a focus. 9. Prove that the tangents to two conjugate hyperbolas at the ends of a pair of conjugate diameters meet on an asymptote. 10. Prove that if any pair of conjugate diameters of a hyperbola are equal, the hyperbola is equilateral. 11. I'rove that in an equilateral hyperbola the asymptotes bisect the angle between any pair of conjugate diameters. ^ CHAPTER XI SOLID ANALYTIC GEOMETRY 146. Plane Analytic Geometry deals with the properties of figures which are wholly contained within a single plane. In Solid Analytic Geometry this restriction is removed. The effect of the addition of one dimension is to generalize the work of the previous chapters, the number of variables being increased to three, the loci of equations being surfaces, etc. The student should note carefully the correspondences between the two systems. 147. Coordinates. — The position of a point is determined with reference to three mutually perpendicular planes, XO F, YOZ, and XOZ. These are called the coordinate jolanes, and designated as the xy-, yz-, and xz-planes respec- tively. Their lines of inter- section, OX, OY, and OZ, are called coordinate axes and designated as the x-, y-, and z-axes respectively ; and their common point, 0, is called the origin. The coordinates of a point P are its distances from the re- spective coordinate planes measured parallel to the coordinate axes. The x-coordinate is the distance from the ^/^-plane parallel to the a;-axis, the y-coordinate that from the a'Z-plane parallel to the 2/-axis, and the z-coordinate that from the 181 Z + c D / / -t p + , __ 4. X^ B / • / / / E 182 ANALYTIC GEOMETRY icy-plane parallel to the 2;-axis. For the point P in the figure, x = FP=BE=CD= OA, y = DP = AE = CF = OB, z = EP = BF = AD = OC. Corresponding to each point there are evidently always three coordinates. Conversely, a point is completely located by its coordinates. For the ic-coordinate fixes it in a plane parallel to the yz-iplsme (or perpendicular to the x- axis), and similarly for the other coordinates ; and these three planes meet in but one point. In plotting a point, we usually take ^+ x= OA, y = AE, z = EP. Z ■+ / c D / // F / / • / / / / / / / P + . B. i x' E The positive directions are as indicated in the figure. The coordinate planes divide space into eight octants, which are distinguished by the signs of the coordinates of the points within them. For example the octant containing P in the figure has all the coordinates positive in sign. 148. Radius Vector and Direction Cosines. — TJie distance of a point from the origin is called its radius vector, and is de- noted by p. For the point P in the figure p = OP. It is evi- dent that p'-^OW + eP = 02' + 'aW + EP". Hence p2 = x2 + y^ + z\ (54) SOLID ANALYTIC GEOMETRY 183 The angles between the line OP and the positive halves of the X-, y- and 2;-axes are called the direction angles of the line OP and are denoted by a, ^, and y, respectively. The cosines of these angles are called the direction cosines of the line. They evidently fix its direction. From the figure it is evident that X = p cos a, 1/ = p cos p, z = p cos 7. (55) Formulas (54) and (55) give at once the important relation cos^ a 4- COS" p -h cos- y = I. (56) It is sometimes convenient to locate a point by means of its radius vector and its direction angles. In this case we P(x, y, z) = P{p, a, 13, y). This mode of representation is analogous to the polar coor- dinate system in plane analytic geometry. 149. Distance between Two Points. — The distance between any two points Pi(xi, y^, z{) and P2(o^2, 2/2? ^2) ^^ given by the formula, d = V(x, - x^y + {y, - yO' + (2i - 22)-. z To prove this, pass planes through the given points Pi and P^, j)arallel to the coordinate planes. These will form a rec- tangular parallelopiped, of } which P1P2 is the diagonal. By elementary geometry, P1P2 = P.A -f- AW + DP(. But P^A = iCi — x^, AD = yx — y-i, and DPj = Zi — z^. Hence we have the formula. (57) 184 ANALYTIC GEOMETRY 150. Direction of a Line. — The direction angles of a line not passing through the origin are defined as the direction angles of a jKirallel line through the origin icith the same positive direction. The positive direction on a line is arbitrary and indicated by the order in which its end-points are read. It is evident that if the direction is changed, the direction angles are re- placed by their supplements. Thus, if the direction angles of PoA are a, /?, and y, those of P^Pn are tt — a, tt — ^, and TT — y. In all cases the direction angles are in value between and TT. In the above figure the edges of the parallelepiped are parallel to the coordinate axes ; hence the direction angles of the line PoPi are : a = AP^P^, /? = BP^P,, y = CP.Pi- Then we have at once cos a = ^ , - , cos 6 = ^^— ^- , cos v = , • (^") a ^ d 'a Since the direction cosines are connected by formula (56) , they may be found if three numbers to which they are pro- portional are known. For if cos a : cos ^ : cos y = a : b : c, then cos- a _ cos^ ^ _ cos'^y _ cos- a + cos'^ ^ -f- cos^ y _ 1 a" ~ U" ~ & " a' + 62_|_c2 ~ a'' ^h^ -\-c^' whence a ^ cos a = , cos p = Va2 4-62_|_c2 V«2 + 62 _^ c* c cos y Va2 + 62+c2 Any three numbers proportional to the direction cosines of a line are called direction numbers of the line. / SOLID ANALYTIC GEOMETRY 185 151. Angle between Two Lines. — - Tlie angle between two lines which do not meet is defined as the angle between two intersecting lines parallel to the given lines and having the same positive directions. If the lines are parallel, the angle between them is or tt, according to their directions. We now derive a formula for the angle between two lines in terms of their direction cosines. Let the lines through the origin parallel to the given lines be OPi and OP2, where the coordi- nates of Pi and P2 ^1*6 (Xi, y^, z^) and (x2, 2/2? ^2) respectively. Let d be the distance between these points ; let pi and pj ^6 their radius vectors ; and let 6 be the angle between the lines. Then 2 _i_ ^ 2 ^2 cos 6 = ^^ ' ^■' 2 pxp2 z '^ 7 / \ A ■^ i<^ ' But Pi' = ^1' + 2/1' + ^x\ P2^=^X2^ + y.^-\-Z2\ ^2 = {x, - x^y + (2/1 - 2/2)' + (2!i - 22)'. .-. cos ^=:Ei^2_+ML±iii2. P1P2 Now from (55) — = cos «!, etc. Hence this last equation becomes ^^ cos 9 = cos ai cos a2 + cos pi cos P2 + cos 71 cos 72- (5^) When the two lines are parallel, we have DT Wl = TT- Cto ft=7r-^2, and yi = 72, and yi = TT — y2. When the two lines are perpendicular, = '^ and (59) becomes Li cos tti cos a2 + cos pi cos p2 4- cos ^i cos 72 = 0. (60) ^3 ^•^-^ b Wr- ■'. 186 ANALYTIC GEOMETRY PROBLEMS 1. Plot the points : {a) (3,4,5), (6,0,0), (-6,1,3); j^) (-2,3,-1), (7,-4,5), (1,0, 1); -^c) (0,0,2), (0,3,-4), (-2, -2, 1); 2. For each of the above points find the radius vector and its direction cosines. 3. Generalize the definition of symmetry with respect to a line so that it will apply to symmetry with respect to a plane. Show that the point (a*, y, s) is symmetrical to {—x^y^z) with respect to the ?/2;-plane ; to {x^ — ?/, 2) with respect to the cc2-plane ; to (a:, ?/, — z) with respect to the a:2/-plane. 4. Show that the point {x, ?/, z) is symmetrical to (— x, — ?/, z) with respect to the ^-axis ; to (— a;, y, — z) with respect to the ?/-axis ; to (.r, — y^— z) with respect to the x-axis. 5. Show that the following points are vertices of a regular tetrahedron : - ^^ ^ >.. (a) (4,0,0), (0, 4,'0), (0,^4), (4,^4,4); (6) (6, - 2, 1), (3, 1, 1), (3, - 2, 4), (2, - 3, 0). 6. Show that the three points (1, 4, - 2), (4, 7, 1), (— 2, 1, - 5) are m a straight line. ' ^^ ^ ^ ^-^ iW Show by tw^o methods that the following are the vertices of a rignt triangle : (a) (- 2,3,5), (-5,-3,3), (0,0, 11); (b) (2,6,-5), (-1,0,-7), (4,3, 1). //^ Find the angles of the triangles whose vertices are given in Pr^Jolem 1. 9. Find the coordinates of the point determined as follows : /) = 8, C0Sa=:f, C0S7=— A. 10. Show that the coordinates of the point Pq dividing the line P1P2 in the ratio n : r-z are _ r2Xi + riXo _ T'^yi 4- ri?/2 _ ^2^1 + r\Z2 '''- n + r, ' y'- ri + r2 ' '' ' n + r^ ' 11. What is the locus of points for which (a) 2 = 0; (e) x = 3, 2=- 2; \b) y = & constant ; (/) p = 6; (c) y z=z x = ; ■ (g) cos a = ; (d) xz=y; ' (h) cosa = cos/3 = ? SOLID ANALYTIC GEOMETRY 187 12. "What is the projection of P( - 2, 3, — 4) on (a) the icy-plane ; >-^ (^ (6) the 0-axis ? >-^r 152. The Locus in Solid Geometry. — Tlie locus of a point in space satisfying a given set of conditions is in general a surface or group of surfaces. To illustrate, consider the locus of a point at a given distance a from a given fixed point C. This is evidently the definition of the surface of a sphere of radius a and center C. Again the locus of a point at a constant distance a from a given straight line I is evidently a circular cylindrical surface having I as an axis and a as a radius. Extending the definition of the equation of a locus given in Chapter I, this shows that the locus of a single equation in the three variables, x, y, and z, is in general a surface. If the coordinates of a point satisfy tico equations simultaneously, its locus must be common to the two surfaces which are the loci of the equations, i.e. it must be their intersection, which is a curve or set of curves. Finally the points determined by a set of three equations will be the intersections of the surface determined by the third with the curves determined by the first two, and hence will be isolated points. Note the analogy to plane analytic geometry, where one equation determines a curve, and tico equations the isolated points in which the curves meet. 153. The Normal Equation of the Plane. — The perpen- dicular upon a plane from the origin is known as the normal axis and the distance of the plane from the origin is known as the normal inter cej^t, A plane is completely determined if the length of the normal intercept p, and the direction angles, «, /3, and y, of the normal axis are known. 188 ANALYTIC GEOMETRY To find the normal equa- tion of a plane, take any point P{x, y, z) in the plane and join it to the origin. Draw the normal axis ON. Let p^ be the radius vector of P, and «i, ySi,. yi, its direction angles ; let «, /8, y be the direc- tion angles of ON \ and let B be the angle between OP and ON. Then by (59), cos Q = cos «! cos a + cos /S^ cos /5 -f cos y, cos y. But GOS0 = —=^- OP Pi Eliminating 6 between these equations, Pi[cos «i cos a + cos /5i cos /3 + cos yi cos y] = p. This becomes on applying (55), X cos a + y cos p + z cos y = p, (61) which is the normal equation. ^ ^ Exercise. — Show that for any point P(x, ?/, z) not in the plane X cos a -\- y cos /3 + <: cos y^ p. 154. Plane Parallel to One or More Coordinate Axes. — Sup- pose the plane is given parallel to the 2;-axis. In this case the normal axis will lie in the .ry-plane and y = -• Hence (61) reduces to the form V-- r X cos a + !/ cos p = />. (61a) Similar equations are obtained for planes parallel to either of the other axes. ^ ^ , d-- 6/) ^ SOLID ANALYTIC GEOMETRY 189' In case the plane is parallel to both the y- and 2-axes, we TT 2 have the normal along the a;-axis. Then y8=y = |, and a = 0. For this case (61) reduces to the form x=p, a result obvious from the definition of coordinates. 155. The General Equation of the First Degree. — The nor- mal equation (61) of the plane is of the first degree. We now wish to prove conversely that the locus of the general first- J^ degree equation Ax + By -\- Cz + D = (i (62) ^5 is a plane. To do this we simply show that (62) can al- ways be reduced to the form (61). Dividing both sides of the equation by ± V^^ ■{- B^-\- C^^ we have J^ d^^. ^■■- :. + - ^ - .■ ' -D By § 150 t he coefficie nts of .^', y, and z a're^irection cosines of a line. Hence this equation has the same form as (61) and its locus is a plane. The direction cosines of its normal are A B ^ C ± V^42 ^B'+ C' ' ± V^l^+5-+C2 ' _t ^j2 + ^^ + (72 and the length of its normal intercept is -D ± -Va" + b'+c^' The sign of the radical is taken opposite to that of D so that the normal distance p shall be positive. Comparison with (61a) shows that if any one of the three variables is missing the locus is a plane parallel to the corresponding axis. 190 ANALYTIC GEOMETRY 156. Angle between Two Planes. — The angle between two planes is readily seen to be the same as that between their normal axes. Hence the angle between two planes can be found by formula (o9). If we wish to express cos d in terms of the coefficients of (62), substitution of the above values of the direction cosines gives AA' + BB' -I- Ca ,no. cos = — -— . (63) V^- + 52 + C2 . \A'' + 5'2 + C Parallel Planes. — If two planes are parallel, their normal axes are the same. Hence their direction cosines are the same or numerically equal with unlike signs. This gives A ±A' B . = etc. By alternation we have A V^2 + 52 ^ (72 VA'^ + B'^+ C""' B _ V^42H-52 + C2 A' B' V^'2_|_^/2_^(7/2 C V^2 + 52 _^ (12 C V^'2 _^ ^/2 ^ (^/2 This gives as the condition for parallelism, ^ = ^ = ^ A' B' C (64) Perpendicular Planes. — If the planes are perpendicular, cos ^ = 0. Then from (63) we have at once AA' + BB' + CC = 0. (60) 157. The Intercept Equation. — The intercepts of a plane are the distances from the origin to the points in which it meets the x-, y-, and 2;-axes. They are denoted by a, b, and c, respectively. If in equation (62) we set y = z = 0, we find that the \ SOLID ANALYTIC GEOMETRY 191 i»-intercept, a, is . Similarly 6 = — —, and c = ^ A ^ B C By transposing D and dividing both sides by — D, (62) may be reduced at once to the form - + l + ~ = l, (66) a b c which is knoT^Ti as the intercept form of the equation. PROBLEMS 1. Write the equations of the following planes in the intercept and normal forms and determine which octant contains the foot of the normal axis. (a)3a:-4?/+82;-5 = 0; (d)4x+5?/-20 = 0; (&)4x+?/ + 2-8 = 0; (e)x + s-l = 0; (c) 3 X - y - 8 + 10 = ; (/) X + ?/ + 2 = 12. ^ " 2. Write the equations of the planes determined by the following data : (a) a = 3, 6 = 4, c = — ^ ; (6) p = 5, cos a = 4, cos ^ = — ^; (c) through the points (1, 1, 1), (2, - 1, - 3), (4, 3, - 2) ; {d) parallel to the plane 3x— 4?/ + 8^— 5 = and containing the point (2, 1, - 4) ; (e) perpendicular to the Diane 3.r — 4?/ +82 — 5 = and contain- ing the points (2, 1, — 4) and the origin ; (/) having the foot of the normal axis at the point (—2, — 4, 1) ; (g) parallel to the ^r-ayis and containing the points (2, — 4, 1) and (3,5, -2). 3. Find the angle between each of the planes of Problem 1 and the plane 8x + 2?/-f22-5 = 0. 4. Find the angle? at which each of the planes of Problem 1 meets the coordinate planes. *^5. Show that the distance from the plane x cos «+ y cos ^ -^ z cos y/TTi 5 — j9 = to the point Cxi, yi, zi) is Xi cos « + 2/1 cos ^ -}- zi cos y — p. ^ 6. Using the result of Problem 5 find the distance from each of the planes of Problem 1 to the point (3, 0, — 2). 7. Find the locus of points equidistant from the points (—2 -3, 1) and (4, 5, ^ 7). 192 ANALYTIC GEOMETRY \^e^ ' ^' ^^^^ *^® point of intersection of the planes 2x -Sy + 2z = 5; Sx + by-3z = 6; 4:X — 2y-\-z = 9. 9. Show that the planes bisecting and perpendicular to the edges of the tetrahedron whose vertices are (3, 1,-2), (4, 2, — 1), (6, 1, —5), (1, 3, 2) meet in a point. 10. Find the locus of points equidistant from the planes 2x — By + 0=4 and Sx — 2y-\-s = 6. 11. Find the volume of the tetrahedron whose vertices are given in Problem 9. i 12. What is the locus of each of the following equations : / (a) x2-3x + 2 .-=0; . (b)x^-y^ = 0', (c) x^ -\- 2 xy + 7j^ - z^ = 0? ilZ. Name two of the following planes which are (a) parallel; (5) perpendicular to each other : 2x-Sy + bz = S; 4:X + Qy + 22=1; 6 a: — 9 V + 15 = 5. 158. The Equations of the Line. — Since any straight line may be regarded as the intersection of two planes, it will be seen from §§ 152 and loo that it requires two equations of the form Ax+By + Cz-hD = to determine a straight line. It is more convenient, ho^^^ver, to determine a line from two given points upon it, or from one point and its direction angles. From these we derive equations for the line. It should be noted that in each case two equations are required. 159. The Symmetrical Equations. — Let the line pass through the point Piix^, ih, z^ and have direction angles a, 3, and y. If P(x, y, z) is any point on the line and d denotes the distance PiP, we have from (58) X — X, n V — Vi z — z, cos a = ^, cos y8 = -' -^S cos y = i- d ' "^ d ' ' d SOLID ANALYTIC GEOMETRY 193 These relations give cos a cos p cos *Y which are the symmetrical equations. If a point Pi and the direction numbers of the line a, h, and c are knowTi, a more convenient form to use is (67a) 160. The Two-point Equations. — Let the line be deter- mined by the points Pi(a?i, y^, z^) and F^ix^ yo z^. Then from (58) the difference of the respective coordinates are direction numbers. Hence the two-point equations may be obtained from (67a) by substituting x^ — x^, 2/1 — 2/2> and 2^ _ z^, for a, b, and c, respectively. They are ^ . x — x^ _ y — y^ __ z — z^ (676) ^ ' X1-X2 1/1-1/2 ^l--^2 161. The Projection Forms. — Let the equations of the projections of the given line on the xy- and ct'2-planes respec- tively be . , ^ y = kx-{-m, ^Q^^ z = Ix -\- n. These equations taken together determine the line. For each equation may be regarded as the equation of the projecting plane. For example, the first by (61a) is the equation of a plane parallel to the 2-axis and hence perpen- dicular to the a;2/-plane. But it contains all points satisfy- ing the relation y = kx -\- m and so must contain the given projection. Similarly, the other equation is that of the plane projecting the line on the a;;2-plane. Thus the line is determined as the intersection of its projecting planes. 194 ANALYTIC GEOMETRY 162. Reduction of the General Equations of a Line to the Standard Forms. — Consider the line determined by the equations 4a; -h?/- 2 + 2 = 0, ic + 4 2/ + 22 — 1 = 0. To reduce these to the projection forms, eliminate first z and then y between the equations. We have at once, 3x' + 2^ + l=0, 5a;-22 + 3 = 0. That the locus of these equations is the same as the locus of those given is obvious, since coordinates satisfying the first pair of equations must satisfy the last. To reduce the given equations to the symmetrical form solve the projection forms for x. We have, 2^+1 2 2 - 3 *< Equating these, we have 1—3 5 * + 22 The denominators are direction numbers ; and if we divide each one by Vl^ +(— f)^ + {W = '^^^^ ^^^J become direc- tion cosines by § 150. Doing so, we get -3 V38 V38 V38 which are the symmetrical equations of a line having direc- tion cosines -7=, -==:, -7=. and passing through the point V3S V3S V3S (0, — i, 4). But these are merely transformations of the original aquations, and so are the symmetrical forms re- quired. SOLID AXALYTIC GEOMETRY 195 PROBLEMS 1. Find the points at which the following lines cut the coordinate planes and draw the lines : (a) 2x-Sy + 2z-6 = 0,Sx-y-z + e=0; (6) 2x-4?/ = 5, Sy-z = 2; , . y-2 ?/ - 3 g+ 1 (c) = - = • ^ ^ 6 2 3 2. Eeduce the equations in Problem 1 to the symmetrical form. 3. Reduce the equations in Problem 1 to the projection form, the projecting planes being perpendicular to the xz- and yz--p\&nes. 4. Find the equations of the straight lines determined as follows : (a) through the points (1, 0, — 5) and (— 2, 3, 1) ; (6) through the point (1, — 1, 2) and parallel to the s-axis; (c) through the point (1, — 1, 2) and perpendicular to the 2- axis ; (d) through the point (1, - 1, 2) and having cos a = i, cos /3 = \. (e) through the point (1, — 1, 2) and parallel to the line of Prob- lem 1 (a). 5. Do the lines of Problem 1 (a) and (6) meet ? 6. Show that if a line is perpendicular to a plane, it has the same direction cosines as the normal to the plane. 7. Show that if a line is parallel to a plane its direction cosmes and those of the normal to the plane satisfy the relation cos a cos a' + cos /3 cos ,3' + cos y cos 7' = 0. 8. Find the angle between the lines x-S ^y-\-2^ z-o ^^^ o- + S ^y-2^ z-4: 2 3 6 6 3 1 9. Show that the line ^^^ = ^^ = ^^ (a) is parallel to the plane 6x — Sy -{- 12 z - 60 = ; (6) lies in the plane Sx-iy-\-Qz-19 = 0; (c) is perpendicular to the plane 2 x -\- ?> y -{- z = 6. 10. Find in the symmetrical form the equations of the locus of points (a) equidistant from the points (3, -1, 2), (4, —6, -5), and (0, 0, - 3) ; (6) equidistant from the planes 2x -3y=6, Sx -2z = 8, and 2x + 32=:5. 196 ANALYTIC GEOMETRY 163. Cylindrical Surfaces. — A cylindrical surface is a sur- face generated by a moving straight line which constantly in- tersects a given fixed curve and remains parallel to a fixed stnttght line. The fixed curve is called the directrix and the generating line the generatrix. Let the directrix of a cylindrical surface be an ellipse in the ir?/-plane sj^mmetrical with respect to the coordinate axes, and let the generatrix remain parallel to the z-Sixis. Then the equation of the ellipse in its plane will be b'^x- + aY = a^W. Let P{x^ y, z) be any point on the surface, and Q the corresponding point on the ellipse. Evidently the coordinates of Q will be {x, y, 0) ; i.e. the x- and y- coordinates of every point on the line PQ are the same as those of Q. But the co- ordinates of Q satisfy the equation of the ellipse. Hence for any point on the cylin- drical surface, 52^^,2 _^ ,^2^2 ^ ^252. The preceding is perfectly general, and we see that it leads at once to the theorem : TJie locus of a cylindrical surface ivhich has for its directrix a curve in the xy-plane and ivhose generatrix moves parallel to the z-axis, lias the same equation as the directrix. The converse, which is readily established, is i Hie locus of an equation in two variables, x and y, is a cylindrical surface, ivhose directrix is the curve in the xy-plane which has the same equation, and whose generatrix moves parallel to the z-axis. SOLID ANALYTIC GEOMETRY 197 Similar theorems hold of course for cylinders whose gen- eratrices are parallel to either the oc- or ^/-axes. Exercise. What is the locus of any equation in one variable ? 164. Surfaces of Revolution. — A surface generated by re- volvinq a plane curve about a fixed line in its plane as an axis is called a surface of revolution. The curve is called the generatrix ; its position with ref- erence to the axis is unchanged during the revolution. Sections of the surface made by planes through the axis are called meridional sections. Erom the definition of such a surface it is evident that (a) sections made by planes perpendicular to the axis are circles ; (h) any meridional section is the generatrix itself. Let us first consider a conical surface generated loy re- volving a straight line about the 2;-axis. Let the position of the generatrix in the x2;-plane be AB, of which the equation is 2 a; + 2 = 5. Let P{x, y, z) be any point of the locus. As the line turns about the 2;-axis, P describes a circle of radius r = LP in a plane parallel to the a'2/-plane and distant from it KP = z. Now when the line is in the x2:-plane, x=r', hence for all posi- tions of the line, 2r + z = 5. But, as P describes a circle of radius r in a plane parallel to the xy'-plsine, we have at once b<^ x^ + y 2 ^ ^ 198 ANALYTIC GEOMETRY Substituting above, we have 2 V.^c^TF -{-z = 5. Simplifying, we have the equation 4(a;2+ 7/) = (z-5y. Consider the surface generated by revolving about the ic-axis a circle of radius a whose center is the origin. The equation of the generatrix in the a.'2;-plane is x2 +z'' = a\ Evidently the point P describes a circle with its center on the a>axis and of radius r ; hence X^ + 7'2 _ ^j2^ But 2/- + 2^ = r\ Hence we have at once, x2 + 2/2 + 22 = a\ (69) This equation is important, as it is the equation of a sphere of radius a and having its center at the origin. PROBLEMS 1. Identify and sketch the following surfaces : (a) a:2-4^2 = 4; {d) xz = 15 ; (6) a:2 + 2 s2 = 2 ; (e) ^2 _,_ 4 ^ _ 5 ^ ; (c) y2^z2_^2y + 62; = 15 ; (/) x^ = 16 z. 2. Find the equation of the surface generated by revolving the curve about the axis indicated : (a) Ho' -{-12y = 3, cc-axis (6) z^ = 2px, a:-axis (c) h-^x'^ + a~z- = a'^b^, a:-axis (d) b-x^ + a-z^ = a'^b-\ z-axis (e) b^oo^ —arz^ = a^b"^, a;-axis (f) b-^x'^ - a^z^=a^b% ^-axis SOLID ANALYTIC GEOMETRY 199 3. Find the equation of the surface of a cone whose vertex is at the origin, with the 0-axis for its axis of revolution and opposite elements perpendicular to each other. 4. Find the equation of the locus of a point equidistant from a given plane and a given line parallel to the plane. 5. Show that the locus of a point equidistant from the point (p, 0, 0) and the ?/2-plane is the paraboloid of revolution y- + z^ = '2px —p^. 6. Show that the locus of a point in space the sum of whose distances from the points ( ± c, 0, 0) is 2 a is the prolate spheroid a^ 62 52 r- )/■- z- 7. How mav the hyperboloid — _^ — — = 1 be defined as a locus ? a- 6- C- 165. Discussion of Surfaces. — The discussion of a sur- face is ill general a much more complicated matter than that of a curve. The notions of intercepts, symmetry, and ex- tent are readily generalized. Besides these, however, we have two other notions, which are of help in determining the nature of a surface. The first is the section of the sur- face by any plane parallel to one of the coordinate planes ; the second is the section of the surface by any of the coor- dinate planes. The latter is called a trace of the surface. To illustrate these, consider the sphere, with equation x" + 2/- + ^- = a\ To find the equations of the traces, set x, y, and z succes- sively equal to 0. When x = 0, we have 2/2 + ^2 = a\ Thus the 2/^-trace is a circle of radius a. The xy- and xz- traces are also circles of the same radius, with equations .T^ -f- ?/2 = a^ and Qt? -f z- = a- respectively. 200 ANALYTIC GEOMETRY The section made by a plane parallel to the ?/2;-plane and at a distance c from it is the curve determined by the given equation and the equation x = c. Eliminating x between the two equations, we have ?/2 -}- 2;2 = a^ — c^, which shows that the section is a circle having the point (c, 0, 0) as a center and Va^ — c^ as the radius. If c'> a, this is imaginary, indicating that the surface lies wholly within the planes x = ± a. A similar discussion holds for sections parallel to the other coordinate planes. To sketch a surface when only two variables of its equa- tion are of the same degree, or of the same degree and sign, first draw sections parallel to the plane of the two variables. These sections and the traces give the best representation of the locus. 166. Quadric Surfaces. — By analogy to the conies, the locus of the general equation of the second degree Ax^ + By^ + Cz^ + Dxy + Eyz + Fxz+Gx -\-Hy+Kz+L = is called a quadric surface. It can be shown that these sur- faces have properties analogous to those of the conies ; and in particular that every plane section of such a surface is a conic. We shall not go into these details, but will confine our discussion to the simpler forms to which the general equation may be reduced. 167. The Ellipsoid. — The locus of the equation x2 y2 22 — h — H — a2 bi c2 r2 1/2 72 -, + r-, + T, = l ('0) is called an ellipsoid. From the form of the equation we see at once that the SOLID ANALYTIC GEOMETRY 201 surface is symmetrical with respect to all of the coordinate planes and coordinate axes. The intercepts on the axes are X = ±a, 2/ = ± ^ z = ± c. Setting x = 0, the equa- tion of the trace on the ?/2-plane is hence it is an ellipse with semi-axes b and c. In like manner the traces on the other coordinate planes are ellipses. Settinsf x = k, the equation becomes ^-f — = — -, a form which shows that there is no section for k numerically greater than a. Dividing by the right-hand member, this becomes r = 1. Hence the various sections are ellipses symmetrical with respect to the ic-axis. The semi-axes, - Va^ _ k"^ and a £. V<^2_;^2^ ^^Q^ smaller as k increases, until the ellipse be- et comes a point when k = a. Similarly the sections parallel to the other coordinate planes are ellipses. The values a, b, and c are called the semi-axes of the ellipsoid. When a = 5 = c, the ellipsoid is evidently a sphere with the center at the origin. Ordinarily the semi-axes are unequal, and in this form of the equation it is usually as- sumed that a > 6 > c. When 6 = c, but a>b or c, the 202 ANALYTIC GEOMETRY ellipsoid is called a prolate spheroid. In this case the sec- tions parallel to the 2/2;-plane are circles. The locus is then a surface of revolution since it is generated by revolving the a;2;-trace (or the x7j-tTSice) about the a:-axis. When a = h but c X / -|-V 1 / 1 / -\ V. •■ - ^ In this case sections parallel to the a;?/-plane, made by the planes 2 = A:, have equations y W- a^ IP- (^ and hence are ellipses which increase in size as the numerical value of k increases. SOLID ANALYTIC GEOMETRY 203 The cc^-trace and the yz-tmce are the hyperbolas />'2 y2 9/2 J.2 __^ = 1 and •'^ — - = 1 respectively. a" e fr & If a = 6, the locus is a surface of revolution about the 2;-axis. The locus of the my 169. Hyperboloid of Two Sheets. equation x^ y- ^' ^-t is called the hyperboloid of two sheets. In this case we consider sections parallel to the 2/2-plane made by the planes x = k. These are the ellipses If k^ < o? the right-hand member is negative. Hence no part of the locus is between the two planes x = ± a. If k = ±a, each corresponding section is a point ellipse. As k increases numerically from a to go the section increases indefinitely. The .T?/-trace and the ajg-trace are the hyperbolas ^ _ ^ = 1 and - — - = 1 respectively, a- 52 ^2 c2 If 6 = c, the locus is a surface of revolution. 204 ANALYTIC GEOMETRY 170. Elliptic Paraboloid. — This is the locus of the equa- tion ^ ^ :>-tAr^^ - + ^' = 2 cz. (73) Proceeding as before, we observe that sections made by the planes z = 'k are el- lipses whose axes increase indefinitely as A; increases. The surface lies wholly above or wholly below the a'2/-plane, according as c is positive or negative. The xz- and 2/2^-traces are parab- olas. If a = 5, the locus is a surface of revolution. 171. Hyperbolic Paraboloid. — This is the locus of the equation 2 ,,9 ^-^ = 2cz. (74) Consider c positive. Sections made by the planes 2; = A: are the hyperbolas As Tc increases from to 00 the vertices of the corresponding sec- tions lie in the .T2:-plane and recede indefinitely from the 2-axis. As k decreases from to — 00 the vertices of the sec- tions are in the ?/2;-plane and recede indefinitely from the z-axis. The xz- and ?/2-traces are parabolas, each having its SOLID ANALYTIC GEOMETRY 205 vertex at the origin, the former extending above the a-^z-plane, the latter below. The xy-tmee is a pair of lines intersecting at the origin. PROBLEMS 1. Find the traces of the following surfaces on each coordinate plane : (a) y^-\-z^ = l2z; (e) x^ - 4:y^ - Az^ = 4: ; (6) x^-y^ = 0; (/) «- + y-=Qz-; (c) x2 + 4?/2 + 4^2 = 4 ; (g) x^ + 4y^ -{-2z^ = W ; id) x^ + 42/-^ - 4^2 = 4 ; (h) x^ = y^ + 9 z\ 2. Find the intersection of each of the above surfaces and the plane .r = 4. 3. Discuss the traces of the following surfaces : a- 0- 4. Show that the sections of —-{-^-~=z are parabolas if perpendic- a^ 0- ular to the x- or the ?/-axis, and ellipses if perpendicular to the ^-axis and 5 > 0. 5. Show that the sections of the surface ^- ^ = 2 perpendicular to the X- or the ?/-axis are parabolas. How do those perpendicular to the :f-axis differ from those perpendicular to the ?/-axis ? 6. Discuss and sketch the following surfaces : (a) 4a:-^+25?/2 + 16 2-^rr 100; (6) 28 x2 + 196 = 9 ?/-2 + 16 z^ ; (c) 4.r2 + 9!/2_^2 + 36 = 0; (fO 16 .r2 + ?/2 ^ 04 ^ 33 ; (e) 100^2 + .36?/2-81x = 0; (/) ?/2_ 22 + 9a; = 0. 206 ANALYTIC GEOMETRY FORMULAS AND EQUATIONS Formulas of Distaxce and Direction Xo. Page Distance: d = -^{x^- x^f + iy^-y^f (1) 12 Point dividing line in the ratio i\ : r^ : r^x^±r^^ y^^nill±Ml. (2) 13 ri 4- r2 Vi H- rg Mid-point : a^^ = J (x^ + a-g), ?/o = 2 (2/1 + 2/2). (2 a) 13 Slope: m= -^/-y^ (^) ^' X-[ iCo Test for parallelism : mi = m2. (4) 18 Test for perpendicularity : mi7?i2 = — 1. (5) 18 Angle beween two lines : tan /? = '^ ~ — ^. (6) 18 1 + mim.2 The Straight Lixe Equations : Point slope form : y ^ y^= m(x — .^2). (7) 40 Two point form : ^ — ^ = "^^ ~'^^ • (7 a) ■ 41 X — X^ X^ — X2 Slope intercept form : y = 7nx + b. (8) 41 Intercept form : ^ + -^ = L (9) 41 a b General form : Ax + By-\-C = 0. (10) 44 Test for parallelism : A.A' = B:B'. 45 Test for perpendicularity : AA' = — BE'. 46 Test for identity : A : .4' = B:B'=C:C'. 46 Normal form : x cos y + y''=e^x\ (29) 110 Generalized standard equations of the conies : '^ihirabola: {y -kf = 2 p{x-h). (30) 117 {x-liy = 2p{y-k), (30a) 117 FORMULAS AND EQUATIONS 209 No. Page Ellipse: (^sd!Y + (jL=^ = l, (31) 118 (y-ky _^(x-hy^^^ (31a) ng Hyperbola: (^^-(^^ = 1. (32) 118 Angle of rotation for eliminating tlie xy term : tan 2^ = -^. (33) 120 A — G Polar Coordinates Eelation to rectangular coordinates : x = pcos^, 7/ = psin^. (34) 129 ^2 = 0^2+2/2, ^ = arc tan '^. (35) 129 Equations of the straight line : pcos^ = a. (36) 131 p sin ^ = a. (36 a) 131 e=c. (37) 131 Equations of the circle : ^ p=r. (38) 131 p = 2rcos^. (39) 131 p = 2r sin 6. (39 a) 131 p = acos^ + 6sin(9. (40) 132 Formula of rotation : = 6' + «. (41) 136 Equations of the conies : p^^lL^. (42) 138 1 — e cos 6 p = ^-E (42 a) 138 1 — e sin 6 210 ANALYTIC GEOMETRY Tangents and Xoemals Slope of a tangent : No. Page To the circle : m = -^- (43) 167 yi To the ellipse : m = -—'' (44) 168 To the parabola : m = i-- (45) 168 2/1 To the hyperbola : m = ^. (46) 168 Length of the subtangent : ^- (47) 171 m Length of the subnormal : myi. (48) 171 Equation of the diameter bisecting chords of slope m : In the ellipse : y = x. (49) 177 In the parabola : yz=i^. (51) 177 m In the hyperbola : y = x. (52) 177 ahn Relation between the slopes of conjugate diameters : ? - In the ellipse : nwi' = (50) 177 In the hyperbola : mm' = —' (53) 178 Solid Analytic Geometry Radius vector and direction cosines : p = a.'2 + 2/2_^22, (54) 182 x = p cos a, y = p cos ft, z = p cos y. (55) 183 cos'^ a + cos^ /? + cos^ 7 = 1- C^^) *■ ^^^ FORMULAS AND EQUATIONS 211 Distance : No page d = VCx-i - x,y +(2/i - y,y + (^1 - z,)\ (57)^183 Direction cosines : cos« = ?i.=^=, cos ;8=?-^!-=^S cosy = ?L=L^. (58)^184 d d cl Angle between two lines : cos ^=cos «! cos cco+cos ^1 cos /32+cos yi cos y2- (59) //185 Condition for parallelism : «i = «2j P\ = ^-2^y\ = 7-2, or «^ = TT — tto, /?! = TT — /?.2J yi = -n- — 72- Condition for perpendicularity : cos «i cos «2 + cos ^1 cos ^2 + cos yi COS yo = 0. (60) *^85 Equations of the plane : Normal form : a; cos « + ?/ cos /? + 2; cos y = p. (61) »^188 General form : Ax + By + Cz + D = 0. (62) ^lB9 Angle between two planes : Test for parallelism : A:A' = B:B'=C:C'. (64) v. , ^ i'-- UNIVERSITY OF CALIFORNIA LIBRARY Y-4