B M EMT bflM IN MEMORIAM FLOR1AN CAJORI X if- PLANE AND SOLID ANALYTIC GEOMETRY AN ELEMENTARY TEXT-BOOK BY CHARLES H. ASHTON, A.M. ASSISTANT PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF KANSAS RE riSED ~ EDITION - ' ' NEW YORK CHARLES SCRIBNER'S SONS 1904 Wof COPYRIGHT, 1900, BY CHARLES SCRIBNER'S SONS CAJORI Norruoofc $rrss J. S. Cushing & Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE The present work is intended as a text-book for class- room use, and not as an exhaustive treatise on the sub- ject. This object has been kept constantly in mind in writing the book, and every subject has been treated from this point of view. A large part of the book was mimeo- graphed and tested by use for several years in the author's classes in Harvard University. The author has tried to meet the needs of a class which occupies from sixty to seventy recitation hours upon the subject, and it is thought that the book ought to be com- pleted by the average class in that time. Necessarily some subjects which usually find a place in books on Analytic Geometry have been omitted ; but it is thought that nothing has been omitted which has an important bearing on future mathematical study. The conies have been treated from their ratio defini- tion, and much space and time have been gained by not repeating proofs which are identical, or very similar, for the three forms of the conic. Analytic methods are used throughout the book, and the author has attempted to give proofs which are concise and easily understood by the average student, but, at the same time, mathemati- cally rigorous. In this connection he would call atten- tion to the proofs in oblique coordinates (Arts. 12, 28), which are usually given without reference to the direc- tions of the lines, and, therefore, do not hold if the positions of the points are changed. v 911318 yi PREFACE Numerous problems, which have been selected with great care, have been inserted after nearly every article. In the early part of the book these are mainly numerical, but later the student has been asked to prove a large number of theorems. A considerable number of theorems which are usually proved in the text are here inserted as problems, and in many places the student has been asked to derive formulas for two of the conies, after the corresponding formulas for one of the conies have been obtained. It is only by solving such problems that the student can acquire any real grasp of the subject. The attention of the teacher is also called to the two chapters on loci (Chaps. VIII and XIV), in which a large number of problems are given and the methods of solving them discussed ; to the treatment of poles and polars by the aid of harmonic division (Chap. XII); and to the system of polar coordinates used in the Solid Geometry. The author desires to acknowledge gratefully the assist- ance of Mr. B. E. Carter of the Massachusetts Institute of Technology, who has read with great care both manu- script and proof ; of Mr. E. V. Huntington, who made many valuable suggestions during the early stages of the work ; and of Mr. W. R. Marsh, his colleague in the preparation of the series, of which this volume is the first to appear. Cambridge, November, 1900. CONTENTS PART I PLANE ANALYTIC GEOMETRY CHAPTER I Introduction ART. PAGE 1. Directed lines 1 2. Addition of directed lines 1 3. Directed angles 2 4. Addition of directed angles 2 5. Measurement of lines and angles 3 6. Angles between two lines 4 7. Law of sines and law of cosines 5 8. The quadratic equation 6 CHAPTER II The Point 9. Cartesian coordinate systems 7 10. Notation 11 11. Distance between two points in rectangular coordinates . 11 12. Distance between two points in oblique coordinates . . 13 13. Points dividing a line in a given ratio 14 14. Harmonic division . . 16 CHAPTER III Loci 15. Equation of a locus 20 16. Locus of an equation 24 17. Plotting the locus of an equation 25 vii Vlll CONTENTS ART. PAGE 18. Symmetry 27 19. Intercepts 31 20. Intersection of two curves 31 21. Locus of u + kv = 0, and uv = 31 CHAPTER IV The Straight Line 22. Introduction . . 23. Line through two points 24. Line determined by its intercepts .... 25. Oblique coordinates 26. Line determined by a point and its direction 27. Line determined by its slope and its intercept on the ' 28. Oblique coordinates 29. General equation of the first degree 30. Two equations representing the same line cannot except by a constant factor . . ... 31. The angle which one line makes with another 32. Perpendicular and parallel lines .... 33. Line making a given angle with a given line 34. Normal form of the equation of a straight line 35. Reduction of the general equation to the normal form 36. Distance of a point from a line .... 37. Oblique coordinates 38. Bisector of the angle between two lines 39. Lines through the intersection of two given lines . 40. Area of a triangle differ ■axis 34 34 36 36 37 38 39 40 42 44 46 48 49 51 53 55 56 57 59 CHAPTER V Polar Coordinates 41. Introduction 63 42. Equation of a locus 64 43. Plotting in polar coordinates 65 44. Natural values of the sines, cosines, tangents, and cotangents 66 CONTENTS IX CHAPTER VI Transformation of Coordinates ART. PAGE 45. Introduction 68 46. Transformation to axes parallel to the original axes . . 68 47. Transformation from one set of rectangular axes to another, having the same origin and making an angle with the first set 70 48. Transformation in which both the position of the origin and the direction of the axes are changed . . . .71 49. Transformation from any Cartesian system to any other Cartesian system, having the same origin .... 72 50. Degree of an equation not changed by transformation of coordinates • • .72 51. Transformation from rectangular to polar coordinates . . 73 CHAPTER VII The Circle 52. Equation 75 53. General form of the equation 76 54. Circle through three points 77 55. Tangent 79 56. Normal 81 57. Tangents from an exterior point 82 58. Tangent in terms of its slope 84 59. Chord of contact 85 CHAPTER VIII Loci 60. Problems 88 61. Problems 91 CONTENTS CHAPTER IX Conic Sections ART. . PAGE 62. Definition and equation 100 63. Parabola 101 64. Central conies 103 65. Ellipse 106 66. Hyperbola 110 67. Asymptotes 114 68. Conjugate hyperbolas 116 69. Equilateral or rectangular hyperbola 117 70. Focal radii of a central conic 118 71. Mechanical construction of the conies 120 72. Auxiliary circles 121 73. General equation of conies when axes are parallel to the coordinate axes 123 CHAPTER X Tangents 74. Equations of tangents 126 75. Normals 128 76. Subtangents and subnormals 129 77. Slope form of the equations of tangents .... 131 78. Theorems concerning tangents and normals .... 133 CHAPTER XI Diameters 79. Equations of diameters 142 80. Conjugate diameters 144 81. Equation of conjugate diameter 147 82. Theorems concerning diameters 148 CONTENTS XI CHAPTER XII Poles and Polars ART. PAGE 83. Harmonic division 154 84. Polar of a point 155 85. Position of the polar 157 86. Theorems concerning poles and polars 159 CHAPTER XIII General Equation of the Second Degree 87. Introduction 166 88. Two straight lines 166 B* - 4 .4 C ^ 0. 89. Removal of the terms of the first degree . . . .168 90. Removal of the term in ^ 169 91. Determination of the coefficients A', C, and F' . . . 170 92. Nature of the locus 172 B* - 4 A C = 0. 93. Removal of the term in xy 175 94. Removal of the term my 177 95. Nature of the locus 177 96. Second method of reducing the general equation to a simple form, when£-4.4C = 178 97. Summary 182 98. General equation in oblique coordinates .... 183 99. Conic through five points 183 CHAPTER XIV Problems in Loci • 185 Xll CONTENTS CHAPTER XV Higher Plane Curves ART. PAGE 100. Introduction 195 101. The parabolas 195 102. The Cassinian oval 196 103. The cissoid 198 104. The conchoid 198 105. The cycloid 200 106. The hypocycloid 201 107. The epicycloid 203 108. The cardioid 204 109. The catenary 205 110. The spirals 205 PART II ANALYTIC GEOMETRY OF SPACE CHAPTER I Coordinate Systems. The Point 1. Introductory 2. Rectangular coordinates 3. Distance between two points 4. To divide a line in any given ratio 5. Projection of a given line on a given axis 6. Polar coordinates 7. Spherical coordinates .... 8. Angle between two lines 9. Transformation of coordinates. Parallel axes 10. Transformation of coordinates from one set of rectangular axes to another which has the same origin 207 208 209 210 211 212 215 216 218 218 CONTENTS X11L CHAPTER II Loci ART. PAGB 11. Equation of a locus 220 12. Cylindrical surfaces 221 13. Surfaces of revolution 221 11. Locus of an equation 223 CHAPTER. Ill The Plane 15. Normal form of the equation of a plane .... 227 16. Reduction of the general equation A x + By + Cz + D = to the normal form 228 Equation of a plane in terms of its intercepts . . . 229 Distance of a point from a plane 229 The angle between two planes 231 Perpendicular and parallel planes ....... 232 Equation of a plane satisfying three conditions . . . 232 CHAPTER IV The Straight Line 22. Equations 235 23. The equations of a line in terms of its direction cosines and the coordinates of a point through which it passes . . 237 24. Given the equations of a line, to find its direction cosines . 238 25. Equations of a line through two points .... 240 CHAPTER V Quadric Surfaces* 26. The sphere 242 27. Conicoids 244 28. The ellipsoid . 245 XIV CONTENTS 29. The imparted hyperboloid 30. The biparted hyperboloid 31. The cone 32. Asymptotic cones . 33. The paraboloids 34. Ruled surfaces 35. Tangent planes 36. Normals . 37. Diametrical planes 38. Polar plane PAGE 247 250 252 253 254 257 259 261 262 264 Answers PART I PLANE ANALYTIC GEOMETRY CHAPTER I INTRODUCTION 1. Directed lines. — If a point moves from A to B in a straight line, Ave shall say that it generates the line AB ; if it moves from B to A, it generates the line BA. In our study of Geometry, AB ^ ^ - c and BA meant the same thing, — j ^ B the line joining A and B with- c j B out regard to its direction. But FlG - 1 « we shall now find it convenient to distinguish between AB and BA as if they were separate lines. The position from which the generating point starts is called the initial point of the line ; the point where it stops, the terminal point. 2. Addition of directed lines. — If a point moves in a straight line from A to B (on any one of the lines in Fig. 1) and then moves in that line, or in that line pro- duced, to (7, the position which it finally reaches is evi- dently the same as if, starting from A, it had moved along the single line AC. The line AC is called the sum of the lines AB and BO; that is, AB + BO=AO. Evi- dently AB + BA = AA = 0, and hence AB = - BA. l 2' ANALYTIC GEOMETRY [Ch. I, §§ 3, 4 3. Directed angles. — If a line starts from the position OA and rotates in a fixed plane about the point into the position OB, it is said to generate the angle A OB. If it rotates from OB to OA, it generates the angle BOA. We shall find it convenient to ^" ^-^B distinguish between the angles / ^ ^ / \ A OB and BOA as if they were / / jP^\ f\ separate angles. The position \f f \ s qZ- _J \a from which the moving line \ / / starts is called the initial side ^t ~S / of the angle ; the position where ^S it stops, the terminal side. There is no limit to the pos- Eig 2 sible amount of rotation of the moving line ; after performing a complete revolution in either direction, it may continue to rotate as many times as we please, generating angles of any magnitude in either direction. Angles which are not equal, but have the same initial and terminal sides (1 and 3, or 2 and 4, Fig. 2) are called congruent angles. In reading the angle AOB, it is not possible to dis- tinguish between the various congruent angles which have OA and OB for their initial and terminal lines, but we shall understand that the smallest of the congruent angles is meant, unless another angle is indicated by an arrow in the figure. 4. Addition of directed angles. — If the moving line starts from OA (in any one of these figures) and rotates first through the angle A OB, and then through the angle BOO, it is evident that the position 00, which the line C.i. I, § 5] INTRODUCTION 3 finally reaches, is the same as if, starting from OA, it had rotated through the single angle A 00. The angle -^J3 Fig. 3. A 00 is called the sum of the angles A OB and BOO; that is, ZAOB +ZB0O=ZA0O. Evidently ZAOB + ZBOA = 0, and hence ZAOB = -ZBOA. 5. Measurement of lines and angles. — The length of a line, or the magnitude of an angle, may be represented by a number, by the familiar process of measurement. That is, the number of times which the given line or angle contains an arbitrarily chosen unit may be used to represent the length of the line or the magnitude of the angle. But we have seen that it is necessary to distinguish between the lines AB and BA, and that it has followed from our definition of addition of lines that AB = — BA. Hence, if the line AB is represented by a positive number, the line BA ivill be represented by the same number with a negative sign. In like manner, if the angle A OB is represented by a positive number, the angle BOA will be represented by the same number with a negative sign. It follows, therefore, that opposite direc- tions are indicated by opposite signs ; that is, if the length of a line or the magnitude of an angle, generated in one direc- tion, is represented by a positive number, then the length of a line or the magnitude of an angle generated in the opposite ANALYTIC GEOMETRY [Ch. I, § 6 direction, is represented by a negative number. Either of two opposite directions may be chosen as the positive direction ; then the other must be taken as the negative. As all our work will be concerned with the algebraic number rather than the geometric line which it represents, it will not be necessary to distinguish between the line AB and the number which represents its length. We shall let AB stand for the number which represents the length of the line from A to B. It is easily shown that the length of the sum of two or more lines that run in the same or in opposite directions is the algebraic sum of the lengths of the separate lines. Hence it is still true that AB+BC= AC, when AB, BO, and AC stand for the lengths of the lines AB, BO, and AC. Since these are now algebraic numbers, it follows that AB = AC— BO. In like manner A OB will be used to represent the magnitude of the angle instead of the angle itself. With this meaning it will still be true that ZAOB + ZBOC=ZAOC. Also, ZAOB=ZAOC-ZBOC. 6. Angles between two lines. — When two lines intersect at a point, they form several angles at that point. To avoid ambiguity, if the lines are directed lines, we shall define the angle between them as the angle from the positive direction of the first line to the positive direction of the second line, the smallest of the congruent angles being chosen. Cn. I, § 7] INTRODUCTION 5 We shall adopt the following notation : Denoting the intersecting lines by single letters, as a and b, the symbol (a, b) shall indicate the angle from the positive direction of the line a to the positive direction of the line 6, to be read, "the angle from a to b." It will sometimes be inconvenient to choose either direc- tion of an unlimited line as positive. (As when the line is given by its equation.) We shall then define the angle which one line makes with another as the angle formed in going from the second to the first in the positive direc- tion of rotation. It is customary to call the angle from a to b positive if its rotation is opposite to that of the hands of a clock; nega- tive if in the same direction as the hands. 7. Law of sines and law of cosines. — The two laws concerning the sines and the cosines of the angles of a triangle are often stated in trigonometry without regard to the direction of the sides of the triangle. But for our work these must be stated in a more accurate form. Let the posi- tive direction of each side of the triangle ABC be fixed. It " FlG 5 can be easily shown that these two laws take the following form, when the directions of the lines and angles are considered. T p . • AB sin (a, b) Law ot sines : — - = . BO sin (6, c) Law of cosines : (AB) 2 = (BC) 2 + (CA) 2 + 2(BC)(CA) cos(a, b). 6 ANALYTIC GEOMETRY [Ch. I, § 8 8. The quadratic equation. — We shall have occasion to use a few theorems in quadratic equations which it seems advisable to reproduce here. Any quadratic equation may be written in the form ax 2 -f bx + c = 0. The two roots of this equation are 2 a , an a x 2 = By addition, X \ ~1" r 2 ~ b a By multiplication, x^x 2 = c a b-Vb 2 -4: ac 2a The sum and the product of the roots can therefore be found directly from the equation without solving. The character of the roots depends on the quantity under the radical, b 2 — 4 ac. If b 2 — 4 ac > 0, the roots are real and unequal, if b 2 — 4 ac = 0, the roots are real and equal, if b 2 — 4 ac < 0, the roots are imaginary. This quantity, b 2 — 4 ac, is called the discriminant of the equation, and when placed equal to zero expresses the condition which must hold between the coefficients, if the two roots of the equation are equal. CHAPTER II THE POINT 9. Cartesian coordinate systems. — The subject of Ana- lytic Geometry is, as its name implies, a treatment of Geometry by analytic or algebraic methods. It is then essential to have the means of translating geometric statements into algebraic and the reverse. Geometric theorems involve the ideas of magnitude, position, and direction. Algebraic methods of representing magnitude and direction have been considered in the previous chapter. The idea of position may be expressed algebraically in many ways. But at present we shall confine ourselves to two methods used in ordinary life. If we wish to locate a town, Ave usually speak of it as being a certain distance in a certain direction from some well-known location. In the plane we must have a fixed point A from which to measure the distance, and a fixed line AB from which to measure the direction of any point P. The point P is com- pletely determined when the angle BAP and the distance AP are given. This system of locating points in a plane is 1G ' called the Polar System, and will be discussed fully later. Another method of fixing the position of a point on the earth's surface is to give its latitude and longitude, or its 7 8 ANALYTIC GEOMETRY [Oh. II, § 9 AY IT X III A IV Y' Fig. 7. distance north or south and its distance east or west from a given pair of perpendicular lines. Constructing a pair of perpendicular lines X'X and Y' Y in the plane, we may locate a point by saying that it is m units above or below X'X and n units to the right or left of Y'Y. If instead of using the words above or below, right or left, we understand that all distances measured up- ivard or to the right are positive, and those meas- ured downward or to the left are negative, two num- bers with the proper signs attached will represent the distances of the point from the two lines, and these two numbers taken together will locate absolutely the position of any point in the plane. These numbers, representing the distances of the point from the two lines, with their proper signs attached, are called the coordinates of the point. The distance NP, measured from Y' Y, parallel to X r X, is called the abscissa, or jr-cobrdinate, and the distance MP, measured from X'X, parallel to Y' Y, is called the ordinate, or /-coordinate, of the point. The line X'X is called the axis of abscissas, or Jf-axis, and Y'Y the axis of ordinates, or X-axis. The two lines to- gether are called the axes of coordinates, or coordinate axes, and their intersection the origin of coordinates, or simply the origin. The abscissa of a point is denoted by the letter x, the ordinate by y, and the two coordinates are oi. ii, § yj tiie roiMT 9 written in a parenthesis (x, y), the abscissa being always written first. It will be seen at once that any point in the plane can be located by means of its coordinates, and that there will always be a point which will correspond to any pair of values we may choose, and that there will be only one such point. We have then a simple means of representing position in a plane by algebraic symbols. This system is called the rectangular, and is a particular case of Cartesian coordinates. In the general Cartesian system the axes are not necessarily perpendicular to each other. In case they are not perpendicular, the system is called oblique. All the definitions given above hold for the oblique system. In Fig. 8, NP is the abscissa of P and MP is its ordi- nate. While rectangular coordinates are more often used because their formulas are simpler, yet it will occa- sionally be desirable to use the more general system. But rectangular coordi- nates will always be un- derstood unless another system is distinctly speci- fied. In locating or 'plotting a point whose coordinates are given, some convenient unit of measure must first be chosen. Then measure off the proper number of these units from the origin along each axis in the direction indicated by the sign of the coordinate. Through the points thus determined draw lines parallel Fig. 8. 10 ANALYTIC GEOMETRY [Ch. II, § to the axes, and their intersection will locate the point whose coordinates were given. The following figures illustrate the method. Coordinate paper having two per- pendicular sets of parallel lines is very useful, and should be obtained by the student. C-A, 7) (-2,-1 XS,8) , and the generalized form of the law of the cosines (Art. 7) gives P X P 2 = Vp 2 jf 2 + jzpf + 2 P 2 K- KP X cos *», where not only the magnitudes, but also the directions of the lines are considered. But P 2 K= M 2 M X = 0M 1 - 0M 2 = x x - x 2 , and KP^M^-M^M^-MzP^y^yv Substituting these values, we have P\P% = V(a?! - a? 2 ) 2 + (2/i - Z/ 2 )' 2 + 2(«i - a? 2 )(2/i - y-i) cos «, [2] as the distance between two points in oblique coordinates. 14 ANALYTIC GEOMETRY [Ch. II, § 13 PROBLEMS 1. Find the distance between the two points whose rec- tangular coordinates are (— 2, 6) and (1, 5). Solution. — In using formulas [1] and [2] we may choose either of the points for Pi and the other for P 2 . Let (-2, 6) he the coordinates of Pi, and (1, 5) the coordinates of P 2 . Then Pi P 2 = V(- 2 - 1)* +(tt - 5)* = VlO. 2. Find the lengths of the sides of a triangle if the rec- tangular coordinates of its vertices are (— 3, 4), (— G,— 1), and (4,-5). 3. Find the lengths of the sides of the triangle, the coordi- nates of whose vertices, referred to axes making an angle of 60° with each other, are (0, 0), (- 5, - 5), and (1, - 3). 4. What is the distance from the origin to the point (a, b) in rectangular coordinates ? In oblique coordinates, if the angle between the axes is 45° ? 5. Show that the points (G, 4), (2, 8), (3, - 2), and (- 1, 2) are the vertices of a parallelogram. 6. Show that the lengths of the diagonals of any rectangle are equal. Note. — Take the two adjacent sides as axes and call the opposite vertex («, b). 13. Points dividing a line in a given ratio. — The next question to be discussed is that of finding the coordinates of a point which will divide the line joining two given points in any given ratio. We must first define what we mean by "dividing the line joining two points in any given ratio " ; for it has a larger meaning here than we have been accustomed to give it. If C is any point on the line AB, it is said to divide the Cm. II, J 13] THE TOIXT 15 lint' AB into the two parts AC and CB (care being taken to read the two parts in just this way) whether the point C lies between A and B or beyond either. It will be — seen that if the point lies between A and £, the ratio, 77^, of the parts into which it divides the line is positive ; while if it lies on AB pro ducecl, the ratio is negative Fig. 15. AC If has a value between CB AC 1 and — 1, C is nearer A, while if ■-— is greater than 1 or less than — 1, C is nearer B. We shall now obtain the formulas for finding the coor- dinates of the point P which divides the line P X P 2 * n ^ ne ratio m 1 : m y V X M M f \ v' Fig. 16. Fig. 17. Draw the ordinates M t P v M 2 P 2 , and MP. Also draw the lines P X K X and PiT parallel to the X-axis. The triangles PP X K X and PKP 2 are evidently similar, and P^ = K,P_P,P PK KPo PP. rn ■ 16 ANALYTIC GEOMETRY [Ch. II, § 14 But P X K X = 0M - 0M X = x - x v PK = 0M 2 -0M =x 2 -z, K X P =MP -MK 1= y ~y v KP 2 = M 2 P 2 - M 2 K=y 2 - y. Substituting these values, we have x-x 1 =d m^ and y-y 1 = wi x 2 - x m a y 2 — y m 2 Solving, » = ""» + *"*, and ,= —* + »■* . [3] If the point P bisects the line P±P V m 1 — ra 2 , and the formulas become « = &±S, and „ = Ki±ife. [4] Let the student go over the demonstration carefully, using the second figure, and assure himself that every step holds as well for that as for the first. Let him also construct other figures with the points in different posi- tions, but using the same letters for corresponding points. Since the demonstration depends only upon the simi- larity of triangles, it will hold also in oblique coordinates. The results are therefore general, and will apply to either system of Cartesian coordinates. 14. Harmonic division. — If the line A C is divided by the points B and 7), internally — ' ' . and externally, in the same Fig. 18. • , , . ,, numerical ratio, or so that — - = — — — , the line .AC is said to be divided har- Jo O AJ o monically. Ch. II, § 14] THE POINT 17 Let the student show that the line BD will then be divided harmonically by the points C and A, or so that BC = BA CD AD The four points A, B, (7, and D are said to form a harmonic range. If parallel lines are drawn through the points A, B, C, and D of a harmonic range, their intersections A', B' , C\ and D', with any transversal, will also be a harmonic range. For, from plane geometry, AB BO A'B' B'C Hence and AD A'D' A B C A^- — * B^. 9^-^ DC D'C A'B' B'C A'D' D'C Fig. 19. D PROBLEMS 1. Find the points of trisection of the line joining (_ 3, _ 4) and (5, 2). Solution. — If we wish to find P, the point of trisection nearest P>, m\X 2 + m^xi mjy 2 + nioyx mi + wi2 2 5 + l(- -3) 1+2 2 2 + l(- -4) <). 1 + 2 and the point of trisection is (|, 0). But if we wish to find P', mi = 1, m® = 2, _ 1 3' a , = 1.5 + 2(-3) 1.2 + 2(-4) y 1 + 2 and the other point of trisection is (— |, — 2). 18 ANALYTIC GEOMETRY [Ch. II, § 14 2. Find the point which divides the line through (—3, —4) and (5, 2) in the ratio — }. 3. Extend the line through (1, 5) and (—3, 4) beyond the latter point until it is three times its original length. Find the coordinates of its extremity. 4. In the triangle whose vertices are (0, 0), (0, 6), (5, 8), find the point on each median which is two-thirds of the dis- tance from the vertex to the middle point of the opposite side, and show that these points coincide. 5. Show that the medians of any triangle meet in a point, choosing the axes so that the vertices may be represented by (0, 0), (a, 0), and (b, c). 6. In the right triangle whose vertices are (0, 0), (0, 6), and (8, 0), show that the distance from the vertex of the right angle to the middle point of the opposite side is equal to one- half of the hypotenuse. 7. Prove that the theorem of problem 6 holds for any right triangle. Note. — Take the legs of the triangle as axes. 8. In the triangle whose vertices are A (— 1, 2), B (4, 5), and (7(3, —.4), a line DE is drawn through the middle points of the sides AB and AC. Show that BC = 2 DE. 9. Prove that the line joining the middle points of the sides of a triangle is equal to one-half of the third side, using the points (x lf 2/j), (x 2 y 2 ), and (x 3 , y s ) as the vertices of the triangle. 10. If the coordinates of three oMhe vertices of a parallelo- gram are (0, 0), (8, 0), and (3, 5), find the coordinates of the fourth vertex, which lies in the first quadrant. 11. Prove that the diagonals of any parallelogram bisect each other. 12. In what ratio is the line joining the points (— 1, 6) and (7, - 2) divided by the point (2, 3) ? by the point (10, - 5)? Ch. II, § 14] THE POINT 19 13. The line joining the points (0, 3) and (9, 0) is divided internally by the point (3, 2). Find the coordinates of the point which divides it externally in the same numerical ratio. 14. Find the coordinates of the point P which forms, with the points A (4, 1), B (2, - 2), and C (- 2, - 8), a harmonic range, if (a) P is between A and 72; (b) P is between B and C. CHAPTER III LOCI 15. Equation of a locus. — In the previous chapter we have considered fixed points only. If a point is made to move in the plane according to some definite law, a curve or locus is generated. (The term " curve " in Analytic Geometry is applied to any locus, including straight lines.) As, for example, a point which remains at a fixed distance from a given fixed point generates a locus called a circle ; a point which is always equally distant from two intersecting lines generates a locus which is the bisector of the angle between those lines ; a point which is always equally distant from the ends of a line generates the perpendicular bisector of that line, etc. If now we can translate the statement of the law govern- ing the movement of a point into an algebraic relation or equation between the coordinates of the points which satisfy the law, we shall have an equation which may be used to represent the curve. For, if our translation is correct, every point whose coordinates satisfy the equation will occupy a position on the path generated by the moving point, since the equation is only a restatement of the law itself in algebraic language. There will be, moreover, no position of the moving point whose coordinates do not satisfy the equation. We shall then have obtained an equation which is satisfied by the coordinates of every 20 Jf Ch. Ill, § 15] LOCI 21 point on the locus, and by the coordinates of no point not on the locus. In the first example given above, if the fixed point is taken as the origin, and if the moving point P remains at a distance a from the fixed point, the relation between the coordinates x and y of every position of P is x 2 + y 2 = a 2 . For (Fig. 21) Oil 2 + MP 2 = OP 2 . We see, moreover, that this equation cannot be satisfied by any point which is not at a dis- tance a from 0. We have then X- translated the given condition into algebraic language. The equation and the curve bear to each other the following recip- [y' rocal relation : The coordinates Fig. 21. of every point on the circle satisfy the equation, and conversely, every point whose coordinates satisfy the equation lies on the circle. When an equation and a curve are connected by this relation, the equation is spoken of as the equation of the curve, and the curve as the locus of the equation. Again, let a point move so as to remain equally distant from the two axes. What is the algebraic translation of this law, or, in other words, what is the algebraic equation which must be satisfied by the coordinates of every point governed by the law ? It is evidently x = y, and this is, therefore, the equation of the bisector of the angle be- tween OX and OY. What is the equation of the bisector of the angle between OY and OX' 1 oo ANALYTIC GEOMETRY [Ch. Ill, § 15 If a point moves so as to be always three units above the X-axis, the ordinate of every point must be three, while no restriction is placed on the abscissa of the point. This law, translated into algebraic language, is, therefore, y = 3 ; for this equation makes just the same statement in regard to the position of every point which satisfies it. What is the equation of the locus of points two units to the left of the Y"-axis ? What are the equations of the axes ? The third illustration was the locus of a point which moves so as always to be equally distant from two fixed points. Place the axes with the origin at one of the points, and the X-axis coincident with the line joining the two points. Let .the distance OA (Fig. 22) between the two points be rep- resented by a. Then the co5r- dinates^ of the two points are (0, 0) and (a, 0). We can translate into an algebraic equa- tion the statement that a point P, coordinates (#, ?/), shall be equally distant from the two fixed points and J., by ex- pressing the distances of P from each of the two points, and equating these two expressions. In Fig. 22, OP = V* 2 + y\ By [i] and AP = VO - a) 2 + f. Bj[l] Equating, Vz 2 + */ 2 = V<>-«) 2 + ?/ 2 . Ch. Ill, § 15] Loci 23 Squaring and reducing, we have, as the equation of the desired locus, a x = ~i Here the final result does not express so clearly as in the previous cases that it is simply a translation of the state- ment of the law. But it has been obtained by simple algebraic reductions from this exact statement. The result is, as we should expect, the perpendicular bisector of the line joining the two fixed points. We have in these simple cases been able to translate the law governing the movement of a point in the plane into an algebraic equation. There are many loci for which this is possible. But the law may be stated in such a way as to require other than algebraic symbols to represent it. For example, the path of any fixed point on the circumference of a wheel rolling on a straight line in a plane is a perfectly definite curve. But the relation between the coordinates cannot be expressed in a single algebraic equation. It requires the introduction of trigo- nometric functions. If a point moves at random, no equation connecting the coordinates of its different positions can be found ; for an equation imposes a law upon the movement of the point. PROBLEMS 1. Find the equation of the locus of points which are equally distant from the points (1, 3) and (—2, 5). 2. Find the equation of the locus of points which are three times as far from the X-axis as from the l^axis. 3. Find the equation of the locus of points which are five units from the point (—3, 4). 24 ANALYTIC GEOMETRY [Ch. Ill, § 16 4. A point moves so as to be always five times as far from the r"-axis as from the point (5, 0). Find the equation of its locus. 5. A point moves so that the sum of the squares of its distances from the points (0, 0) and (5, — 5) is always equal to 40. Find the equation of its locus. 6. A point moves so as to be always three times as far from the point (1, — 2) as from the point (—3, 4). Find the equation of its locus. 7. A point moves so that the sum of its distances from the two axes is always equal to 10. Find the equation of its locus. 8. A point moves so that its distance from the X-axis is always one-half its distance from the origin. Find the equation of its locus. 9. A point moves so that its distance from the point (—4, 1) is always equal to its distance from the origin. Find the equation of its locus. 10. A point moves so that the square of its distance from the origin is always equal to the sum of its distances from the axes. Find the equation of its locus. 16. Locus of an equation. — Looking at the question from the other side, let us consider what will be the geometric interpretation of any given equation in x and y. It is at once evident that only the coordinates of certain points in the plane will satisfy the equation ; for, if we give any particular value to x, one or more values of y will be determined. The point, then, cannot occupy any position at random in the plane, yet it is not confined to a finite number of positions. For, since any value we please may be assigned to x, there will be an indefinite number of positions whose coordinates will satisfy the Ch. Ill, § 17] LOCI 25 equation. Moreover, it appears that these points are not scattered indiscriminately over the plane, since random values of x and y will not satisfy the equation. Small changes in the value of x will in general produce small changes in the value of y. Points may therefore be found as close as we please to each other, and from this we may infer that they are situated on some curve. This curve which contains all the points which satisfy the equation and no others is called the locus of the equation. 17. Plotting the locus of an equation. — How shall we determine the locus of any given equation ? Sometimes the locus is at once evident. For example, what is the geometric interpretation of the equation y = 3 ? The equation says nothing concerning the abscissas of points on the locus, but fixes the ordinate of every point. All points which satisfy it must therefore lie at a distance of three units above the X-axis. Hence the locus is a line parallel to the X-axis, and three units above it. Again, consider the equation x — y. It states in alge- braic language that a point moves so as to remain equally distant from the two axes. Its locus is therefore the line which bisects the angle between the two axes. Sometimes it is easy, as in these cases, to translate the algebraic equation into the law which governs the move- ment of the point, and hence determine the exact form and position of the locus. But this is often difficult, and we must have other means of determining the curve. We can always determine as many points as we please on the locus by giving to one of the coordinates a series of values and determining the corresponding values of the other. 26 ANALYTIC GEOMETRY [Ch. Ill, § 17 Place these points in their proper positions in the plane, and when a sufficient number has been obtained, a smooth curve passed through them will show approximately the form of the curve. The points can be determined as near to each other as we please, and the approximation can be carried to any required degree of accuracy. This is called plotting the curve. We shall plot the locus of the equation Give consecutive values to x, and find the corresponding values of y. If x = 0, y = 10, x=l, y= 8, x=% 9f= 6, x=8, y= 4, z = 4, y = 2, x = 5, y= o, x = - i, y = i2, x = - 2, y = 14, x— — 3, 2/ = 16, etc. etc. Plotting the points (0, 10), (1, 8), (2, 6), etc., and passing a curve through the points, we see that they all appear to lie on a straight line. This method, however, does not assure us that the locus is a straight line. It only shows that, so far as our construction is accurate, it appears to be a straight line. We shall show later that every equation of the first degree represents a straight line. Ch. Ill, § 18] LOCI 27 Again, let us plot the locus of the equation 25. x*-y* Solving the equation for y, we have y = ± Va? 2 — 25, from which it appears that y is imaginary, so long as — 5 < x < + 5. There will therefore be no points on the locus for which x is numerically less than 5. If x — 5, y — 1 x = — 5, y — ; x = 6, y = ±VIlj x = -6, ?/ = ±Vll; # = 7, y == ± V24 etc. Plotting the points (5, 0), ((3, + Vll), (6, - Vll), etc., and passing a smooth curve through them, we y have the curve in Fig. 24. It can be seen from the equation that each branch goes off indefi- nitely, never again turn- ing toward either axis ; for as x increases, y in- creases indefinitely. 18. Symmetry. — A curve is said to be symmetrical with respect to one of two axes (rectangular or oblique) when that axis bisects every chord parallel to the other. A curve is said to be symmetrical with respect to a point when that point bisects every chord drawn through it. It is easily proved that if a curve is symmetrical with respect to two axes, it is symmetrical with respect to 28 ANALYTIC GEOMETRY [Ch. Ill, § 18 their point of intersection. Now, if, upon substituting any value for x in an equation, we find two values of ?/, equal numerically but with opposite signs, the curve is evidently symmetrical with respect to the X-axis. Or, if, for every value of ?/, we find two values of x, equal numerically but with opposite signs, the curve is evi- dently symmetrical with respect to the T^-axis. If both these occur, the curve must be symmetrical with respect to the origin. It appears that the first of these conditions can be satisfied when y occurs in the equation in even powers only, and the second when x occurs in even powers only. A curve is therefore symmetrical with respect to the X-axis tvhen its equation does not contain odd powers of y ; it is symmetrical with respect to the Y-axis ivhen its equation does not contain odd powers of x. It is symmetrical with respect to the origin if its equation contains no term of an odd degree in x and y. We can therefore tell at once whether a curve is sym- metrical with respect to either or both axes. This is useful in plotting ; for if a curve is symmetrical with respect to the Jf-axis, it is only necessary to plot the part above that axis and form the same curve below ; if sym- metrical with respect to the I^-axis, to plot the part at the right of that axis and form the same curve at the left. The curve which we have just plotted, x 2 — y 2 = 25, is evidently symmetrical with respect to both axes. It would have been sufficient to have plotted that part which lies in the first quadrant and determined the rest of the curve from this, Ch. Ill, § 18] LOCI 29 PROBLEMS 1. Plot the loci of the following equations : (a) x 2 + y 2 = 9- (/) 4 a? + 9 ?/ 2 = 0. (b) x 2 -\-7f = 0. (g) 4or-9/ = 0. ( C ) ^-^ = 0. (/t) i^ = 4a;. (d) 4 a 2 + 9 f = 36. (i) a 2 = 4 y. (e) 4ar , -9, and speak of the equation as ?/ = or v = 0. The letters u and v are simply used as ab- breviations for expres- sions in x and y of any degree. Then u = will represent some curve, and v = another curve. Let us consider what will be represented by the equation u -h kv = 0, where Fig. 26. 32 ANALYTIC GEOMETRY [Ch. Ill, § 21 k is any constant quantity, positive or negative. Let (x v y x ) be any point of intersection of the two curves u — and v = 0. Its coordinates will satisfy both these equations, and hence will satisfy the equation u + kv = 0. The locus of u -f kv = must therefore pass through all the points common to the two curves u = and v = 0. More- over, it will not pass through any other point of either curve. For the coordinates of any such point will cause one of the expressions u or v to vanish, but not the other, and therefore cannot satisfy the equation u -f- kv = 0. Again, let us consider what will be represented by the equation uv = 0. It is evident that the coordinates of every point which cause either u or v to vanish will satisfy this equation, and that the coordinates of no other point can satisfy it. uv = must therefore represent the loci of the two equations u = and v = 0, taken together. For example, xy — represents both coordinate axes. PROBLEMS 1. Find the intercepts of the curves whose equations are given on page 29. 2. Find the points of intersection of the following curves : (a) x 2 -f- if = 25 and x + y = 4. (6) x~ + y 2 = 25 and 3x-4t/=25. ( C ) ar + y 2 = 25 and x + 2 y = 10. (d) 3.i- 9 + 4?/ 2 = 24 and a 2 -?/ 2 = 4. (e) y 2 = 4:X and # — ?/ -f- 1 = 0. (/) a- 9 -f 4?/ 2 = 16 and 6y = x 2 . 3. If the equations of the sides of a triangle are x + 7 y + 11 = 0, 3 x + ?/ — 7 = 0, and # — 3y + l = 0, find the length of each of the medians. Ch. Ill, § 21] LOCI 33 4. Which of the points (3, -1), (7, 2), (0, -2), and (8, ,3) are on the locus of the equation 4 x — 7 y = 14. 5. Find the length of the chord of intersection of the loci of 7? + y- = 13 and y- = 3 x 4- 3. 6. For what values of b are the two intersections of the loci of y = 2 x + b and \f = 4 a; real and distinct ? imaginary ? coincident ? 7. Write a single equation which will represent the two bisectors of the angles between the axes. 8. Plot the two lines which are represented by each of the following equations : (a) x 2 + xy = 0. (c) 2x> 4- 5xy - 3y 2 = 0. (b) x 2 -ox = -6. (d) 2if-xy + ±x-9y = -4:. CHAPTER IV THE STRAIGHT LINE 22. We have seen that, if we know the law of the move- ment of a point, we can often determine the equation of its locus. We shall now proceed to the systematic study of a few such loci, beginning with the straight line. The two most common ways of determining the position of a line are to give either two points on it, or a single point and the direction of the line. If either of these sets of conditions is given, the line is fully determined, and we should be able to find the algebraic relation which must be satisfied by every point on it. 23. Line through two points. — Let the line pass through the two points P v (x v y^), and P 2> (z 2 , y 2 ), and let P, (#, y), be any point on the line. Draw the or din ate s 3I 1 P V MP, and M 2 P V and the line P X K parallel to OX. Then from the similarity of the two triangles P X LP and Y P 1 KP 2 we have LP KP 2.. Fig. 27. P X L P X K But LP = y- y x P X L = x — x v KP 2 = y 2 - y v P 1 K=x 2 -x v 34 Ch. IV. §23] THE STRAIGHT LINE 35 Substituting these values, we have y -V\ = Vi-y\ t 1-5-1 This is then the algebraic relation between the coordi- nates x and y of any point on the line and the constants x v y v x 2 , and y 2 , and is therefore the equation of the line. It is called the two-point form of the equation of the straight line. Let the student show that this equation cannot be satisfied by the coor- dinates of any point not on the line. The student should here, and in all the fol- lowing demonstrations, assure himself that the proof is perfectly general. Place the lines and points in different positions, being careful to give the same letter to corresponding points, and the demonstrations ought to hold, letter for letter. For example, try Fig. 28 with the above demonstration, being careful to note that P X L = M x + OM = OM - 0M V LP = LM + MP =MP - ML, P 1 K=3I 1 0-{- 0M 2 = 0M 2 - 0M V KP, = KM, + M 9 P K M 2 P 2 - M 2 K. Equation [5] may be written in the determinate form x % y 1 = o. 3G ANALYTIC GEOMETRY [Ch. IV, §§ 24, 25 24. Line determined by its intercepts. — If the two given points should be, in particular, the points where the line cuts the axes, or if, in other words, the intercepts a and b are given, the equation can be found easily by sub- stituting («, 0) for (x v y^) and (0, 5) for (# 2 , y^) in equation [5]. It becomes y _ Q = b - x — a — a or reducing, — + ? = 1. [6] This is called the intercept form of the equation of the straight line. . Let the student derive equation [6] geometrically with- out using equation [5]. 25. Oblique coordinates. — In obtaining these equations of the straight line we have made no use of the fact that the axes are perpendicular. The 011I3- idea used was the similarity of triangles, which will be true in oblique as well as rectangular coordinates. The results will hold therefore for both systems of Cartesian coordinates. PROBLEMS 1. Find the equation of the straight line through the points (- 1, 5) and (6, 0). Solution. — In applying formula [5] either point may be chosen as Pi and the other as P 2 . Here let (6, 0) be Pi and (- 1, 5) be P 2 . Sub- stituting in [5], we have as the equation of the line 5 x + 7 y = 30. 2. Find the equations of the lines through the following points and find the intercepts of these lines on the axes : (a) (- 5, 4) and (S, - 1). (c) (4, 2) and (4, - 2). (6) (0 ; 0) and (4, 3). (d) (3, 5) and (- 7, 5). Ch. IV, § 26] THE STRAIGHT LINE 37 3. Find the equation of the line whose intercepts are 3 and — 1. 4. Does the line joining the two points (6, 0) and (0, 4) pass through the point (3, 2) ? the point (—4, 5) ? 5. What condition must be satisfied if the point (x lf y{) lies on the line joining the points (x 2 , y.,) and (x 3y y 3 ) ? 6. The line joining the points (6, 2) and (7, — 3) is divided in the ratio of 2 to 5. Find the equation of the line joining the point (— 5, — 5) to the point of division. 7. The coordinates of the vertices of a triangle are (2, 1), (3, —2), and (—4, — 1). Find the equation of the medians, and show that the coordinates of the point of intersection of any two medians satisfy the equation of the third, and that the three medians therefore meet in a point. 8. What are the equations of the diagonals of the rectangle whose vertices are (0, 0), (a, 0), (0, 6), and (a, b) ? Find the point of intersection, and show that they bisect each other. 9. What system of lines is represented by the equation y X -+. a o 1, if we keep a constant and allow b to vary ? if we keep b constant and allow a to vary ? 26. Line determined by a point and its direction. — If the second condi- tion mentioned in Art. 22 be given, — a point on the line and the direc- tion of the line, — we can obtain its equation as follows : Let (x v y{) be the given point, and let the direction of the line be determined by the angle — 7) or sni (« — 7) [9] If the coordinates of the given point are (0, b), [9] reduces to y= . ^ ^ +&- [10] Sill (ft) - y) L J When ay = 90°, these two forms will be seen to reduce to the equations [7] and [8]. 40 ANALYTIC GEOMETRY [Ch. IV, § 29 PROBLEMS 1. What is the equation of the line which passes through the point (—6, 6) and makes an angle of 60° with the X-axis ? 2. Find the equation of a straight line if (a) b = 6 and y = 30°, (b) b = -5 and y = tan" 1 f , (c) b = S, y = 30°, and -7)J and b is the intercept on the !F-axis, they can have any real value whatever. AVe have then reduced the general equation Ax + By + O = to the slope form of the equation of a line, and it must represent that line for which 1 = and b = — If B B B 0, the general equation reduces at once to a X = — which we know to be the equation of a line parallel to the y-axis. We have then shown that the general equation of the first degree always represents a straight line. Another method of showing that the locus ■ x_ of any equation of the first degree is a straight line is as follows : ■ FlG * • Let (x v ?/ 1 ), (x T y 2 ), and (# 3 , y%) be the coordinates of any three points on the locus of the equation Az + By + C=Q, 42 ANALYTIC GEOMETRY [Ch. IV, § 30 These coordinates must satisfy the equation. Substi tuting, we have (1) Ax 1 + By 1 + C=Q, (2) Ax 2 + By 2 + C=0, (3) ^ % +% 8 + (7=0. Subtracting (2) from (1), we have -A(x 2 - x x ) = B(y 2 - yi ), x n — x, B or i - y\-y2 * 2 constant factor. The converse is easily seen to be true : if tiro equation* of the first degree differ only by a constant factor^ they rep- resent the same straight line, PROBLEMS 1. Find the values of a, b, and I for the line whose equation is 2 the given angle, there can be only a single line which passes through P x and makes the angle with MN, where is measured in the positive direction of rotation. Ch. IV, § 34] THE STRAIGHT LINE 49 Let ES be this line. Let the inclination of MN be y v and of BS be 7. Then from [11, a], tan d> = — L . * 1 + «, Solving for 7, we have , _ / 1 4- tan (f) 1 — Zj tan (/> The equation of BS will therefore be If MS is parallel to J£ZV", tan = 0, and the equation becomes If BS is perpendicular to ilifiV, tan = go, and the equation becomes l \ These formulas might be used to write the equations of parallels and perpendiculars in place of the methods given in the previous section. PROBLEMS 1. Find the equation of the line through the origin which makes an angle of 60° with the line x — 3 y = 10. 2. Find the equation of the line through (1, 4) which makes an angle of 135° with the line joining (1, 4) with the intersec- tion of 5 x — 2 y = 17 and 3 x + 4 y = 5. 34. Normal form of the equation of a straight line. — If we have given the length of the perpendicular or nor- mal from the origin on a line, together with the angle 50 ANALYTIC GEOMETRY [Ch. IV, § 34 which this normal makes with the positive direction of the .X-axis, the line is completely determined. The perpen- dicular distance is represented by jt?, and the angle by a. Through draw a line making an angle a with OX. If any distance Off is laid off on this line either in the positive di- rection (along the termi- nal line of the angle), or in the negative direction, and through H a line AB, perpendicular to Off, is drawn, that line is com- pletely determined. It is convenient to restrict a to positive values from 0° to 360°. In case we wish to speak of a complete set of parallel lines without changing «, it will be necessary to allow p to be either positive or negative, but every line in the plane can be determined by positive values of both a and p, and this will always be understood unless otherwise stated. We have seen that the equation of the line AB in terms of its intercepts is - + %- = 1. a o line Fig. 34. But for all positions of the V P and -=cos«, or a = cos a p f- = Sill «, or sin a Substituting these values of a and 5, the equation of AB becomes x cos a + y sin a = p. [15] Ch. IV, § 35] THE STRAIGHT LINE 51 This is called the normal form of the equation of a straight line. Let the student show that the equation of a straight line in oblique coordinates in terms of a and p is x cos a -f- y cos (o> — a) = p. Note. — The equations - = cos a and - = sin a are true for all cases, a b ' since if p is positive, a and cos a have the same sign, and also b and sin a. While if p is negative, they have the opposite signs. PROBLEMS 1. What is the equation of the straight line in which (a) a = 60°, and p = 5 ? (d) a = 225°, and p = ? (6) a = 120°, and j> = 5 ? (e) a = 45°, u> = 60°, and p = 1 ? (c) « = 330°, and p= - 5 ? (/) a= - 60°, shall always be positive, then in any numerical case that sign must be given to the C radical which will make . a negative number ± VA 2 + B 2 to correspond to — p. Hence the sign of the radical mast be chosen opposite to the sign of O. This will always be understood unless the contrary is stated. PROBLEMS 1. Reduce the equation 3 x -f- 4 y = 10 to the normal form. Solution. — Here ± VA 2 + B 2 = ± 5, and since C is negative, we must divide by + 5, and the equation becomes f x + $ y = 2. Hence cos a = f, sin a = f, and p = 2. The line can be easily plotted. What would have been the values of a and p, if — 5 had been chosen ? 2. Reduce the following equations to the normal form and plot the lines which they represent : (a) lx-3y = 25, (d) aj + 4 = 0, (6) ». + 2y = -8, (e) 5y-3 = 0, (c) 2x-2/ = 0, (/) aj-3y + 4 = 0. Ch. IV, § 36] THE STRAIGHT LINE 53 3. What system of lines is represented by the equation x cos a -f y sin a — p = 0, if we keep a constant and allow p to vary ? If we keep j> constant and allow a to vary ? 36. Distance of a point from a line. — Let it be required to find the distance of the point P 1 from the line AB when the equation of AB is given in the form x cos a + y sin a — p = 0. Draw MJST through P l parallel to AB and continue the perpendicular OH to meet it at K. The equation of MN will be x cos a + y sin a — p 1 = 0, where p x may be either positive or negative. For, as the value of a is fixed and as MN can be any line parallel to AB, it may be on the opposite side of the origin from AB, and in this case p x will be negative. (See Art. 34.) Since P x lies on MN, its coordinates (x v y-^) must satisfy the equation of MN, 54 ANALYTIC GEOMETRY [Ch. IV, § 36 Hence x x cos a + y x sin a = p v Now wherever P x may lie, RP X = HK= OK- 011= p x - p. Hence BPi = a?icosa + yisma-p, [17, a] If the equation is given in the form Ax + By +(7=0, it is necessary first to reduce it to the normal form and then substitute x x for x and y x for y. Hence MP, = ±VJFTW • [IT, 8] TAe radical must be given the sign opposite to that of C. It appears from the way RP X has been chosen that the result will be positive when the point and the origin are on opposite sides of the line ; negative, when they are on the same side of the line. PROBLEMS 1. Find the distance of the point (3, 5) from the line 2 X -3y + 6 = Q. 2. Find the distance of the origin from the line 3aj + 4y-5 = 0. 3. Find the area of the triangle whose vertices are (0, 3), (4, 0), and (5, 5) by calculating the length of one side and the distance of the opposite vertex from that side. 4. Given the line Sx — 4y = 10 and the point (—3,5). Find the equation of the line through the point perpendicular to the given line ; find the point of intersection of this perpen- dicular with the given line ; find the distance of the given point from this point of intersection. 5. Use the method indicated above to find the distance from the point (a^, y x ) to the line Ax + By + C = 0, Ch. IV, § :J7J THE STRAIGHT LINE 55 6. Find the distance bet ween the two parallel lines 7 x — 8 y = 15, and 7 x — 8 y -— 40. Which line is nearer the origin? 7. Show that the point (3, 1) is on the same side of the line x -+- 4 ?/ = 8 as the origin. 37. Oblique coordinates. — It will be noticed that sec- tions 31-36 have reference to rectangular coordinates only. The corresponding formulas in oblique coordinates are rather complicated and seldom used. We shall simply state what they are without obtaining them. To reduce Ax -f By + '= to the normal form, % cos a -+- y cos (to — a) = p, multiply the equation by sin (o ^A 2 + B*-2ABcosc0 The angle between two lines whose equations in oblique coordinates are A l x + B l y + C 1 =0 and A 2 x -f B 2 y + C 2 = is tan 6 = (A X B 2 - A 2 B,~) sin <* A X A 2 H- B X B 2 — {A 1 B 2 -f- A^B^) cos A' /0 38. Bisector of the angle between two lines. — Let the equations of the two lines AB and MN be (1) A x x + ^ + Cj = 0, and (2) ^ + ## + C 2 = 0, and let (V, ?/) be any point on the bisector of the angle between them. Since every point in the bisector of an angle is equally distant from the sides, HP' and KP 1 are numerically equal. But KP i = a j + By + c 1 and Hpt = A y + B y + ^ ±Vyl 1 2 +^ 1 2 ' ±Vvl 2 2 + ^ 2 2 Hence the relation which must exist between a;' and y in order that P' may be a point on the bisector is Y' Fig. 37. Aix' + Bip' +Ci _ A 2 x' + B 2 y' + C 2 ( ± \/^i a + JSi 2 ± V.4 2 2 T#? [18] If the signs of the denominators have been chosen in accordance with the rule given in Art. 36, the positive siern in the second member indicates that P' and the Ch. IV, §39] THE STRAIGHT LINE 57 origin are either on the same side or on opposite sides of each of the lines, and that therefore the equation repre- sents the bisector of the angle in which the origin lies; while if the minus sign is chosen, it represents the bisector of the angle in which the origin does not lie. If either C x or C 2 is zero, one or both of the lines pass through the origin, and this test cannot be used. PROBLEMS 1. Find the equations of the bisectors of the angles between the two lines 3 x — 4 y = 10 and 4 x -f 3 y — 7. Show that the two bisectors are perpendicular. 2. Show that the bisectors of any pair of supplementary adjacent angles are perpendicular to each other, using the two lines in Art. 38. 3. The equations of the sides of a triangle are 3a? = 4y, 4 x = — 3 y, and y = 6. Show that the bisectors of the interior angles meet in a point. Show also that the bisector of the interior angle at one vertex and the two bisectors of the exterior angles at the other vertices meet in a point. 39. Lines through the intersection of two given lines. — If the equations of two given lines are (1) Ap + Btf+0^% and (2) A 2 x + B$ -+- 2 = 0, and we form the equation (3) A x x + B x y +C X + k(A 2 x^ + B 2 y + C 2 ) = 0, where k can have any value, it will represent for every value of k some line through the intersection of .the first two. For the coordinates of the point of intersection of the loci of (1) and (2), which must satisfy both of 58 ANALYTIC GEOMETRY [Ch. IV, § 39 these equations, must satisfy (3) also. Moreover, it represents any line through their intersection, for k can always be chosen so as to make the locus of (3) pass through any given point. It is only necessary to substitute the coordinates of the point in the equation and determine k so that the equation is satisfied. In this wa} r the equation of the line through the intersection of two lines and any other point may be obtained without actually finding the coordinates of the point of intersection. If any other condition sufficient to determine the line is given (for example, its slope), k can always be deter- mined so that the line will satisfy the condition. PROBLEMS 1 . What is the equation of the line through the intersection of 2 x + 3 y - 4 = and x + 2 y - 5 = 0, and the point (2, 3) ? Solution. — The equation of any line through the intersection of the given lines is 2x + 3y-4 + k(x + 2y- 5)= 0. Since the line is to pass through the point (2, 3), these coordinates must satisfy the equation. Hence k = — 3, Substituting this value, we have x + 3y-ll=0, as the equation desired. 2. What is the equation of the line passing through the origin and the intersection of the lines x + oy — 8 = and 3. In the triangle whose sides are 5x-6y = 16, 4z + 5?/=20, and x + 2y = 0, find the lines through the vertices and parallel to the opposite sides without finding the coordinates of the vertices. Ch. IV, § 40] THE STRAIGHT LINE 59 4. Find the equation of a line through the intersection of the lines 2x — 3y + l = and x + 5 y -f 6 = 0, which is per- pendicular to the first of these lines. 5. Find the equation of the line through the intersection of the lines y = 7 x — 4 and y = — 2 x + 5, which makes an angle of 60° with the X-axis. 6. Find the equation of the line through the intersection of the lines 5y — 2x — 10 = and y-f4a? — 3 = 0, and also through the intersection of the lines 10 y -f- x -f- 21 = and 3y-5x -fl = 0. 40. Area of a triangle. — If the coordinates of the vertices of a triangle are given, the area of the triangle may be found in the fol- lowing manner : The area is equal to the numerical value of \HP^P X P V Fig. 38. 0. ^ 2 ) 2 + (^l-y 2 ) 2 - JIP S is the distance of — P 3 from the line P X P T The equation of P X P 2 is 2 - Vi) x ~ 2 - x \)y - *iy* + xtf x Hence HP = ^ 2 ~ yi)x * ~ ^ 2 ~ X ^ y * ~ X ^ 2 + X ^K and the area = \ [(y 2 -y{)x z ~(x % -x{)y z - x x y 2 +a^yj, = K («i - ^2)2/3 + (^2-^3)2/1 + (a? 3 -a?i)2/ 2 ]. [19] The form of the result is easily remembered since the subscripts follow the cyclic order. The sign of the result 60 ANALYTIC GEOMETRY [Cii. IV, § 40 may be disregarded, since it is only the numerical value of the area we wish. Formula [19] may be written in the determinate form A = i \ PROBLEMS 1. Find the area of the triangle whose vertices are (1, — 3), (-4, 3), and (5,5). 2. Show that the area of any quadrilateral is [Oi?/2 - a^/O + (x 2 y 3 - x 3 y 2 ) + (x& 4 - x^) + (x A y x - x l y,)\ 3. What is the area of the quadrilateral the equations of whose sides are x = 0, x+y=Q, x+2y=5, and 6x+y-\-58=0? 4. Obtain the formula for the area of a triangle by drop- ping perpendiculars from each of the three vertices upon the X-axis, and considering the trapezoids formed. GENERAL PROBLEMS 1. Show that the triangle whose vertices are (3, 2), (— 1, —3), and (—6, 1) is a right triangle. 2. An isosceles right triangle is constructed with the hy- potenuse on the line x -f 4 y = 10, and the vertex of the right angle at the point (3, 4). Find the coordinates of the other vertices. 3. Find the equation of the line through the point (5, fi) which forms with the axes a triangle whose area is 80. Four solutions. 4. Find the equation of a line through the point (—1, 5) such that the given point bisects that portion of the line be- tween the axes. 5. Find the equation of a line through the point (3, — 6) such that the given point divides that portion of the line be- tween the axes in the ratio 3 : — 1. Ch. IV, §40] THE STRAIGHT LINE 61 6. Find the equation of the line passing through the point (8, 2) such that the portion of it included between the lines x — 2y = 6 and x + y = 5 shall be bisected at the given point. 7. On the line y — 5 = a segment is laid off, having for the abscissas of its extremities 2 and 5, and upon this segment an equilateral triangle is constructed. What are the coordinates of the third vertex? 8. Find the point on the line 4?/ — 5x + 28 = which is equidistant from the points (1, 5) and (7, — 3). 9. Find the points which are equidistant from the points (4, — 3) and (7, 1), and at a distance 3 from the line 15 x -f- 8 y = 120. 10. The coordinates of the vertices of a triangle are (5, 2), (4, — 7), and (3, 7). The side joining the first two points is ilivided in the ratio 4 : 7, and through this point lines are drawn parallel to the other sides. Find their points of intersection with the other sides. 11. On each side of the triangle in problem 10, find the point which is equidistant from the other sides of the triangle. 12. The equations of the sides of a complete quadrilateral are 2y + 7 x=l4, x—2y=l, x+ky=— 4, and 7 x— ky=— 28. Show that the middle points of the three diagonals lie on a straight line. 13. Show that the perpendiculars let fall from any point of the line 2 x + 11 y = 5 upon the two lines 24 x + 7 y = 20 and 4 x — 3 y = 2 are equal to each other. 14. Perpendiculars are dropped from the point (0, 4) to the sides of a -triangle whose vertices are (1, 5), (5, — 1), and (6, 0). Show that the feet of these perpendiculars lie on a line. 15. Find the equation of a line through the intersection of the lines 2 x — 7 y = 3 and x + 3 y = 8, which is perpendicular to the line joining the origin to the intersection of these lin»s 62 ANALYTIC GEOMETRY [Ch. IV, § 40 16. Show that the four points (2, 1), (5, 4), (4, 7), and (1, 4) are the vertices of a parallelogram. 17. Find the area of the triangle formed bylthe three lines y = m x x + (?!, y = m with OX. sin w r/ sinO-4>) 3 sinw sin w y'Sin*. [23] sin « What do these formulas becotne when co — 90° ? When co = 90° and $ = 6? 50. Degree of an equation not changed by transformation of coordinates. — The degree of an equation cannot be changed by transformation from one system of Cartesian coordinates to any other. For we have seen that in each case we replace x and y by expressions of the first degree Ch. VI, §51] TRANSFORMATION OF COORDINATES 73 in x f and y\ and that therefore the degree of the equation cannot be raised. Neither can it be lowered, for it would then be necessary to raise the degree in transforming back to the original axes, since we must obtain the origi- nal equation. 51. Transformation from rectangular to polar coordi- nates. — Let it be required to find the equations of transformation for transforming from a given set of rec- tangular axes, OX and OY, to a polar system having as its origin and OX as its initial line. The relations between ^, y, />, and 6 are seen at once from the triangle OMP; for sin0 = f^, , n OM U± ^ and cos u — —— , or *_ , (cf) (ar + ?/ 2 ) 2 = o 2 (.r 3 — ?/ 2 ), (/t) x 2 + y 2 + 2 ax = aVx^ + tf. 3. Obtain the rectangular equation of the curve whose polar equation is p = a cos 0. We might make this transformation by using the two for- mulas [25], but it will be found to be easier first to multiply both members of the equation by p, giving p- = ap cos 6. Using the formulas x = p cos and p 2 = x 2 + y 2 , this reduces at once to x 2 + y 2 = ax. This is then the rectangular equation of the curve whose polar equation is p = a cos 0. 4. Obtain the rectangular equations of the curves whose polar equations are (a) p = a sin 6, (/) p — a sin 2 6, (b) p = a + ——, (g) p 2 cos2 = a 2 , (c) p = a (1 + cos 0), (li) P = a (cos 20 + sin 2 (9), (d) p 2 = a 2 cos 2 6, (i) p = a (1 + cos 2 0), (e) p = a — b cos 0, (,;') p = 2 a tan • sin 0. CHAPTER VII THE CIRCLE 52. Equation. — The locus of points equidistant from any fixed point is called a circle. Hence, to find the equation of a circle, it is necessary to express the algebraic relation be- tween the coordinates of such points. If the origin is taken at the centre of the circle, the equation is evidently z 2 + * 2 , where Fig. 46. the radius of the circle. For the distance of any point (#, y) from the origin is Vx 2 + y 2 . If the centre be taken at any point C, whose coordinates are (a, /3), the distance CP from the centre to any variable point P is w(x — a) 2 + (y — /3) 2 . Hence the equation of the circle is If the centre is on the -X"-axis, /3 = 0, reduces to (x — «) 2 + y 2 = r 2 ; [26] and the equation if on the P"-axis, a = 0, and the equation reduces to x 2 + iy - /3) 2 = r 2 . 75 76 ANALYTIC GEOMETRY [Ch. VII, § 53 Problem. — Find the equation of a circle, (a) tangent to both axes ; (b) passing through the origin and having its centre on the X-axis. 53. General form of the equation. — Expanding [26], we have x 2 + y i _ 2 ax - 2 /3y + a 2 + /3 2 - r 2 = 0. a and ft can have any value, positive or negative, and r can have any positive value. Hence the equation is in the general form of # 2 + 2/ 2 + Zto + Ey + F=0, [27] where D = -2a, U = - 2 j3, and F = a 2 + /S 2 - r 2 . And if an equation is to represent a circle, it must be in the form of [27]. It will be noted that this is not the most general form of the equation of the second degree. For tins IS Ax 2 + Bxy + C> 2 + Bx + Fy + F = 0. When the two equations are compared, it will be seen that the term in xy is wanting in [27], and that the coeffi- cients of x 2 and y 2 are equal, or B=0, and A = C. Hence both these conditions must be satisfied in order that the general equation of the second degree may repre- sent a circle. But will it always represent a circle when these condi- tions are satisfied ? It will be necessary to determine whether there are always values of a, /3, and r which cor- respond to all values of D, F, and F. Solving the equa- tions given above for «, /3, and r, we have -B n - B /3 = —^, and r = l^B 2 + F 2 -±F. Ch. VII. § 54] THE CIRCLK 77 Hence there will always be real values for a and /3 for all values of D, K and F. But if D 2 + F 2 - 4 F < 0, the value of r is imaginary, and there will be no point in the plane which will satisfy the equation. But since it has the form of the equation of a circle, it is said to represent an imaginary circle. Again, if D 2 + F 2 — 4 F= 0, r = 0, and the equation represents the point («, /3) onl} r . It is called a null circle. We see then that we shall have a real circle only in case D 2 + F 2 — 4jP> 0. But no equation in the form of [27] can represent any other locus. Hence it is said to represent a circle, real, if I) 2 + F 2 - 4 F > ; null, if D 2 + F 2 -4F=Q ; imaginary, if D 2 + F 2 - 4 F < 0. 54. Circle through three points. — We know from plane geometry that three points not in a straight line determine a circle. It ought therefore to be possible to find the equation of the circle passing through three such points, (x v 3/j), (x 2 , # 2 ), and 3 , y 3 ). This may be done by determining 2), F, and F of the general equation [27] so that these coordinates will satisfy that equation. Substituting these coordinates successively in equation [27], we have x 2 + y 2 + D Xl + E yi + F = Q, x 2 + y 2 + Dx 2 + Fy 2 + F=0, From these three equations it is always possible to determine D, F, and F (if the three points do not lie on a 78 ANALYTIC GEOMETRY [Ch. VII, § 54 line), and their values substituted in the general equation [27] will give the equation of the circle through the points. PROBLEMS 1. What is the equation of a circle, if («) its centre is at the point (—2, 3), and r = G, (b) its centre is at the point (— 3, —4), and r = 5, (c) its centre is at the point (5, 3), and it is tangent to the line 3 a,- — 2y = 10, (d) its radius is 10, and it is tangent to the line 4 x+oy = 70 at the point (10, 10), (e) it passes through the three points (4, 0), ( — 2, 5), (0, —3), (/) it circumscribes the triangle, the equations of whose sides are x -+- 2 y — 5 = 0, 2 x + y — 7 = 0, and x — y -f 1 = 0, (r/) it has the line joining the points (3, 4) and (— 2, 0) as a diameter, (h) it passes through the points (5, — 3) and (0, G) and has its centre on the line 2 x — 3 y = 6, (i) it passes through the points (5, —3) and (0, 6) and r=6? 2. Find the coordinates of the centre and the radius of each of the following circles : (a) x 2 + y 2 + $x-6y-10 = 0, (6) x 2 + y~ + 8 x — 6 y -f- 50 = 0, (c) ar + ?/ 2 + 6?/-16 = 0, (d) 3ar + 3?/ 2 -7a:-8 = 0. 3. Show that if the equations of two circles differ only in the constant term, they represent concentric circles. 4. Show that the equation of a circle in oblique coordinates is in the form of xr + 2 cos o) • xy + if + Dx + Ey + F=0. What conditions must be satisfied by the general equation of the second degree that it may represent a circle when referred to any particular set of oblique coordinates ? Ch. VII, §55] THE CIRCLE 7 ( .» 5. Show that the equation of any circle through the points of intersection of two given circles, * 2 + //'"' + Dp 4- E& + F, = 0, and x 2 -f }r + D. 2 x + E,y + F, = 0, can be expressed in the form x- + if + D x x + E& + F, + % (.r + // 2 + XV? + Eojy + F 2 ) = 0. What is the locus of this equation when k = — 1 ? 6. Obtain the equation of the common chord of the two circles, x- + y- + 6x-y = 0, and ar + y- — 4 y + 10 = 0, and show that it is perpendicular to their line of centres. 7. Prove that the common chord of any pair of intersecting circles is perpendicular to their line of centres. 8. What would be the statement of problems 5 and 7, if the two circles do not intersect ? 55. Tangent. — A tangent to any curve is denned as follows : Let a secant through a fixed point P 1 of the curve intersect the curve again at P 2 . Let P 2 move along the curve toward P v The secant will revolve about P v and as P 2 approaches P x the secant will ap- proach a certain limiting position. This line, which is the limiting position approached by the secant as P 2 ap- proaches P v is called the tangent to the curve at P v The method of finding the equation of the tangent to any curve of the second degree is the same for all. The demonstration should, therefore, be studied carefully in the case of the circle where the work is the simplest. 80 ANALYTIC GEOMETRY [Ch. VII, § 55 According to the definition we must first write the equation of a secant through two points, and then find the limiting form which this equation approaches when the two points approach coincidence. Let (x v y{) and (x 1 + \ y^ + &) be the coordinates of P x and P 2 , adjacent points on the circle x 2 + y 2 = r 2 . The equation of the line through these two points is (by [5]) y — y\ = k. x — x x h If we let P 2 approach P v h and k will approach zero, and the limit of the second member will be indeterminate. This would be neces- sary since we have made no use of the fact that P 2 must ap- proach P x along the circle. Unless P 2 ap- proaches P x along some curve, P X P 2 will have no limiting posi- tion. It will there- fore be necessary to determine in the case of each curve the value of the k expression -■ In the case of the circle about the origin, ri the coordinates of the points P x and P 2 must satisfy the equation x 2 + y 2 = r 2 . We have, therefore, (1) rr x 2 + y^ = r 2 , Fig. 47. and (2) x 2 + 2hx 1 + h 2 + y 2 + 2 ky l + k 2 = r 2 . Ch. VII, § 50 J THE CIRCLE 81 Subtracting (1) from (2), we have 2 hx x + A 2 + 2 ky t + k 2 = 0, k or, transposing and solving for -, ill h_ 2 x x + h t h~ 2 yi + k Substituting in the former equation of the secant P X P V we see that y _ , fi _ 2x^ + h x — x x 2y 1 -\-k is another form of its equation in the circle x 2 -f y 2 = r 2 . If now we let h and k decrease, the limit of the second member is no longer indeterminate, but becomes K The equation of the tangent is therefore * V-Vx = \ x - x x yl which by the aid of (1) reduces to scias + y\V = r 2 . [28] Let the student show by the same method that the equation of the tangent to the circle a* + yZ + Dx + IJy + F=0 is xix + 2/12/ +?(<*> + ^) + f (V + 2/i)+ F = 0. [29] 56. Normal. — The normal at any point of a curve is the line through the point, perpendicular to the tangent at the point. Its equation can be obtained by first writing the equation of the tangent at the point, and then that of a perpendicular to it through the point of contact. The equation of the normal to the circle x 2 + y 2 = r 2 , at the point (x x y^)^ is seen to be y x x — x x y = 0. 82 ANALYTIC GEOMETRY [Ch. VII, § 57 PROBLEMS 1. Obtain the equations of the tangents and normals to the following circles, and show that in each case the normal passes through the centre of the circle : (a) x 2 + f = 25, at (3, 4), (6) x> -jly + 2*- 4y + 5 = 0, at (- 1, 2), (c) x 2 + y 2 — 14 x — 4 ?/ — 5 = 0, at the points whose abscissas are 10. (d) x 2 -f- y 2 — 6 a* — 14 ?/ — 3 = 0, at the points whose abscissas are 9. 2. Find the angle in which the two circles x 2 + ?/ 2 — 4 x = 1 and a* 2 + y 2 — 2 ?/ = 9 intersect. Note. — The angle between two curves is the angle between their tangents at the point of intersection. 3. Show that the following circles cut each other orthogo- nally (or intersect at right angles) : tf + y 2 - 8a? + 4y+ 7 = 0, x 2 + y 2 - 10 x - 6 y + 21 = 0. 4. Show that the length of the tangent from the point ( x d Vi) to the circle x 2 + y 2 -\- Dx + Ey -+- F = is VV + 2/f + Bx x + Jg?y, + if. Note. — Use the right triangle having for its legs the tangent and the radius to the point of contact. The length of the hypotenuse is the dis- tance from the point (a?i, y{) to the centre of the circle. 5. What is the length of the tangent from the point (- 2, 6) to the circle x 2 + y 2 + 2y = 5? 57. Tangents from an exterior point. — The equation of the tangent which we have obtained can be applied only when we know the coordinates of the point of contact Ch. VII, § 57] THE CIRCLE 83 ( x v !/{). There are other conditions which will determine the tangent. Consider first the tangent from a given exterior point. The method of procedure may here be best shown by an illustration. Let it be required to find the equation of a tangent from the point (5, 10) to the circle whose equation is .^ + i/ 2 = 1 oo. Let the coordinates of the unknown point of contact be (x v y^). Then the equation of the tangent will be x x x + y x y = 100. Now this tangent is to pass through the point (5, 10), and therefore these coordinates must satisfy its equation, or 5^ + 10^ = 100. This is one equation connecting x 1 and y v and the fact that the point (x v y x } lies on the circle gives another, ^2 + ^2 = 100. The algebraic solution of these equations gives 2-1 = 0, 10, or ^=10, ^ = 6. There are, therefore, as we should expect, two points of contact of tangents from the given exterior point, viz. : (0, 10) and (8, 6). Substituting these values in the equation of the tangent, we have and 10 # = 100, 8 x + 6 y = 100, as the equations of the tangents through the point (5, 10). 84 ANALYTIC GEOMETRY [Ch. VII, § 58 PROBLEMS 1. Obtain the equations of the tangents to the following circles : (a) x- + y~ = 49, from (6, 8). (b) x 2 + y 2 -±x-22 = 0, from (- 2, 6). (c) x 2 + f + 5 y = 25, from (7, - 1). 2. Obtain in each of the problems the equation of the line joining the points of contact of the two tangents. 3. Obtain in this way the equation of the chord of contact of tangents from the exterior point (x x , y{) to the circle x 2 + y 2 = r 2 . 58. Tangent in terms of its slope. — When the slope of the tangent is given, we might proceed as in Art. 57, for we could obtain one equation by placing the slope of the X-i tangent, -, equal to the given slope. Solving this with xf + y^ = 100, we could find x x and y x just as before. But another method is more important. The equation of any line which has the given slope I may be written in the form 7 , , y = lx + b. It is then only necessary to find what value of b will make it a tangent to the circle x 2 + y 2 = r 2 . Every line of the system will cut the circle in two points, real, imaginary, or coincident. If the points are coincident, the line is a tangent. Starting the solution of y = lx + b and x 2 + y 2 = r 2 , we have at once by substitution (l + /2) a a + 2 lbx + b 2 - 7^ = for determining the abscissas of the points of inter- section. tin. VII, § 59] THE CIRCLE 85 This Avill in general have two distinct roots, but (by Art ' 8)if (2 lb)* =4(1 + P)(J 2 -r»), these roots are equal. This equation therefore gives the value of £, which makes the line a tangent. Solving for b, we have b = ± rvr+?. There are then two tangents to the circle which have any given slope. Their equations are y=lx± rVTTW. [30] PROBLEMS 1. Obtain the equations of the tangents to the circle x 2 + y 2 = 49, which are (a) parallel to the line 3 x — 2 y = 10 ; (b) perpendicular to the same line. 2. Obtain the equations of the tangents to the circle x 2 -|- if -f. 6 x = 0, which are perpendicular to the line x-3y.+ 4 = 0. 3. Determine the relation between a, b, and r if the line - ' + -- = 1 is tangent to the circle ar + y 2 = r 2 . a b 4. Determine the value of k if the line 3x — Ay — k is tangent to the circle x 2 -+- y 2 — 8 x + 12 y — 44 = 0. 5. Find the condition which must be satisfied if the line Ax + By + C = is tangent to the circle x* + y 2 + Dx + Ey + F=0. 59. Chord of contact. — We have seen that, from any point P x outside the circle x 2 + y' z = r 2 , two tangents can be drawn to the circle. Let it be, required to find the ANALYTIC GEOMETRY [Ch. VII, § 59 equation of the chord P 2 P 3 through the two points of contact of these tangents. The equations of the tangents P 2 P X and P 3 Px are (by [28]) and x 3 x + y$ = r 2 . Both these equations X must be satisfied by On Vi)- Hence x 2 x x -f y 2 y x = r 2 , and x 3 x t + y 3 y 1 = r 2 . But these are just the conditions which must be satisfied if the points P 2 and P 3 are on the line ocioc + y\\f = r 2 . [31] This is, therefore, the equation of the line P 2 P 3 which is the chord of contact. It Avill be noted that this equation has the same form as the equation of the tangent. It represents the tangent if the point P x is on the circle ; but if P x is outside the circle, it is the equation of the chord of contact. Let the student show that the equation of the chord of contact of tangents from an exterior point to the circle is 0. [32] PROBLEMS 1. Find the length of the chord of contact of tangents from the point (3, 4) to the circle $ + y 2 = 4. Oh. VII, §69] THE CIRCLE 87 2. Find the equation of the circle which touches the line 2x — ?/ = 10 at the point (3, —4) and passes through the point (5, 1). 3. Find the equation of the circle which passes through the point (1, 1) and also through the intersections of the circles '<•' + .'/" — •> a;+4y=10, and x 2 +y 2 =5x. [See prob. 5, page 79.] 4. Find the equations of the three common chords of the three circles in problem 1, page 84, and show that they inter- sect in a point. 5. Find the equation of the circle inscribed in the triangle whose sides are represented by the equations 4 x -f- 3 y — 10, x — 5 y = 15, and 3 x — 4 y = 8. 6. Find the area of the triangle formed by the axes of coordinates and the tangent to the circle x 2 + y 2 = r 2 at the point (i\, y x ). 7. Construct the circles x 2 + y- — x -+- 2 y, and x 2 + y 2 = 2x. Find the equations of their line of centres, their common chord, and points of intersection. Show that their common chord is perpendicular to their line of centres. At what angles do the circles intersect ? 8. Show that in any circle a line perpendicular to the tan- gent at the point of contact passes through the centre. 9. Show that an angle inscribed in a semicircle is a right angle. 10. Show that the perpendicular from any point of a circle on a chord is a mean proportional between the perpendiculars from the same point on the tangents at the extremities of the chord. 11. Show that the chord of contact of tangents from an exterior point is perpendicular to the line joining that point to the centre of the circle. CHAPTER VIII LOCI 60. We have seen that when a property common to all points of a locus is given, the translation of this property into an algebraic equation between the coordinates of the points gives the equation of the locus ; for this is just what is meant by the equation of a locus, — an equation which is satisfied by the coordinates of every point which satisfies the given conditions, and by no other points. The actual Avork then always consists in this translation of a condition expressed in language into a relation between the coordinates expressed in an algebraic equation. Any method which enables us to do this may be employed. The simple methods have already been exemplified in the previous chapters. In these cases the law may be expressed as an equation in x and y at once by the aid only of a sim- ple geometrical construction. There are many problems which may be solved in this way. PROBLEMS 1 . The sum of the squares of the distances of a moving point from two fixed points is constant. Find the locus of the moving point. Let the .X-axis pass through the fixed points, with the origin midway between them. Then («, 0) and (— a, 0) will represent the points. Let (x, y) be any position of 88 Ch. VIII, § 60] LOCI 8y tlic moving point. Then placing the sum of the squares of the two distances equal to a constant, Jc, we have + [(.* + a) 2 + # + [(c — a)6 + «&]# = kcb. Now since these two lines both pass through P\ its coordinates (V, y' ) must satisfy both equations, or kbx' + [(« — c)6 + e&],/ = &a&, and kbx' 4- [0? — &)& + ak~\y' = kcb. Here then are two equations between x\ y\ and k. The elimination of k will give a single equation in x' Ch. VIII, § 01] LOCI 93 and y' which must be the equation of the locus of P'. For it will be the algebraic expression of the relation which must exist between the coordinates of P\ that it may be 'the intersection of the two diagonals. The elimination is here easily performed. For, adding, we have 2 kbx r + k(a + e)y' = kb(a + c). Dividing by k and dropping the primes, we have as the equation of the locus, 2 bx + (a + c*)y = b(a + c'). Let the student find from the conditions of the problem two points through which the curve must pass, and test the result obtained above by substituting in it the coor- dinates of these points. 18. Find the locus of the intersection of the diagonals of rectangles inscribed in a given triangle. 19. On the sides of a given triangle measure off equal distances from the extremities of the base, and at these points erect perpendiculars to the sides. Find the locus of the point of intersection of these perpendiculars. 20. The ends of the hypotenuse of a given right tri- angle touch the coordinate axes. Find the locus of the vertex of the right angle. 21. Parallel lines are drawn with their ends on the two axes. Find the locus of the point which divides them in the ratio of m : n. 22. One side and the opposite angle of a triangle are fixed. Find the locus of the centre of the inscribed circle. 94 ANALYTIC GEOMETRY [Ch. VIII, § 61 23. Each radius of the circle, x 2 -f y 2 = r 2 , is extended a distance equal to the ordinate of its extremity. Find the locus of its terminal point. 24. In a rectangle, ABCD, let EF and GcH be drawn parallel respectively to AB and BO. Find the locus of the intersection of HF and EG-. 25. In the previous problem let ABCD be any paral- lelogram and solve with the aid of oblique coordinates. 26. Find the locus of the middle point of a system of parallel chords of the circle x 2 -f y 2 — r 2 . Let y = Ix + b be the equation of any one of the parallel chords ; let (x v y^) and (z 2 , y 2 ) be the coordinates of the points where it cuts the circle, and (x\ y') the coor- dinates of the point midway between these points. It is required to find an equation connecting x' and y' which may contain I but must not contain b. Starting the solution of the two equations, y = lx + b and x 2 -h y 2 = r 2 , we have (1 + l 2 )x 2 + 2lbx + b 2 -r 2 = 0, the two roots of which must be x 1 and x 2 . But x' = ^±^2- Hence (1) x' = - -^~- (See Art. 8.) Since the point (V, y'} lies on the line y = Ix + b, its •coordinates must satisfy that equation, or (2) y f = Ix' + b. Ch. VIII, § LOCI 95 We have then two equations in a/, y\ Z, and Z>, from which b must be eliminated. From (2) b = y' — lx'. Sub- stituting this value in (1) and reducing, we have as the equa- tion of the desired locus x -f ly = 0. Since this equation is of the first degree and contains no constant term, it represents a straight line through the centre of the circle, and con- forms to the ordinary definition of a diameter, evidently perpendicular to the parallel chords. It is 27. Find the Iogus of the middle points of chords which pass through a fixed point (x v y x ) of the circle x 2 + y 2 = r 2 . Let P\ (V, ?/ ), be the middle point of any chord through P v (x v y{). Let (x v y^) be the coordinates of P 2 , the other extremity of the chord. From the formulas for bisecting a line [4], we have (1) x' = x, 4- x, and (2) y' = U±±l2. And, since P 2 is a point on the circle, (3) + Here are three equations between the variables x f and y\ the constants x v y v and r, and the parameters x 2 and y v It is therefore possible !»»; ANALYTIC GEOMETRY [Ch. VIII, § 61 to eliminate the parameters and obtain a single equation in terms of the variables and constants only. Solving (1) and (2) for x 2 and y v we have x 2 = 2 x' - x v and y 2 = 2 y f -, y v Substituting these values in (3), we have 4 a/» + 4 y' 2 - 4 a^' - 4 y t y' + z x 2 + y x 2 = r 2 . But a^ 2 + j/j 2 = r 2 , and, dropping primes, the equation reduces to x 2 -{-y 2 -x 1 x-y 1 y = 0. This is the equation of the locus of P' . It is a circle on 0P 1 as a diameter, since its centre is at the point I -J, 2l J, and it passes through the origin. When, as in the above problem, we have to determine the locus of a point situated on a moving line which revolves about some fixed point in it, polar coordinates are often convenient. The fixed point is taken as the pole, and the distance from it to any position of the Y ^ moving point becomes " A the radius vector. The following prob- lem will illustrate the method : 28. Find the locus of the middle points of chords of the circle, ;r r\ which pass through a fixed point, (x v t^), not on the circle. Ch. VIII, §61] LOCI 97 Le1 J\ be a fixed point through which the secant P X P S passes, and let it be required to find the locus of _P', the middle point of P 2 Ps- Transform the equation of the circle to polar coordinates, with P 1 as origin. The equa- tions of transformation are (by [20] and [24]), x = X-, 4- p cos 6, (1) 1 y = Vi + p sin 0, and the equation of the circle becomes (2) p 2 + 2 (x x cos 6 + y x sin 6) p + x x 2 + y 2 - r 2 = 0. Let p' and 6' be the polar coordinates of P' . The vec- torial angles of P 2 , P' ', and P 3 are evidently the same, and if 6' be substituted for 6 in (2), the solution of the resulting equation, p 2 + 2 Oj cos 0' + y. x sin 6") p + ^ 2 + y x 2 - r 2 = 0, for ^ will give p 2 and p 3 , the two values of p for the points P 2 and P 3 . But / = ^, and p., + p 3 = — 2 (a^ cos r + ^ sin 0'). (See Art. 8.) ence p' = — (x x cos 6' + y 1 sin0'). This equation expresses the relation which must exist between the polar coordinates of P' ', and, dropping primes, we have as the polar equation of the locus, referred to P x as origin, p = — x x cos 6 — y x sin 6. The equations for transforming back to rectangular coordinates, obtained from (1), are p cos 6 = x — x v and p sin 6 = y — y r 98 ANALYTIC GEOMETRY [Ch. VIII, § 61 From these we see that If the polar equation of the loeus is multiplied by p, and these values substituted, it becomes (x - x x y + (y- 2/j) 2 = - x x x + x 2 - y x y + y 2 , or x 2 + y 2 — x x x — y x y = 0. This is the rectangular equation of the locus referred to the original origin, and is seen to represent a circle on OP 1 as diameter. 29. Solve problem 27 by means of polar coordinates, and problem 28 by means of rectangular coordinates. 30. Find the locus of the points which divide in the ratio m : n chords through a fixed point (x v y-^) of the circle x 2 + y 2 = r 2 . 31. Lines through a fixed point P 1 cut the circle x 2 + y 2 = r 2 in the points P 2 and P g . Find the locus of a point P of this line, if P p = ^ -Pl-**2 X ™1™3 , 1 " P l P i + P l P i 32. Chords through a fixed point of a circle are extended their own length. Find the locus of their extremity. 33. Lines are drawn from a fixed point P v meeting a fixed circle in P 2 . On P X P 2 a point P is taken so that P X P x P X P 2 = k 2 . Find the locus of P. 34. Lines are drawn from a fixed point P v meeting a fixed line in P 2 . Find the locus of the point which divides P X P 2 in the ratio m : n. Cn. vin, §oi] loci 99 35. Lines are drawn from a fixed point P v meeting a fixed line in P v Find the locus of a point P on these lines if P X P x P Y P 2 = P. 36. Find the locus of points from which tangents to two given fixed circles are equal. (See problem 4, page 82.) Show that the locus is a line perpendicular to the line joining the centres of the two circles. CHAPTER IX M CONIC SECTIONS 62. Definition and equation. — If a point moves so that the numerical ratio of its distance from a fixed point to its distance from a fixed line remains constant^ its locus is called a conic. Let the fixed line be taken as the Z"-axis, and X a perpendicular through the fixed point F as the X-axis. Let the perpen- dicular distance OF of the fixed point from the fixed line be represented by m. Let P be any position of the moving point. Then Ave are to find the equation of the locus of P when FP MP F Fig. 55. G) (any constant) = e. But FP = V(x - m) 2 + y\ and MP == x. Then (1) becomes — ^ ) ~r V _ e and y = y ' Substituting these values, the equation becomes y 2 = 2 moc. [34] From this equation we see that the curve passes through the origin; that it is symmetrical with respect to the X-axis ; that it is real only to the right of the y-axis ; and that as x increases, y increases, — at first more rapidly than x, until x = — , then more and more slowly. It has, therefore, the form shown in Fig. 56, [02 ANALYTIC GEOMETRY [Ch. IX, But, for the study of the distant points, polar coordinates are better adapted. Transforming to polar coordinates with as origin, equa- tion [34] becomes _ 2 m cos 6 p ~ sin 2 e ' "When 6 = 0, p is infinite, and the curve, therefore, does not cut the .X-axis a second time. But if we give to 6 any finite value, however small, p will have a finite value, which will be very large for small values of 0, and will decrease as 6 increases, until for 6 = — , p — 0. We see, then, that every line through except the X-ax's cuts the curve a second tune, a fact which does not appear from the rectangular form of the equation. Yet the discussion of that form showed that the curve constantly recedes from the X-axis. It can be shown by the aid of the equation of the tangent that the curve approaches parallelism with the X-axis. This particular species of conic is called the parabola. We have already defined the line iLCVas the directrix ; the point F as the focus ; OX as the principal axis ; and as the vertex of the curve. We saw that Fig. 56. DO \DF \m. The coordinates of the focus, referred to as origin, are therefore f — , 771 The equation of the directrix h x— — — , Ch. IX, §64] CONIC SECTION'S 103 The line LI! through 1\ perpendicular to the -3T-axis and terminated by the curve, is called the latus rectum. The abscissa of 11 is seen to be — , and by substituting this value for x in the equation of the parabola, its ordinate is found to be m. The length of the latus rectum is there- fore 2 m. PROBLEMS 1. "What is the equation of the parabola having its vertex at the origin, and its focus (a) on the X-axis, at a distance — to the left of the origin ; (b) on the I'-axis, above the origin ; (c) on the I"-axis, below the origin ? 2. What is the equation of the parabola if the focus is at the origin and the vertex at a distance — to the left of the origin ? 3. What is the equation of the parabola, if its vertex is at the point («, ft) and its axis is parallel to the X-axis ? 4. What is the equation of the parabola which has its ver- tex at the origin and passes through the points (3, — 4) and (-3, -4)? 5. Obtain the equation of the directrix, the coordinates of the focus, and the length of the latus rectum in the parabola 64. Central conies. e^l. We see from the form of the equation of a conic, (1 - e 2 )x 2 - 2 mx + y 2 + m 2 = 0, [33] that it always represents a curve symmetrical with respect to the .X-axis. When e = l, we have seen that there is but one value of x for each value of y. But when e =^= 1, there will be, in general, two numerically unequal values 104 ANALYTIC GEOMETRY [Ch. IX, § G4 of x for any given value of y. The curve is therefore not symmetrical with respect to the Y"-axis. But it will be shown to be symmetrical with respect to a line parallel to that axis. Transform the equation to a new ori- gin midway between the points where the curve — «= — X cuts the X-axis. The Z-axis will then be found to be an axis of symmetry. Placing y = in [33], we have Fig. 57. (1 — e 2 )x 2 — 2 mx + m 2 = 0. The two solutions of this equation will give the inter- cepts, OA and 0A f , on the X-axis. Let these he denoted by x 1 and x 2 . But we wish to know 00 (=#), being the middle point of A' A. Hence - x x + x^ But we know that the sum of the roots of a quadratic is , where a and b are the coefficients of x 2 and x a respectively. (Art. 8.) Hence x x + x 2 = 1-e' , and x = 9/1 The equations for transforming from to C as origin will then be (by [20]) x — x' + m -, and y = y' '. Ch. IX, §64] CONIC SECTIONS 105 Substituting in [33], it becomes ^ 1 + l-* 2+ (l-, 2 J) 2 J + y' 2 - 2 w* ( x' 4- r-^ ) + m 2 = Reducing and dropping primes, Q.-W + 9 2 _ e 1 m 1 Dividing by ,- hn 2 1- -e 2 x 2 e 2 m 2 +-£-=1. (1 - e 2 ) 2 1 e 2 ??? 2 Let — — - = r« 2 , and the equation becomes (1 - e') 2 2 2 l + ^b] =1 - [35] If then this transformation is possible, we have reduced [33] to a form which represents a curve symmetrical with respect to both axes. It is always possible except for the case when e — L But if e = 1, no value could be obtained for 5?, and the point O would not exist. This has been discussed in Art. 63. All other cases are in- cluded in equation [35]. The intercepts of the curve on the new axes, obtained from equation [35], are ± a and ± aVl — e 2 . This equa- tion must therefore represent two classes of curves quite dissimilar in form ; for while all intercepts are real when e < 1, we see that the intercepts on the I^-axis will be imaginary when e > 1. If, when e < 1, we let the inter- 10(3 ANALYTIC GEOMETRY [Ch. IX, § 65 cepts on the P"-axis, ± a Vl — e 2 , be represented by ± b, equation [35] becomes 5+5- 1 - ^ But since we wish to Avork with equations having only real coefficients, b cannot represent the same expression when e > 1, for Vl — e 2 would be imaginary. We then let ± a Ve 2 — 1 = ± b' , and equation [35] becomes %-&-*• P"] The b used in the first case is the actual intercept. In the second case b' is the real coefficient of the imagi- nary intercept, and b 2 = - b'\ We see then that there are three distinct forms which the locus may take. If e = 1, the conic has been called a parabola ; if e < 1, it is called an ellipse ; and if e > 1, an hyperbola. The ellipse and hyperbola are called central conies to distinguish them from the non-central conic, the parabola. They may be treated together from the single equation [35], or from their separate equations. Let the student show that, if the directrix is taken as the X-axis, and a perpendicular to it through the focus as the J^-axis, the simplest equation of the central conies is — — — + '^-==1. What is its form for the « 2 (1 — e l ) a 2, ellipse ? hyperbola ? 65. Ellipse. e0F\ and the O ,A ,F' Fig. 1-e 2 points must take the po- sitions indicated on the figure. We have shown that, when e < 1, the equation of the conic referred to the new axes (see Fig. 58) is ^ + #1=1 a 2 b* ' [36] From the form of this equation we see that the curve is symmetrical with respect to both axes, and hence to their intersection ; that it cuts the X-axis at the points (± a, 0), and the F-axis at (0, ± b) ; that the values of x are real only for values of y from — b to + b ; and that the values of y are real only for values of x from — a to -f a. A more careful plotting of the points will show that it has the form shown in Fig. 59. The line D'lP has been called the directrix, and the point F f the focus. Place the points F and D on the X-axis so that CF=F'Q and CD = D'C, and draw DR perpendicular to the X-axis. The symmetry of the curve 108 ANALYTIC GEOMETRY [Ch. IX, § 65 shows that if we had used the line BH and the point F as directrix and focus, and the same value of e, the same curve would have been found as the locus. The curve can be said therefore to have these two lines BR and B'R' as directrices, and the two points F and F' as foci. We can now obtain the relations between the various lines in the figure. We have seen that FB = D'F' = m, CB = B'C = 1-e 5 CA = A'C =a em CB = B'0 =b = em VI It follows that CB = ~, and that the equations of the directrices are v = % and x=-^. [38] e e Ch. IX, § 66) CONIC SK(TI()XS 1U ( .) Also that CF=CD-FI) = — ^— - wi = -^- = ae. 1 — £ 2 l — ^ 2 It is convenient to let CF be represented by a single letter c. Then ae, or ^ In obtaining equation [36], we let b 2 = a 2 (l — e 2 ~). Solving for e 2 , we have * = «L=JL [39] a 2 u J Comparing these two values of e, we have a 2 — b 2 = c 2 . [40] From this we see that BF, being the hypotenuse of a right triangle whose legs are c and 5, is equal to a. It also shows that a is always larger than b, or that i'i(=2a), the axis perpendicular to the directrices, is larger than B'B(=2b). A' A has been called the trans- x ~ verse or principal axis ; B'B is called the conju- gate or minor axis of the curve. If the foci of the ellipse are on the I^-axis, the vertex A also lies on that axis, and B on the .X-axis (Fig. 60). Its equation is (see end of Art. 64) - + ^=1. [41] b 2 a 2 L J 110 ANALYTIC GEOMETRY [Ch. IX, § GO All the formulas found above hold for [41], except the equations of the directrices, which are , a y = ±- e PROBLEMS 1. Find a, b, c, e, and the equations of the directrices in the ellipse, (a) 4a 2 + 9?/ 2 = 36, (b) 9x* + 4:y 2 = ?>6, (c) 3ar° + 8>/ 2 =10. 2. Find the equation of the ellipse having its centre at the origin and its foci on the AT-axis, if (a) a = 3 and b = 2, (d) b = 4 and c = 3, (&) b = 3 and e = i, (e) a = 5 and c = 3, (c) a = 6 and e = f, (/) c = 4 and e = J. 3. Show that the length of the latus rectum (line through 2 b- the focus perpendicular to the axis) of the ellipse is 4. Show that the circle is the limiting form of the ellipse as a and b approach equality. What is the eccentricity of the circle, and where are its foci and directrices ? 5. What is the equation of the ellipse which has its centre at the origin and its axes coincident with the coordinate axes, and which passes through the points (4, 1) and ( — 3, 2) ? 6. What is the equation of an ellipse if its centre is at the point (a, /3) and its axes are parallel to the coordinate axes ? 66. Hyperbola. e>l. When e > 1, one of the intercepts, -, is positive, 1 + e 071 and less than ??i, while the other, , is negative. OC 1 — e ° will also be negative. The points, A, A', (7, and .F, will, therefore, take the positions indicated in Fig. 61. Cii. IX, § 66] ('••NIC SECTIONS 111 We have shown that, when e > 1, the equation of the conic reduces to A' AF --^- = 1. r37i a 2 b' 2 L J Again we see that the - curve is symmetrical with respect to hoth axes, and hence with respect to the origin; that it cuts the X-axis at the points (± a, 0), and does not cut the !F-axis ; and that the values of y are real only for values of x numerically equal to or greater than a. The exact form can be obtained more readily from the polar Fig. 61. equation. Transforming — a 1 with C as origin, we have 22— = 1 to polar coordinates b' 2 l P 2 = a 2 b' 2 &'2 CO s 2 0-a 2 sin 2 When 6 = 0, p = ± a, and as 6 increases, the denomi- nator decreases, the fraction increases, and the point recedes from the origin. This will continue as long as the denominator remains positive. As soon as the de- nominator becomes negative, the value of p becomes imaginary. There is then a value of beyond which the curve does not exist. This value of 6 is that which makes the denominator, b' 2 cos 2 6 — a 2 sin 2 0, zero, or a For every value of 6 between tan 6 = tan" tan" 1 ! and there will be a real value of p, these val- ues growing larger as 6 approaches tan _1 (H — j 112 ANALYTIC GEOMETRY [Ch. IX, § 66 tan -1 ( ). The lines then which pass through the V aJ f V\ f V\ origin, making the angles tan M H J and tan _1 ( J with the JT-axis, do not cut the curve, while every line lying between these lines cuts the curve in two real points. The curve must therefore approach parallelism with these lines as the point recedes from the origin, and it will be shown in the next article that the curve approaches coinci- dence with these lines. Such a line is called an asymptote. If we continue to increase 0, we see that there will be no real value oi.p until tan 6 again becomes numerically b> less than — Then p goes through the same changes a in value, decreasing until it equals ± a. But we have shown that the curve is symmetrical with respect to both axes, and there is therefore no need of discussion beyond the first quadrant. The following is the form of the hyperbola : 7 Place the points F' and D' on the X-axis so that F'C=OF md D'C=CD, and draw D'H' perpendicular Ch. IX, § GG] CONIC SECTIONS 113 to the X-axis. The symmetry of the curve again shows, as in the ellipse, that the hyperbola may be said to have two foci, J 7 and F\ and two directrices, DR and D' W '. We can now <>l>!lfc^tln' relations between the various lines in the figure. ^Bliavc seen that DF=FD' = m, e*-r CA = A'C =a = em e 2 -l CB = B'0 =b' =— — VeP-1 It follows that CD = -, and that the equations of the directrices are a t a ac = Also that * = 2, and * = -?. , [42] 0F= CD + DF=^-+m = -pL = ae. e z — 1 e & — 1 It is convenient to let OF be represented by a single letter c. Then c = ae, e or e = . a In obtaining equation [37], we let V 2 = a 2 (e 2 — 1). Solving for e 2 , we have Comparing the two values of e, we have a? + b' 2 = c 2 . [44] 114 ANALYTIC GEOMETRY [Ch. IX, § G7 There is, in the hyperbola, no restriction on the relative values of a and b' . Note. — In the following articles we shall follow the ordinary custom, and use b in place of b'. s f i til 67. Asymptotes. — The slopes of the asymptotes were seen [Art. 60] to be ± -. Hence their equations are b . b y=~ x -> and y = ~ a x > or written as a single equation, a 2 b 2 They are evidently the diagonals of the rectangle formed by drawing lines parallel to the axes through A, A', B, and B'. It remains to be shown that the curve not only approaches parallelism with these lines, but actually approaches coinci- dence with them ; or that the perpendicular distance P X M from any point P x on the hyperbola to the asymptote decreases indefinitely, as P x recedes from the origin along the curve. (See Fig. 62.) Since the equation of the asjmiptote is bx — ay = 0, P 1 M= bx * - ay ± - (By [17]) V6 2 + a 2 But, since P x is a point on the curve, b 2 x x 2 — a 2 y x 2 = a 2 b 2 , or, factoring, bx x - ay x bx x + ay x Ch. IX, § G7] CONIC SECTIONS 115 Hence 1\M (^ 1 + a?/ 1 )V^ + a 2 This expression evidently decreases as x 1 and y x in- crease, approaching zero as a limit. The curve therefore a pp roaches its asymptote indefinitely, PROBLEMS 1. Find a, b, c, e, and the equations of the directrices and asymptotes of the hyperbola, (a) a*-25tf = 25, 0) 9x 2 --if- = 3(j, (c) 2x 2 -5f- = 20. 2. Find the equation of the hyperbola having its centre at the origin and its foci on the X-axis, if (a) a = 3 and b = 2, (d) b = 4 and c = 5, (b) b = 3 and e = 2, (e) a = 4 and c = 5, (c) a — 5 aild e = f , (/) c = 10 and e = 3. 3. What is the equation of an hyperbola, if its centre is at the point («, /?) and its axes are parallel to the coordinate axes ? 4. What is the equation of the hyperbola which has its centre at the origin and its foci on the X-axis, and which passes through the points (5, 3) and (—3, 2) ? 5. Show that the latus rectum of the hyperbola is - — • 6. Show that the foot of the perpendicular from the focus of an hyperbola on an asymptote is at the distance a from the centre and b from the focus. 7. Show that the circle of radius b, whose centre is at the focus of an hyperbola, is tangent to the asymptote at the point where it is cut by the directrix. 8. Show that the product of the two perpendiculars let fall from any point of an hyperbola on the asymptotes is constant. 116 ANALYTIC GEOMETRY [Ch. IX, § 68 68. Conjugate hyperbolas. — Tf, in deriving the equa- tion of the conic, the directrix is taken as the X-axis, and a perpendicular to it through the focus as the JT-axis, its Fig. 03. simplest form, in the case of the hyperbola, is = 1. # 2 y 2 __ If the definitions of a and b are interchanged, using b to represent the semi-transverse axis (which is here the real intercept of the hyperbola on the !F-axis), the equation becomes -«). If the axis of the parabola is parallel to the Y"-axis, its equation is * 0-«) 2 =±2m(j/-/3). In each case the term in xy is wanting, and all of the equations are seen to be special cases of the general equa ion ^ + ^ + J)x + ^ + F = ^ If neither A nor C is zero in this equation, it may be written in the form aU + v-rr-^icfv* + E v+ :E2 V m + & - f- A \ X+ A X + ±A*) + t V + C y + 4C r >riA + IC F ' or, if we represent the second member by K, (•+n) (» + 2 + v " =1. K K A O Ch. IX, §73] CONIC SECTIONS 125 If A and C have the same sign, this takes the form of the equation of the ellipse whose centre is at the point and in which a — \ — , and b=\—-. The -1A 26V _^ *0 ellipse wiH be real, null, or imaginary, according as a and b are real, zero, or imaginary. Let the student show that if A and have opposite signs, the equation represents an hyperbola, or (if K = 0) two intersecting lines. Also that, if either A or is zero, the equation repre- sents a parabola, or a pair of parallel lines. PROBLEMS 1. Determine the nature and position of the locus of 2x 2 + 3y 2 -6x + 4:y = l0. This equation may be written in the form 2(^-3x + f) + 3(. ! /+| 2 / + |) = 10 + f + |=^ ) (*-f) 2 , (y + f) 2 ,-. 9 5 T 9 5 12 18 The locus is an ellipse, having its centre at the point (f , — -§), and in which a = Vff , and b = V-f-f • 2. Determine the nature and position of the locus of the following equations : (a) x> + 2y 2 -6x + y = 10, (d) 3ar - y 2 + Gy = 0, 0) a? + 4x-2y = W, (e) f + 2x-±y = 6 (c) ±x 2 -3if-±x + S = 0, 3. Obtain the polar equation of each of the conies, the focus being used as the origin and the transverse axis as the initial line. CHAPTER X TANGENTS 74. The method of finding the equation of a tangent to any conic at a given point is the same as that used in the case of the circle (Art. 55}. The equation of a secant through the given point P v (x v y±), and an adjacent point P 2 , (x x + h, y x + &), on the curve is V ~ V\ _ * x — x 1 h It is necessary termine in each to de- case a Fig. 69. value of - which will h not be indeterminate when h and k approach zero. We shall give the work in detail for the ellipse, b 2 x 2 -\- a 2 y 2 = a 2 b 2 . Since the points P x and P 2 lie on the ellipse, their coordinates must satisfy its equation, or (1) b 2 x 2 + a 2 y 2 =a 2 b\ (2) b 2 x 2 + 2 bVix^ + b 2 h 2 + a 2 y 2 + 2 a 2 ky x + a 2 k* = a 2 b 2 . Subtracting (1) from (2), we have 2 b 2 hx x + b 2 h 2 + 2 a 2 ky x + a 2 k 2 = 0, Jc = 2 b 2 x, + b 2 h h 2 a 2 y x + a 2 k 126 or Ch. X, § 74] TANGENTS 127 The equation of the secant may therefore be written // - ,y t = 2 b 2 x x + b 2 h x — x 1 2 cfiy l + a 2 k Now, if we let P 2 approach P v h and & will approach zero, and the limit of the second member is no longer b 2 x indeterminate, but becomes 1 - The equation of the tangent is therefore y-y\ = ft 2 *i 7 x - x x a 2 y x or clearing of fractions and transposing, b 2 x x x + a 2 y^y = b 2 x 2 + a 2 y 2 . But b 2 x 2 + a 2 y 2 = a 2 b\ and the equation of the tangent reduces to b-xix + ary x y = a 2 b 2 . [47] Let the student show that the equation of the tangent to the hyperbola, b 2 x 2 - ahf = a 2 b 2 , is bx x x - cpyxy = aW-, [48] the parabola, y 2 =2mx, is y\y = m(x + xi), [^9] the locus of the general equation of the second degree, Ax 2 + Bxy + Cy 2 + Dx + Ey + F= 0, is Axix+~-(x { y + yix)+ Cy x y [ 50 ] + ~(x + x l ) + §(y + yO+F=o. These formulas can be most easily remembered and applied if we notice that they may be obtained from the 128 ANALYTIC GEOMETRY [Ch. X, § 75 equation of the conic by replacing x 2 and y 2 by x x x and constant quantities being unchanged. The method of finding the equations of the tangents from an exterior point is the same as that given for the circle. (See Art. 57.) Let the student show that the equation of the chord of contact of tangents from an exterior point will, in each case, take the same form as the equation of a tangent at the point of contact. (See Art. 59.) 75. Normals. — The normal at any point of a conic is the line through the point, perpendicular to the tangent at the point. It can be found in any case by writing the equation of a tangent, and then writing the equation of a perpendicu- lar to the tangent through the point of contact. For example, the tangent to the ellipse has been found to be b 2 x x x + a 2 y x y = a 2 b 2 . A perpendicular to this line will have the form a 2 y x x — b 2 x Y y = k. Since the normal passes through P v k = cfiy^^ — b 2 x^j v and the equation of the normal becomes a 2 y x x - b 2 x x y = (a 2 - b 2 )x 1 y v In like manner, the equation of the normal to the hyperbola is ^ + ^ = ^ + ^ and to the parabola is y x x + my = x 1 y l + my v The student should note that these formulas apply only when the equations of the curves are in the simplest form. Cn. X, § 7G] TANGENTS 129 He is advised not to use them in solving problems, as they are not easily remembered, but in each case to write the equation of the tangent and then that of a perpen- dicular to the tangent at the point of contact. 76. Subtangents and subnormals. — The projections on the X-axis of those parts of the tangent and normal included between the point of contact and the X-axis are called the subtangent and subnormal. Eig. 70. In the ellipse (Fig. 70) M X T is the subtangent and JfjiV is the subnormal for the point P v The equation of the tangent at P 1 is b 2 x x x -f a 2 y x y = a 2 b 2 , and the equa- tion of the normal is cPy 1 x—b 2 x x y — (cP' — b 2: )x 1 y l . Find- ing the intercepts on the X-axis, we have CT=~ and CN= ^ ~ ^ x v But M X T= CT- CM X = - ocf OCi and M^^Ctf-CM^ (a 2 - b 2 )x^ V Xv [51] [52] 130 ANALYTIC GEOMETRY [Ch. X, § 70 2 2 Let the student show that for the hyperbola — — &- = 1, a 2 b 2 the subtangent equals a ~ a?1 > [53] the subnormal equals -^«i, [54] and that for the parabola y 2 = 2 mx, the subtangent equals - 2 xi, L^li the subnormal equals in, [5t>] PROBLEMS 1. Find the equations of the tangents and normals to (a) 3ar 9 + 4?/ 2 = l9 at (1, 2), (6) 2ar*-2/ 2 = 14 at (3,-2), (c) ?r = 6 ;i' at (6, — 6), (t?) 2^ 2 -3a7/ + 6.T-2 = at (2,3). 2. Find the lengths of the snbtangents and subnormals in (a), (6), and (c), problem 1. 3. Find the equations of the tangents to (a) 16 x 2 + 25 y 2 = 400 from (3, 4), (b) y- = ±x from (-3, -2), (c) ar- 3/ +2 a- + 19 = from (-1,2). 4. Find the chords of contact of the tangents in problem 3. 5. Find the lengths of the tangents and normals in («), (Jj), and (c), problem 1. Note. — The terms "length of tangent" and " length of normal 1 ' are used to indicate the distances on the tangent and normal from the point of contact to the points where they cut the X-axis. 6. Find the angles between the ellipse 4 x 2 + y 2 = 5, and the parabola y 2 = 8 x, at their points of intersection. 7. Show how the subtangent or subnormal in the parabola may be used to construct the tangent at any point of the curve. Ch. X, § 77] TANGENTS 131 8. From the fact that the subnormal in the parabola is constant, show that the tangent approaches parallelism with the axis as the point of contact recedes from the origin. 9. Show that if the normals of an ellipse pass through the centre, the ellipse is a circle. 10. Show that the distance from the focus of a parabola to any tangent is one-half the length of the corresponding normal. 11. Show that the focus of a parabola bisects the portion of the axis intercepted by a tangent and the corresponding normal. 77. Slope form of the equations of tangents. — For many problems it is convenient to have the equation of the tan- gent in terms of its slope only. This can be found for each of the conies just as it was found for the circle in Art. 58. We shall give the outline of the work for the ellipse. Starting the solution of the equation of any line, y= lx + /3, with the equation of the ellipse, b 2 x 2 -f- a 2 y 2 — a 2 b 2 , we have, for obtaining the abscissas of the points of intersection, the equation (52 + a 2 ?2 ) 3,2 + 9 a H$x + a 2 (/3 2 - b 2 ) = 0. There will be two solutions of this equation, and hence two points where the line cuts the ellipse. But if (see Art. 8) 4 aH 2 ^ = 4 (b 2 + a 2 l 2 ~)Qa 2 P - a 2 b 2 ), or /3 = ±Va 2 Z 2 + 6 2 , the two solutions of this equation are equal, the two points of intersection of the line with the ellipse have become coincident, and the line is tangent to the ellipse. 132 ANALYTIC GEOMETRY [Ch. X, § 77 The equation of a tangent having a given slope I is therefore ____. y = lx± VaH 2 + V 2 . [57] Let the student show that the equation of the tangent having a given slope I is for the hyperbola, b 2 x 2 ~ a 2 y 2 = a 2 b 2 , y = lx± y/aH 2 - b 2 , [58] for the parabola, y 2 = 2 mx, y = i* + f v [59] PROBLEMS 1. Find the equations of the tangents to the ellipse 4 x 2 -f y 2 = 4 which are parallel to the line 2a? — 4^ + 5 = 0. 2. Find the equation of the normal to the parabola y 2 = Sx, which is parallel to the line 2 x -f 3 y = 10. 3. Find the equations of the tangents to the ellipse x 2 + 2 y 2 — x + y = 0, which are perpendicular to the line x — 5 y = 6. 4. Find the condition which must be satisfied if the line x 2 v 2 y = lx + fl is tangent to the hyperbola — — *- = — 1. 5. Find the points on each of the conies where the tangents are equally inclined to the axes. For what case is the solution impossible ? 6. Find the points on the ellipse and hyperbola where the tangents are parallel to the line joining the positive extremi- ties of the axes. x 1/ 7. Show that the line - + \ = 1 is tangent to a p X 2 V 2 (a) the ellipse — + fr = 1, if or b~ « 2 + ^ = l. « 2 + /? 2 ' o 2 (b) the hyperbola — — ^ = 1, if a 2 b 2 1 . a 2 y3 2 (c) the parabola y 2 = 2 mx, if ma + 2/3 2 = 0. Sm X, § 78 j TANGENTS 133 8. Find the equations of the common tangents to the ellipse ar + 9 y 2 = 9, and the circle x 2 + y 2 = 4. Note. — Write the equation of the tangent to each curve in the slope form and determine the value of I which will make the two equations identical. 9. Find the equations of the common tangents to the ellipse 4 ar + 9 y 2 = .'56, and the hyperbola x 2 — y 2 = 16. Show that there would be no common tangent, if the second member of the equation of the hyperbola had any value less than 9. Why ? Construct the figure. 10. Find the equations of the common tangents to the circle ar + y 2 = 9, and the parabola y 2 = Sx. How many solutions are there ? Why ? Construct the figure. 11. Show that two tangents can be drawn to any conic from an exterior point. 12. Through any given point how many normals can be drawn to (a) an ellipse, (b) a parabola ? 13. Obtain the equation of the tangent at the point P 1 of the parabola y 2 = 2 mx, by determining I in the slope form of the equation of the tangent in terms of x x and y x . 78. Theorems concerning tangents and normals. — 1. The tangent and normal at any point of an ellipse bisect the exterior and interior angles respectively between the focal radii drawn to the point of contact. Let P^T and P X N be the tangent and normal to the ellipse —--\-^-=\ at the point P v We wish to show that a 2 b 2 P X T bisects the angle FP X K, and that P X N bisects the angle F'P^F. It is a well-known theorem of elementary geometry that the bisector of an interior angle of a triangle divides the opposite side into segments which are proportional to the 134 ANALYTIC GEOMETRY [Ch. X, § 78 adjacent sides of the triangle. The converse theorem is also true. It is therefore sufficient to show that F'P X = F'JV FP X JVF' The equation of the normal P X N is ahj^x - b\y = O 2 - b 2 )x l y v Fig. 71. a 2 — b 2 Its intercept CN on the X-axis is — x v or, since in a 2 — b 2 a the ellipse — = e 2 , a 2 CJSr=e 2 x v Also, F f O=CF=ae. Hence F , N=F , C+CN= ae + e 2 x x = e(a + ex x ), and NF=CF - CN= ae - e 2 x x = e(a - ex{). But (by [45]) F , P 1 = a -f- ex v and FP X = a — &r r Ch. X, § 78] TANGENTS 135 Hence and the normal bisects the angle F'P, = F'N FP X NF' F' P X F. Since the tangent is perpendicular to the normal, it bisects the supplementary angle FP X K. Note. — It is upon this principle that whispering galleries are con- structed. If the whole or part of the sides of a room is a surface formed by revolving an ellipse about its major axis, all waves of sound, light, or heat starting from one focus and striking this surface will be reflected to the other focus. 2. In an hyperbola the tangent and normal at any point bisect the interior and exterior angles respectively between the focal radii. 3. If an ellipse and hyperbola are confocal (or have the same foci), they intersect orthogonally (or at right angles). Fig. 72. Since the direction of a curve at any point is along the tangent at that point, two curves intersect orthogonally, if their tangents at the point of intersection are perpen- dicular to each other. This is evidently the case here, since the tangent to the ellipse bisects the exterior angle 136 ANALYTIC GEOMETRY [Ch. X, § 78 between the focal radii, and the tangent to the hvperbola bisects the interior angle. The curves therefore intersect orthogonally. 4. The tangent at any point of a parabola makes equal angles with the focal radius drawn to the point of contact, and with the axis of the curve. In the parabola y 2 = 2 mx, let P X T and P X N be the tan- gent and normal at P v Join P X F and draw P^K parallel Fig. 73. to OX. We wish to prove that the angles TP X F and FTP X are equal. If we let the abscissa of P x be x v its ordinate will be ± V2 mx v since P x is a point on the parabola. Then the equation of the tangent at P x is ± ^/2mx 1 • y = m (x + x^) . If in this equation we let y = 0, we find the intercept OT to be —x,. Hence TO m x v and TF=x 1 + ^. But FP X = \ (x x - |Y + 2 mx 1 = x x + Cii. X, § 78] TANGENTS 137 Hence TF=FP V and the angles TP X F and FTP 1 are equal. What other angles are also equal in the figure ? Note. — Parabolic reflectors depend on this principle. If a surface is formed by revolving a parabola about its axis, all waves of light, etc., which start from the focus will be reflected in lines parallel to the axis of the parabola. 5. Two parabolas which have the same focus and axis, but which are turned in opposite directions, cut each other orthogonally. 6. The chord of contact of tangents to a parabola from any point on the directrix passes through the focus. Fig. 74. The coordinates of any point L on the directrix may be represented by ( — — , yA. The equation of P X P V the chord of contact of tangents from this point, is y x y = mx m? y = lx + m 21 X ml y = ~T ~ ~2 138 ANALYTIC GEOMETRY [Ch. X, § 78 Since the coordinates of the focus (— , J satisfy this equation, the chord of contact must pass through the focus. Let the student prove the converse theorem, viz. : Tan- gents at the ends of any focal chord meet on the directrix. 7. Prove that the same theorems hold for the ellipse and hyperbola. 8. Any two -perpendicular tangents to the parabola meet on the directrix. Two perpendicular tangents may be represented by the equations and By solving these equations simultaneously, the point of intersection of the two tangents is found to be r_m m(\ - P)- | which is evidently a point on the directrix. Let the student prove the converse theorem, viz. : Two tangents to a parabola from any point on the directrix are perpendicular to each other. 9. Tangents to a parabola at the ends of any focal chord are perpendicular to each other. 10. Show that theorems 8 and 9 do not hold for cen- tral conies, but that perpendicular tangents (a) to an ellipse meet on the circle x 2 + y 2 = a 2 -j- b 2 ; (£>) to an hyper- bola meet on the circle x 2 + y 2 = a 2 — b 2 . (See Chap. 14, Prob. 1.) 11. The line joining any point on the directrix of a pa- Ch. X, § 78] TANGENTS 139 rabola to the focus is perpendicular to the chord of contact of tangents from the point. Take the coordinates of the point L (Fig. 74) on the directrix as ( — — , y x ), and show that the line LF which joins this point to the focus is perpendicular to the chord mx m- of contact y x y 12. Prove the same theorem for the central conies. 13. The two tangents which may be drawn from an exterior point to any conic subtend equal angles at the focus. 14. In the parabola the perpendicular from the focus on any tangent meets it on the tangent at the vertex ; the per- pendicular meets the directrix on the line through the point parallel to the axis of the parabola. The equation of the tangent at any point (x v yd is y x y = mx 4- mx v The equation of a per- pendicular to the tangent through the focus is y x x + my = ^1- The coordinates of the point of intersection of these two lines are (». t> Fig. 75. They therefore meet on the F'-axis, which is the tan- gent at the vertex. Let the student prove the second part of the theorem, 140 ANALYTIC GEOMETRY [Ch. X, § 78 15. Show that theorem 14 does not hold for central conies, but that the perpendiculars from the foci of a cen- tral conic on any tangent meet the tangent on the circle x* + y*= a 2 . (See Chap. 14, Prob. 5.) 16. The perpendicular from a focus on any tangent to a central conic meets the corresponding directrix on the line joining the centre to the point of contact of the tangent. 17. In any conic, tangents at the ends of the latus rectum meet the X-axis on the directrix. 18. The tangent at any point of the parabola meets the directrix and latus rectum produced at points equally dis- ta n t from the focus. 19. The product of the perpendiculars from the foci of a central conic on any tangent is constant and equal to P. 20. The semi-minor axis b of a central conic is a mean proportional between a normal and the distance from the centre to the corresponding tangent. 21. The tangents at any point of an ellipse and the cor- responding point on the auxiliary circle pass through the same point on the axis. PROBLEMS 1. Show that, if the point of contact of a tangent to an hyperbola moves off along the curve, the tangent approaches the asymptote as its limiting position. 2. Find the equations of the tangents to the hyperbola which pass through the centre. (Use slope form of the equa- tion of the tangent.) 3. Show that the portion of any tangent to an hyperbola included between the asymptotes is bisected at the point of contact. Ch. X. § 78 ] TANGENTS 141 4 Show that the area of the triangle formed by any tan- gent to an hyperbola and the asymptotes is constant. 5. Through any point of an hyperbola parallels to the asymptotes are drawn. Show that the area of the parallelo- gram formed by these lines and the asymptotes is constant. 6. Show that the normal at one extremity of the latus rectum of the parabola is parallel to the tangent at the other extremity of the latus rectum. 7. Obtain the equation of the parabola referred to tangents at the ends of the latus rectum as coordinate axes. 8. Show that the distances of the vertex and focus of a parabola from the tangent at one end of the latus rectum are in the ratio of 1 : 2. 9. Show that the directrix of a parabola is tangent to the circle described on any focal chord as a diameter. 10. Show that the tangent at the vertex of a parabola is tangent to the circle described on any focal radius of the parabola as a diameter. 11. The tangent and normal at a point of the ellipse form an isosceles triangle with the X-axis. Find the coordinates of the point. 12. Prove that the angle between two tangents to a parabola is one-half of the angle between the focal chords drawn to the points of contact. 13. Show that the length of a normal in an equilateral hyperbola is equal to the distance of the point of contact from the centre. 14. Find the points on the conjugate axis of an hyperbola from which tangents to the hyperbola are perpendicular to each other. CHAPTER XI DIAMETERS 79. The diameter of a conic may be denned as the locus of the middle points of a system of parallel chords. The method of finding this locus has already been illustrated for the circle on page 94. We shall repeat the work for the ellipse -- -+- ^ = 1. Let y = l x x -f- /3 be the equation of any one of the parallel chords ; let (x v y^) and (z 2 , y 2 ) be the coordinates of the Fig. 76. points where it cuts the ellipse, and (V, y r% ) the coordinates of the point midway between these two points. 142 Ch. XI, § 79] DIAMETERS 143 Starting the solution of the two equations, y — l x x-\-^ and b 2 x 2 + a 2 y 2 = a 2 b 2 , we have (J) 2 + a 2 ! 2 ) x 2 + 2 a\px + a 2 (/& - b 2 ) = 0, the two roots of which must be x 1 and x T But Hence x f = x 1 ±x. x>= *M- b 2 + a\ 2 [By Art. 8] Since (V, y 1 ) is on the line y = l x x 4- A y r may be found by substituting the value of x f in that equation. This gives Hence y cfil, - * 2 ' or, dropping primes, we have as the equation of the diameter, b 2 x + a-hy = 0. [60] Fig. 78. 144 ANALYTIC GEOMETRY [Ch. XI, § 80 Let the student show that the equation of a diameter oi ^2 2 the hyperbola, ^ — %- = 1, is b 2 x - a?hy = 0, T611 a 2, ¥ the parabola, y 2 '=2mx, is l^y-m. [62] The form of the equation of a diameter of an ellipse or hyperbola shoivs that it passes through the centre of the conic, and that it therefore conforms to the ordinary definition of a diameter. But all the diameters of a parabola are seen to be parallel to the axis. Since any value may be given to l v all lines through the centre of an ellipse or hyperbola and all lines parallel to the axis of a parabola are diameters. Let the student obtain the equation of the diameter of each conic by the following method : Transform to polar coordinates, with the middle point (V, y'j of any one of the parallel chords as origin. If tan -1 l x is substituted for 6 in this equation of the conic, the resulting values of p should be equal in magnitude, but have opposite signs. The necessaiy condition for this will be an equation be- tween x\ y\ and l v which will therefore be the equation of the diameter. 80. Conjugate diameters. — If we let l 2 be the slope of the diameter which bisects a system of chords of slope l v we see that for the ellipse 7 b2 7 7 & and for the hyperbola 7 b 2 77 b* Ch. XI, § 80] DIAMETERS 145 Since in these expressions l x and l 2 are interchangeable, it is evident that, if we started out with a system of chords of slope Z 2 , the corresponding diameter would have l x for its slope. Hence the two diameters which have Z x and l 2 for their slopes are so related to each other that each bisects all chords parallel to the other. Such diameters are said to be conjugate to eacli other. Their equations are y = l x x, and y = l 2 x, where Z x and l 2 are connected by the relations given above. In the ellipse, l x and l 2 have opposite signs, and the diameters must pass through different quadrants. But in the hyperbola, l x and l 2 have the same sign, and the diameters must pass through the same quadrant. In either case, as l x decreases in numerical value, l 2 increases, and as one diameter approaches the major axis, the other will approach the minor axis from one side or the other. The limiting case is seen to be the two axes. They con- form to the definition of conjugate diameters, since every line paral- lel to one is bisected by the other. They are the only conjugate diameters which are perpendicular to each other. If in the ellipse one diameter, P X K V starts coincident with the major axis and revolves in the posi- tive direction, its conjugate diameter, P 2 K 2 , will start Fig. 79. 146 ANALYTIC GEOMETRY [Ch. XI, § 80 coincident with the minor axis and also revolve in the positive direction, since we have seen that the two must pass through different quadrants. Let the student show that the angle P X CP V in which the minor axis lies, will always be obtuse. If, in the hyperbola, J > 1 K 1 starts coincident with the major axis and revolves in the positive direction, its conjugate diameter, P 2 K 2 , will start coincident with the minor axis and revolve in the negative direction. For the two diameters must remain in the same quadrant. Since b 2 b the product of the two slopes is — , if L is less than -, /„ , a* a must be greater than -, and the two diameters must there- a fore rernain on opposite sides of the asymptote. As l x approaches -, L also approaches -, and the asymptote a a is therefore the limiting position of both diameters. It Ch. XI, § 81] DIAMETERS 147 is evident that only one of two conjugate diameters can cut the hyperbola. But in speaking of the length of the other diameter, we shall mean the distance, P 2 iT 2 , between the points where it cuts the conjugate hyperbola. 81. Equation of conjugate diameter. — If one diameter is n'iven in any way, as by means of its slope or the coor- dinates of its extremity, it is easy to determine the con- jugate diameter. For it is only necessary to write the equation of a line through the centre whose slope bears the required relation to the slope of the given line. If the coordinates of P l (Fig. 79) are (x v y x ), the equa- tion of CP l is x x y — y x x = 0, and l x = — . Then for the ellipse, * I = h2 = ^ , 2 a\ a^yl and the equation of the conjugate diameter P 2 ^2 * s 9^ + mii = . T631 a 2 b~ L J The solution of this equation with the equation of the ellipse gives for the coordinates of P 9 ( f*i — l ), and (ay bx \ " ^ " a ' for the coordinates of K -f 1 l • 2 \ b a J Let the student show that in the hyperbola the equa- tion of a diameter conjugate to the diameter through (a\, vO is £!£_ 2i2 = 0, [64] a 2 b- L J and that the coordinates of its extremities are fay, bx,\ if ay, bx,\ 148 ANALYTIC GEOMETRY [Ch. XI, § 82 PROBLEMS 1. Find the equations of a pair of conjugate diameters of the hyperbola x 2 — 8 y 2 = 96, one of which bisects the chord whose equation is 3 x — 8 y = 10. 2. Find the equation of the diameter of the parabola y 2 = 6x, which bisects all chords parallel to the line x+3y = 8. 3. Find the equation of a diameter of the ellipse £x 2 + 9y 2 = 36, if one end of its conjugate diameter is (f V3, 1). 4. What is the equation of the chord of the ellipse 9 x 2 + 36 f = 324, which is bisected by the point (4, 2) ? 5. Find the equation of a chord of the ellipse 13ar 9 + ll?/ 2 = 113 through the point (1, 3), which is bisected by the diameter 2y = 3x. 6. A diameter of the ellipse 15 1/ 2 + 4 x 2 = 60 is drawn through the point (1, f). Find the equation of the conjugate diameter and its points of intersection with the ellipse. 7. Find the length of the diameter of the hyperbola 9 x 2 — 4 y 2 = 36, which is conjugate to the diameter y = 3 x. 8. What is the equation of the chord of the parabola y 2 = 6 x, which is bisected by the point (4, 3) ? 9. A diameter of the hyperbola 4 x 2 — 16 y 2 = 25 passes through the point (1, — 2). Find its extremities and the extremities of its conjugate diameter. 10. What is the relation between the slopes of the con- jugate diameters of the equilateral hyperbola xy = k? 82. Theorems concerning diameters. — 1. In the ellipse the sum of the squares of any tivo conjugate semi-diameters is equal to the sum of the squares of the semi-axes. Ch. XI, § 82] diameters 149 In Fig. 81, let CP t = a! and 0P 2 = b'. Then (by [1]) and .12 — >. 2 b z a 2 Fig. 81. Adding, a' 2 + b' 2 = (a 2 + 6 2 ) ^ + (a 2 + & 2 ) But since P x is on the ellipse, %+^r = l, and a' 2 + b' 2 = a 2 + b 2 . a z b z 2. In the hyperbola the difference of the squares of any two conjugate semi- diameters is equal to the difference of the squares of the semi-axes. 3. The product of the focal distances of any point on a central conic is equal to the square of the semi-diameter conjugate to the diameter through the point. 150 ANALYTIC GEOMETRY [Ch. XI, § 82 Since the focal distances of the point (x v y^) have been shown (Art. 70) to be a -f- ex 1 and a — ex v it is necessary to show that b' 2 = a 2 — e 2 x x 2 . From theorem 1 we have b 2 + a 2 ' But since P x is a point on the ellipse, W _ J2 -- "-I" tfy* = a 2 b 2 -b 2 x 2 , or ^- = a 2 b 2 x 2 Hence 5' 2 = a 2 — x? -\ ^-, 1 a 2 ai-f'sL^)^, or (by [39]), = a 2 — A-, 2 . 4. Prove the same theorem for the hyperbola. 5. TAe tangents at the ends of a diameter of a central conic are parallel to the conjugate diameter. In the ellipse the equation of the tangent at P 1 is a 2 ^ b 2 ~~ This is seen at once to be parallel to the conjugate diameter -^ + ¥j¥- = 0. In the same way the tangent at P 2 can be shown to be parallel to P V K V This theorem appears also from the fact that the tan- gents are special cases of the system of parallel chords. 6. The tangent at the end of a diameter of the parabola is parallel to the system of chords which the diameter bisects. 7. The area of the parallelogram formed by tangents at the ends of conjugate diameters of a central conic equals the area of the rectangle on the principal axes. Oh. XI, § 82] DIAMETERS 151 Let ABED be the parallelogram formed by the tangents at the ends of the conjugate diameters P X K X and P 2 K 2 . B^^ Y Ji sl — - — y^ \ ^\T^ \ T \ \ y^^^ A D z^~~K s Fig. 82. The sides of the parallelogram are evidently 2 a 1 and 2 V '. From (7 drop a perpendicular (7M" on AB. Its length is the distance from the origin to the line b 2 x x x + a 2 ^?/ = a 2 b 2 , or (by [17]) a% 2 CM= ■Vb% 2 + a 4 ^ 2 But the area of the parallelogram = 2 OMx AB=2 CMxP 2 K 2 , 2 a*b* -J 152 ANALYTIC GEOMETRY [Ch. XI, § Let the student give the proof for the hyperbola. r 8. In the hyperbola the parallelogram formed by the tangents at the ends of conjugate diameters has its vertices on the asymptotes. 9. In the hyperbola, the line joining the ends of conju- gate diameters is parallel to one asymptote and is bisected by the other. 10. Shoiv that the angle between two conjugate diameters i ab sin ']' a'b 11. Conjugate diameters of the rectangular hyperbola are equal. 12. The ellipse has a pair of equal conjugate diameters which coincide with the asymptotes of the hyperbola, which has the same axes as the ellipse. Ch. XI, § 82] DIAMETERS 153 PROBLEMS 1. Prove that conjugate diameters of an equilateral hyper- bola are equally inclined to the asymptotes. 2. Prove that any two perpendicular diameters of an equi- lateral hyperbola are equal. 3. Prove that the straight lines drawn from any point in an equilateral hyperbola to the extremities of any diameter are inclined at equal angles to the asymptotes. 4. Prove that the points on either the major or minor auxiliary circle, which correspond to the extremities of a pair of conjugate diameters, subtend a right angle at the centre of the ellipse. 5. Prove that chords drawn from any point of a central conic to the extremities of a diameter (called supplemental chords) are always parallel to a pair of conjugate diameters. 6. If Pi and P 2 are the extremities of a pair of conjugate diameters of a central conic, prove that the normals at P x and P 2 , and the perpendicular from the centre on P 2 P 2 meet in a point. 7. If a perpendicular is drawn from the focus of a central conic to a diameter, show that it meets the conjugate diameter on the corresponding directrix. 8. Tangents at the extremities of a pair of conjugate diam- eters of an ellipse form a parallelogram (Fig. 82). Show that the diagonals of this parallelogram are also a pair of conjugate diameters. CHAPTER XII POLES AND POLARS 83. Harmonic division. — In Art. 14 we have said that four points A, B, C, and D on a line form a harmonic ., AB AD A ,, , 4 b c p range, if — = - — , and that FlG - 84 - the pairs of points A and C, and B and D are then called conjugate harmonic points. From this definition it is easily seen that if we keep A and C fixed and allow B and D to move, as B approaches C, B will also approach C, and as B approaches the middle point of AC, B will recede indefinitely. When B coin- cides with the middle point of AC, it has no conjugate harmonic point. When B moves from the centre toward A, B comes in from the left toward A. It is desirable for our purposes to express the relation between these points in terms of distances from A only. From the definition, AB x CD = AD x BC. Substituting CD = AD -AC and BC= A C - AB, this becomes ABxAD-ABx AC= AD x AC - AD x AB, 2AB xAD or AC = AB + AD Let the student show that BD = -— — —- — . Connect BA + BC these results with harmonic progression in algebra by show- ing that A C is a harmonic mean between AB and AD, 154 Ch. XII, § 84] POLES AND POLARS 155 84. Polar of a point. — The polar of a point with respect to any conic is defined as the locus of points which divide harmonically secants through the fixed point. The methods of finding this locus are the same for all the conies. In problem 31, Chapter VIII, the student has been asked to find it for the circle by the aid of polar — P, Fig. 85. coordinates. The same method might be employed here, but it is thought best to use a very similar one, into which, however, polar coordinates do not enter. We shall find the polar of the point P x with respect to the ellipse ^ + a y = a ^ Transform to the point P x as origin by the aid of the equations x = x f + X, — />/' y + Vv 156 ANALYTIC GEOMETRY [Ch. XII, § 84 The equation of the ellipse becomes b 2 x 2 + a 2 y 2 + 2 b 2 x x x + 2 a 2 y x y + b 2 x 2 + a 2 ?^ 2 - a 2 6 2 = 0. Let any line, y — Ix, through P 1 cut the ellipse in the points P 2 and P 3 . We wish to find the locus of a point P' on this line, so situated that P v P 2 , P', and P 3 form a harmonic range. By the theorem of Art. 14, P v M 2 , M, and M z will also form an harmonic range, and hence If we start the solution of the equation of the line, y = lx, with the equation of the ellipse, we have (b 2 + a 2 / 2 ) x 2 + 2 \b 2 x x + aHyJ x + b 2 x 2 + a 2 y 2 - a 2 b 2 = 0, rom which t (see Art. 8) from which to determine the values of x 2 and x s . Hence _ b 2 x 2 + a 2 y, 2 - a 2 b 2 X ^- & + <£!* A , 2(b 2 x l + a 2 ly 1 ) and x 2 + x s = - V+« 2 Z 2 Hence ,^, ^4-^-^ o 1 x l + « t^i Since P' lies on the line y = Ix, its coordinates satisfy the equation, and y' = lx\ or l = ^-f Substituting this SB value of I in the equation above, and reducing, we have as the equation of the locus, referred to the point P x as origin, b 2 x x x + (Py^y 4- b 2 x? + a 2 y x 2 — a 2 b 2 = 0. Cii. XII, § 85] POLES AND POLARS 157 When transformed back to the original origin by the aid of the equations x=x' — x x and y = y' — y v this equation becomes b 2 xix + oryx y= a 2 b 2 , [65] Since this equation is of the first degree, we have arrived at the singular result that the locus is a straight line. It is called the polar of the point P l with respect to the ellipse, while the point P x is called the pole of the line. The theory of poles and polars is of great impor- tance in some branches of geometry. Let the student show that the polar of the point (x v y^) with respect to (a) the circle, x 2 -f y 2 = r 2 , is x x x + y x y = r 2 , [Q6] (b) the hyperbola, b 2 x> - a 2 y 2 = a 2 b 2 , is b 2 x x x - a 2 y x y = a 2 b 2 , [67] (e) the parabola, y 2 = 2 mx, is y x y = mx + mx\, [68] (c?) the locus of the general equation, Ax 2 + Bxy + Cy 2 + Dx + Uy + F=0, is Axxx + — (y x x + x x y) + Cy x y + ¥>(x + x l ) + ^(y + y l )+F [69] 85. Position of the polar. — From what we have said about harmonic ranges, it is evident that the polar of every point outside the conic cuts the conic, and that the polar of every point within the conic does not cut it. The form of the equation shows that, when the point is outside the conic, the polar coincides with the chord of contact 158 ANALYTIC GEOMETRY [Ch. XII, § 85 and, when the point is on the conic, the polar becomes the tangent. Again, as P x recedes from the conic, its con- jugate harmonic point P' approaches the middle of the chord, and its polar, therefore, approaches coincidence with a diameter. If the point P l is inside a central conic and approaches the centre, the polar evidently recedes indefinitely. PROBLEMS 1. What is the equation of the polar of (a) (1, — 2) with respect to the conic of' + 4 y 2 = 16 ? (6) (6, — 4) with respect to the conic y 2 = 4 # ? (c) (— 3, 2) with respect to the conic 5 x 2 — 8 y 2 = 24 ? (d) (0, 0) with respect to the conic x 2 +2xy + 3y 2 -4 : x-l0 = 0? 2 . What is the pole of the line 3 x — 2 y = 5 with respect to the circle x 2 + y 2 = 25 ? Solution. — The polar of the point (xi, y{) with respect to the circle is X\X + y\y = 25. We wish to find the values of Xi and y x which will make this line coincident with the line 3 x — 2 y = 5, or 15 x — 10 y = 25. They are evidently xi = 15 and y± = — 10. 3. What is the pole of the line 5 x + ky = 7 with respect to the ellipse x 2 + 2y 2 =10? 4. What is the pole of the line x — y = 10 with respect to the parabola y 2 = Sx? 5. What is the pole of the line Ax+By+ C=0 with respect to the hyperbola b 2 x 2 — a 2 y 2 — a 2 b 2 ? 6. Tangents are drawn to the circle y 2 = 10x — x 2 at the points where it is cut by the line y = 4 x — 7. What is their point of intersection ? 7. Through the point (# 1? y{) a line is drawn parallel to the polar of the point with respect to the ellipse b 2 x 2 + a 2 y 2 = a 2 b' 2 . Find the coordinates of the pole of this parallel. Ch. XII, § 86] POLES AND TOLA US 159 86. Theorems concerning poles and polars. — 1. If a set of points lie on a line, their polars all p> a8S through the pole of that line ; and conversely, if a set of lines pass through a point, their poles lie on the polar of that point. Let P 2 be the pole of the line MN, and P 1 any point on MN We wish to show that MS, the polar of P v will pass through P 2 . If the coordinates of P x and P 2 are (x v y x ) and (x 2 , y 2 ), the equation of PS is b 2 x x x + a 2 y x y = a 2 b 2 , and of MN is b 2 x 2 x + a 2 y 2 y = a 2 b 2 . But we know that the coordinates of P 1 must satisfy the equation of MN, or b 2 x 2 x x + a%y 1 = a 2 b 2 . Now this is just the condition which must be satisfied, if P 2 lies on MS. Hence P 2 lies on MS, and as the point P x moves along the line MN, its polar will revolve about P 2 , the pole of MN Let the student prove the converse theorem, and also both theorems for the hyperbola and parabola. It follows from this theorem that tangents at the extremities of any chord through P x meet on MS. For the pole of every chord through P x lies on MS, and we have seen (Art. 85) that tangents at the extremities of a chord intersect at the pole of that chord. From this property the polar may be defined as the locus of the 160 ANALYTIC GEOMETRY [Ch. XII, § 86 intersection of tangents at the extremities of chords through any fixed point. This property enables us to construct the polar of any point ; for any number of points on the polar may be Fig. 87. determined by finding the intersections of tangents at the extremities of chords through the point. 2. The polar of any point P 1 with respect to a central conic is parallel to the tangent at the point where the diameter through P 1 cuts the conic. Y Fig. 88. Ch. XII, § 86] POLES AND POLARS 161 Let the coordinates of the point P 2 where CP 1 cuts the hyperbola be (x v y a ). Then the equation of the tangent 2 k b 2 x 2 x — a 2 y 2 y = a 2 b 2 , and the equation of the polar of P x is b 2 x x x — a 2 y x y = a 2 b 2 . But since P x and P 2 are on the same line through the x x origin, — = — -, and these lines are evidently parallel. Let the student prove the same theorem for the ellipse. 3. The polar of any point P x with respect to a parabola is parallel to the tangent at the point where a diameter through P 1 cuts the parabola. Y Fig. 89. We may let the coordinates of P 2 be (x v y-[)> Then the equation of the tangent at P 2 is y x y = mx + mx v and the equation of the polar of P 1 is y x y = mx + mx v These equations are seen at once to represent parallel lines. 162 ANALYTIC GEOMETRY [Ch. XII, § 86 These two theorems show that the polar of a point on a diameter is one of the system of parallel chords bisected by that diameter. 4. If the line joining the centre C of any central conic to any point P x cuts the conic in P 2 and the polar of P 1 in P 3 , then CP 1 x CP 3 = OP*. We shall give the proof for the hyperbola, using Fig. 88. The equation of QP X is y = — x. The coordinates of P 2 , where this line cuts the hyperbola are found to be ahx l ~~A <%1 and -\Jb 2 x^ — a 2 y-f ^/b 2 x x 2 — a 2 y^ and the coordinates of P 3 , where it cuts the polar, b 2 x x x — a 2 y x y — a 2 b 2 , a 2 b 2 x x , a 2 b 2 y 1 b 2 x x 2 — a 2 y x 2 b 2 x x 2 — a 2 y x 2 Hence CP X = ^x 2 + y 2 , ^ 2 ~ V b 2 x 2 -a 2 y 2 _ g%^ X * + y* 3 " b 2 x 2 - a 2 y 2 ' From these values we see at once that CP X x CP 3 = OP 2 . Let the student prove the same theorem for the ellipse. Show that in the parabola (Fig. 89) P 2 bisects the line P X P V Ch. XII, § 80] POLES AND POLARS 163 5. The line which joins any point to the centre of a circle is perpendicular to the polar of the point with respect to the circle. The proof of this theorem appears at once as soon as the equations of the lines are written. This theorem enables us to state theorem 4 for the circle as follows : 6. The radius of a circle is a mean proportional between the distance from the centre to any point and the distance from the centre to the polar of that point. 7. The polar of the focus is the directrix in (a) the ellipse, (b) the hyperbola, (c) the parabola. The proof of this theorem appears at once in each case when the coordinates of the focus are substituted in the equation of the polar. This theorem is evidently equiva- lent to theorems 6 and 7 on tangents. 8. Any chord through the focus of a conic is perpendicular to the line joining its pole ivith the focus. This theorem is equivalent to theorems 11 and 12 on tangents and is proved in the same manner. 9. The line joining the centre of a central conic to any point P 1 cuts the directrix in K. Show that the line KF is perpendicular to the polar of P v 10. Two triangles are so related that the vertices of the first are the poles of the sides of the second, with respect to a conic. Prove that the vertices of the second are also poles of the sides of the first. Two such triangles are said to be conjugate to each other. If in any triangle the vertices are the poles of the opposite 104 ANALYTIC GEOMETRY [Oh. XII, § 86 sides, the triangle and its conjugate coincide, and it is called a self-conjugate triangle. 11. If a line is drawn through a point parallel to the axis of a parabola, that portion of it included between the point and its polar is bisected by the parabola. How does this conform to the definition of harmonic division ? 12. The two lines, which join the focus of a conic to any point ctnd to the intersection of the polar of that point ivith the corresponding directrix, are perpendicular to each other. 13. Write the equation of the polar of a point P 1 on a diameter of a central conic. Let P x recede indefinitely along the diameter and shoiv that the polar approaches, as its limiting position, the diameter conjugate to the given diameter. Show that this would be true, if the point receded along any line parallel to the given diameter. How must this theorem be stated for the parabola ? PROBLEMS 1. Show that the polars of the same point, with respect to two conjugate hyperbolas, are parallel. 2. Show that the four points, in which any line is cut by the asymptotes of an hyperbola and by a pair of conjugate diameters, form a harmonic range. 3. What is the polar of the focus of an ellipse, with respect to the major auxiliary circle ? 4. Obtain the equation of the polar of the point P x with respect to the rectangular hyperbola xy = k. What are the coordinates of the foci and the equations of the directrices of this hyperbola? Prove that your results are correct by showing that the directrix is the polar of the focus. Ch. XII, § 86] POLES AND POLARS 165 5. Show that the polar of one extremity of a diameter of an hyperbola, with respect to its conjugate hyperbola, is the tangent at the other extremity of the given diameter. 6. If a perpendicular is let fall from any point P x upon its polar, prove that the distance of the foot of this per- pendicular from the focus is equal to the distance of the point P x from the directrix of the parabola. 7. An ellipse and an hyperbola have the same transverse and conjugate axes. Show that the polar of any point on either curve, with respect to the other, is tangent to the first curve. CHAPTER XIII GENERAL EQUATION OF THE SECOND DEGREE 87. We have seen that the equations of all the conies are of the second degree. We shall now prove that an equation of the second degree must always represent a conic, either in one of the ordinary forms or in one of the limiting cases, and show how to reduce any given equation to the simplest equation of one of these conic sections. 88. Two straight lines. — We have seen that there are certain equations of the second degree which can be factored, and hence represent two straight lines. Let us determine what condition must be satisfied by the coefficients of the general equation, (1) Ax 2 + Bxy + Cy 2 + Bx + Ey + F= 0, when it can be separated into two linear factors. Arrang- ing it according to the powers of x and solving, we have (2) x = - (By + i>) ± ^/(B 2 - 4 AO)y 2 + ( 2 BB - ±AE)y + B 2 - 4 AF. 2A If the general equation is to be factored into two linear factors, the quantity under the radical must be a perfect square. The condition for this is (3) (2BB-4AB) 2 - ±(B 2 - ± AC)(B 2 - 4 AF) = 0, or (4) 4ACF+BBE-AE 2 -CB 2 -FB 2 = Q. 166 Ch. XIII, § 88] EQUATION OF THE SECOND DEGREE 167 This is, then, the condition which must be satisfied by the coefficients of the general equation, when it can be separated into two linear factors. The first member is called the discriminant of the equation. It is usually represented by the letter A. Note. — If A = 0, the work will have to be changed somewhat, but the same form will always be obtained for the discriminant. When this condition is satisfied, the equation can always be factored, but it is not necessary that the factors should be real. For if B 2 - 4 A C is negative, from (3), B 2 - 4 AF must also be negative, and while the expression under the radical is a perfect square, its square root will contain imaginary coefficients. The equation will in this case break up into a pair of factors with imaginary coefficients, and we speak of it as representing a pair of imaginary lines. There will be, however, one real point on the locus ; for the intersection of the two imaginary lines will always be a real point. If B 2 — 4 AC is positive, the factors represent real and intersecting lines. If B 2 -±AC=0, 2 BD- ±AE must also reduce to zero, and the quantity under the radical is reduced to D 2 — 4 AF. The lines are therefore parallel. They are real and distinct if D 2 - 4 AF> 0, real and coincident if D 2 — 4 AF = 0, imaginary if D 2 — 4 AF < 0. PROBLEMS 1. Obtain the discriminant by the following method: Let the two factors be x -f b x y -f- y 2 + F = 0. 91. Determination of the coefficients A', C, and F' . — We have shown how to determine F' ; and since tan 2 6 is known, the values of A' and C may be found, and the result fully determined. But much of the labor involved may, in practice, be avoided by the following method : Adding equations (13) and (15), we have (18) A' + C = A + a Subtracting the same equations, we have (19) A'-C'=(A- 6 7 ) cos 2 (9-f-^ sin 2 (9. Squaring (19) and (14), and adding, we have (20) (A! - C r ) 2 + B 12 = (A- O) 2 + B 2 . Squaring (18) and subtracting from (20), we have (21) B' 2 -4A r C' = B 2 -4,AO. These results hold good for all transformations from one system of rectangular axes to another. Ch. XIII, § 91] EQUATION OF THE SECOND DEGREE 171 If the general equation has been reduced to equation (IT), B' = 0, and (21) reduces to (22) 4A'C f =4tAC-B*. From the two equations (18) and (22), A' and C can be found. But there will be two values of each, corre- sponding to the two possible values of 0, and it will be necessary to be able to choose the proper values. We have let (C- A) sin 2 6 + £cos 2 6 = 0. Multiplying this equation by (^4. — (7) and equation (19) by B and subtracting, we have (23) (B 2 + (A- C) 2 ) sin 2 = B(A' - C). If now the acute value of 6 be chosen, the first member will always be positive, and the factors of the second member, B and A' — C, must have the same sign. It will be easy then to choose the proper values for A' and C '. The determination of F' may also be considerably sim- plified. Multiply equation (9) by x and (10) by y and add. The sum is 2 Ax* + 2 Bx y + 2 Cy* + Dx + Ey n = 0. Combining this with (8), we have ■*" = f*o + fs'o + -* r - Substituting the values of x and y , CD 2 + AE 2 - BDE+ B*F- 4 A OF = -A B 2 -±AC tf-lAQ (24) F' = 172 ANALYTIC GEOMETRY [Ch. XIII, § 92 92. Nature of the locus. — The general equation has now been reduced by transformation of coordinates to the form (IT) A'x 2 + C'y 2 + F' = 0. Neither of the coefficients A ! or C can be zero, for they must satisfy equation (22), and we are only con- sidering the case where B 2 — -±AC^0. If F' =£ 0, (17) can be written in the form -F' "*" -F'~ ~aJ~ a The nature of the curve evidently depends on the rela- tive signs of A', C' ', and F'. If A' and C have the same sign and F' has the opposite sign, equation (17) will represent a real ellipse; for it can be written in the form ^ + £=1 a 2 b 2 If A', C 1 ', and F' all have the same sign, equation (17) can be written in the form ^ + ^ = - 1 a 2 ^b 2 This equation has no real locus, but is said to represent an imaginary ellipse. Again, if A 1 and C have opposite signs, the equations will take one of the two forms ^_^ 2 -1 or ^-t- \ a 2 b 2 ~ ' a 2 6 2 ~ ' Ch. XIII, §92] EQUATION ()K THE SECOND DEGREE 173 according 1 as F' has the same sign as C or as A'. These are both real hyperbolas. From equation (22), 4 A' C = 4 A C — B 2 , we see that A' and C have the same or opposite signs according as B 2 — 4: AC is negative or positive. If then F' is not zero, or what is the same tiling, if the discriminant does not vanish, and if B 2 — \ AC ^ 0, the general equation has been shown to represent an ellipse, real or imaginary, when B 2 — 4 AC < 0, an hyperbola, always real, when B 2 — 4 AC > 0. If F' = 0, equation (17) reduces to (25) A'x 2 + C'y 2 = 0. When A' and C have the same sign, the equation may be looked upon as representing a pair of imaginary lines, since the equation can be separated into a pair of linear factors with imaginary coefficients. These lines have a real point of intersection, the origin. Or it may be looked upon as the equation of an ellipse in which the axes have become zero. From this point of view, it is spoken of as representing a null ellipse. • When A! and C have opposite signs, the equation can always be separated into two real factors, representing two real and intersecting lines. These results will be seen to agree with those obtained in Art. 88. PROBLEMS 1. Determine the character of the locus of the following equation, reduce it to its simplest form, and plot : 5x 2 + 2xy + 5y 2 -12x-l2y = 0. 174 ANALYTIC GEOMETRY [Ch. XIII, § 92 Substituting these values for the coefficients in (4), we obtain A = -1152. Also B 2 -4AC = -96. The locus is therefore an ellipse, real or imaginary. The simplest method for determining the coordinates of the centre is to write the equations (9) and (10) and solve for x and y . The first of these may be obtained by multiplying the coeffi- cient of every term which contains x by the exponent of x, decreasing that exponent by unity, and leaving out all terms which do not contain- x. The second may be formed in a similar way, using y. In this case they are 10 x + 2y - 12 = 0, and 2 x + 10y Q - 12 = 0. From these the coordinates of the centre are found to be (1, 1). From equation (24), F' = — 12. The equation, referred to the point (1, 1) as origin, is then Sx 2 + 2xy + 5y 2 -12 = 0. Next revolve the axes through an angle 6, such that tan 2, = -^ = ^. We have decided to use the acute value of 6, which is here -• 4 To determine A! and C, we use the equations (18) and (22) or A r +C' = A + C = 10, and 4 A'C = 4 AC - B 2 = 96. Solving, we have A' = 6 or 4, and C = 4 or G. But since we chose the acute value of 6, we must choose A' and C so that A' — C has the same sign as B. This is positive. Hence the final form of the equation is 6x 2 +4y 2 = 12. Ch. XIII, § 03] EQUATION OF THE SECOND DEGREE 175 But this is the equation of the curve referred to axes with origin at the point (1, 1), and making an angle of - with the original axes. 4 Constructing these axes and plotting the equation Gar 2 + 4/ = 12 with respect to them, we have the locus of the origi- nal equation, referred to the original axes. 2. Determine the char- FlG - 90 - acter of the loci of the following equations, reduce them to their simplest forms, and plot : (a) 2a 2 + 2f- 4^-47/4-1 = 0, (b) x* + tf + 2x+2 = 0, ( c ) ±xy- 2x + 2=0, (d) y 2 -5xy + 6x 2 -Ux + 5y + 4: = 0. Case II. B 2 -AAC = 0. 93. Removal of the term in xy. — We have seen that, if B 2 — 4 A = 0, it is not possible to transform to a new origin such that the terms in x and y shall disappear. In this case we shall first revolve the axes through an angle 0. Proceeding as in Art. 90, we obtain the equation A'x 2 4- B'xy 4- O'f 4- D'x + E'y + F=0, where (13) A f = A cos 2 6 4- B sin 6 cos 6 4- tfsin 2 0, (14) B' = (C - A) sin20 + Bcos20, (15) C = Asin 2 -Bsin0cos0 + Ccos 2 0, (26) B' = Bcos0 + Esin0, (27) E' = - 2) sin + E cos 0, 176 ANALYTIC GEOMETRY [Ch. XIII, § 93 The values of A', B\ and C are the. same as those used in Art. 91. The results there obtained will therefore apply here. These were, (18) A' + C = A+ C, and (21) B't-lA'C'^B't-'iAa Let 6 be so chosen that B' = 0, or tan 20 = — -• Then A — C since B 2 — \AQ = 0, it appears from (21) that either A! or C must reduce to zero at the same time. It can easily be shown that one of the two values of 6 will give A' = 0, and the other, C = 0. Let that value be chosen which will make A' = A cos 2 6 + B sin 6 cos + C sin 2 = 0. Solving, we have A + B tan 6 + C tan 2 (9 = 0, The general equation will be reduced by this transforma- tion to (29) O'f + D'x + B'y + F=Q, where (30) C = A + C, (3n^= B D-2AH % ±V^ 2 + 4A 2 and (32)^= ^B + 2AD ± V£ 2 + 4 A 2 It appears then that C cannot vanish, since A and Q have the same sign ; that D' or E' may vanish, but since BD — 2 A E is the value of the discriminant when B 2 — 4 J. (7= 0, I)' cannot be zero unless A = 0. Ch. XIII, § 95] EQUATION OF THE SECOND DEGREE 177 94. Removal of the term in /. — Transform equation (29) to a new origin (:r , ?/ ). It becomes (33) O'f + D'x + E"y + F' = 0, where (34) E" = 2 O'y + E\ and (35) F> = C'y* + D'x Q + E'y + F. We can then, in general, choose such values for x and y Q that E" = F' = 0. Solving the two equations and C'y* + D% + E'i, + F=0, , E' , J?' 2 - 4 6 7 '^ we have y = - — , and x = ^ QJJ)] If D' =£ 0, these values are always finite, and the trans- formation is possible. The equation will be reduced by it to (36) C'f + D'x = 0. If D' = 0, no value can be obtained for x which will make F' = 0. But if we transform to the point (0, ?/ ), the equation will be reduced to (37) C'y 2 + F' = 0. 95. Nature of the locus. — When B 2 -4AO=0, the general equation has been reduced by transformation of coordinates to one of the two forms (36) Cy + 3'x=0,OTtf=--%;X % D[ C (37) C'f + F f = 0,oTf = --^ 178 ANALYTIC GEOMETRY [Ch. XIII, § 90 The first of these equations always represents a real parabola. The second is obtained only when A = 0, and represents, as we should expect, a pair of lines. In this case the lines are evidently parallel and real and distinct, if C and F ! have opposite signs, real and coincident, if F' = 0, imaginary, if (7' and F' have the same sign. It has been shown that (30) C' = A + C, and (31) D>=-Z V- 2A E , and when A = 0, it can be shown that 4 AF - D 2 (38) F' = ±A From these values the reduced form of the equation can be determined. But in any numerical problem the method of the following section will be found to be simpler. PROBLEM 1. Show that the above conditions which determine the nature of the parallel lines are the same as those given at the end of Art. 88. 96. Second method of reducing the general equation to a simple form, when B 2 - 4 AC = 0. — When B 1 - 4 AO= 0, the terms of the second degree in the general equation Ch. XIII, §9C] EQUATION OF THE SECOND DEGREE 179 form a perfect square, and the equation can be written in the form (ax + cy) 2 + Dx + Ey + F = 0, where a = V^L and e = V(7. Introduce arbitrarily the quantity k inside the paren- thesis, and subtract from the rest of the equation whatever has been added by this introduction. It becomes (ax + cy + 7c) 2 + (D - 2 ak)x + (E-2 ck)y + F-k 2 = 0. Now choose such a value for k that the two lines repre- sented by the equations ax -f- cy + k = and (D- 2 a¥)x + (E- 2 ck)y + F-k 2 = shall be perpendicular to each other. Let I be such a value of k. The equation will then take the form (ax + cy + 2 = D'x + E'y + F f . Divide both members of this equation by a 2 + c 2 , and both divide and multiply the second member by VD /2 + E' 2 , and write the result in the form f ax + cy + W = Vi>' 2 + E' 2 ( D'x + E'y + F' \ \ ^a^+7 2 J <# + = and ax + cy -f I = 0. It is therefore the equation of the curve referred to these lines as Y and Jf-axes respectively. The positive direction of the X-axis can be fixed by finding the inter- cepts of the curve on the original axes, and determining by inspection which way the parabola is turned. PROBLEMS 1. Plot the locus of the equation tf _ 2 xy + y 2 - 8 x + 16 = 0. Following the method described above, write the equation in the form (x — y + k) 2 -(8 -f- 2 k)x + 2 ky + 16 - k 2 = 0. If the two lines represented by the equations x — y + k = and _(8 + 2 ft) a; -f 2 % + 16 - A: 2 = Ch. XIII, §96] EQUATION OF THE SECOND DEGREE 181 are perpendicular to each other, k = — 2. Substituting this value in the equation and transposing, it becomes (x-y-2y = 4:(x + y-3). Dividing both members by a 2 + c 2 , and both dividing and multiplying the second member by V ' D n -f E' 2 , it becomes (^)'= 2v2 (^) V2 { X + y -° \ or y' 2 =2V2x', where y' is the perpendicular distance of any point (#, y) of the locus from the line x — y — 2 = 0, and where x' is the distance from x-\-y—S=0. It is therefore the equation of the locus referred to these lines as X and Y- axes. Construct the two lines. From the original equation we see that the curve touches the X-axis at the point (4, 0), and does not cut the F-axis. It is then easily seen which FlG - 91 - is the positive direction of the axis O'X', and the curve can be plotted as in Fig. 91. 2. Plot by this method the locus of the following equations: (a) x 2 - 2 xy + y 2 - 6 x - 6 y + 9 = 0, (6) x 2 + 6 xy -f- 9 y 2 + x - 6 y - 9 = 0, (c) 2x 2 + $y 2 + 8xy + x + y + 3 = 0, (d) f - 2 x - 8 y + 10 = 0, (e) 4:X 2 + 4xy + y 2 + 6 = 0. 183 ANALYTIC GEOMETRY [Ch. XIII, § 97 when A = 0, • 97. Summary. — It has been shown in this chapter that the general equation of the second degree represents, ' and when B 2 — 4 AC < 0, an ellipse (real or imaginary), when A * 0, j ftnd when & _ 4 A Q = ^ ft parabola? and when B 1 — 4 A C > 0, an hyperbola ; and when B 2 — 4 AC < 0, a null ellipse (two imaginary lines), and when B 2 — 4 A C = 0, two parallel lines (real, coincident, or imaginary), and when B 2 — 4 A C > 0, two real intersect- ing lines. All of these forms may be obtained as plane sections of a right circular cone, and are all included under the term "conies." An equation of the second degree must therefore represent some conic either in its regular or degenerate form. PROBLEMS Determine the nature of the locus of each of the following equations : 1. 3x?-2xy + y 2 + 2x + 2y + 5 = 0. 2. x 2 + xy + y 2 + 2 x + 3y -3 = 0. 3. 2X 2 - 5xy- 3/4-9 x- 13 y + 10 = 0. 4. 4z 2 + 2xy-\f + 6x + 2y + 3 = 0. 5. 9x 2 - 12xy + ±tf- 24 a; + 16 y- 9 = 0. 6. 9ar J -6a-?/ + ?/ 2 + 4a; + 3?/ + 16 = 0. 7. 25 x 2 + 40 xy + 16 y 2 + 70 x + 56 y + 49 = 0. 8. 13 a; 2 + 14 ajy + 5 y 2 + 14 x + 10 y + 5 = 0. 9. 4 x 2 + 9 t/ 2 - 8 a? + 54 y + 85 = 0. 10. 3a: 2 + 10a^ + 7?/ 2 + 4x-f-2?/ + l = 0. Cm. XIII, §99] EQUATION OF THE SECOND DEGREE 183 98. General equation in oblique coordinates. — If the general equation is referred to axes which are oblique, we can first transform to rectangular axes with the same origin. The resulting equation will be in the form A'x 2 + B'xy + Cif + D'x + E'y + F' = 0. This can then be treated by the methods of this chapter. It must, therefore, represent a conic. 99. Conic through five points. — The general equation of the second degree contains six constants, but only five of these are independent, since any one we please may be reduced to unity by division. Five conditions are therefore sufficient to determine the conic. For example, it can be made to pass through five points, and in general no more than five. For, if the coordinates of the five points are substituted in turn in the general equation, there will be five equations from which, in general, we can determine five coefficients in terms of the sixth, which will divide out after substitution. If a sixth point were given, there would be six simultaneous equations in five variables, which is not possible unless some of the equations are not independent. This will only happen when the sixth point lies on the conic through the other five. If three of the points lie on a line, the conic evidently breaks up into this line and another line through the other two. If four points lie on a line, the solution is indeter- minate ; for this line and any other through the fifth point will be a conic through the five given points. Other conditions may be given, as in the case of the circle, where A = C and B= 0. These two conditions 184 ANALYTIC GEOMETRY [Ch. XIII, §99 restrict the number of points through which a circle can be passed to three. Similarly, a parabola can be passed through only four points, since the condition B 1 — 4 AC=Q must be satisfied. But here, since the condition is a quad- ratic, there may be two parabolas which pass through the four points ; or imaginary solutions may be obtained, and four points may therefore be chosen through which no real parabola can be drawn. PROBLEMS 1. Find the equation of a conic through the points (a) (2, 3), (0, - 3), (2, 0), (5, 5), (- 5, - 5). (6) (5, 3), (4, 4), (2, 6), (7, 1), (0, 0). (o) (2, 4), (4, 3), (6, 2), (0, - 1), (1, 0). 2. Find the equation of a parabola through the points (a) (0,0), (8,8), (4,2), (-4,2). (b) (0, 0), (1, 0), (- 1, 1), (- 1, - 1). (c) (4,3), (0,-4), (6,1), (-6,2). (d) (12,-6), (3,0), (0,2), (-3,4). 3. Determine the nature of the conies obtained in problems 1 and 2. CHAPTER XIV PROBLEMS IN LOCI 1. Find the locus of the vertex of a right angle whose 1. or ij' sides are tangent to the ellipse — + *- <(- The equations of any two perpendicular tangents P' K and P'L may be written in the form y Ip + 'y/lfcP + lPi and t/ = ljc + ^l 2 *a 2 + P, where l 1 l 2 = — 1. If JP f is their point of intersection, its coordinates (V, y') must satisfy both equations. Substituting these co- ordinates, and replacing l 2 by , we have y = ^+VZ 1 2 a 2 -h^ 2 , Fig. 92. two equations in x\ y\ and the variable param- eter l v By eliminating l v we shall obtain a single equa- tion in x r and y f . Clearing of fractions, transposing, and squaring, 185 186 ANALYTIC GEOMETRY [Ch. XIV y' 2 - 2 l x x'y' + l 2 x' 2 = l 2 a 2 + b 2 l 2 y' 2 + 2 Z^y 4- x' 2 = a 2 + ^ 2 6 2 Adding, (l + Z 1 2 )y 2 +(l + ? 1 2 )^ 2 = (l + ? 1 2 )a 2 + (l + Z 1 2 )6 2 . Dividing by (1 + I 2 ), y' 2 + a:' 2 = a 2 + 6 2 , or x 2 -+- ?/ 2 = a 2 + 6 2 . The locus is the director circle, a circle having the same centre and Va 2 + b 2 as radius. 2. Find the locus of the intersection of perpendicular tangents to a parabola. 3. Find the locus of the intersection of tangents to the ellipse if the product of their slopes is constant. As in problem 1, the equations connecting x\ y\ l v and Z 2 are (i) / = z 1 ^+VZ 1 2 ^ 2 + P, (2) y' = l 2 x'+Vl 2 2 a 2 + b 2 , (3) 1,1, = k. But the method of elimination used in that problem will not apply here. Transpose and square (1) and (2), y' 2 - 2 l^'y* + l 2 x' 2 = l 2 a 2 + b 2 , y' 2 - 2 l 2 x'y f + ? 2 V 2 = J 2 a 2 + b 2 . Write these as affected quadratic equations in l x and Z 2 , (4) (a 2 - x f2 ) I 2 + 2 x'y\ + b 2 - y' 2 = 0, (5) (a 2 - x' 2 ) I 2 + 2 x'y'l 2 + b 2 -y' 2 = Q If now we write the equation (6) (a 2 - x' 2 ) z 2 + 2 x'y'z + b 2 -y' 2 =0 Ch. XIV] PROBLEMS IN LOCI 187 (an affected quadratic in 2), it appears from (4) and (5) that l x and l 2 are the two roots of (6), and hence that b 2 — y' 2 b 2 — y' 2 l ih = n2 U ' But l lh = k- Hence ^-^T 2 = k is the equation of the desired locus. Dropping primes and reducing, we have kx 2 — y 2 = ka 2 — b 2 . If k = — 1, it becomes x 2 -h y 2 = a 2 + b 2 , as in problem 1. 4. Find the locus of the intersection of tangents to the parabola if the product of their slopes is constant. 5. Find the locus of the feet of perpendiculars from a focus on tangents in the (a) ellipse, (5) hyperbola, (c) parabola. 6. Find the locus of the intersection of tangents at the ends of conjugate diameters of an ellipse. Note. — Solve this as a special case of problem 3. 7. Find the locus of the intersection of tangents at the ends of conjugate diameters of an hyperbola. 8. Radii vectores are drawn at right angles from the centre of an ellipse. Find the locus of the intersection of tangents at their extremities. 9. Find the locus of the middle point of chords joining the ends of conjugate diameters of an ellipse. Let (V, y) be the middle point of any such chord. If (rr^j) are the coordinates of P v ( ~~, ? -— 1 ) will be the coordinates of P 2 . ay. bx* x *~b y ^^r Then x' = ^ » and V* = 2 ' or (1) 2 bx' = bx x — ay v and (2) 2 ay' = ay x + bx v 188 ANALYTIC GEOMETRY [Ch. XIV Since (x x y^) lies on the ellipse, (3) b 2 x 2 +a 2 y 2 = a 2 b 2 . From these three equations we can obtain a single equation in x' and y' by eliminating x 1 and y v From (1) and (2), „. _ hx' + ay'_ *-* b ' -bx' a Fig. 93. Substituting these val- ues in (3), it reduces to a? ,y^_ 1 'a 2 b 2 ~2 10. Find the locus of the vertex of a triangle whose base is a line joining the foci and whose sides are parallel to two conjugate diameters. 11. Find the locus of the middle point of chords drawn through a fixed point in the (a) ellipse, (£>) parabola. 12. Tangents are drawn to the parabola y 2 =2mx. Find the locus of their pole with respect to the circle x 2 + y 2 = r 2 . 13. The two circles x 2 + y 2 = a 2 and x 2 + y 2 — ax = are tangent internally. Find the locus of the centres of circles which are tangent to both the given circles. Ch. XIV] PROBLEMS IN LOCI 189 Let the two circles be drawn, and let (V, y') be the centre of any circle which is tangent to both circles. Then the lines O'P' must pass through B, the point of contact of the two circles, and OP' must pass through 0. Hence, P' 0=00 -OP' = r- OP', and P'B = P'0'-BO' = P'0' --> But P' C and P'B are radii of the same circle. Hence, r-OP' =P'0' Fig. 94. Or f-v. x' 2 4- y' 2 4 + y '2 Squaring and reducing, the equation of the locus re- duces to 8 x 2 + 9 y 2 — 4 rx — 4 r 2 = 0. What curve is this and how is it situated? 14. Find the locus of the centres of all circles which pass through the point (0, 3) and are tangent internally to x 2 + y 2 = 25. 15. Find the locus of the centres of circles which are tangent to a given circle and pass through a fixed point outside of that circle. 16. Lines are drawn from the point (1, 1) to the hyperbola x 2 — y 2 = 1. Find the locus of the points which divide these lines in the ratio of 2 to 1. 190 ANALYTIC GEOMETRY [Ch. XIV 17. Lines are drawn from the centre of the circle x 2 + y 2 = r 2 , cutting the circle in A and the line, x = a, in B. Find the locus of P, if 0, A, B, and P form a harmonic range. Show that the result will represent an ellipse, hyperbola, or parabola, according as 4 r 2 < a 2 , 4 r 2 > a 2 , 4 r 2 = a 2 . 18. Find the locus of the vertex of a triangle if the length of the base is c, and the product of the tangents of the base angles is k. Let P r be any position of the vertex of the triangle, and 00 the base. We know that tan OOP' • tanP'C0 = &. But tan OOP = ^, x' and tan P' 6 7 = -^— -• c — x Hence the condition which must be satisfied is J2 Fig. 95. tf = Jc. x\c - x f ) Dropping primes and reducing, we have Jcx 2 + y 2 — hex = 0. This will be an ellipse or hyperbola, according as h is positive or negative. In either case the coordinates of the centre will be & »)• and the semi-axes will be - and 19. Find the locus of the vertex of a triangle if the length of the base is c, and the product of the tangents Ch. XIV] PROBLEMS IN LOCI 191 of the half base angles is k. Show that the locus is an ellipse with the extremities of the base as foci. Note. — Express the tangents of half the base angles in terms of the ft)(* ~ O three sides by the aid of the formula tan \ y s(s — a) 20. Find the locus of the vertex of a triangle if the length of the base is c, and one of the base angles is twice the other. 21. Find the locus of the intersection of tangents to the (a) parabola, (5) ellipse, ( 2 Zy 2 4-2Zrry+ z' 2 = a^ + V^ 2 Adding, (1 + P)y' 2 + (1 + Z 2 >' 2 = a x 2 + a 2 Z 2 + b 2 + V? 2 Ch. XIV] PROBLEMS IN LOCI 193 J kit from (3), af + h 2 = « 2 + b 2 . Substituting and factoring, we have (1 + l 2 )ij' 2 + (1 + P)x' 2 = a\l + I 2 ) + b 2 (\ + Z 2 ), ij 2 + s'2 = a 2 + ^ 2 , or, dropping the primes, we have for the equation of the locus x 2 + y 2 = a 2 + b 2 . 30. Find the locus of the intersection of two perpen- dicular lines which are tangent respectively to two con- focal parabolas. Note. — Write the equation of the parabola referred to the focus as origin, y 1 — 2 mx + w 2 , and obtain the equation of the tangent to it in terms of the slope, y = Ix + m ^ + l ^ > 31. Find the locus of the points of contact of tangents drawn from a fixed point on the principal axis to a set of confocal ellipses. 32. Find the locus of the middle points of chords in a circle, which are tangent to an internal concentric ellipse. Let (V, y') be the middle point of the chord, and (x v 2/j) and (# 2 , 3/ 2 ) its extremities. Then the equation of the chord will be The condition which makes this line tangent to the ellipse W'x 2 + a 2 y 2 = a 2 b 2 is ( 1) (**i-**Jfj9*-*X# + 52. \ x 2 — x x J \x 2 — xj Since (x x y^) and Qr 2 y^) are on the circle, 194 ANALYTIC GEOMETRY [Cii. XIV (2) x* + y* = r\ and (8) = y -l±Ms. From these five equations Ave can eliminate x v y v x v and y v and obtain a single equation in x' and y\ which will be the equation of the locus. 33. Given two concentric ellipses, one within the other, on the same axes. Find the locus of the pole of tangents to the inner with respect to the outer. 34. Find the locus of the middle points of a set of parallel chords intercepted between an hyperbola and its conjugate. 35. Normals are drawn to an ellipse and the circum- scribing circle at corresponding points. Find the locus of their point of intersection. 36. A perpendicular is drawn from a focus of an ellipse to any diameter. Find the locus of its intersection with the conjugate diameter. 37. Find the locus of the middle point of all chords of an ellipse of the same length 2 c. Note. — Find the polar equation of the ellipse referred to the point (sc', y') as origin. Then express the conditions that the two values of p are each equal numerically to c, but opposite in sign. Eliminate 0. 38. Find the locus of the intersection of the ordinate of any point of an ellipse, produced, with the perpendicu- lar from the centre to the tangent at that point. CHAPTER XV HIGHER PLANE CURVES 100. Introduction. — Any locus which lies wholly in a single plane, and which cannot be represented by an algebraic equation of the first or second degree, is spoken of as a higher plane curve. Their equations may be purely algebraic, or they may involve functions other than alge- braic, when they are spoken of as transcendental. There are an infinite number of such curves ; but only a few of those which are of importance in y the study of the Calculus will be discussed here. 101. The parabolas. — The locus of any equa- tion of the form y = ax n is called a parabola of the T > /7th degree. It is usual to restrict n to values greater than unity. If n = 2, we have the ordi- nary parabola along the F-axis. If n= 3, the locus is called the cubi- cal parabola, and has the form shown in Fig. 97. If n Fig. 97. |, the locus is called the semicubical parabola, and has the form shown in Fig. 98. 105 19G ANALYTIC GEOMETRY [Ch. XV, § 102 All parabolas are similar to one of these three forms according to the value given to n. Let the student plot the locus for various values of n. Let him also show that, if n is an even integer, or a fraction with an even numerator and an odd denomina- tor, the curve is similar to the ordinary parabola ; if n is an odd integer, or a fraction with an odd numerator and an odd denomina- tor, the curve is similar to Fig. 97 ; while, if n is a fraction with an odd numerator and an even de- nominator, the curve is similar to Fig. 98. 102. The Cassinian oval. —The locus of a point, the product of whose distances from two fixed points is constant, is called a Cassinian oval. The two fixed points are called the foci of the oval. To find its rectangular equation, let the X-axis go through the two foci, F and F, and let the T-axis bisect FF . Take OF equal to e, and let (x, y) be the coordi- nates of P, any point on the locus. Fig. 98. Ch. XV, § 102] HIGHER PLANE CURVES 197 Then FP = V(z - c) 2 + ~y\ and F'P = V(x + c) 2 + f. But FP.F'P = m 2 . Hence [0 - c) 2 + f] [O + c) 2 + y 2 ] = m 4 , or (a? + y 2 + c 2 ) 2 - 4 c 2 x 2 =m* is the equation of the Cassinian oval. The intercepts of the curve on the axes are ± Vc 2 ± m 2 and ± Vm 2 — c 2 . Hence if cra, the curve cuts the X-axis in four real points but does not cut the I^-axis. It must, therefore, consist of two distinct ovals, as shown in Fig. 100. If c = m, all the intercepts are zero and the curve Fig. 100. goes through the origin. In this case the equation re- duces to x 2 + y 2 = 2 c 2 (x> - y 2 ), or in polar coordinates, p 2 =2c 2 cos 2 6. This special form of the Cassinian oval is called the lemniscate. It has already been discussed on page 67. 198 ANALYTIC GEOMETRY [Ch. XV, § 103 103. The cissoid. — The cissoid may be defined as follows : on any diameter OA of a circle, lay off equal dis- tances CM and CN on each side of the centre, and at the points M and N erect MK and NL perpendicular to the diameter. Draw OK and OL. The locus of the intersection of OK with NL and OL with MK is the cissoid of Diodes. To obtain its rectangular equa- tion, let OA be the X-axis and the origin. Then 031= NA = x, and NL = VON- NA = V(2 a - x)x. From the similarity of the tri- angles OMP and ONL, OM: ON:: MP : NL, or x : : 2 a — x : : y : V(2 a — x): Fig. 101. Hence y 1 = - za — x which is the rectangular equation of the cissoid. It is evidently symmetrical with respect to the X-axis, and has the line x = 2a as an asymptote. 104. The conchoid. — Let A be a fixed point at a dis- tance a from a fixed line OX. Draw the line AP through A cutting OX at B, and on this line lay off a constant distance BP(=b) both ways from B. The locus of P is called the conchoid of Nicomedes. To find its rectangular equation, take the fixed line Ch. XV, §104] HIGHER PLANE CURVES 199 OX as the X-axis, and OA as the F"-axis. Drop a per- pendicular from P on the X-axis and continue it to meet AK, drawn parallel to the same axis. Then But Hence or AP = V* 2 + (# + a) 2 , and #P = £. AP = KP BP MP g 2 -K,y + fl) 2 = Q/ + a) 2 6 2 y 2 Y The fixed point ^1 is called the pole, and the fixed line OX the directrix of the conchoid. If a < b, the curve has the form shown in Fig. 102. If a — 5, there is no loop, but the curve has a cusp at A. If a > 5, the lower branch of the curve cuts the F-axis in a single point above A. Note. — Among the most noted problems of the ancient mathema- ticians were the Trisection of an Angle and the Duplication of the Cube by the aid of ruler and compass alone. It has lately been shown that the solution of these problems in this way is impossible. Both problems involve the solution of a cubic equation, and both may be made to depend upon the construction of two mean proportionals between two straight lines. This has been accomplished in various ways by aid of higher plane curves, and it was for this purpose that both the Conchoid and Cissoid were invented. 200 ANALYTIC GEOMETRY [Ch. XV, § 10/ 105. The cycloid. — The path described by a point on the circumference of a circle which rolls on a straight line is called a cycloid. Let C be the centre of the moving circle of radius a, and let P be the fixed point on its circumference. To find the rectangular equation of the curve, let the X-axis coincide with the fixed line, and choose as the origin of coordinates one of the points where P coincides with that line. Draw CM perpendicular to OX, and PK Fig. 103. parallel to the same line. Let 6 represent the circular measure of the angle MCP through which the radius CP has revolved. Let the coordinates of P be (x, y). Then x = OM- PK, and y = MC - KC. But 0M= arc MP = a0, PK= PC sin 0, and KC = PC cos d. Hence x — a (6 — sin 0), y = a(l — cos 0). If is eliminated from these two equations by finding Ch. XV, § 106] HIGHER PLANE CURVES 201 its value in the second equation and substituting in the first, we have x = a vers -1 ( - ) — V2 ay — y 2 . But this single equation is not so convenient to use as the pair of equations from which it was obtained. These two equations, containing a third variable, are equivalent to the single equation from which 6 has been eliminated. The locus consists of an infinite number of branches, similar to the one shown in Fig. 103, extending both to the right and to the left of the origin. 106. The hypocycloid. — The path described by a point on the circumference of a circle which rolls on the inside of a fixed circle is called a hypocycloid. Let a be the radius of the fixed circle, and b the radius of the rolling cir- cle. Let P be the fixed point on the rolling circle. Take the centre of the fixed circle as the ori- gin of coordinates and let the X-axis pass through A, one of the points where P coin- FlG - 104 - cides with the fixed circle. Consider the rectangular coordinates (x, y) of any position of P. Drop perpen- a x 202 ANALYTIC GEOMETRY [Ch. XV, § 106 diculars from C and P to the X-axis, and through P draw LR parallel to that axis. The radius OK of the fixed circle, drawn through the point of contact K, passes through the centre C of the rolling circle. Let Z A0K= $, and ZPCK= 6. Then, since the arcs AK and PK are equal, acf) = bd, or 6 = - . o Now x = 031= ON- PL, y = MP = NC-LC. But ON =00 cos = (a - 5) cos <£, iV r C= 0(7 sin (j> = (a-b) sin <£, PZ = PC 7 cos RPC=b cos [180° -(0- £)] = — b cos (6 — $)= — b cos f — — — j c/>, and LC= b sin (0 — <£)= 5 sin (— -jr— - )<£• Then x = (a — 5) cos $ + 6 cos f — — — ] <£, y = (a — b~) sin $ — b sin ( ~~ ) . As in the cycloid, these two equations, containing a third variable , may be used in place of a single equa- tion in x and y to represent the curve. The most important special case is the four-cusped hypo- cycloid, in which a = 4 b. The equations here reduce to x = | a cos + \a cos 3 <£, y = | a sin — | a sin 3 (/>. But sin 3 <£ = 3 sin — 4 sin 3 c/>, cos 3 <£ = 4 cos 3 — 3 cos . Ch. XV, § 107] Hence x = a cos 3 , and y = a sin 3 (f>. Raising to the § power and adding-, we have X s + y % = «* as the rectangular equa- tion of the four-cuspecl hypocycloid. The form of the curve is shown in Fig. 105. HIGHER PLANE CURVES 203 107. The epicycloid. — The path described by a point Fig. 106. 204 ANALYTIC GEOMETRY [Ch. XV, § 108 on the circumference of a circle which rolls on the out- side of a fixed circle is called an epicycloid. Let the student show that the equations of the epi- cycloid are 'a + b s x — (a-\-b~) cos (j> — b cos y = (« + b) sin <\> — b sin b a + b * 108. The cardioid. — An important special case of the epicycloid is the cardioid, in which a = b; but instead of Fig. 107. obtaining its equation as a special case of that curve, we shall find it easier to obtain its polar equation at once from the definition. Let 0, the original point of contact of the two circles, Ch. XV, § 110] HIGHER PLANE CURVES 205 be chosen as the polar origin, and the diameter CA con- tinued through this point as the polar axis. Let the second circle roll on the first until the line of centres CC makes an angle 0G 7 O"(= ) with its original position, and let P be the new position of the point of contact. Then OP = p, and the angle AOP = 0. From and P drop the perpendiculars OM and PN o\\ CC. Evidently the arc OK equals the arc PK, and the angles PC C and OCC are therefore equal. Then OM=PN, and OP is parallel to CC. Then 6 = <£. Now CC = CM + MN + NC = 2 a, or 2 CM+ OP = 2 a, or 2 a cos 6 -f p = 2 a, or jo = 2 a (1 - cos 0). 109. The catenary. — The curve assumed by a perfectly flexible chain of uniform weight per linear unit, when suspended at its ends, is called a catenary. Its equation may be obtained in the form where e is the base of the Naperian system of logarithms, and the origin of coordinates is a units below the lowest point of the curve. 110. The spirals. — The curve traced by a point which revolves about a fixed point, and, at the same time, re- cedes from or approaches this point according to some definite law, is called a spiral. The fixed point is called 206 ANALYTIC GEOMETRY [Ch. XV, § 110 the centre, and the curve traced daring one revolution is called a spire. If any two radii vectores have the same ratio as the angles they make with the initial line, the equation of the spiral is evidently p = kO. The form of the curve is shown in Fig. 108. The dotted line indicates the portion of the locus obtained by giving negative values. Let the student plot the spirals whose equations are The curve whose equation is log p = Jc0, or p = a 9 , is called the logarithmic spiral. When = 0, p = 1. For increasing positive values of 0, p increases very rapidly ; while for decreas- ing negative values of 0, p decreases more and more slowly, and approaches zero as a limit. There are, there- fore, an indefinite number of spires, growing smaller as they wind about the origin, but never passing through that point. Fig. 108. PART II ANALYTIC GEOMETRY OF SPACE CHAPTER I COORDINATE SYSTEMS. THE POINT 1. In the following chapters on Analytic Geometry of Space, a knowledge of the methods and results of Solid Geometry and of Plane Analytic Geometry is presumed. Many of the methods and formulas to be given for three dimensions are closely analogous to methods and formulas in two dimensions, with which the student is already familiar ; and in all such cases the discussion will be condensed into as brief a form as possible. For convenience of reference, the following theorems and definitions from solid geometry are cited : If a straight line is perpendicular to a plane, it is per- pendicular to every line through its foot in the plane. If a straight line is perpendicular to any two straight lines through its foot in a plane, it is perpendicular to the plane. The angle between two lines not in the same plane is the same as the angle between two intersecting lines parallel respectively to the given lines. 207 208 ANALYTIC GEOMETRY OF SPACE [Ch. 1, § 2 The orthogonal projection of a point on a plane (o:* an axis) is the foot of the perpendicular from the point to the plane (or the axis). The projection of a portion of a line or curve on a plane (or an axis) is the locus of the projections of all its points. The angle which a line makes with a plane is the angle which it makes with its projection on the plane. The angle between two planes is measured by the angle between two lines, one in each plane, drawn perpendicular to their intersection at the same point. 2. Rectangular coordinates. — In applying algebra to the geometry of space, we must first devise some method of representing the position of a point in space by numbers. Construct three mutually perpendicular planes, X-Y, Y-Z, and Z-X, dividing all space into eight compart- ments, called octants. These planes are spoken of as coordinate planes, their point of intersection, 0, as the origin, and their lines of intersection, OX, OY, and OZ, as coordinate axes. A point in space is located by means of its distances, AP, BP, and CP, from the coordinate planes, measured parallel to the coordinate axes. The three numbers which represent these distances are called the rectangular coor- dinates of the point, and are always written in the order O* y, z). We shall consider distances as positive when measured to the right, forward, or upward ; that is, parallel to OX, OY, and OZ. Distances measured in the opposite directions will then be negative, The octant 0-XYZ is Ch. 1, § 3] COORDINATE SYSTEMS. THE POINT 209 V called the first, and the others may be numbered in any convenient way. The position of any point (#, y, z) may be determined by taking on the axes the distances OL, OM, and 6W, equal to these coordinates, and through the points X, M, iV, passing planes parallel to the coordinate planes, forming a rectangular par- allelopiped ; the point of intersection of these planes _ will be the point required. It is evident that rectan- gular coordinates in a plane is a special case of this more general system, in which one of the coordinates has become zero. We ought therefore to be able to reduce all of the formulas in three dimensions to the corresponding formulas in two dimensions by plac- ing z equal to zero. - p <' i — i— > i j i J' Fig. 1. PROBLEMS 1. Plot the following points : (5, 4, 3), (- 3, 4, 1), (- 3, - 1, 2), (2, - 3, 1), (1, 1, - 2), (-1, 4, -2), (-3, -2, -1), (4, -1, -2); (3, 4, 0), (-2, 0, 1), (0, -1, 3); (5, 0, 0), (0, 3, 0), (0, 0, -2). 3. Distance between two points. — Let P x and P 2 be any two points in space, and through each of them pass three planes parallel to the coordinate planes, form- ing a rectangular parallelopiped. 210 ANALYTIC GEOMETRY OF SPACE [Ch. I, §4 Since the square of the diagonal of a rectangular parallelopiped equals the sum of the squares of its edges, iyy = />,*,' + JW +P 1 T 1 '. But P 1 B 1 = x 2 — x v A#i = #2 - VV and P\T X = z 9 — z v Fig. 2. Hence P 1 P 2 = V(*2 - #i) 2 + (2/2 - 2/O 2 + (s 2 - »i) 2 . [1] The distance, /a, of any point from the origin is P = V^ + 2/ 2 + z 2 . [2] evidently 4. To divide a line in any given ratio. — Let the point P P m P divide the line P X P 2 so that —J— = PP Project the line P X P 2 on the X- ^-plane, form- ing the trapezoid P 1 C 2 , in which 1 P 1 = z v C 2 P 2 = z 2 , and OP = z. It will be noticed that this is the same figure used in Art. 13, Part I. Hence z- iniZ\ + m,\Zi Ch. I, §5] COORDINATE SYSTEMS. THE POINT 211 If P x P, 0) are called the spherical coordinates of P. The arrows indicate the usual choice of positive direction. 216 ANALYTIC GEOMETRY OF SPACE [Ch. I, § 8 Let the student show that the relations between rec- tangular and spherical coordinates are oc = p sin <|> cos 6, y = p sin sin 0, [8] z — p cos <|>. Note. — Spherical coordinates have usually been called polar coordi- nates. But the application of the system described in Art. 6 is more nearly analogous to the uses of polar coordinates in two dimensions. 8. Angle between two lines. — Let u v /3 r 7 r and « 2 , /3 2 , y 2 be the direction angles of two lines, and let 6 be the angle between them. Draw parallels to these lines through the origin, and on each of these parallels take a point, as P x and P . Then by [1] Fig. 7. *VY = Oi - ^) 2 + (jfi - y*y + Oi - * 2 ) 2 > or by [6] = (p x cos a x — p 2 cos « 2 ) 2 ~l~G i cos ^i~p2 cos ^) 2 + 0>icos7 1 - / ? 2 cos7 2 ) 2 , or by [7] = Pi+p^ — 2 p x p 2 (cos a x cos a 2 + cos /3 X cos fi 2 -f- cosyj cos7 2 ). Ch. I, § 8] COORDINATE SYSTEMS. THE POINT 217 But by the law of the cosines Hence cos = cos a x cos a 2 + cos pi cos p 2 + cos 71 cos 72. [9] If the lines are perpendicular cos 6 = 0, and the condi- tion for perpendicularity is cos ai cos a 2 4- cos Pi cos p 3 + cos -yi cos Y2 = 0. [10] If the lines are parallel, they must make the same angles with the axes, and the conditions for parallelism are 01 =o 2 , Pi = p 2 , and 71 = Y2. [11] PROBLEMS 1. Show that the three lines whose direction cosines are 12-3-4. 4 12 3 . orirl 3 -4 1 2 T5> T^> T 3 J T¥> T 3 > T3 J djllu T3 J T 3 > T 3 are mutually perpendicular. 2. Show that (3, 30°, 60°, 90°), and (5, 30°, 90°, 60°) are possible polar coordinates of two points, and find the angle they subtend at the origin. 3. Show that the conditions for parallelism are consistent with [9] when = 0°. 4. Find the rectangular coordinates of the points in problem 2. 5. Find the polar coordinates of the point (3, — 6, 2). . 6. Find the angle subtended at the point (1, 2, 3) by the points (2, 3, 4) and (5, 4, 3). 218 ANALYTIC GEOMETRY OF SPACE [Ch. I, § 10 9. Transformation of coordinates. Parallel axes. — If the new axes are parallel to the old, and the coordinates of the new origin, re- ferred to the old axes, are (# , y , 2 ), the equations of transformation are easily seen (see Fig. 8) x to be ac = x + oc' f V = 2/o + V, [12] Z = Z + Z . 10. Transformation of coordinates from one set of rec- tangular axes to another which has the same origin. — Let («ii fi v Yj), (« 2 , £ 2 , 7 2 ), and (« 3 , /3 3 , y 3 ) be the direction angles of OX\ OY\ and OZ' with respect to the original axes. The coor- dinates (#, y, 2) of any point P are the projec- tions of OP on OX, OF, and OZ. But the broken line made up of x\ y ! , and z' extends from to P, and will therefore have the same projections on Y the axes as OP. Hence (by Art. 5) Fig. 9. ac = x' COS ai + y' cos a 2 + Z' COS a 3 , y = 05' cos Pi + y' cos p 2 + s' cos p 3 , 3 = 05' cos 71 + 2/' cos 72 + z' COS 73. [13] Ch. I, § 10] COORDINATE SYSTEMS. THE POINT 219 Let the student show that the transformation of coordi- nates cannot alter the degree of an equation. (See Art. 50 Part I.) PROBLEMS 1. What will be the direction cosines of OX, OY, and OZ referred to the new axes in Art. 10 ? 2. What six relations hold between a lf ft, y 1} u 2 , ft, etc., from [7] ? 3. What six relations hold between u 1} ft, y l9 a 2 , (3 2 , etc., from [10] ? 4. Show that the twelve relations obtained in problems 2 and 3 are equivalent to only six independent conditions. How many of the coefficients in equations [13] are independent ? CHAPTER II LOCI 11. Equation of a locus. — If a point moves in space according to some law, it will generate some locus. As, for example, a point keeping at a fixed distance from a fixed point will generate the surface of a sphere. If we can translate the statement of the law into an algebraic relation between the coordinates of the points which satisfy the law, we shall have, as in plane analytic geometry, an equation which can be used to represent the locus. In the above example, if the origin is at the centre, the equa- tion of the surface will be x 2 + y 2 + z 2 = r 2 ; for this states that the point (x, y, z), which satisfies it, must remain at the distance r from the origin. The planes parallel to the coordinate planes are evi- dently represented by x = k v y = & 2 , and z = Jc s ; for these equations state that the points which satisfy them are at a fixed distance from the coordinate planes. PROBLEMS 1 . What are the equations of the coordinate planes ? 2. What are the equations of the planes bisecting the angles between the X-Y and 5 -Z-planes ? Between the Y-Z and Z-X-planes ? 3. What equation must be satisfied by the coordinates of a point which remains at a distance of 5 units from the X-axis ? 5 units from the F-axis ? What is the locus in each case ? 220 Ch. II, §§ 12, 13] LOCI 221 4. Find the equation of the locus of a point which is 5 units from the point (3, 2, 5). 5. What equations must be satisfied by the coordinates of a point which is equidistant from the three points (1, 3, 8), (- 6, - 4, 2), and (3, 2, 1) ? 12. Cylindrical surfaces. — If a cylindrical surface is formed by the movement of a line, which remains parallel to one of the axes, while moving along a directing curve in the plane of the remaining axes, its equation in three dimensions will be the same as the equation in two dimen- sions of the directing curve, and will contain only two variables. For, suppose the line remains parallel to the Z-axis and the directing curve lies in the X-Z-plane ; then, for any position of the line, the relation between the x and y coordinates of any point on it will be the same as the relation between the x and y coordinates of the point where the line touches the directing curve, while the z coordi- nate may have any value whatever. The equation in x and y of the directing curve is, therefore, the only necessary relation between the coordinates of any point on the sur- face, and as it is not satisfied by any point not on the surface, it is (when interpreted as an equation in three dimensions) the equation of the surface. In a similar manner, it may be shown that the equations of cylindrical surfaces, whose elements are parallel to the JT-axis, contain only y and z ; parallel to the !F-axis, only x and z. 13. Surfaces of revolution. — Surfaces generated by the revolution of a plane curve about one of the coordinate axes form another class of surfaces whose equations can be determined easily. 222 ANALYTIC GEOMETRY OF SPACE [Ch. II, § 13 For example, let it be required to determine the equation of the surface generated by the revolution of the ellipse 1 about the X-axis. Let P' (V, y\ z') be any z a? r a 2_h 6 2 Fig. 10. point on the surface, and through P' pass a plane perpen- dicular to the X-axis. The section of the surface made by this plane is evidently a circle. Hence LP' = LK. But LP' = V/M^T 2 " and OL = x\ The coordinates of iT in the X-P-plane are, therefore, x' and Vy 2 + z' 2 , and since K is a point on the ellipse 1, these coordinates must satisfy that equation, ^2 f a 2_h 6 2 or + y"> + z> = 1. Dropping primes, we have as the equation of an ellipsoid of revolution about the X-axis, a 2 "^ 2 "^ 2 " ' Ch. II, § 14] LOCI 223 A general rule for finding the equation of a surface of revolution, formed by revolving a plane curve about one of the coordinate axes, may be stated thus: Replace in the equation of the plane curve the coordinate perpendicular to the axis of revolution by the square root of the sum of the squares of itself and of the third coordinate. PROBLEMS 1. Find the equation of the surface generated by a line moving parallel to the Z-axis along x 2 if (a) the ellipse - + |- 2 = 1, (b) the parabola y 2 = 2 mx, (c) the line x + 3 y = 6. 2. What is the equation of a circular cylinder whose axis is parallel to the F-axis and passes through the point (3, 0, 5), and whose radius is 5. 3. Find the equation of the surface of revolution, formed by revolving about the X-axis (a) the line y = 4, (a cylinder) (6) the line x = y, (a cone) (c) the circle x 2 -\- y 2 = r 2 , (a sphere) (cZ) the parabola y 2 = 2 mx, (a paraboloid of revolution). 4. Obtain the equations of the hyperboloids of revolution formed by revolving the hyperbola about (a) its transverse axis ; (b) its conjugate axis. 14. Locus of an equation. — Again, as in plane analytic geometry, an equation between x, y, and z expresses a necessary relation between the coordinates of every point which satisfies it, and hence cannot be satisfied by points taken at random in space. It is easy to see that the 224 ANALYTIC GEOMETRY OF SPACE [Ch. II, § 14 points which satisfy it may be taken as near to each other as we please. Moreover, any such equation represents a surface of some kind, as we shall now prove. Let/(:r, y, z)= Q be an equation of any degree between x, y, and z. If we substitute x — k (any constant), the resulting equation, f(y, z} = 0, must represent the rela- tion between y and z for all points of the locus for which x = k, or which lie in a plane at distance k from the Y-Z-plane. But since the locus of /(?/, z) = lies wholly in this plane, it is a plane curve. Hence the intersection of any plane parallel to the Y-Z-plnne with the locus of f(x, y, z) = is a plane curve. This can be proved in like manner for all planes parallel to the X-Y and X-Z- planes. If the axes are revolved through any angle, the equation of the locus will be of the same general form and every plane parallel to the new axes will cut it in a plane curve. Hence all planes cut the locus in a plane curve, and the locus is therefore a surface. If, in particular, the equation is of the first degree, its intersection with any of these planes will be a straight line. An equation of the first degree therefore always represents a plane. If an equation does not contain a term in z, the relation between x and y will not be changed by a change in z. The sections of the locus parallel to the X-F-plane are therefore all alike, and the locus is a cylindrical surface, having all its elements parallel to the Z-axis. In like manner, if an equation does not contain a term in y, it represents a cylindrical surface parallel to the y-axis; if it contains no term in x, a cylindrical surface parallel to the X-axis. Ch. II, § 14] LOCI 225 If in particular the equation is of the first degree, the surface becomes a plane parallel to one of the axes. If two equations are simultaneously satisfied by the coordinates of points on a locus, that locus must consist of the points common to the loci of the two equations. Hence two equations of the form f x (#, y, z) = and f 2 (x, y, z) = 0, taken togetlier, represent a curve in space, the intersection of the surfaces which they represent. In particular, if these tw T o equations are of the first degree, this locus will be the intersection of the two planes which they represent. Hence two equations of the first degree, used simultaneously, represent a straight line. Three equations used simultaneously are satisfied by the coordinates of a finite number of points only, — the points of intersection of the curve represented by two of the equa- tions with the surface represented by the third. The curves of intersection of any surface with the coordinate planes are called the traces of the surface. Their equations may be found from the equation of the surface by placing each of the coordinates in turn equal to zero. The general method of determining the form of the surface represented by any given equation will be taken up in the chapter on quadric surfaces. PROBLEMS 1. What surface is represented by the equations ? (a) x = y, (d) x 2 + y 2 = 25, (!>) y = z, (e) x 2 + f- + z 2 = 25, (c) x-y=5, (f)x*-2y = 0. 226 ANALYTIC GEOMETRY OF SPACE [Ch. II, § 14 2. Obtain the traces on each of the coordinate planes of the loci of the following equations, and from these traces deter- mine roughly the nature of the surface: (a) x? + tf = 9, (d) y 2 = ±z, (b) x-y + 2z = 10, (e) J + J + g^l, (c)a? = 2y, (f)x> + tf-2z = 0. 3. What is the equation of the surface generated by the revolution of the hyperbola xy = k about the X-axis. 4. What is the position of a line whose equations are x + 3y = 10 and Sx — 4,y = S? 5. The equations of any two surfaces may be represented by U= and V= 0, where U and Fare abbreviations for algebraic expressions of any degree in x, y, and z. Show that lU+kV=0 will represent a surface which passes through all the points common to the loci of U = and V= 0, and which meets neither of these surfaces at any other points. Show also that the locus of UV= will consist of the loci of TJ— and V= 0. CHAPTER III THE PLANE 15. Normal form of the equation of a plane. — Let ON be the normal to the plane (a straight line of indefinite extent perpendicular to the plane), and let <*, /3, and 7 be the angles which this normal makes with the axes. Let p be the perpendicular distance OK from the origin to Fig. 11. the plane, measured along the normal. Let P(x,y,z) be any point in the plane. The line PK will be perpen- dicular to ON, and the projection of OP on ON will be OK or p. But the projection of OP on ON is the same as the projection on ON of the broken line OL, LC, CP, 227 228 ANALYTIC GEOMETRY OF SPACE [Ch. Ill, § 16 or #, ?/, z. From [5] the projection of OL on OJV is #cos«; of L C, y cos /3 ; of CP, 2 cos 7. Hence oc cos a -f y cos p + 2 cos y - p = 0. [14] This is called the normal form of the equation of a plane. The distance p is measured from the origin to the plane, and is positive or negative according as it runs in the positive or negative direction of the normal. It is usually possible to choose the direction from the origin to the plane as the positive direction of the normal, so that p will usually be a positive number. The angles «, /3, and 7 are measured from the positive directions of the axes to the positive direction of the normal. 16. Reduction of the general equation Ax + By + Cz + D = to the normal form. — It lias been shown in the previous chapter that every equation of the first degree represents a plane. Let the general equation of the first degree, Ax + By + Cz + D = 0, be the equation of a plane, and let x cos a + y cos /3 -f z cos 7 — p = be the equation of the same plane in the normal form. Then, since the two equations represent the same plane, they can differ only by a common factor. Then JcA = cos a, kB = cos /3, and JcC= cos 7. Hence h = , and the ± V J 2 + B 2 + O 2 equation ^ T a? + ' — 1/ + + , n = = [151 Ch. Ill, § 18] THE PLANE 229 is in the normal form. If we wish to keep p positive, it is necessary to choose the sign of the. radical opposite to the sign of D. Then the coefficient of x is cos «, etc. 17. Equation of a plane in terms of its intercepts. — If the intercepts of a plane on the axes are a, 5, and c, the coordinates of the points where it cuts the axes are (a, 0, 0), (0, 5, 0), and (0, 0, c). If these coordinates are substituted successively in the general equation Ax + By + Cz + D = 0, we have a = -, 6 = — -, and c = -• A B V But the general equation may be written in the form x y z _, + 7T=L A J5 (7 From this we have, by substitution, °° + V + s = l [16] a b c L J as the equation of a plane in terms of its intercepts. 18. Distance of a point from a plane. — Let it be re- quired to find the distance of the point P x from the plane UK, when the equation of UK is given in the form x cos « 4- y cos /3 -f- z cos 7 — p = 0. Pass a plane RS through P v parallel to UK. Its equation will be x cos a + y cos /3 -f- 2 cos 7 — jt^ = 0, Fig. 12. 230 ANALYTIC GEOMETRY OF SPACE [Ch. Ill, § 18 where p x can be either positive or negative, since it is the distance from the origin to the plane BS, measured along the nor- mal to UK. The coordi- nates of P x must satisfy the equation of BS. Hence x x cos a -f y x cos ft + z x cos 7 = p v Now, wherever P x may lie, MP X = NN X = ON x - ON=p x -p = x x cos a + y 1 cos ft + z 1 cos 7 — p. If the equation is given in the form Ax + By + Cz + D = 0, MP, = Ax * + ^ + Cz i+M , r 171 ± V^ + B* + C 2 L J where the sign of the radical, is chosen opposite to that of D. The distance MP 1 is positive when the point and the origin are on opposite sides of the plane ; negative when they are on the same side of the plane. PROBLEMS 1. Given the plane 3x — Sy -\- z = 12, find (a) the direction cosines of a normal, (b) its distance from the origin, (c) its distance from the point (3, — 2, 6), (d) its intercepts. Ch. Ill, § 10] THE PLANE 231 2. Find the equation of a plane, if the foot of the perpen- dicular from the origin on it is the point (3, 1, — 5). 3. On which side of the plane 7 x-\-Ay = 5 is the point (0, 7, 3) ? How is this plane situated ? What are its traces on the coordinate planes ? 4. Show that the three planes 2x + 5y -\- 3 2 = 0, x — y + 42 = 2, and 1 y — 524-4 = intersect in a straight line. 5. Find the equation of the plane which bisects the angle between the two planes A^x 4- B^ + C x z + D x = and Afc + 5^/4- C# 4- A = 0. 6. Find the equation of a plane through the origin and the line of intersection of the planes x 4- 3 y — 4 z = 10 and 5y— 624-3 = 0. (See problem 5, page 214.) 19. The angle between two planes. — The angle between two planes is easily seen to be equal to the angle between their normals. If the two planes are x cos «! -f y cos /3 X 4- z cos y x — p 1 = 0, and x cos « 2 4- y cos /3 2 + 2 cos y 2 — /? 2 = 0, the angle between them is given by cos 9 = cos ai cos a 2 + cos Pi cos p 2 4- cos 71 cos 72. [18] If the two planes are A x x 4- B x y 4- C x z + D l = 0, and J. 2 z -f- B 2 y 4- 2 = 0, the angle between them is given by cose = AlA * + B lB * + Cl ^ 2 ["191 ± V^ + .B^ + Cl 2 ' V^ 2 2 4" ^2 2 + C2 2 If the sign of the first radical is chosen opposite to the sign of D v and the sign of the second opposite to the sign 232 ANALYTIC GEOMETRY OF SrACE [Ch. Ill, § 21 of D 2 , 6 will be the angle between the positive directions of normals to the planes. 20. Perpendicular and parallel planes. — If two planes are perpendicular, cos 6 = 0, and AiA 2 + BiB 2 + dC 2 = 0. [20] If two planes are parallel, the direction cosines of their normals must be equal, or and 4 A V^ 2 + B? Br + 0* VA 2 + A 2 B 2 + c 2 2 V^! 2 + B* Or + c* V^ 2 S % '+i = 0, and A 2 x + B 2 y + C 2 z + D 2 = will in general represent a line, the only exception being when — J = — 1 = —J., and the planes are parallel. A 2 B 2 C 2 But the line may be determined by any pair of planes which pass through it, and it is convenient to pick out those planes which have the simplest form. The equa- tion of any plane through the line can be written in the form A x x + B lV + C x z + J) 1 + k(A 2 x + B 2 y + 2 z + D 2 )= 0. When none of the coefficients A v B v etc., are zero, it will always be possible to choose k in such a way as to eliminate y and reduce the equation to the form x = mz + a. Again, k may be so chosen as to eliminate x and reduce the equation to the form y = nz -f- b. Then the equations x = mz + a, and y = nz -f 6, 235 236 ANALYTIC GEOMETRY OF SPACE [Ch. IV, § 22 each determine a plane through the line, and hence may be used as the equations of the line. These planes are seen to be the projecting planes of the line, perpendicular to the X-Z and Y-Z-planes. The equations of any two of the three projecting planes may be chosen as the equations of the line. In practice, to reduce the equations of a line to their simplest form, we simply eliminate one of the variables and then another from the two equations. Indeed, it is evident algebraically that any set of values which satisfy a pair of equations must also satisfy any equation which can be deduced from them. If some of the coefficients A v B v etc., are zero, it will always be possible by elimination to reduce the equations to one of the three forms x = mz + #> y = qx + c, x = e, or y = nz + 5, z = d, z =f. The first form includes all lines not parallel to the X-Y- plane ; the second, lines parallel to the X-!F-plane, but not parallel to the y-axis ; the third, lines parallel to the P-axis. PROBLEMS 1. Write the equations of each of the coordinate axes. 2. Write the most general form of the equations of a line in each of the coordinate planes ; parallel to each of the coordi- nate planes ; parallel to each of the coordinate axes. 3. Show how to find the points where a given line pierces the coordinate planes, and by this means plot the lines in problem 4. Ch. IV, § 23] THE STRAIGHT LINE 237 4. Reduce these equations to their simplest forms (a) 2a?-3# + z - 6 = 0, (6) 2x + 3y x + 7/ — 3 z — 1 = 0. (c) 2a? + 4y + 3« + 6 = 0, 3a? + 6y + 22-l = 0. 3t/ (e) 2a>-3y- z + 2 = 0, (/) 4. y 4a?-6y + 32J-l = 0. 2t/ 6 z - 12 = 0, 4# — t/ + 12z + 4 = 0. (rt) 4 y + 3*+ 1=0, — 2 12 = 0. 3z- 2 = 0, 2 + 4 = 0. 5. Find the equations of the line of intersection of the plane 2x — 3y -\- z — 6 = with the coordinate planes. 23. The equations of a line in terms of its direction cosines and the coordinates of a point through which it passes. — Let «, /3, and 7 be the direction angles of the Fig. 13. c line and P t a point through which it passes. Let P be any point on the line. Then from the figure x — x x = P X P cos a, y-y l = P X P cos ft z — z 1 = P X P cos 7. 238 ANALYTIC GEOMETRY OF SPACE [Ch. IV, § 24 Solving these for P X P, and equating the values, we have aj - asi y - y\ z - Z\ cos a cos COS X PROBLEMS [22] 1 . What form will these equations take when a = 90° ? when a = 90°, and £ = 90° ? 2. Find the equations of a line through the point ( — 1, 2, -3) if (a) a = 60°, /? = 60°, y = 45°; (6) a = 120°, (3 = 60°, y = 135°j (c) cos a = | V3, cos /3 = i, cos y = 0. Show that the given values are possible in each case and plot the line. 3. Find the equations of a line through the origin, equally inclined to the axes. 24. Given the equations of a line, to find its direction cosines. — The method is best shown by an example. Let the equations of a line, reduced to their simplest form, be x = 5 z — 6, and y = 2 z + 3, or Let cos « cos p cos y be the equation of the same line. These equations are of the same form and, since they represent the same line, X + 6 5 -V 2 3_ z -0 1 X -x x _ -V — Vi _ z ~ z i x 1 = — 6, y x = 3, and z 1 = 0, Ch. IV, § 24] THE STRAIGHT LINE 239 and the denominators, 5, 2, and 1, are proportional to cos a, cos /3, and cos 7. They can be made identical with them by multiplying by a suitable factor R. Then cos « = 5 R, cos /3 = 2 R, and cos y=R. 1 Then by [7] 25 R 2 + 4 R? + R 2 = 1, and i? = Hence cos a = , cos /3 = t__ , cos 7 = 30 "30 V30' V30' and the equation can be written in the form a? + 6 _ y~3 _ g-0 ~5~" ~~T~~~T~' V30 V30 V30 PROBLEMS 1. Show that, if the equations of a line can be written in the form x = mz -\- a, and y = nz + &> they may be changed into the form # — a 2/ — & 2 Vm 2 + w 2 4- 1 Vm 2 + w 2 + 1 Vm 2 + n 2 + 1 where — m — = cos a, n = cos /3, Vm 2 + w 2 + 1 Vm 2 + ri 2 + 1 and — — = cos y. Vm* + rr + 1 2. What form will the equations take, if their simplest forms are y = qx + c, aj = e, 240 ANALYTIC GEOMETRY OF SPACE [Ch. IV, § 25 3. Find the direction cosines of the lines whose equations are (a) 2x + 3y-2z-13 = 0, 3x + 6y-2z-24: = 0. (b) 2x + 2y-3z- 2 = 0, 4ic— y — z — 6 = 0. (c) 2x + 4y + 3z + G = 0, 3x + 6y + 2z- 1 = 0. (d) 4y + 3z + 1= 0, 3y-2z-12 = 0. 25. Equations of a line through two points. — Let (x v y v Zj) and (:r 2 , y v z 2 ~) be the two points. The equation of any line through the first point is (by [22]), x-x x _ y-y x _ g- ^ cos a cos /3 cos 7 If the second point lies on this line, x i ~ x \ I/2 - V\ Z 2~ z \ cos a cos /3 cos 7 Dividing, we have, as the equations of a line through the two points, as-asi _y-yi_z-z\ 1-23-1 0C2 -vc\ 2/2 - 2/1 22 - Si PROBLEMS 1. Establish equation [23] from an independent figure without using equation [22]. 2. Discuss the special cases of [23], when x 2 = x x , y 2 = yi, or z 2 = Zj. Ch. IV, § 25] THE STRAIGHT LINE 241 3. Find the equations of a line passing through the points (a) (0, 0, - 2) and (3, - 1, 0), (b) (- 1, 3, 2) and (2, - 2, 4), (c) (2, - 3, 1) and (2, - 3, - 1). 4. Find the equations of the line joining the origin with the intersection of the planes 3 x - 2 y + z + 4 = 0, a- + 4?/ + 2z = 0, y-3z-7 = 0. 5. Are the three points (1, - 1, 2), (2, 3, - 1), and (3, 2, 2) in a straight line ? 6. Show that the two lines x-2 = 2 y-G = 3z, and ±x -11 = ±y -13 = 3z meet in a point, and that the equation of the plane in which they lie is 2x-6y + 3z + U = 0. 7. Show that the line 4:X = 3y = — z is perpendicular to the line 3 x = — y = — 4z. 8. Find the point of intersection of the line 2a -4 = 3?/ + 1 = 2 + 6 with the plane x -{- G y — o z = 16. 9. What is the equation of the plane determined by the point (3, 2, — 1) and the line 2x — 5 = 5y + l = z? CHAPTER V QUADRIC SURFACES 26. The sphere. — A sphere maybe defined as the locus of a point whose distance from a fixed point is constant. If (# , y , Zq) is the centre and r the radius, the equa- tion of the sphere is evidently (x - aco) 2 + (y - 2/0) 2 + (s - zo) 2 = r 2 . [24] If the centre is at the origin, the equation becomes a? 2 + y 2 + z 2 = r 2 . [25] Expanding [24], we see that the equation of every sphere is of the form oc 2 + y 2 + z 2 + Gx + Hy + Iz + K = 0, [26] where a _h _i 2 ' #o _ 2 ' z ° ~~ 2' and r = J V£ 2 + H 2 + I 2 - 4JT. Every equation in the form of [26] will therefore repre- sent a sphere, real, if G 2 + H 2 + I 2 - 4 K> 0, null, if G 2 + H 2 + I 2 - 4 K= 0, imaginary, if G* + ff* + J* - 4 ^T< 0. Comparing [26] with the general equation of the second degree, Ax 2 + By 2 + Cz 2 + Dyz + Ezx + Fxy+Gx + iry + Iz + K=0, 242 Ch. V, § 26] QUADRIC SURFACES 243 we see that the general equation will represent a sphere, if D = U=F=Q, and A = B = 0. A sphere may, in general, be passed through any four points ; for the substitution of their coordinates in [26] will give four equations which will, in general, determine G, J5T, 7, and if. PROBLEMS 1. Find the equation of a sphere with (a) centre at (5, — 2, 3), radius equal to 1. (b) centre at (2, — 3, — 6), passing through the origin. (c) centre on the Z-axis, radius a, passing through the origin. 2. Find the centre and radius of each of the following spheres, when real : (a) ar + 2/ 2 + z 2 -2a + G?/-8z + 22 = 0. (b) x 2 + y 2 + z 2 + 10x-4 : y + 2z + 5 = 0. (c) 3a? + 3tf + 3z 2 + 12x + 12 y + 18 2 + 3 = 0. (d) x 2 4-2,2 + 32 + 60;= 0. (e) ar° + 2/ 2 + z 2 + 4.T + 2/ + 5z + 21 = 0. 3. Find the equation of the sphere passing through the four points, (a) (2, 5, 14), (2, 10, 11), (2, 5, - 14), (2, - 10, - 11), (6) (0, 0, 0), (2, 8, 0), (5, 0, 15), (- 3, 8, 1). 4. Find the equation of a sphere passing through the origin and concentric with the sphere through the points (7, 7, 8), (_ i, _ 5, - 8), (- 5, 7, - 6), (3, - 5, 10). 5. Find the equation of a sphere with its centre at the origin and touching the sphere & + f + z 2 - 8 x - 6y + 2Az + 48 = 0. . 244 ANALYTIC GEOMETRY OF SPACE [Ch. V, § 27 6. Show that the equation of the sphere whose diameter is the line joining the points (a^ y x , z^ and (x 2 , y 2 , z 2 ) may be put in the form (x - x{) (x - x 2 ) + (y- 2/0 (y - y 2 ) + (z - z Y ) (z - z 2 ) = 0. 7. Show that the equation x 2 + y 2 + z 2 = r 2 will have the same form, if the axes are turned through any angle without changing the origin. 27. Conicoids. — Any surface whose equation is of the second degree in x, y, and z is called a quadric surface or conicoid. The sphere is a special case of such a surface. It is possible, by suitable transformation of coordinates, to reduce the general equation of the second degree in x, y, and z to one or other of these two forms, (1) Ax 2 + By 2 + Cz 2 = D, (2) Ax 2 + By 2 = Cz, where ^4, B, C, and D may be any quantities, positive, negative, or zero. But for our present discussion, let neither A, B, nor vanish. The locus of equation (1) is evidently symmetrical with respect to each of the coordinate planes, and hence with respect to the origin. Such surfaces are therefore called central quadrics. If D =£ 0, equation (1) may be written in the form ±^±g±*?=l. [27] a 2 b 2 c 2 u J If D = 0, it may be written Ch. V, § 28] QUADRIC SURFACES 245 Non-central quadrics arc included under equation (2). It may be written in the form g ± g=2~ [20] We shall now investigate the forms of the surfaces represented by these equations. 28. The ellipsoid. ^ + £? + ^=l._ The surface is a 1 ¥ c l symmetrical with respect to each of the coordinate planes. Its intercepts on the X, Y, and Z-axes are ± a, ± b, and ± c. The section of the surface made by the X-Y- plane is obtained by putting 2 = 0, and its equation is -5+^ = 1, which represents an ellipse with semi-axes a and b. The section made by a plane parallel to this coordinate plane is found by putting z = z v This gives ^ + t„ = l- Z 4, or t + t =1 a 2 b 2 c 2 of -j which represents an ellipse, in the plane z = z v with semi-axes a^l \ and byll — -L, the centre lying on the Z-axis. As z 1 increases numerically from to ± c, the section diminishes in size, until when z x = c it shrinks to a null ellipse, the single point (0, 0, a, the equation again represents an hyperbola, but the transverse axis is now parallel to the Z-axis. As x x increases, the semi-axes increase without limit. Ch. V, § 29] QUADRIC SURFACES 249 Similarly, the sections made by planes parallel to the Z-X-plane will be found to be hyperbolas, the transverse axis being parallel to the X-axis, when the distance of the cutting plane is less than b, and parallel to the Z-axis, when the cutting plane is beyond y = b ; the transition from one set of hyperbolas to the other being a pair of intersecting lines in the plane y = b. wm^mmmmmmmmmmmmmmmmmmmmmmmmtmmm Fig. 17. The surface is called the hyperboloid of one sheet, or the unparted hyperboloid, extending along the Z-axis. The equations -+ \- + % = 1 and -r- — &- + \ — 1 represent a 2 b l £ yyw y L If b = c, the equation becomes -- — &- — - = 1, which is a 1 c 2 2 n \lil nted by the equation '— — ^- = 2cz J i a 2 b 2 hyperbolic paraboloid. (See Fig. 22.) -v>2 n \lil sented by the equation '— — ^- = 2cz. It is called an J L a 2 b 2 PROBLEMS 1. Prove that in both the elliptic and hyperbolic parabo- loids the sections parallel to the X-Z-plane are equal parab- olas ; also that the sections parallel to the F-Z-plane are equal parabolas. 2. Show from the results of problem 1 that a paraboloid may be generated by the motion of a parabola, whose vertex moves along a parabola lying in a plane, to which the plane of the moving parabola is perpendicular ; the axes of the two parabolas being parallel, and (a) in the elliptic paraboloid, their concavities turned in the same direction ; (b) in the hyperbolic paraboloid, their concavities turned in opposite directions. 3. Show that an ellipsoid may be generated by the motion of a variable ellipse, whose plane is always parallel to a fixed plane, and which changes its form in such a manner that the extremities of its axes lie in two ellipses, which have a com- mon axis, and whose planes are perpendicular to each other and to the plane of the moving ellipse. 4. Find the equation of the cone, whose vertex is at the centre of an ellipsoid, and which passes through all the points of intersection of the ellipsoid and a given plane. 5. Find the equation of the cone, whose vertex is at the centre of an ellipsoid, and which passes through all the points common to the ellipsoid and a concentric sphere. 6. If a, b, c is the order of magnitude of the semi-axes of the ellipsoid in problem 5, and if the radius of the sphere is b, show that the cone breaks up into a pair of planes, whose intersections with the ellipsoid are circles. Ch. V, § 34] QUADUIC SURFACES 257 34. Ruled surfaces. — A surface, through every point of which a straight line may be drawn so as to lie entirely in the surface, is called a ruled surface. Any one of these lines which lie on the surface is called a generating line of the surface. The cylinder and cone are familiar examples of such surfaces. We shall now show that the imparted hyper- boloid and the hyperbolic paraboloid are also ruled surfaces. The equation of the unparted hyperboloid may be written in the form ^_z2 _y* a* c 2 6 2 ' or e+3(H)=K)M> If now we write the two equations a c k x \ oj in which k x may have any value, it appears that every point, whose coordinates simultaneously satisfy these equations, will satisfy the equation of the hyperboloid, and will therefore lie on the surface. But these two equations, used simultaneously, are the equations of a line, and, from what we have shown, that line must lie wholly in the surface. But since h l may have any value, there will be an indefinite number of such lines, and it may be easily shown that one of them passes through each point of the surface. 258 ANALYTIC GEOMETRY OF SPACE [Ch .V, § 34 In the same manner it may be shown that there is another set of lines whose equations are =K 1+ f) which lie wholly in the surface. A line of this set may also be passed through any point of the surface. Hence, through any point on this ruled surface, there may be passed two lines which lie wholly in the surface. Each line of one set cuts every line of the other set, but does not cut any line of the same set. Let the student show that the hyperbolic paraboloid is also a ruled surface. Figures 17 and 23 show the two sets of generating lines on both these surfaces. None of the other conicoids are ruled surfaces. PROBLEMS 1. Prove that, if a plane is passed through a generating line of a conicoid, it will also cut it in another generating line. Will the two generating lines belong to the same set ? 2. Prove that every generating line of the ruled paraboloid is parallel to one of the planes - ± *- = 0. a b 3. Obtain the equations of the generating lines which pass through the point (x lf y lf z^) of (a) the ruled paraboloid, (6) the ruled hyperboloid. 4. Prove that the plane, which is determined by the centre and any generating line of a ruled hyperboloid, cuts the sur- face in a parallel generating line, and touches the asymptotic cone in an element. 5. Show that, in both the ruled hyperboloid and the ruled paraboloid, the projections of the generating lines on the prin- cipal planes are tangent to the principal sections. Ch. V, § 35] QUADRIC SURFACES 259 35. Tangent planes. — A tangent line to a surface may be defined as follows : Through P x and P 2 , two adjacent points on the surface, draw a secant line. The limiting position, which this secant approaches as P 2 approaches P v is called a tangent line to the surface at the point P v Since P 2 may approach P x along the surface in an indefinite number of ways, there will be, in general, an indefinite number of tangent lines at any point of a sur- face. These will, in general, lie in a plane which is called the tangent plane at the point P v We shall obtain the equation of the tangent plane at the point P x of the ellipsoid ^-|_£? + ^ = l. a 2 b 2 c 2 Transforming this equation to parallel axes with the origin at P x (by [12]), and then to polar coordinates (by [6]), we have as the equation of the ellipsoid in polar coordinates (origin at P 2 ) 2 / cos 2 a cos 2 /3 cos 2 7 9 \ a 2 b 2 ~J~ l2p ( x i cosa , fficosff ( giCQS7 \ =0> V a 2 b 2 c 2 J For every set of values of a, /3, 7 in this equation there will correspond two values of p ; one value will always be zero, which agrees with the fact that the origin is a point on the surface : the other value is — _ 9. f x x cos a y x cos ft z l cos 7 V a 2 b 2 c 2 . cos 2 /3 cos 2 7 b 2 c 2 260 ANALYTIC GEOMETRY OF SPACE [Ch. V, § 35 which gives the distance from P x to any second point P of the ellipsoid, measured along the secant whose direc- tion angles are a, /3, 7. Now let P 2 approach P x along the ellipsoid ; then the secant line through P 1 and P 2 will approach as its limit- ing position a tangent line at P v whose direction angles we shall call «', fi\ 7'. That is, as P 2 approaches P v , -. x, cos a , y, cos 6 , z, cos 7 approaches zero, and — l — h ,«, + - 1 — 5 — - ap- Hence, by the theory of limits, x x cos (t! y x cos ft z x cos y r _ ~ a*~ ¥~ c 2 " * If (/o', «', ft, 7') are the polar coordinates of any point on any one of the tangent lines through P v this equation expresses the only relation which must hold between those coordinates, and is therefore the polar equation (referred to P 1 as origin) of the locus of the tangent lines through P r Multiplying by p' and transforming to rectangular coordinates (by [6]) we have ^ + M + fb| = 0. Again a 2 b 2 c 2 transforming to the original origin (by [12]), we have a 2 b 2 c 2 L J as the required equation of the tangent plane. Let the student show that the equations of the tangent planes to the hyperboloids, a 2 b 2 c 2 ' are «*g ± tm-*& = 1 . [31] a 2 b 2 c 2 L J Ch. V, § 3G] QUADRIC SURFACES 201 the paraboloids, — ± ^- = 2 cz, a 1 ¥ are *& ± U& = e (.* + *d* [32] 36. Normals. — The line perpendicular to the tangent at the point of contact is called the normal to the surface at that point. Its equation for any particular surface can be easily obtained from the definition. PROBLEMS 1. Prove that every tangent plane to a cone passes through the vertex. 2. Prove that all the normal lines of a sphere pass through the centre of the sphere. 3. Show that the length of a tangent to a sphere from the point (#!, yn z x ) is the square root of the quantity obtained by substituting (a^, y x , z{) for (x f y, z) in the equation of the sphere. 4. Show that the locus of points from which equal tangents may be drawn to a sphere is a plane. This plane is called the radical plane of the two spheres. 5. Prove that the radical planes of three spheres meet in a line. This line is called the radical axis of the three spheres. 6. Prove that the radical plane of two spheres is perpen- dicular to their line of centres. 7. Prove that the radical axis of three spheres is perpen- dicular to the plane of their centres. 8. Show (from its definition) that the tangent plane at a point P x of a ruled surface contains the two generating lines of the surface which pass through P v 262 ANALYTIC GEOMETRY OF SPACE [Ch. V, § 37 9. Prove that every plane which contains a generating line of a ruled surface is tangent to the surface at some point on the generating line. 37. Diametral planes. — The locus of the middle points of a set of parallel chords of a quadric surface will be found to be a plane. This plane is called a diametral plane. Let a v fi v y x be the direction angles of a set of parallel chords in the ellipsoid, and let (V, y f ', z') be the coordi- nates of the middle point P' of any one of these chords. Transform the equation of the ellipsoid to polar coor- dinates with P' as origin. Its equation (by [12] and [6]) is 2 / cos 2 ft cos 2 /3 cos 2 7\ 9 / Vcos a ?/'cos/3 g'cos y P \~aT + b* + c 2 J P \ a 2 V 3 , «', /3', 7'), and on it locate a point P' (//, a', /?', 7') such that p pi _ 2P \ P 2 X P\ P * or ,/ _ 1m.. P 1 P* + P 1 P Z ' P R 2 + p s Then /? 3 and /9 3 are evidently the roots of the equation /cos 2 «' cos 2 8' cos 2 7 f \ Hence (by Introduction, Art. 8) a 2 "*" 6 2 "*" c 2 P '=- x x cos a' y x cos /3' 2; x cos 7' - + This is an equation connecting the polar coordinates of P\ and is, therefore, the polar equation of the desired locus. Transforming to rectangular coordinates and to the original origin, Ave have, as the equation of the polar plane, a Z + b l + C 2 ~ 1- L dt) J Let the student obtain the equation of the polar plane of a point with respect to each of the quadric surfaces. 266 ANALYTIC GEOMETRY OF ST ACE [Ch. V, § 38 PROBLEMS 1. Prove that the polar plane of 1\ with respect to any quadric surface passes through the points of contact of all the tangent lines from P x to the surface. 2. Prove that the polar planes of all points in a given plane pass through the pole of that plane ; and, conversely, the poles of all planes passing through a given point lie on the polar plane of that point. 3. Prove that the polar planes of all points on a given diameter of a quadric surface are parallel to the tangent plane at the extremity of the diameter. 4. Prove that the polar plane of P x with respect to a sphere is perpendicular to the diameter through P v 5. Prove that in the sphere the product of the distance of the pole from the centre and the distance of the polar plane from the centre is equal to the square of the radius. 6. Prove that the distances of two points from the centre of a sphere are proportional to the distances of each from the polar plane of the other. LOCI PROBLEMS 1. Find the locus of points which are equally distant from two intersecting planes. Show that it consists of two planes which are perpendicular to each other. 2. Show that the locus of a point, the sum of the squares of whose distances from any number of points is constant, is a sphere. 3. A point moves so that the sum of the squares of its dis- tances from the six faces of a cube is constant; show that its locus is a sphere, Ch. V, § 38] QUADRIC SURFACES 267 4. ^l and B are two fixed points, and P moves so that PA = nPB-, show that the locus of P is a sphere. Show also that all such spheres, for different values of n, have a common radical axis. 5. Show that the locus of the point of intersection of three mutually perpendicular tangent planes to an ellipsoid is a sphere about the centre of the ellipsoid, whose radius is Va 2 + b 2 + c 2 . 6. Show that the locus of the point of intersection of three mutually perpendicular tangent planes to a paraboloid is a plane. 7. Find the locus of a point whose distance from a given point bears a constant ratio to its distance from a fixed plane. 8. Three fixed points on a line lie, one in each coordinate plane ; find the locus of any fourth fixed point of the line. 9. Show that the locus of the points, which divide in any g.ven ratio all straight lines terminated by two fixed straight lines, is a plane. 10. A line of constant length has its extremities on two fixed straight lines j show that the locus of its middle point is an ellipse. ANSWERS, Page 10 *■ « (!• §)• (-!• I)- (-!• -!)• (I- -|> |V2,o). (o, |V5). (~|V2, o)» (o, -|V2~)' (ft) 5. (a) (a, 0), (6, c), (a + b,c). (b) (a, 0), (0, c), (a, c). 6. (a) (0,0), (0,0), (|, ^ a y w (1- o). (-1- o), (o, *.). (0 (f V3, 0), (-gvj. |), (~§A -I)- Page 14 2. \'34> Vl30, 2\/29. 3. 5V3, 2Vl3, V7. 4. Va 2 + 6 2 , Va 2 + 6 2 + a&V2. Page 18 2. (-19, -16). 12. f, --V-. 3. (-11,2). 13. (-9,6). 10. (11,5). 14. (a) (V, -|). (6) (1, -1). Page 23 1. 6x-4?/ + 19 = 0. 6. 2 x 2 + 2 ?/ 2 + 14 x - 19 y +55 = 0. 2. y = 3x. 7. x + ?/- 10 = 0. 3. x 2 + ?/ 2 + 6 x - 8 y = 0. 8. x 2 - 3 ?/ 2 = 0. 4. 24x 2 + 25 ?/ 2 - 250 x + 625 = 0. 9. 8x-2?/+17=0. 5. x 2 + ?/ 2 - 5 a: + 5 y + 5 = 0. 10. x 2 + y 2 - x - y = 0. Page 32 3. 2V5; |V2; jVl70, 6. &<£; 6>i; & = fr 5. 6, 7. x 2 - i/ 2 = 0. 269 270 ANSWERS Page 36 2. (a) 5x + 8y = 7. (c) x - 4 = 0. (6) 3x-4y = 0. (tf)y-5 = 0. 3. x~Sy = S. 4. Yes. No. 5. 2/i (Xa - £3) +2/2 (£3 - »i)+ 2/3 (xi - x 2 ) = 0. 6. 39 x - 79 y = 200. 7. Equations of medians, x — y — 1 = 0, x + 2 y + 1 = 0, x - 13 y - 9 = 0. Point of intersection, (i, — f). 8. Equations of diagonals, bx + ay = ab, bx — ay — 0. Point of intersection, [-» - V V2 2J Page 40 1. \/3x--j/=-6(V3 + l). 2. (a) >/3x-3y = - 18. (6) 3x - 5 y = 25. (c) x-y=-3 3. 7V3x + 7y = llV3-2. 4. -2. Page 43 2. (a) a = 10, b = - f , I = J. 4. (a) x - 8 y + 5 = 0. (6) « = -*, b = l,l = l (6)2*- y-2 = 0. (c) a = 0, 6 = 0, Z = - 4. (c) 3 x + 5 y = 0. (d) a = — 4, & = 00, Z = ao Page 46 2. tan- 1 (2^|). 3. 135°, tan-!(7), tan-i(6$), tan-i(lf). 4. Exterior angle between first two lines, tan _1 (— |f). Opposite inte- rior angles, tan" 1 (||), tan-!(- - 2 ^). Page 47 2. 7x-3y = ll. 3. (a) y = 0, 4 x + y - 24 = 0, 2 x - y = 0. (&) x - 4 y = 0, x + 2y-6 = 0, x - 4 = 0. (c) x - 3 = 0, x - 4 y + 11 = 0, x + 2 y - 10 = 0. (d) 4 x - 5 y = 0, x + y - 6 = 0, 8 x - y - 24 = 0. Page 49 \. (6 + 5V'3)x-.-3y = 0. g. 7 x - 3 y + 5 = 0. ANSWERS 271 Page 51 1. (a) x + y/3 y - 10 = 0. (6) x - V3 ?/ + 10 = 0. (c) VSx-y + 10 = 0. (d) a + y = 0. 0) 4 x + (2 + 2 V3)y - 4 V2 = 0. (/) 2 x - ( V2 -f V'0)y - 24 = 0. 1. T 3 3\/l3. 4. 4z + 3?/-3 = Page 54 2. - 1. (H. -tt); 7|. 3. 11$. 6. T 2 T 5 3 \/lT3 ; the first. Page 57 1. x + 7y + 3 = 0',7x—y 17=0. 2. llz-35y=0. 3. 49 x + 98?/- 272 : 4. 3 x + 2 ?/ + 7 = 0. Page 58 ; 15 x - 18 y - 320 = 5. V3z-i/ + 3-V3 ; 4 z + 5 ?/ :0. 6. 10 X- 3=0. 3 y + 4 = 0. Page 60 1. 32. 3. 3263 Page 60. General Problems 2( 6 41"\ . /"78 23\ • VT7> T7J> UT» IT) 3. 4. 6. 7. 9. 11. 15. 2 x + 5 y = 40 ; 18 a + 5 y = 120 ; 6(2 + V7> + 5(2 ± y/l)y = 120. 5x-y+10 = 0. 5. 2x + 3?/ + 12 = 0;6x + ?/-12 = 0. (a) x — 4 y = 0. (31, 5±|V3). 8. (10,51). (16f, -0&); (4|, -jf). 10. (8A* 1H)J 4 tt, 3&). /5VI97 ± 3V82 4V197 ± 14 V82\ V Vl97 +V82 / 4V29 + 3V82 t * V29+V82 / 4V29 + 5V197 V V29 + VI97 5 x + y = 26. Vl97±\/82 / 7V29 + 7V82 \. V29 ± V82 / ' - 7\Z29±2Vl97 V29+VI97 (Cl - C 2 )2 2 (mi — m 2 ) 17 18. 5. 21. (5,5). Page 70 2. 4 x - 3 y + 15 = 0. 272 ANSWERS Page 71 1. Sx+ Ty - 10 = :0. 5. (- -4, 2). 3. xy = 5- 6. 11 tan-i(- ■2). 4. tan- 1 f. Page 74 2. (a) (&) p2 : « 2 &2 4 . (a) x 2 + y 2 - ay (b) (x-b)Vx* (c) x 2 +y 2 -ax- = 0. + *■ = ax. a 2 cos 2 tf + & 2 sin 2 2 m cos sin 2 - a Vx 2 + y~ = 0. («) P 2 -- a 2 (d) (x* + y*y = (e) x 2 + y 2 + bx- a 2 (x 2 - -aVs 2 ■ y 2 )- + 2/ 2 cos 2 = 0. w (/) (so P 2 -- P = P = P = = a 2 cos 2 0. = a sin 2 0. = — a cos 0. 2 a - cot 2 COS0 (/) (* 2 + ? 2 ) 8 = fa) x 2 - ?/ 2 = a 2 . (0 (x2 + y 2 )3 = 4 a 2 xV 2 (x 2 + 2 4a¥. 2 xy — */ 2 ) 2 . (/O P = = a(l — 2cos0). ( j) z 3 + ^ 2 - 2 a*/ 2 = Page 78 1. (a) x 2 + 2/ 2 + 4 x - 6 */ - 23 = 0. (b) x 2 + 2/ 2 + 6x + 8y = o. (c) x 2 + */ 2 - 10 x - 6 ?/ + 33J-f = 0. (d) x 2 + y 2 - 36 x - 32 -y + 480 = 0, or x 2 + i/ 2 - 4 x - 8 y - 80 = 0. (e) 19 x 2 + 19 y 1 + 2 x - 47 */ - 312 = 0. (/) 3 x 2 + 3 y* - 13 x - 11 y + 20 = 0. (flO x 2 +«/ 2 -x-42/-6 = 0. (h) 3x 2 + 3y 2 - 114 x- 64 y + 276 = 0. (*) x2 + 2 /2_ (5± 9VT|)x-(3±5VH)2/-18±30VH: 2. (a) (-4, 3); V35. (c) (0, -3); 5. (6) Imaginary. (d) (£, 0); |Vl45. 6. 6 x + 3 y - 10 = 0. Page 82 1. (a) 3 x + 4 y = 25 ; 4 x - 3 y = 0. (6) Indeterminate. (c) 3 x + 7 y = 93 ; 7 x - 3 y = 43. 3 x - 7 */ = 65; 7 x + 3 */ = 55. (d) 6 x + 5 */ = 114 ; 5 x - 6 y = - 27. 6 x - 5 y = 44; 5 x + 6 ?/ = 57. 2. 45°. 5. V4L ANSWERS 273 Page 84 1. (a) (21±4V51)x+(28^3\/51)y = 350. (b) x + 5 y - 28 = j 5 x - y + 10 = 0. (c) 2 x - // = 15 ; 58 x + 71 */ = 335. 2. (a) 6 x + 8 ?/ - 49 = 0. (c) 14 x + 3 y = 55. (6) 2 x - 3 ?/ + 9 = 0. 3. xi x + yi y = r 2 . Page 85 1. (a) 3a:-2y±7Vl3 = 0. (6) 2x + 3 y ± 7V13 = 0. 2. 3x + ?/ + 9±3VT0 = 0. 3. 1 + 1 = 1. r- a 2 b~ r 1 4. A: = 3G±20v / G. 5. A 2 E 2 + B 2 D 2 + 4BCE-2ABDE-4A 2 F-4 C 2 + 4 ACD-4 B 2 F=0. Page 86 1. f>/2T. 2. x 2 + ?/ 2 -f 52 x - 21 y- 265 = 0. 3. 2 x 2 + 2 ?/ 2 - 13 x - 6 ?/ + 15 = 0. 64V26 + 375\ 2 , / 2 V26 -f 435 \ 2 _ / 301 no/ V: 25V26+170' \ 25V26 + 170/ \25>/26 + 170/ 6. -^-. 7. 2x + y = 2; x - 2 y = ; (0,0), (f, |); tan"* 2. 2x01 Page 89 3. Perpendicular bisector of the line joining the two points. 5. Circle of radius r about (xi, y\). 6. Perpendicular bisector of the line joining the two points. 7. Bisector of the angle between the lines. 8. Circle about the centre of the square. 9. Circle whose centre is on the line through the fixed point, perpendic- ular to the fixed line. 11. Circle whose centre is on the base of the triangle, extended. 12. Circle whose centre is at the centre of the triangle. 14. Circle whose centre is at the intersection of the two lines. 15. Circle whose centre is on the line OX. 18. Line through the centre of the base and the centre of the altitude of the triangle. 19. A straight line. 20. Two lines through the origin. 21. A line through the origin. 22. A circle. 23. x 2 + y 2 - rVx 2 + y 2 = ry. 274 ANSWERS 24. A diagonal of the rectangle. 25. A diagonal of the parallelogram. 30. x 2 + y -2 _ _1«_ {XlX + yiy) = ( tULz m + n \m + n, 31. xix + y x y = r 2 . 32. An equal circle tangent to the given circle at the fixed point. 33. (x 1 2 +y l 2 -r 2 )[(x-xO 2 +(y-y l ) 2 ] + 2k 2 (x l x + y 1 y-x 1 2 -y{ 2 ) + tf = 0. 34. A line parallel to the fixed line. 35 A circle. Page 103 1. (a) y 2 = - 2 mx. (6) x 2 = 2 my. (c) x 2 = — 2 my. 2. y 2 = 2 mx + m 2 . 4. 4 x 2 = — 9 y. 3. G/-/3)2 = 2m(x-a). 5. x = -2 ; (2, 0); 8. Page 110 1. (a) a = 3, & = 2, c = V5, e = |V5, x = ± — -• V5 (6) a = 3, 6 = 2, c=V5, e = | V5, y=±-iL v5 (c) a = |V30, 6 = 1V5, c = | ^3, e = \ VTO, * = ± f V3. 2. ( a ) 4 *2 + 9 y 2 = 36. (d) 16 x 2 + 25 y 2 = 400. (6) 3 x 2 + 4 ?/ 2 = 36. (e) 16 x 2 + 25 i/ 2 = 400. (c) 5 x 2 + 9 ?/ 2 = 180. (/) 8as* + 9y 2 = U52. 5. 3x 2 + 7 1 / 2 = 55. 6. il^)l + iyj=jy = l. a 2 b 2 Page 115 1. (a) a = 5, 6 = 1, c = V26, e = | V20, & = ± — , a; ± 5 y = 0. V26 (6) a = 2, 6 = 3, c=Vl3, e = |Vl3, ac=±-4=, 3 x ± 2 y = 0. Vl3 (c) a=Vl0, 6=2, c=VH, e = iV35, x=±fVl4, 2x±VlO>=0. 2. («) 4 x 2 - 9 i/ 2 = 36. (d) 16 x 2 - 9 y 2 = 144. (6) 3 x 2 - y 2 = 9. O) 9 x 2 - 16 y 2 = 144. (c) 5 x 2 - 4 y 2 = 125. (/) 72 x 2 - 9 y 2 = 800. 3 (s-«) a _(y-fl) 8 = ;L 4 Impossible . a 2 6 2 Page 118 1. 4 x 2 — y 2 = - 4 ; a = 1, 6=2, e = |V5; latus rectum = 1 ; foci, (0, ± V5) ; directrices, y = ± % V5. 2. V2. 4. 2xy = a 2 . ANSWERS 275 Page 123 1. -y-v^; f^- 2 - |Vl3±|V66. Page 130 1. (a) 3 ac + 8 y = 19 ; 8x-3y = 2. (b) 3x + y = 7 ; x - 3 y = 9. fr) x+2y = -6; 2»-y = 18. (d) 5 x - y = - 8 ; 6 & + 5 y = 27. 2- (a) ¥; -I- (&) -l! 6. (c) -12; 3. 8. (a) y = 4; 3& + 2y = 17. 4. (a) 12 x + 25 y = 100 (b) x-y=-l; x + Sy=-9. (6)x4^3. (c)» + 3y = 5;as-3y = -7. (c) y = 3. 5. (a) |\/73; ^V73. (6) fVTO; 2vT6. (c) 6 a/5; 3V5. 6 tan-i(± 3). Page 132 1. x-22/iVTT^O. 5 . (™, ±ml; 2. 18 x + 27 y = 88. V 2 ' 3. 5x+2/_ V Va 2 +& 2 vW& 2 / 4. /3 = ± V& 2 - a 2 Z 2 . ( x « 2 , ± ^ \ V Va 2 - & 2 Va 2 - bV Impossible in hyperbola when b > a. V V2 V2/ 8. 5 y ± xVlE ± 4\/l0 = 0. Four tangents. 9. 1 y ± 2 x V35 ± 4V91 = 0. Four tangents. 10. x ± yy/S + 6 = 0. Two tangents.. 12. (a) 4. (&) 3. Page 148 1. 3 x - 8 y = ; x - 3 y = 0. 5. 20 x + 33 y = 125. 2. */ + 9 = 0. 6. 8x + 45y = 0;( 45 _ , § — V 3. 2*V§ + 3y = 0. 7. ^V3. ^±Vl51 T Vl61/ 4. x + 2 y = 8. 8x-y = l. 9. (± i Vl5, T i\/l5); (± § Vl5, T t^ Vl5). 10. h=- 12- Page 158 1. (a) x - 8 y = 16. (c) 15 x + 16 y = - 24. (b) x + 2 g = - 6. Cd) x + 5 = 0. 276 6. (-If, H). ANSWERS 4. (-10,4). 5. (-^, W\. 7 / a 2 b 2 Xi a*b 2 y\ \ \6 2 a;i 2 + a-y{ 2 b' 2 x{ 2 + cPyi 2 ) 1. Imaginary ellipse 2. Real ellipse. 3. Two intersecting lines. 4. Hyperbola. 5. Two parallel lines. Page 182 6. Parabola. 7. Two coincident lines. 8. Point. 9. Point. 10. Hyperbola. Page 186 4. x-. 2k 2. The directrix. 5. (a) x 2 + y 2 = a 2 ; (6) x 2 + y 2 = a 2 ; (c) x = 0. 7. The asymptotes. v 2 ' 6* e. ^+r-2. 10. ft 2 * 2 + a-Y 2 = 6 2 c 2 . 12. 2/^-— x. a 4 6* a 2 6 2 11. («) An ellipse ; (6) A parabola. 14. 25 x 2 + 16 y 2 - 48 y - 64 = 0. 15. 2 r Vx J + y l — 2 x x x - 2 ?/ii/ + Xi 2 + y{ 2 — r 2 = 0. 16. x 2 x + fy-| = 0. 3 x 2 — ?/ 2 — 2 ex = 0. (Take the origin at the vertex of the smaller angle. ) 22. A parabola. 24. 2 xy — y\X — Xiy = 0. 27. y 2 = -2 mx. 30. x =-r>L±m. 2 32. (x 2 + y 2 ) 2 = d 2 x 2 + b 2 y 2 . 34. 4 &! 2 xV 2 - 4 rtiV = «i 2 &i 4 - 36. A directrix. 23. x 2 + ?/ 2 =-. 26. (x 2 + y 2 -2 ax) 2 = a 2 (a-x) 2 +a'V 2 28. ab' 2 Vx 2 + 2/ 2 = 6 2 x 2 + d 2 y 2 . 31. x 2 + y 2 -( c2 + m ' 2 )x + c 2 = 0. a 4 6 4 35. Circle with radius a + b. 37. *f^+ W 1 + * + * = !■ 38. *- + ^=l. a 2 a 4 . ft 2 ANSWERS Page 211 V^O; (6, -10, 20). 2. (*L±**±** Vi + ** + &, *i-r*2 + z*\ \ 3 3 3 / COS" a i V3. Page 214 COS ■/s _ 3 Vn _ -4 , cos 7 1 VTT 7 a = P = 7 = cos « = = , Vll cos « = p , cos /3 = — —-, cos 7 = Voo VUG Voo 1 2 3 cos « = — — , cos /3 = - — , cos 7 = — — « Vl4 Vl4 Vl4 60°. Page 217 COS -1 -, COS" 6 -, COS ■'!)■ 4. (fV3, |, 0); (|V3, 6. cos" 1 £V15. o, I). Page 220 3. y 2 + z 2 = 25 ; x 2 + z 2 = 25. 4. a 2 + ?/ 2 + z 2 - 6 x - 4 y - 10 z + 13 = 0. 5. 7s + 7y+10z = 9;2a;-y-72=-30. Page 223 2. x 2 + z 2 - 6 x - 10 2 + 9 = 0. 3. (a) y 2 + z 2 = 16. (c) x 2 + y 2 + z 2 = r 2 . (6) x 2 - y 2 - z 2 = 0. (cZ) y 2 + z 2 = 2 ma;. 4. & 2 x 2 - a 2 y 2 - a 2 z 2 = a 2 b 2 ; 6 2 x 2 - a 2 */ 2 + b 2 z = a 2 b 2 . Page 230 1. (a) JL, -=L*, _1_. (6) J§_. V74 V74 V74 V74 2. 3 x + y - 5 2 = 35. , . 19 V74 00 4, - 6. 3 x + 59 y — 72 z = 0. Page 233 1. (a) 11 » - 17 y - 13 2 + 3 = 0. (&)4s-7y-5s-£ 2. 12 x- 17?/ + 42 -3 = 0. 5. 2se + z-8 = 0. 3. y + 3 z + 3 = 0. 6. 10 x - 19 y - 32 z = 0. 4. 7x + 4y-4z-22=0. = 0. 278 ANSWERS Page 236 1. x = y = 0; y = z = 0; z = x = 0. 2. -s = 0, Ax + 2ty + G = 0, etc. ; 2 = &, ^4x + #y + C = 0, etc. ; s = A*i, ?/ = &2, etc. 4. (a) x = $z + l y = iz-i. (d) y = 2, z = - 3. (?)) a: = -■¥*, y = ^2 + 4. (e) * = fy--l, 2 = 1. (c) e = - 4, a = - 2 y + 3. (f) y = -l, z = 2. 5. s = 0, « = f y + 3 ; x = 0, z = 3y + 6; y = 0,z=-2x + fi. Page 238 2. (a) re = */ - 3, x = \\/2z + |V2 - 1. (ft) se = -y+l, x = \\'2 z + l\/~2 - \. (r) x = V3 y - 2 V3 - 1, z = - 3. 3. x = y = z. Page 240 3- (a) ?, =A jj. (c) 4=i =1. 0. 7 y/b V5 3' 3' 3* (&) It % % (<0 1> 0, Page 241 3. (a) x = -3y, (b) x = f z - 4, (c) a; = 2, e = - 2 y - 2. y = - f * + 8. y = - 3. 4. x = 0,2 y + z = 0. 8. (4,1,-2). 9. 12 x - by - bz - 31 = Page 243 1. (a) x 2 + y* + z 2 - 10 x + 4 y - 6 z + 37 = 0. (6) x' 2 + */ 2 + 2 2 - 4 x + 6 y + 12 z = 0. (c) x 2 + if- + 2 2 ± 2 ax = 0. 2. (a) (1,-3, 4), 2. (d) (-3, 0,0), 3. (b) (-5, 2, 1), 5. (e) Imaginary. (c) (-2, -2, -3), 4. 3. (a) Indeterminate ; points lie in a plane. (b) x 2 + y 2 + z 2 -2x-8y-16z = 0. 4. x 2 + y 2 + z*-2x-2y -2z = 0. 5. x 2 + y 2 + s 2 = 4. 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