<^ 
 
 \I^ 
 
 V 
 
ELEMENTS 
 
 .Analytic Geometry. 
 
 BY 
 
 G. A. WENTWORTH, A.M., 
 
 PROFESSOR OF MATUE.MATKS IN VUILLU'S EXETER ACADEMY. 
 
 I 
 
 BOSTON: 
 
 PUBLISHED BY GINN & COMPANY. 
 
 188G. 
 
Entered, according to the Act of Congress, in the year 1886, by 
 
 G. A. WEXTWORTH, 
 
 in the Office of the Librarian of Congress, at Washington. 
 
 J. S. CusuiNG & Co., Pkinteks, Boston. 
 
PREFACE. 
 
 rpHIS book is intended for beginners. As beginners generally find 
 great difficulty in comprehending the connection between a locus 
 and its equation, the opening chapter is devoted mainly to an attempt, 
 by means of easy illustrations and examples, to make this connection 
 clear. 
 
 Each chapter abounds in exercises ; for it is only by solving 
 problems which require some degree of original thought that any 
 real mastery of the study can be gained. 
 
 The more difficult propositions have been put at the ends of the 
 chapters, under the heading of "Supplementary Propositions." This 
 arrangement makes it possible for every teacher to mark out his own 
 course. The simplest course will be Chapters I.-III. and Chapters 
 V.-VII., with Review Exercises and Supplementary Propositions left 
 out. Between this course and the entire work the teacher can exer- 
 cise his choice, and take just so much as time and circumstances will 
 allow. 
 
 The author has gathered his materials from many sources, but he 
 is particularly indebted to the English treatise of Charles Smith. 
 Special acknowledgment is due to G. A. Hill, A.M., of Cambridge, 
 Mass., for assistance in the preparation of the book. 
 
 Corrections and suggestions will be thankfully received. 
 
 G. A. WENTWORTH. 
 Phillips Exeter Academy, 
 July, 1886. 
 
 184022 
 
CONTENTS. 
 
 CHAPTER I. Loci and their Equations. 
 
 PAOB. 
 
 1-4. Preliminary Notions 1 
 
 5. Circular Measure of an Angle 5 
 
 6-7. Distance between Two Points 6 
 
 8-9. Division of a Line 8 
 
 10-16. Constants and Variables 10 
 
 17-24. Locus of an Equation 14 
 
 25-29. Intersections of Loci 22 
 
 30-34. Construction of Loci 25 
 
 35. Equation of a Curve 31 
 
 Review Exercises .• . 33 
 
 CHAPTER n. The Straight Line. 
 
 36-41. Equations of the Straight Line 36 
 
 42-47. General Equation of the First Degree 43 
 
 48-51. Parallels and Perpendiculars 47 
 
 52-53. Angles 50 
 
 54-56. Distances 54 
 
 57-58. Areas 59 
 
 Review Exercises 62 
 
 59-67. Supplementary Propositions 67 
 
 CHAPTER in. The Circle. 
 
 68-70. Equations of the Circle 77 
 
 71-75. Tangents and Normals 83 
 
 Review Exercises 90 
 
 76-87. Supplementary Propositions 95 
 
VI CONTENTS. 
 
 CHAPTER IV. Different Systems of Co-ordinates. 
 
 SKCTIONS. p^eH 
 
 88-90. Oblique Co-ordinates 105 
 
 91-92. Polar Co-ordinates 107 
 
 93-100. Transformation of Co-ordinates HI 
 
 Review Exercises 117 
 
 CHAPTER V. The Parabola. 
 
 101-108. The Equation of the Parabola 119 
 
 109-112. Tangents and Normals 126 
 
 Review Exercises 131 
 
 113-127. Supplementary Propositions 134 
 
 CHAPTER VI. The Ellipse. 
 
 128-140. Simple Properties of the Ellipse 145 
 
 141-147. Tangents and Normals 155 
 
 Review Exercises . . _ 161 
 
 148-164. Supplementary Propositions 163 
 
 CHAPTER VII. The Hyperbola. 
 
 165-173. Simple Properties of the Hyperbola 180 
 
 17-4-179. Tangents and Normals 187 
 
 Review Exercises 189 
 
 180-197. Supplementary Propositions 190 
 
 CHAPTER VIII. Loci of the Second Order. 
 
 198-200. General Equation of the Second Degree 201 
 
 201-202. Loci of the Second Order that have a Finite Centre . . 207 
 
 203. Loci of the Second Order that have no Finite Centre . . 210 
 
 204. Special Cases of Loci 215 
 
 205. Summary 217 
 
 206-208. Examples and Definitions 217 
 
 Exercises •. . . . 221 
 
AJN^ALYTIC GEOMETRY, 
 
 CHAPTER I. 
 
 LOCI AND THEIR EQUATIONS. 
 
 Preliminary Notions. 
 
 1. Let XX' and YY' (Fig. 1) be two fixed lines perpen- 
 dicular to each other, and intersecting in a point 0. These 
 lines divide the plane in which they lie into four similar parts. 
 
 ■n 
 
 E 
 
 P 
 
 
 P, F 
 
 X' 
 
 IN 
 
 
 
 M 
 
 K 
 
 
 IB 
 
 T 
 
 
 3... 
 
 v 
 
 Fig. I. 
 
 A H 
 
 Let these parts be called Quadrants (as in Trigonometry), and 
 ^distinguished by naming the area between OX and OF the 
 \ first quadrant ; that between Y and OX' the second quad- 
 rant ; that between OX' and F' the third quadrant ; and 
 that between OY' and OX i\ie fourth quadrant. 
 
 Suppose the position of a point is described by saying that its 
 
2 ANALYTIC GEOMETRY. 
 
 distance from YY', expressed in terms of some chosen unit 
 of length, is 3, and its distance from JTX' is 4. It is clear 
 that in each quadrant there is one point, and only one, which 
 will satisfy these conditions. The position of the point in 
 each quadrant may be found by drawing parallels to YY' 
 at the distance 3 from YY\ and parallels to XX' at the dis- 
 tance 4 from XX' ; then the intersections Pj, F^, Pz, and P^ 
 satisfy the given conditions. 
 
 
 :l!> , 
 
 T 
 
 ,B 
 
 E 
 
 ■f. 
 
 
 ^F 
 
 X' 
 
 \N 
 
 
 
 M: X 
 
 
 ..i^....„. 
 
 T 
 
 \l\ 
 
 G 
 
 \c 
 
 Y' 
 
 & H 
 
 Fig. I. 
 
 2. In order to determine which one of the four points, 
 Pi, P-i, P3, Pi, is meant, we adopt the rule that opposite 
 directions shall be indicated by unlike signs. As in Trigo- 
 nometry, distances measured from YY' to the right are con- 
 sidered positive ; to the left, negative. Distances measured from 
 XX' upward are positive; downward, negative. Then the 
 position of Pi will be denoted by +3, +4 ; of P„ by -3, +4 ; 
 of P3, by -3, -4; of P„ by +3, -4. 
 
 3. This method of determining the position of a point in a 
 plane is the method commonly employed in Analytic Geom- 
 etry. It enables us to represent position by numbers ; and^ 
 by reasoning with these numbers, to investigate the properties' 
 of geometrical figures. The science of Analytic Geometry 
 consists of investigations of this kind. 
 
LOCI AND THEIR EQUATIONS. 6 
 
 Note. The first man who employed this method successfully in 
 investigating the j>roperties of certain figures was tlie Fre«ch jihiloso- 
 pher Descartes, whose work on Geometry appeared in the year 1(137. 
 
 Co-ordinates ; XX' is called the Axis of Abscissas, or Axis of x ; 
 YY', the Axis of Ordinates, or Axis of y. The intersection is 
 called the Origin. 
 
 The two distances (with signs prefixed) which determine 
 the position of a point are called the Co-ordinates of the point ; 
 the distance of the point from YY' is called its Abscissa; and 
 the distance from XX', its Ordinate. 
 
 The letters x and ?/ are in common use as general symbols 
 or abbreviations for the words "abscissa" and "ordinate" 
 respectively. For the sake of brevity, a point is often rep- 
 resented algebraically by simply writing the values of its 
 co-ordinates within brackets, the value of the abscisea being 
 always written first. 
 
 Thus Pi (Fig. 1) is the point (3, 4), F, the point (-3, 4), 
 Ps the point (— 3,'^^4), and P4 the point (3, —4). In gen- 
 eral the point whose co-ordinates are x and 7/ is the point 
 
 (^,y)- 
 
 / Ex. 1. 
 
 1. "What* are the co-Qi*dinates of the origin ? 
 
 2. In what cjuadrants are the following points (a and b 
 being given lengths) : 
 
 (~-a,-h),: (-a,h), (a,b), {a,-b). 
 
 3. To what quadrants is a point limited if its abscissa is 
 positive ? negative ? ordinate positive ? ordinate negative ? 
 
 *4. In what line does a point lie if its. abscissa = 0? if its 
 ordinate = ? 
 
 5. A point {x,y) moves parallel to the axis of .r ; which one 
 of its co-ordinates remains constant in value ? 
 
4 ANALYTIC GEOMETRY, 
 
 6. Construct or p/onhe points: (2,3), (3,-3), (-1,-3), 
 (-4,4), (3,0), (-3,0), (0,4), (0,-1), (0,0). 
 
 Note. To plot a point is to mark its proper position on paper, when 
 its co-ordinates are given. The first thing to do is to draw the two 
 axes. The rest of the work is obvious after a study of Nos. 1-3. 
 
 7. Construct the triangle whose vertices are the points 
 (2,4), (-2,7), (-6,-8). 
 
 8. Construct the quadrilateral whose vertices are the points 
 (7,2), (0,-9), (-3,-1), (-6,4). 
 
 9. Construct the quadrilateral whose vertices are (—3, 6), 
 (-3, 0), (3, 0), (3, 6). What kind of a quadrilateral is it ? 
 
 10. Mark the four points (2, 1), (4, 3), (2, 5), and (0, 3), 
 and connect them by straight lines. What kind of a figure 
 do these four lines enclose? 
 
 11. The side of a square = 'a; the origin of co-ordinates 
 is the intersection of the diagonals. What are the co-ordinates 
 of the vertices (i.) if the axes are parallel to the sides of the 
 square ? (li.) if 'the axes coincide with the diagonals ? 
 
 ^-(..)(f f) (-1 2} (-J -5} (I -^)^ 
 
 (i,.)(2V2, 0), (0, ?vl), (^-|v/2, 0), (0, -2V2]. 
 
 12. The side of an equilateral triangle = a ; the origin is 
 taken at one vertex and the axis of x coincides with one side. 
 What are the co-ordinates of the three vertices ? 
 
 Ans. (0,0), (a, 0), K f VSY 
 
 13. The line joining two points is bisected at the 
 If the co-ordinates of one of the points are a and b, what 
 the co-ordinates of the other ? 
 
 14. Connect the points (5, 3) and (5, — 3) by a straight 
 line. What is the direction of this line ? *' 
 
 ./|V 
 
 origin, 
 are 
 
 \ 
 
LOCI AND THEIR EQUATIONS. 
 
 Circular Measure. 
 
 5. In Analytic Geometry, angles are often expressed in 
 degrees, minutes, and seconds; but sometimes it is very' con- 
 venient to employ the Circular Pleasure of an angle. 
 
 In* circular measure an angle is defined by the equation 
 
 1 arc 
 
 angle = — - — . 
 radius 
 
 in which the word " arc " denotes the length of the arc corre- 
 sponding to the angle when both arc and radius are expressed 
 in terms of a common linear unit. 
 
 This equation gives us a correct measure of angular magni- 
 tude, because (as shown in Geometry) for a given angle the 
 value of the above ratio of arc and radius is constant for all 
 values of the radius.' 
 
 If the radius = 1, the equation becomes 
 
 angle = arc ; that is, 
 
 In circular measure an angle is measured hy the length of 
 the arc suhtended hy it in a unit circle. 
 
 It is shown in Geometry that the circumference of a unit 
 circle = . 2 tt; as this circumference contains 3G0° common 
 measure, the two measures are easily compared by means 
 of the relation 
 
 360 degrees = 2 TT units, circular measure. 
 
 Ex. 2. 
 
 1. Find the value in circular measure of the angles 1®, 
 45°, 90°, 180°, 270°. 
 
 180' ^''' ^''' ''' 
 
 2. In circular measure, unit angle is that angle whose arc 
 
 is equal to the radius of the circle. What is the value of this 
 
 angle in degrees, etc. ? 
 
 Am. 57° W 45^^ 
 
ANALYTIC GEOMETRY. 
 
 Distance between Two Points. 
 
 6. To find the distance between two given points. 
 
 Let P and Q (Fig. 2) be the given points, x^ and 3/1 the 
 co-ordinates of P, x, and 3/2 those of Q. Also let d= PQ = 
 the required distance. 
 
 3f 
 
 Fig. 2. 
 
 iS^X 
 
 Traw PJ/and Qil^-L to OX, and PP II to OX. 
 
 Then 0J/= .Tj, MP =^ y,, 
 
 ON=x„ m = y2f 
 
 PR = X2 — Xu QR = y., - 2/1. 
 
 By Geometry, 
 
 c^2 == (^x., - x,y + (y, - y,y ; 
 
 whence d = ^{x.^ — a?i)"-^ -ff 2/2 — Z/i)'-^- [1] 
 
 Since (a^i — a'2)^ = (^2 — ^l)^ it makes no difference which 
 point is called (.Ti, 3/1), and which (0:2, 3/2). 
 
 7. Equation [1] is perfectly general, holding true for points 
 situated in any quadrant. Thus, if P be in the second quad- 
 rant and Q in the third quadrant (Fig. 3), x^—x^ is obvi- 
 ously equal to the leg RQ ; and since 3/2 is negative, 3/2 — 2/1 is 
 the sum of two negative numbers, and is equal to the absolute 
 length of the leg RP with the — sign prefixed. 
 
LOCI AND TUEIR EQUATIONS. 7 
 
 Note. The learner should satisfy himself that equation [1] is perfectly- 
 general, by constructing other special cases in which the points Pand Q 
 are in different quadrants. In every case he will find that the numeri- 
 cal values of the expressions (-Tj — a-^) and (ya~2/i) ^-re the legs of the 
 right triangle the hypotenuse of which is the required distance PQ. 
 
 Equation [1] is merely one illustration of a general truth, of which 
 the learner will gradually become convinced as he proceeds with the 
 stud}'- of the subject ; namely, that theorems and formulas deduced by 
 reasoning with points or lines in the first quadrant {where the co-ordinates 
 are always positive) must, from the very nature of the analytic method, 
 hold true when the points or lines are situated in the other quadrants. 
 
 Ex. 3. 
 
 Find the distance z 
 
 1. From the point (—2, 5) to tile point (—8, —3). 
 
 2. From the point (1, 3) to the point (6, 15). 
 
 3. From the point (—4, 5) to the point (0, —2). 
 
 4. From the origin to the point (—6, —8). 
 
 5. From the point (a, b) to the point (—a, —h). 
 
 Find the lengths of the sides of a triangle 
 
 6. If the vertices are the points (15,-4), (-9,3), (11,24). 
 
 7. If the vertices are the points (2, 3), (4, -5), (-3, -6). 
 
 8. If the vertices are the points (0, 0), (3, 4), (-3, 4). 
 
 9. If the vertices are the points (0, 0), (--a, 0), (0, —h). 
 
 10. The vertices of a quadrilateral are (5, 2), (3, 7), (—1, 4), 
 (—3, —2). Find the lengths of the sides and also of the 
 diagonals. 
 
 11. One end of a line whose length is 13 is the point 
 (—4, 8) ; the ordinate of the other end is 3. What is its 
 abscissa? 
 
 12. What equation must the co-ordinates of the point {x, y) 
 satisfy ii'its distance from the point (7, —2) is equal to 11 ? 
 
8 
 
 ANALYTIC GEOMETRY. 
 
 13. What equation expresses algebraically the fact that 
 the point (x, y) is equidistant from the points (2, 3) and 
 (4,5)? 
 
 14. If the value of a quantity depends on the sqiiare of a 
 length, it is immaterial whether the length be considered 
 positive or negative. Why ? 
 
 Division of a Line. 
 8. To bisect the line joining two given points. 
 
 Let P and Q (Fig. 4) be the given points {x^, y^) and {x^.y-^. 
 Let X and y be the co-ordinates of i?, the mid-point of PQ. 
 
 The meaning of the problem is to find the values of x and y 
 in terms of x^^, yi, and x^, y%. 
 
 Y 
 
 pZ 
 
 f( 
 
 I 
 
 : 
 
 1 
 
 
 
 
 
 
 
 M 
 
 a 
 
 .Y 
 
 X 
 
 Y 
 
 
 
 
 
 
 /« 
 
 
 
 / 
 
 i 
 
 
 p/ 
 
 / 
 
 A 
 
 \ 
 
 1 
 
 
 
 
 
 M 
 
 < 
 
 5 NX 
 
 Fig. 4. 
 
 Fig. 5. 
 
 Draw P3f, PS, QN 1. to 0X\ also draw PA, PB \\ to OX. 
 Then rt. A PPA = rt. A PQB (hypotenuse and one acute 
 angle equal). 
 
 Therefore PA = PB, and AP = BQ\ 
 
 also 218 =SN. 
 
 By substitution, x — Ti -— X2 
 
 whence x = — ^-^^ — '- 5 y = 
 
 X, and y — y^ = y^ — y, 
 
 Vx -^1/2 
 
 [2] 
 
LOCI AND THEIR F.QUATIONS. 9 
 
 9. To divide the line joining two given points into two imi is 
 having a given ratio ni : n. 
 
 Let Pand Q (Fig. 5) be the given points {i\, ?/,) and (x^,^^)- 
 Let H be the required point, such that Pli : RQ — ni: n, and 
 let X and y denote the co-ordinates of B. 
 
 Complete the figure by drawing lines as in Fig. 4. 
 
 The rt. A PEA and EQB, being mutually equiangular, 
 are similar ; therefore 
 
 BB BQ n '^"^ 5q BQ^ n 
 Substituting for the lines their values, we have 
 
 ^~^ ^._ ^-^^ and y~y^ __ ^^^ 
 x-i — X n Vi — y ^ 
 
 Solving these equations for a: and y, we obtain 
 
 _ mx, + nx^ _ my, + ny^ 
 
 If m = ?2, we have the special case of bisecting a line already 
 considered ; and it is easy to see that the values of x and y 
 reduce to the forms given in [2]. 
 
 X. 4. 
 
 "What are the co-ordinates of the point 
 
 1. Half-way between, the points (5, 3) and (7, 9) ? 
 
 2. Half-way between the points (-6, 2) and (4, -2) ? 
 
 3. Half-way between the points (5, 0) and (-1, -4) ? 
 
 4. The vertices of a triangle are (2, 3), (4, ~5), (-3, -6) ; 
 find the middle points of its sides. 
 
 5. The middle point of a line is (6, 4), and one end of the 
 lineis (5, 7). What are the co-ordinates of the other end ? 
 
 6. A line is bisected at the origin ; one end of the line is the 
 point {—a, b). What are the co-ordinates of the other end ? 
 
10 ANALYTIC GEOMETRY. 
 
 7. Prove that the middle point of the hypotenuse of a right 
 triangle is equidistant from the three vertices. 
 
 8. Prove that the diagonals of a parallelogram mutually 
 bisect each other. 
 
 9. Show that the values of x and y in [2] hold true when 
 the two given points both lie in the second quadrant. 
 
 10. Solve the problem of § 9 when the line FQ is cut 
 externally instead of internally, in the ratio m : n. 
 
 11. What are the co-ordinates of the point which divides 
 the line joining (3,-1) and (10, 6) in the ratio 3:4? 
 
 12. The line joining (2, 3) and (4, —5) is trisected ; deter- 
 mine the point of trisection nearest (2, 3). 
 
 13. A line AB is produced to a point C, such that i?(7 = 
 ^AB. If A and B are the points (5, 6) and (7, 2), what are 
 the co-ordinates of C? 
 
 14. A line AB is produced to a point C, such that AB : BC 
 = 4:7. If ^ and B are the points (5, 4) and (6, -9), what 
 are the co-ordinates of (7? 
 
 15. Three vertices of a parallelogram are (1, 2), (—5, —3), 
 (7, -6). What is the fourth vertex ? 
 
 Constants and Variables. 
 
 10. In Analytic Geometry a line is regarded as a geometric 
 magnitude traced or generated by a moving point, — just as we 
 trace on paper what serves to represent a line to the eye by 
 moving the point of a pen or pencil over the paper. 
 
 We shall find that great advantages are to be gained by 
 defining a line in this way, but we must be prepared from 
 the outset to make an important distinction in the use of 
 symbols representing lengths. We must distinguish between 
 symbols which denote definite or fixed lengths and those 
 which denote variable lengths. 
 
LOCI AND THEIR EQUATIONS. 
 
 11 
 
 11. A simple example will serve to illustrate this difference. 
 Let A (Fig. 6) be the point (3, 4). Then OA = V9 + 16 = 5. 
 Now let a point F describe the line OA by moving from 
 to A, and let the co-ordinates of P be denoted by x and y ; 
 also let z denote the length OF at any position of F. Then 
 it is clear that the distance OF or z will be equal to 0, to 
 begin with, and will increase in value continuously until it 
 becomes equal to 5. 
 
 F 
 
 M B X 
 
 Fig. 6. 
 
 Here the word continuoiishj deserves special attention. It 
 means that F must pass successively through every conceiv- 
 able position on the line OA from to A; that, therefore, z 
 must have in succession every conceivable value between 
 and 5. There will be one position of F for which z is 
 equal to 2 ; there will be another position of F for which 
 z is equal to 2.000001 ; but before reaching this value it must 
 first pass through all values between 2 and 2.000001. 
 
 In the same way the co-ordinates of F, namely, x and y, 
 both pass through a continuous series of changes in value 
 unlimited- in number, the abscissa x increasing continuously 
 from to 3, and the ordinate y from to 4. 
 
 We may now divide the lengths considered in this example 
 into two classes : 
 
 (1) Lengths supposed to remain constant in value, namely, 
 the co-ordinates of A and the distance OA ; (2) lengths sup- 
 posed to vary continuously in value, namely, the co-ordinates 
 of F, (x and y), and the distance OF or z. 
 
12 ANALYTIC GEOMETRY. 
 
 Quantities of the first kind in any problem are called co7i- 
 stant quantities, or, more briefly, Constants. 
 
 Quantities of the second kind are called variable quantities, 
 or, more briefly. Variables. 
 
 12. Two variables are often so related that if one of them 
 changes in value, the other also changes in value. The second 
 variable is then said to be a function of the first variable. 
 The second variable is also called the dependent variable, 
 while the first is called the independent variable. Usually 
 the relation between two variables is such that either may be 
 treated as the independent variable, and the other as the 
 dependent variable. 
 
 Thus, in § 11, if ^ve suppose z to change, then both x and 
 y will change ; the values of x and y then will depend upon 
 the value given to z ; that is, x and y will be functions of z. 
 But we may also suppose the value of x, the abscissa of P, 
 to change ; then it is clear that the values of both y and z 
 must also change. In this case we take x as the independent 
 variable, and values of y and z will depend upon the value 
 of X ; that is, y and z will hQ functions of x. 
 
 13. The most concise way to express the relations of the 
 constants and variables which enter into a problem is by 
 means of algebraic equations. 
 
 The co-ordinates of P (Fig. 6) throughout its motion are 
 always x and y ; and the triangle 0PM is similar to the 
 triangle OAB. Hence, for any position of P, 
 
 ^ = -, and ^ — x^ -^y"^ \ 
 X 3 * 
 
 whence, by solving, 
 
 4 T 5 
 
 ?/ = -X, and z = -x, 
 
 equations which express the values of y and z, respectively, 
 in terms of x as the independent variable. 
 
LOCI AND THEIR EQUATIONS. 13 
 
 14. In § 11, instead of assuming 3 and 4 as the co-ordi- 
 nates of P, we might have employed two letters, as a and b, 
 with the understanding that these letters should denote two 
 co-ordinates which remain constant in value during the motion 
 of P. If we choose these letters, and then proceed exactly 
 as in § 13, we obtain for the values of y and z, 
 
 •^ a ' a 
 
 15. There is a noteworthy difference between the constants 
 3 and 4 and the constants a and b. The numbers 3 and 4 
 are unalterable in value ; they cannot be supposed to change 
 under any circumstances. The letters a and b are constants 
 in this sense only, that they do not change in value when we 
 suppose X or y or z to change in value ; in other words, they 
 are not functions of a; or y or z in the particular problem, 
 under discussion. In all other respects they are free to rep- 
 resent as many. different values as we choose to assign to them. 
 
 Constants of the first kind (arithmetical numbers) are called 
 absolute constants. Constants of the second kind (letters) are 
 called arbitrary or general constants. 
 
 16. By general agreement, variables are represented by the 
 last letters of the alphabet, as x, y, z; while constants are 
 most commonly represented by the first letters, a, b, c, etc., or 
 by the last letters with subscripts added, as x^, 3/1, x^, y^, etc. 
 
 Ex. 5. 
 
 1. A point P (x, y) revolves about the point Q (.Tj. y^), keep- 
 ing always at the distance a from it. Mention the constants 
 and the variables in this case. What is the total change in 
 the value of each variable ? 
 
 2. A point Q (x, y) moves : first parallel to the axis of y, 
 then parallel to the axis of x, then equally inclined to the 
 axes. Point out in each case the constants and the variables. 
 
14 
 
 ANALYTIC GEOMETRY. 
 
 Locus OF AN Equation. 
 
 17. Let us continue to regard x and y as the co-ordinates of 
 a point, and proceed to illustrate the meaning of an algebraic 
 equation containing one or both of these letters. 
 
 Take as the first case the equation a; — 4 = 0, whence a: = 4. 
 It is clear that this equation is satisfied by the co-ordinates of 
 every point so situated that its abscissa is equal to 4 ; there- 
 fore it is satisfied by the co-ordinate of every point in the line 
 
 Fig. 7. 
 
 AB (Fig. 7), drawn I! to OY, on the right of OF, and at the 
 distance 4 from Y. And it is also clear that this line con- 
 tains all the points whose co-ordinates will satisfy the given 
 equation. 
 
 The line AB, then, may be regarded as the geometric rep- 
 resentation or meaning of the equation x — 4 = 0; and con- 
 versely, the equation x — 4 = may be considered to be the 
 algebraic representative of this particular line. 
 
 In Analytic Geometry the line AB is called the locus of 
 the equation a; — 4 = ; conversely, the equation :r — 4 = is 
 known as the equation of the line AB. 
 
 The line AB is to be regarded as extending indefinitely in 
 both directions. If AB be described by a point P, moving 
 parallel to the axis of y, then at all points x is constant in 
 
LOCI AND THEIR EQUATIONS. 
 
 15 
 
 value and equal to 4, while y (which does not appear in the 
 given equation) is a variable, passing through an unlimited 
 number of values, both positive and negative. 
 
 18. The equation x — y — ^,ox: x = y, states in algebraic 
 language that the abscissa of the point is always equal to the 
 ordinate. 
 
 
 ' 
 
 Y 
 
 
 
 77/ 
 
 Values of X. 
 
 Values of y. 
 
 
 
 p/ 
 
 ^ 
 
 . . . 
 
 . . 0. 
 
 
 
 
 
 1 . . . 
 
 . . 1. 
 
 
 / 
 
 
 
 2 . . . 
 
 . . 2. 
 
 / 
 
 
 
 
 X 
 
 1—1 
 1 
 
 . . -1. 
 
 / 
 
 
 
 
 etc. 
 
 etc. 
 
 / 
 
 
 
 
 Fig. 8. 
 
 If we draw through the origin (Fig. 8) a straight line 
 AB, bisecting the first and third quadrants, then it is easy to 
 see that the given equation is satisfied hy every 'point in this 
 line and hy no other points. If we conceive a point P to 
 move so that its abscissa shall always be equal to its ordinate, 
 then the point must describe the line AB. In other w^ords, 
 if the point P is obliged to move so that its co-ordinates 
 (which of course are variables) shall always satisfy the con- 
 dition expressed by the equation a; — y = ; then the motion 
 of P is confined to the line AB. 
 
 The line AB is the locus of the equation x — y = 0, and 
 this equation represents the line AB. 
 
 19. The equation 20: + ?/— 3 = is satisfied by an unlimited 
 number of values of x and y. We may find as many of them 
 as we please by assuming values for one of the variables, and 
 computing the corresponding values of the other. 
 
16 
 
 ANALYTIC GEOMETRY. 
 
 If we assume for x the values given below, we easily find 
 for y the corresponding values given in the next column. 
 
 Values of x. Values of y. 
 
 3. 
 
 1 1. 
 
 2 -1. 
 
 o o 
 
 O — 'J. 
 
 4 -5. 
 
 -1 5. 
 
 -2 7. 
 
 -3 9. 
 
 -4 11. 
 
 etc. etc. 
 
 Fig. 9. 
 
 Plotting these points (as shown in Fig. 9), we obtain a series 
 of points so placed that their co-ordinates all satisfy the given 
 equation. By assuming for x values between and 1, 1 and 
 2, etc., ^ve might in the same way obtain as many points as 
 we please between A and JB, B and C, etc. In this case, 
 however, the points all lie in a straight line (as will be shown 
 later) ; so that if any tiuo points are found, the straight line 
 drawn through them will include all the points whose co-ordi- 
 nates satisfy the given equation. Now imagine that a point P, 
 the co-ordinates of w^hich are denoted by x and y, is required 
 to move in such a way that the values of x and y shall always 
 satisfy the equation 2x-\-y — ^ = 0-, then P must describe 
 the line AB, and cannot describe any other line. 
 
 The line AB is the locus of the equation 2x-\- y — o = 0. 
 
 20. Thus far we have taken equations of the first degree. 
 Let us now consider the equation x^ — y'^ = 0. By solving for 
 ?/, we obtain y = d= a;. Hence for every value of x there are 
 
LOCI AND THEIR EQUATIONS. 
 
 17 
 
 two values of y, both equal numerically to x, but having unlike 
 signs. Thus, for assumed values of x, we have corresponding 
 values of y given below : 
 
 Vahies of x. Values of y. 
 
 0. 
 
 1 
 o 
 
 3 
 
 1 
 
 o 
 
 Fig. 10. 
 
 1, 
 
 -1. 
 
 o 
 
 — 2. 
 
 3, 
 
 -3. 
 
 1, 
 
 1. 
 
 2 
 
 9 
 
 3, 
 
 3. 
 
 zK 
 
 V 
 
 / 
 
 /b 
 
 / 
 
 \ 
 
 X 
 
 A^ 
 
 
 \ 
 
 By plotting a few points, and comparing this case with the 
 example in § 19, it becomes evident that the locus of the 
 equation consists of two lines, AB, CD (Fig. 10), drawn 
 through the origin so as to bisect the four quadrants. 
 
 21. There is another way of looking at this case. The equa- 
 tion x-—y'^ = 0, by factoring, may be written {x—y) (:r+y) = 0. 
 Now the equation is satisfied if either factor = ; hence, it is 
 satisfied if x — y = Q, and also if x-\-y = 0. We know (see 
 § 19) that the locus of the equation x — y = is the. line 
 AB (Fig. 8). And the locus of the equation x-\-y = (or 
 x — ~y^ is evidently the line CD, since every point in it is 
 so placed that the two co-ordinates are equal numerically but 
 unlike in sign. Therefore the original equation x^ — y'^ = 
 is represented by the pair of lines AB and CD (Fig. 10). 
 
 22. Let us next consider the equation x^ 
 ing for y, we obtain y = ±V25 — :? 
 
 7/^=25. Solv- 
 When rr < 5 there 
 
 are two values of y equal numerically but unlike in sign. 
 When :r = 5, y = 0. When :r > 5 the values of y are imagi- 
 nary ; this last result means that there is no point with an 
 abscissa greater than 5 whose co-ordinates will satisfy the 
 given equation. 
 
18 
 
 ANALYTIC GEOMETRY. 
 
 By assigning values of x differing by unity, we obtain the 
 following sets of values of x and ?/ ; and by plotting the j^^oints, 
 and then drawing through them a continuous curve, we obtain 
 
 the curve shown in Fig. 11. 
 
 Fig. II. 
 
 dues of X. 
 
 Values of y 
 
 . . . 
 
 . . zh5. 
 
 1 . . . 
 
 . . ±4.9. 
 
 9 
 
 . . zt4.6. 
 
 3 . . . 
 
 . . ±4. 
 
 4 . . 
 
 . . ±3. 
 
 5 . . 
 
 . . 0. 
 
 -1 . . 
 
 . . ±4.9. 
 
 — 2 . 
 
 . . ±4.6. 
 
 -3 . . 
 
 . . ±4. 
 
 --4 . . 
 
 . . ±3. 
 
 -5 . . 
 
 . . 0. 
 
 In this case, however, the locus may be found as follows : 
 Let F (Fig. 11) be any point so placed that its co-ordinates, 
 X = 03f, y = MP, satisfy the equation x^ -\-y' = 25. Join 
 OP; then a;' + y' = OP' ; therefore 0P= 5. Hence, if P is 
 anywhere in the circle described with as centre and 5 for 
 radius, its co-ordinates will satisfy the given equation ; and 
 if P is not in this circle, its co-ordinates will not satisfy the 
 equation. This circle, then, is the locus of the equation. 
 
 23, The points whose co-ordinates satisfy the equation 
 y2 — ^x lie neither in a straight line nor in a circle. Never- 
 theless, they do all lie in a certain line, which is, therefore, 
 completely determined by the equation. To construct this 
 line, we first find a number of points which satisfy the equa- 
 tion (the closer the points to one another, the better) and then 
 draw, freehand or with the aid of tracing curves, a continuous 
 curve through the points. 
 
 The co-ordinates of a number of such points are given in 
 
LOCI AND TllElK EQUATIONS. 
 
 19 
 
 the table below. It is evident that for each positive value of 
 X there are two values of y, equal numerically but unlike in 
 sign. If we assume a negative value for x, then the value 
 of y is imaginary ; this result means that there are no 
 points to the left of the axis of y which will satisfy the given 
 equation. 
 
 Values of a-. Values of y. 
 
 0. 
 
 1 ±2. 
 
 2 ±2.83. 
 
 3 ±3.46. 
 
 4 =h4. 
 
 5 ±4.47. 
 
 6 ±4.90. 
 
 7 ±5.29. 
 
 8 ±5.66. 
 
 9 ±6. 
 
 — 1 imaginary. 
 
 Fig. 12. 
 
 In Fig. 12 the several points obtained are plotted, and a 
 smooth curve is then drawn through them. It passes through 
 the origin, is placed symmetrically on both sides of the axis 
 of X, lies w^holly on the right of the axis of y, and extends 
 towards the right without limit. It is the locus of the given 
 equation, and is a curve called the Parabola. 
 
 24. After a study of the foregoing examples, we may lay 
 down the following general principles, which form the foun- 
 dation of the science of Analytic Geometry : 
 
 I. Every algebraic equation involving x and ?/ is satisfied 
 by an unlimited number of sets of values of x and y ; in other 
 words, X and y may be treated as variables, or quantities vary- 
 ing continuously, yet always so related that their values con- 
 stantly satisfy the equation. 
 
20 ANALYTIC GEOMETRY. 
 
 II. The letters x and y may also be regarded as represent- 
 ing the co-ordinates of a point. This point is not fixed in 
 position, because x and y are variables ; but it cannot be 
 placed at random, because x and y can have only such values 
 as will satisfy the equation ; now since these values are con- 
 tinuous, the point may be conceived to Tnove continuoushj , and 
 will therefore describe a definite line, or group of lines. 
 
 III. The line, or group of lines, described by a point 
 moving so that its co-ordinates always satisfy the ec[uation is 
 called the Locus of the Equation ; conversely, the equation sat- 
 isfied by the co-ordinates of every point in a certain line is 
 called the Equation of the Line. 
 
 lY. An equation, therefore, containing the variables x and 
 y is the algebraic representation of a line ; it determines a 
 certain line in the same sense that two co-ordinates determine 
 a certain point. And, conversely, the line (or group of lines) 
 which is the locus of an equation is the geometric representa- 
 tion of the equation. 
 
 Ex. 6. 
 
 Determine and construct the loci of the following equations 
 (the locus in each case being either a straight line or a circle) : 
 
 1. a:- 6 = 0. 9. 9:r--25 = 0. 
 
 2. rr + 5 = 0. 10. ^x--y- = 0. 
 
 3. y = -1. 11. .r^-16/ = 0. 
 
 4. :r=:0. 12. .r^ + /=-36. 
 
 5. y = 0. 13. .T- + y--l = 0. 
 
 6. x + y = 0. 14. :r(y+5) = 0. 
 
 7. x~2y = 0. 15. (^ - 2) (^-- 3) = 0. 
 
 8. 2.r-f3y-fl0-0. 16. (y- 4) (y + 1) = 0. 
 
LOCI AND THEIR FAiUATlONS. 21 
 
 17. AVhat is llie gooinetric meaning of the equation 
 
 Hint. Resolve the equation into two hinoniial factors, 
 
 18. What is the geometric meaning of the equation 
 
 19. "What two lines form the locus of the equation 
 
 20. Is the point (2, — 5) situated in the locus of the equa- 
 tion 4:?;-3y — 22--=:0? 
 
 Hint. See if tlio co-ordinates of the point satisfy tlie equation. 
 
 21. Is the point (4, —6) in the locus of the equation 7/'~9x? 
 
 22. Is the point (—1, —1) in the locus of the equation 
 16^;^ + 9?/^ -f 15a; - Gy - 18 = ? 
 
 23. Does the locus of the equation a;^ + ?/"'^=100 pass through 
 the point (— G, 8) ? 
 
 24. Which of the loci represented by the following equa- 
 tions pass through the origin ? 
 
 (1) 3.^ + 2 = 0. (5) 3:^- = 2y. 
 
 (2) 3x-U7j + 7 = 0. (6) 3^'-lly-=0. 
 
 (3) a;^-lGy2_lO = 0. (7) x' -I67/ = 0. 
 
 (4) ax + h'(/ + c=--0. (8) ax^-hy = 0. 
 
 25. The abscissa of a point in the locus of the equation 
 3x ~ 4:7/ — 7 = is 9 ; what is the value of the ordinate ? 
 
 Ans. 5. 
 
 26. Determine that point in the locus of y^ — 4:r=0 for 
 which the ordinate = - 6. ^.^^ ^j^^ ^^^-^^ ^.^^ _ ^^^ 
 
 27. Determine the point where the line represented by the 
 equation 7x + y -14 — cuts the axis of x. 
 
 A71S. The point (2, 0). 
 
22 analytic geometry. 
 
 Intersections of Loci. 
 
 25. The term Curve, as used in Analytic Geometry, means 
 any geometric locus, including the straight line as well as 
 lines commonly called curves. 
 
 The Intercepts of a curve on the axes are the distances from 
 the origin, measured along the axes, to the points where the 
 curve cuts the axes. 
 
 26. To find the intercept of a curve, having given its equation. 
 
 The intercept of a curve on the axis of x is the abscissa of 
 the point where the curve cuts the axis of x. The ordinate of 
 this point = 0. Therefore, to find this intercept, put y = in 
 the given equation of the curve, and then solve the equation 
 for X ; the resulting value of x will be the intercept required. 
 
 If the equation is of a higher degree than the first there 
 will be more than one value of x ; and the curve will cut the 
 axis of X in as many points as there are o^eal values of x. 
 
 If an iynaginary value of x is found it is to be rejected. 
 But in order to make the language of geometry correspond to 
 that of algebra, we may say in this case that the curve cuts 
 the axis of x in an imaginary point ; that is, a point that has 
 one or both of its co-ordinates imaginary. 
 
 Similarly, to find the intercepts on the axis of y, put a- = 
 in the given equation, and then solve it for y ; the resulting 
 real values of y will be the intercepts required. 
 
 27. To find the points of intersection of two curves, having 
 given their equations. 
 
 Since the points of intersection lie in both curves, their co- 
 ordinates must satisfy both equations. Therefore, to find their 
 co-ordinates, solve the two equations, regarding the variables 
 X and y as unknown quantities. 
 
 If the equations are both of the first degree, there will be 
 
LOCI AND THEIR EQUATIONS. 23 
 
 onl}^ 07ie pair of values of x and y, and one point of inter- 
 section. 
 
 If the equations are, one or both of them, of higher degree 
 than the first, there may be several pairs of values of x and 
 y ; in this case there will be as many points of intersection as 
 there are pairs of real values of x and y. 
 
 If imaginary values of either x or y are obtained, there are 
 no corresponding points of intersection. 
 
 28. If a curve pass through the 07'igin, its equation, reduced 
 to its simplest for VI, cannot have a constant term ; that is, cannot 
 have a teo-ni free from both x and y. 
 
 Since in this case the point (0, 0) is a 2:»oint of the curve, 
 its equation must be satisfied by the values x ^= 0, and y = 0. 
 But it is obvious that these values cannot satisfy the equation 
 if, after reduction to its simplest form, it still contains a con- 
 stant term. Therefore the equation cannot have a constant 
 term. 
 
 29. If an equation has no constant term, its locus must pass 
 through the origiii. 
 
 For, the values x^=0, y = must evidently satisfy the equa- 
 tion, and therefore the point (0, 0) must be a point of the 
 locus. 
 
 Ex. 7. 
 
 Find the intercepts of the following curves : 
 
 1. 
 
 4.1' + 3y- 48 = 0. 
 
 8. 
 
 x-2, = 0. 
 
 
 o 
 
 5y — 3:r — 30 =^ 0. 
 
 9. 
 
 x--9 = 0. 
 
 
 3. 
 
 x' + y' = \Q. 
 
 10. 
 
 x'--y- = 0. 
 
 
 4. 
 
 9x'-\-^f=U. 
 
 11. 
 
 y' = 4:x. 
 
 
 5. 
 
 9.r'-4y^ = 16. 
 
 12. 
 
 x'^y'-^x- 
 
 - 8y = 32. 
 
 6. 
 
 9x^-4y = 16. 
 
 13. 
 
 X- -j-y'^ — ix - 
 
 -8y = 0. 
 
 7. 
 
 a'x' + by = a'b\ 
 
 14. 
 
 (x-5)' + (y- 
 
 ~6)= = 20. 
 
24 ANALYTIC GEOMETRY. 
 
 Find the j^oints of intersection of the following curves : 
 
 15. 3a;-4y + 13 = 0, ll:r + 7y — 104 = 0. 
 
 IG. 2a; + 3y = 7, x — y = l. 
 
 17. :r — 7y+25 = 0, x' + f = 25. 
 
 18. 3:r + 4y = 25, x'-}-f = 25. 
 
 19. ^ + y = 8, :i'' — 3/^ = 34. 
 
 20. 2.r = y, :i;2_|_y2._;L0a7 = 0. 
 
 21. The cfjuations of the sides of a triangle are 2x-j-9y 
 + 17=0, 7.y-y-38 = 0, a:-2y-f 2 = 0. Find the co- 
 ordinates of its three vertices. 
 
 22. The equations of the sides of a triangle are 5a7+6y=12, 
 3.r — 4?/ = 30, ^ + 5y = 10. Find the lengths of its sides. 
 
 23. Find the lengths of the sides of a triangle if the equa- 
 tions of the sides are x= 0, 'i/ ^=0, and 4:X -f- 3y = 12. 
 
 24. What are the vertices of the quadrilateral enclosed by 
 the straight lines x — a = 0, x -}- a = 0, y — b = 0, y -\~ h = ? 
 What kind of a quadrilateral is it ? 
 
 25. Does the straight line 5.r + 4y = 20 cut the circle 
 
 X' + y2 r= 9 ? 
 
 26. Find the length of that part of the straight line 
 3.r — 4y = which is contained within the circle :2;^ + y^ = 25. 
 
 27. Which of the following curves pass through the origin 
 of co-ordinates? 
 
 (1) 7:r-2y + 4 = 0. (4) ax-{-h2j = 0. 
 
 (2) 7a:-2y = 0. (5) ax + 1?/ -}- c = 0. 
 
 (3) f - x' = 4y. (6) x'~i-\-a = a + xy. 
 
 28. Change the equation 4.r + 2y — 7=0 so that its locus 
 shall pass through the origin. 
 
loci and their equations. 25 
 
 Construction of Loci. 
 
 30. If we know that the locus of a given equation is a 
 straight line, the locus is easily constructed ; it is only neces- 
 sary to find any two points in it, plot them, and draw a 
 straight line through them with the aid of a ruler. 
 
 Likewise, if we know that the locus is a circle, and can find 
 its centre and its radius, the entire locus can then be immedi- 
 ately described with the aid of a pair of compasses. 
 
 It will appear later on that the fovTn of the given equation 
 enables us at once to tell whether its locus is a straight line 
 or a circle. 
 
 If the locus of an equation is neither a straight line nor a 
 circle, then the following method of construction, which is 
 applicable to the locus of any equation without regard to the 
 form of the curve, is usually employed. 
 
 31. To construct the locus of a given equation. 
 
 The steps of the process are as follows : 
 
 1. Solve the equation with respect to either x or y. 
 
 2. Assign values to the other variable, differing not much 
 from one another. 
 
 3. Find each corresponding value of the first variable. 
 
 4. Draw two axes, choose a suitable scale of lengths, and 
 plot the points whose co-ordinates have been obtained. 
 
 5. Draw a continuous curve through these points. 
 
 Discussion. An examination of the equation, as shown in 
 the examples given below, enables us to obtain a good general 
 idea of the shape and size of the curve, its position with respect 
 to the axes, etc. ; in this way it serves as an aid in construct- 
 ing the curve, and as a means of detecting numerical errors 
 made in computing the co-ordinates of the points. Such an 
 examination is called a discussion of the equation. 
 
26 
 
 ANALYTIC GEOMETRY. 
 
 Note 1. This method of constructing a locus is from its nature an 
 approximate method. But the nearer the points are to one another, the 
 nearer the curve will approach the exact position of the locus. 
 
 Note 2. In theory, it is immaterial what scale of lengths is used. 
 In practice, the unit of lengths should be determined by the size of the 
 paper compared with the greatest length to be laid off upon it. Paper 
 sold under the name of " co-ordinate paper," ruled in small squares, j\ 
 of an inch long, on each side, will be found very convenient in practice. 
 
 32. Construct the locus of the equation ■ 
 9.1-2 + 4/ -576 = 0. 
 If we solve for both x and y, we obtain the following values : 
 
 (1) 
 
 (2) 
 
 fV64 
 
 y 
 
 a: = ±|Vl44 
 
 r 
 
 By assigning to x values differing by unity, and finding 
 corresponding values of ?/, we obtain the results given below. 
 To each value of x, positive or negative, there correspond two 
 values of ?/, equal numerically and unlike in sign. By plotting 
 the corresponding points, and drawing a continuous curve 
 through them, we obtain the closed curve shown in Fig. 13. 
 
 Values of . 
 
 c. 
 
 
 Values of y. 
 
 ... ±12. 
 
 ±1 
 
 
 
 . =h 11.91. 
 
 ±2 
 
 
 
 . ± 11.62. 
 
 =4=3 
 
 
 
 . ±11.13. 
 
 =±=4 
 
 
 
 ± 10.39. 
 
 =b5 
 
 
 
 . ± 9.36. 
 
 ±6 
 
 
 
 ± 7.93. 
 
 =h7 
 
 
 
 ± 5.80. 
 
 ±8 
 
 
 
 ± 0. 
 
 ±9 
 
 
 
 ± imaginary 
 
LOCI AND TIIEIR EQUATIONS. 27 
 
 Discussion. From equations (1) and (2) we see that if 
 x = 0, y = ± 12, and if y = 0, x = ±8; therefore the inter- 
 cepts of the curve on the axis of x are + 8 and — 8, and those 
 on the axis of y are -f 12 and —12. These intercepts are the 
 lengths OA, OA', and OB, OB', in Fig. 13. 
 
 If we assign to :r a numerical value greater than 8, posi- 
 tive or negative, we find by substitution in equation (1) that 
 the corresponding value ofy will be imaginary. This shows 
 that OA and OA' are the maximum abscissas of the curve. 
 Similarly, equation (2) shows that the curve has no points 
 with ordinates greater than -f-12 and —12. 
 
 The greater the numerical value of x, between the limits 
 and + 8 or and — 8, the less the corresponding value of y 
 numerically ; why ? 
 
 From equation (1) we see that corresponding to each 
 value of X, between the limits and ± 8, there are tiuo real 
 values of y, equal numerically and unlike in sign. Hence, for 
 each value of x between and dr 8 there are two points of the 
 curve placed equally distan|; from the axis of x. Therefore 
 the curve is symmetrical with respect to the axis of ;r ; in 
 other words, if the portion of the curve above the axis of x 
 be revolved about this axis through 180°, it will coincide with 
 the portion below the axis. Similarly, it follows from equa- 
 tion (2) that the curve is also symmetrical with respect to 
 the axis of y. Therefore the entire curve is a closed curve, 
 consisting of four equal quadrantal arcs symmetrically placed 
 about the origin 0. The name of this curve is the Ellipse, 
 
 33. Construct the locus of the equation 
 4.r-y^ + 16 = 0. 
 
 Solving for both a; and y, we obtain 
 
 y = ±2V^+4, (1) 
 
 ■ x = l^:^. (2) 
 
28 
 
 ANALYTIC GEOMETRY. 
 
 We may either assign values to x, and then compute those 
 of ?/ by means of (1), or assign values to y, and compute those 
 of X by means of (2) ; the second course is better, because 
 there is less labor in squaring a number than in extracting 
 Its square root. 
 
 By assigning values to y, differing by unity from to +10, 
 and from to —10, and then proceeding exactly as in the 
 last example, we obtain the series of values given below, and 
 the curve shown in Fig. 14. 
 
 Values of 3/. 
 
 
 ±1 
 
 ±2 
 ±3 
 
 ±4 
 ±5 
 ±6 
 ±7 
 
 Fig. 14. 
 
 d=9 
 ±10 
 
 Values of x. 
 
 — 4. 
 
 -3.75. 
 
 -3. 
 
 -1.75. 
 
 0. 
 
 2.25. 
 
 5. 
 
 8.25. 
 
 12. 
 
 16.25. 
 
 21. 
 
 Discussion. An examination of equations (1) and (2) 
 yields the following results, the reasons for which are left 
 as an exercise for the learner : 
 
 The intercepts on the axes are : 
 
 On the axis of x, OA = — 4. 
 
 On the axis of y, 0B = + 4, and OC^ - 4. 
 
 If we draw through A the line AD 1. to OX, the entire 
 curve lies to the right of AD. 
 
 The curve is situated on both sides of OX, and is sym- 
 metrical with respect to OX. 
 
 The curve extends towards the right without limit. 
 
LOCI AND THEIR EQUATIONS. 
 
 29 
 
 The curve constantly recedes from OX as it extends 
 towards the right. 
 
 This curve is called a Parabola; the point A is called its 
 Vertex, the line ^Xits Axis. 
 
 34. Construct the locus of the equation 
 y =■ sin X. 
 
 If we assume for x the values 0°, 10°, 20°, 30°, etc., the cor- 
 responding values of y are the natural sines of these angles, 
 and are as follows : 
 
 Values of y. 
 0.77. 
 0.87. 
 0.94. 
 0.98. 
 1. 
 
 If we continue the values of x from 90° to 180°, the above 
 values of y repeat themselves in the inverse order (e.g., if 
 X = 100°, y = 0.98, etc.) ; from 180° to 360° the values of y 
 are numerically the same, and occur in the same order as 
 between 0° and 180°, but are negative. 
 
 Values of x. 
 
 Values of ?/. 
 
 Values of x. ^ 
 
 0° . . . 
 
 . 0. 
 
 50° ... . 
 
 10° . . . 
 
 . 0.17. 
 
 60° ... . 
 
 20° . . . 
 
 . 0.34. 
 
 70° ... . 
 
 30° . . . 
 
 . 0.50. 
 
 80° ... . 
 
 40° . . . 
 
 . 0.64. 
 
 90° ... . 
 
 Fig. 15. 
 
 In order to express both x and y in terms of a common 
 linear unit, we ought, in strictness, to use the circular meas- 
 ure of an angle in which the linear unit represents an angle 
 
30 ANALYTIC GEOMETRY. 
 
 of 57.3°, very nearly (see § 5). But it is more convenient, 
 and serves our present purpose equally well, to assume that 
 an angle of 60° =^ the linear unit. This assumption is made 
 in Fig. 15, where the curve is drawn with one centimeter as 
 the linear unit. 
 
 Discussion. The curve passes through the origin, and cuts 
 the axis of .t at points separated by intervals of 180°. Since 
 an angle may have any magnitude, positive or negative, the 
 curve extends on both sides of the origin without limit. The 
 maximum value of the ordinate is alternately +1 and —1 : 
 the former value corresponds to the angle 90°, and repeats 
 itself at intervals of 360° ; the latter value corresponds to 
 the angle 270°, and repeats itself at intervals of 360°. The 
 curve has the form of a wave, and is called the Sinusoid. 
 
 Ex. 8. 
 Construct the loci of the following equations : 
 
 1. Sx-7j-2 = 0. 13. y'~l = 0. 
 
 2. y = 2x. 14. 7/ = x\ 
 
 3. x' = y\%^br,^^iz^UMAj 15. xij = 12. 
 
 4. 
 
 x'Jry' = '^00. 
 
 
 16. 
 
 X = sin y. 
 
 5. 
 
 x'-f^25. 
 
 
 17. 
 
 y = 2sm X. 
 
 6. 
 
 ^x'-y' = 0. 
 
 
 18. 
 
 y = sin 2x. 
 
 7. 
 
 ^x' + 9y' = U4:. 
 
 
 19. 
 
 y = cos X. 
 
 8. 
 
 y^-l6x = 0. 
 
 
 20. 
 
 y = tan x. 
 
 9. 
 
 y2^16:r = 0. 
 
 
 21. 
 
 y = cot X. 
 
 10. 
 
 x'-2x-10y-5-- 
 
 --0. 
 
 22. 
 
 y = sec X. 
 
 11. 
 
 y2_2y_l0a;-=0. 
 
 
 23. 
 
 y = CSC X, 
 
 12. 
 
 ^x-sy+(y-2y= 
 
 = 25. 
 
 24. 
 
 y = sin X + cos x. 
 
loci and their equations. 31 
 
 Equation of a Curve. 
 
 35. From what precedes, we may conclude that every equa- 
 tion involving x and y as variables represents a definite line 
 (or group of lines) known as the locus of the equation. 
 Regarded from this point of view, an equation is the state- 
 ment in algebraic language of a geometric condition which 
 must always be satisfied by a point {x, y), as we imagine it 
 to move in the plane of the axes. For example, the equation 
 x = 2y states the condition that the point must so move that 
 its abscissa shall always be equal to twice its ordinate ; the 
 equation x^ -\- y"^ — A. states the condition that the point must 
 so move that the sum of the squares of its co-ordinates shall 
 always be equal to 4 ; etc. 
 
 Conversely, every geometric condition that a point is 
 required to satisfy must confine the point to a definite line 
 as its locus, and must lead to an equation that is always 
 satisfied by the co-ordinates of the point. 
 
 Hence arises a new problem, and one usually of greater 
 difficulty than any thus far considered, namely : 
 
 Given the geometric condition to he satisfied by a point, to 
 find the equation of its locus. 
 
 The great importance of this problem lies in the fact that 
 in the practical applications of Analytic Geometry the law of 
 a moving point is commonly the one thing known to start 
 with, so that the first step must consist in finding the equa- 
 tion of its locus. 
 
 Ex. 9. 
 
 1. A point moves so that it is always three times as far from 
 the axis of x as from the axis of y. AVhat is the equation of 
 its locus? 
 
 2. What is the equation of the locus of a point which moves 
 so that its abscissa is always equal to-f6? — G? 0? 
 
32 ANALYTIC GEOMETRY. 
 
 3. What is the equation of the locus of a point which moves 
 so that its ordinate is always equal to +4? — 1? 0? 
 
 4. A point so moves that its distance from the straight line 
 :r = 3 is always numerically equal to 2. What is the equation 
 of its locus ? 
 
 5. A point so moves that its distance from the straight line 
 y = 6 is always numerically equal to 3. Find the equation 
 of its locus. Construct the locus. 
 
 6. A point moves so that its distance from the straight 
 line :^ -f 4 = is always numerically equal to 5. Find the 
 equation of its locus. Construct the locus. 
 
 7. What is the equation of the locus of a point equidistant 
 
 (1) from the parallels x = and a; = — 6 ? 
 
 (2) from the parallels y = 7 and ?/ — — 3 ? 
 
 8. What is the equation of the locus of a point always 
 equidistant from the origin and the point (6, 0) ? 
 
 Find the equation of the locus of a point 
 9. Equidistant from the points (4, 0) and (— 2, 0)". 
 
 10. Equidistant from the points (0, — 5) and (0, 9). 
 
 11. Equidistant from the points (3, 4) and (5, —2). 
 
 12. Equidistant from the points (5, 0) and (0, 5). 
 
 13. A point moves so that its distance from the origin is 
 always equal to 10. Find the equation of its locus. 
 
 14. A point moves so that its distance from the point (4, — 3) 
 is always equal to 5. Find the equation of its locus, and con- 
 struct it. What kind of curve is it ? Does it pass through 
 the origin ? Why ? 
 
 15. What is the equation of the locus of a point whose 
 distance from the point (— 4, — 7) is always equal to 8 ? 
 
LOCI AND THEIR EQUATIONS. 33 
 
 16. About the origin of co-ordinates as centre, with a radius 
 equal to 5, a circle is described. A point outside this circle 
 so moves that its distance from the circumference of the circle 
 is always equal to 4. What is the equation of its locus ? 
 
 17. A high rock A, rising out of the water, is 3 miles 
 from a perfectly straight shore BC. A vessel so moves that 
 its distance from the rock is always the same as its distance 
 from the shore. What is the equation of its locus ? 
 
 18. A point A is situated at the distance 6 from the line 
 J^C, A moving point P is always equidistant from A and 
 BC. Find the equation of its locus. 
 
 19. A point moves so that its distance from the axis of x is 
 half its distance from the origin ; find the equation of its locus. 
 
 20. A point moves so that the sum of the squares of its 
 distances from the two fixed points (a, 0) and (—a, 0) is the 
 constant 2^' ; find the equation of its locus. 
 
 21. A j)oint moves so that the difference of the squares of 
 its distances from (a, 0) and (— a, 0) is the constant k^ ; find 
 the equation of its locus. 
 
 Ex. 10. (Review.) 
 
 1. If we should plot all possible points for which x = — 5, 
 how would they be situated ? 
 
 2. Construct the point (x, y) if a:= 2 and 
 
 (1) y = 4^'-3, (2) 3a; -2?/ = 8. 
 
 3. The vertices of a rectangle are the points («, h), (—a, Z>), 
 (—a, — ^), and (a, — i). Find the lengths of its sides, the 
 lengths of its diagonals, and show that the vertices are equi- 
 distant from the origin. 
 
 4. What does equation [1], p. 6, for the distance between 
 two points, become when one of the points is the origin ? 
 
34 ANALYTIC GEOMETRY. > 
 
 5. Express by an equation that the distance of the point 
 (x, y) IVoui the point (4, G) is equal to 8. 
 
 6. Express that the point {x, y) is equidistant from the 
 points (2, 3) and (4, 5). 
 
 7. Find the point equidistant from the points (2, 3), (4, 5), 
 and (6, 1). What is the common distance ? 
 
 8. Prove that the diagonals of a rectangle are equal. 
 
 9. Prove that the diagonals of a parallelogram mutually 
 bisect each other. 
 
 10. The co-ordinates of three vertices of a parallelogram 
 are known : (5, 3), (7, 10), (13, 9). What are the co-ordi- 
 nates of the remaining vertex ? 
 
 11. The co-ordinates of the vertices of a triangle are (3, 5), 
 (7, —9), (2, —4). Find the co-ordinates of the middle points 
 of its sides. 
 
 12. The centre of gravity of a triangle is situated on the 
 line joining any vertex to the middle point of the opposite 
 side, at the point of trisection nearest that side. Find the 
 centre of gravity of the triangle whose vertices are the points 
 (2,3),(4,-5), (-3, -6). 
 
 13. The vertices of a triangle are (5, - 3), (7, 9), (- 9, 6). 
 Find the distance from its centre of gravity to the origin. 
 
 14. The vertices of a quadrilateral are (0, 0), (5, 0), (9, 11), 
 (0, 3). Find the co-ordinates of the intersection of the two 
 straight lines which join the middle points of its opposite sides. 
 
 15. Prove that the two straight lines which join the middle 
 points of the opposite sides of any quadrilateral mutually 
 bisect each other. 
 
 16. A line is divided into three equal parts. One end of 
 the line is the point (3, 8) ; the adjacent point of division is 
 (4, 13). What are the co-ordinates of the other end? 
 
LOCI AND THEIR EQUATIONS. 35 
 
 17. The line joining the jDoints (^'i, 7/1) and (x.^, 7/2) is divided 
 into four equal parts. Find the co-ordinates of the jooints of 
 division. 
 
 18. Explain and illustrate the relation which exists between 
 an equation and its locus. 
 
 19. Construct the two lines which form the locus of the 
 equation x'^ — 7x = 0. 
 
 20. Is the point (2, — 5) in the locus of the equation 
 
 21. The ordinate of a certain point in the locus of the 
 equation :i-- + / + 20a; — 70 = is 1. What is the abscissa 
 of this point ? 
 
 22. Find the intercepts of the curve x^-\-y — 5x—7y-{-6 = 0. 
 
 Find the points common to the curves : 
 
 23. x' + 7/ = 100, and 7/ — 12 x = 0. 
 
 24. x'^ + y' = 5a^ and rc^ — 4 ay. 
 
 25. h'-x"- -{- ahf ^ d'b\ and x- + y-' = a\ 
 
 26. Find the lengths of the sides of a triangle, if its ver- 
 tices are (6, 0), (0, -8), (-4, -2). 
 
 27. A point moves so that it is always six times as far from 
 one of two fixed perpendicular lines as from the other. Find 
 the equation of its locus. 
 
 28. A point so moves that its distance from the fixed 
 point A is always double its distance from the fixed line AB. 
 Find the equation of its locus. 
 
 29. A fixed point is at the distance a from a fixed straight 
 line. A point so moves that its distance from the fixed point 
 is always twice its distance from the fixed line. Find the 
 equation of its locus. 
 
CHAPTER 11. 
 
 THE STRAIGHT LINE. 
 
 Equations of the Straight Line. 
 
 36. Notation. Throughout this chapter, and generally in 
 equations of straight lines, 
 
 a = the intercept on the axis of x. / 
 
 h = the intercept on the axis of y. ^ 
 
 y = the angle between the line and the axis of x. 
 971 = tan y. ' 
 p = the distance of the line from the origin. ' 
 
 a = the angle between p and the axis of x. "^ 
 
 These six quantities are constants for a given straight line, 
 but vary in value for different lines : a, b, m, and p may 
 have any values from -co to -}-co; y and a, any values from 
 0° to 360°. 
 
 The constant m is often called the Slope of the line ; its 
 value determines the direction of the line. 
 
 In order to determine a straight line, two geometric con- 
 ditions must be given. 
 
 37. To find the equation of a straight line passing through 
 two given points {x^, y^ and {x2, yi). 
 
 Let A (Fig. 16) be the point {x^, y^), B the point {x.^, y^) ; 
 and let P be any other point of the line drawn through A 
 and jB, X and y its co-ordinates. Draw AC, J5D, PIT, J. to 
 OX and ^^i^ II to OX. 
 
THE STRAIGHT LINE. 
 
 37 
 
 The triangles APF, ABE are similar ; therefore 
 
 PF _^BE 
 AF AE 
 
 'No^y FF= 7/ — 2/i, AF^-x — Xi, BE=y.j_—y^, AE^x^—x^. 
 
 Therefore 
 
 1 ^2 •*'l 
 
 This is the equation required. 
 
 [4] 
 
 Y 
 
 F 
 
 i \D 
 
 
 
 M 
 
 ^ ^ 
 
 \ X 
 
 Fig. 16. 
 
 Fig. 17. 
 
 It is evident that the angle PAF=y. Therefore each 
 side of equation [4] is equal to tan y or m. The first side 
 CO ntains the two variables x and y, and the equation tells us 
 
 th at they must vary in such a way that the fraction ^ — ^ 
 ( . _ X — x^ 
 
 shall remain constant in value, and always equal to m. 
 
 Note. In Fig. 16 the points A, B, and P are assumed in the first 
 quadrant in order to avoid negative quantities. But the reasoning will 
 le ad to equation [4] whatever be the positions of these points. In 
 Fis. 17 the points are in different quadrants. The triangles APF, 
 /\BE are to be constructed as shown in the figure. They are 
 similar ; and by taking proper account of the algebraic signs of the 
 q uantities, we arrive at equation [4], as before. The learner should 
 sjtudy this case with care, and should study other cases devised by him- 
 stdf, till he is convinced that equation [4] is perfectly general. 
 
 i 
 
38 
 
 ANALYTIC GEOMETRY. 
 
 38. To find the equation of a straight line, given one point 
 (^^\j 1/y) ?^ ^^^ ^^^^'<^ <^^-^ i^^(^ angle y. 
 
 Let the figure be constructed like Fig. 16, omitting the 
 point B and the line BED. Then it is evident that 
 
 whence, 
 
 AF X - X, ' 
 2/ — 2/i = m{oc — Xj), 
 
 Y 
 
 B 
 
 y^... _Jn 
 
 AX 
 
 
 y 
 
 U X 
 
 [5] 
 
 Fis. 18. 
 
 39. To find the equation of a straight li^ie, given the inteTcc]^)^^ 
 h and the angle y. 
 
 Let the line cut the axes in the points A and Jj (Fig. LS). 
 Let F be any point (x, y) in the line. Draw FM 1. to 0\X, 
 and EC II to OX. 
 
 Then OE^h, FEC=y, EC=x, FC=y-h] i 
 
 n 
 
 40. To find the equation of a straight line, given 'its intercept 's 
 a and b. ^ 
 
 Let the line cut the axes in the points^ and E (Fig. 19 c, 
 and let F be any other point {x, y) in the line. Then OA = ^ i, 
 
 therefore 
 
 .7 ^ 
 
 ''' X ' 
 
 whence 
 
 y = inx -\- h. 
 
THE STRAIGHT LINE. 
 
 39 
 
 OB = h. Draw 7\1/_L to OX. The triangles FMA, BOA 
 are similar ; therefore 
 
 PM MA OA-OM 
 
 BO OA 
 
 OA 
 
 r-«---l_ 
 
 X 
 
 h a 
 
 a' 
 
 
 
 or 
 
 whence - + ,, =1. [7] 
 
 ^ This is called the Symmetrical Equation of the straight line. 
 
 41. To find the cquatio7i of a straight line, given its distance 
 p from the origin, and the angle a. 
 
 Y 
 
 B 
 
 s 
 
 
 
 / 
 
 
 \^ . 
 
 
 
 
 M 
 
 Vx 
 
 Fig. 20. 
 
 Let AB (Fig. 20) be the line, F any point in it. Draw 
 OS ± to AB, meeting AB in S ; PM ± to OX; MR II to 
 AB, meeting 08 in R ; and PQ ± to AB. 
 
 Then p = OS =0P+ QP, a = SOM= PMQ. 
 
 By Trigonometry 
 
 OP = Oil/ cos a = X cos a, 
 QP ~ PM sin a = y sin a. 
 Therefore • Oi?4- (?P = ;7 = a: cos a + ?/ sin a. 
 
 Or a? cos a -I- 2/ sin a = p. rg] 
 
 This is called the Normal Equation of the straight line. 
 The quantity p is ahvags j^ositive (like the radius of unit 
 circle in Trigonometry). 
 
40 
 
 ANALYTIC GEOMETRY. 
 
 Note 1. Observe that all the equations of the straight line which 
 have been obtained are of the first degree. 
 
 Note 2. For the value of the sines, cosines, and tangents of the more 
 common angles, see Appendix. 
 
 Ex. 11. 
 
 Find the equation of the straight line passing through the 
 two points : 
 
 1. (2, 3) and (4, 5). 
 
 2. (4, 5) and (7, 11). 
 
 3. (-1,2) and (3,-2). 
 
 4. (-2,-2) and (-3,-3). 
 
 5. (4, 0) and (2, 3). 
 
 6. (0, 2) and (-3,0). 
 
 7. 
 
 (2, 5) and (0, 7). 
 
 8. 
 
 (3, 4) and (0, 0). 
 
 9. 
 
 (3, 0) and (0, 0). 
 
 10. 
 
 (3, 4) and (-2,4). 
 
 11. 
 
 (2, 5) and (-2, -5). 
 
 12. 
 
 (??i, ?2) and (-???, — 7i). 
 
 Find the equation of a straight 
 
 13. (4, 1) andy-r45°. 
 
 14. (2, 7) andy = 60°. 
 (1^ (-3, 11) and y = 45°. 
 
 16. (13,-4) and 7 = 150°. 
 
 (ft. (3, 0) and 7 = 30°. 
 
 18. (0, 3) and 7 = 135°. 
 
 19. (0, 0) and 7 = 120°. 
 
 20. (2,-3) and 7 = 0°. 
 
 21. (2, -3) and 7=90°. 
 (^^ i.= 2and7 = 45°. 
 
 23. ^ = 5,7=45°. 
 
 24. Z» = -4, 7 = 45°. 
 
 25. 5^-4,7 = 30°. 
 2^ ^=:_4, 7 = 0°. 
 
 27. 5 = -4, 7 = 60°. 
 
 28. 5 = -4, 7 = 90°. 
 
 line, given : 
 
 $3': ^ = -4,7=120°. 
 
 30. b = -i, 7 = 135°. 
 
 31. ^> = - 4, 7 = 150°. 
 (B^ ^ = -4,7 = 225°. 
 
 33. a = 4, 6 = 3. 
 
 34. a = -6,b = 2. 
 
 35. a^= — 3, 6 = — 3. 
 ^ a = c>, 5 = -3. 
 
 37. a = -10, h = D. 
 
 38. a = l, b=-l. 
 (39y a^n, h = — n. 
 
 40. a = ?2, h^in. 
 
 41. p = 5, a = 45°. 
 42j p=--^, a = 120°. 
 43"^ ;^ = 5, a=240°. 
 44. » = 5, a = 300°. 
 
THE STRAIGHT LINE. 41 
 
 Write the equations of the sides of a triangle : 
 
 45. If its vertices are the points (2, 1), (3, — 2), (-4, -1). 
 
 4G. If its vertices are the points (2, 3), (4, — 5), (- 3, - G). 
 
 47. Form the equations of the medians of the triangle 
 described in No. 53. 
 
 48. The vertices of a quadrilateral are (0, 0), (1, 5), (7, 0), 
 (4, — 9). Form the equations of its sides, and also of its 
 diagonals. 
 
 Find the equation of a straight line, given : 
 
 49. a = 7i, 7-30°. 51. p = 6, y = 45°. 
 
 50. a = - 3, (x„ y,) = (2, 5). 52. ;? ^ 6, y - 225^ 
 
 Note. The best way, in general, to construct a straight line from its 
 equation is to find its intercepts (f 26), and then lay them off on the 
 axes. If the line passes through the origin, it has no intercepts ; but 
 in this case a second point in the line is easily found by assuming any 
 convenient value (as 1) for x, and then computing the corresponding 
 value of 7/ from the given equation. 
 
 The intersection of two straight lines is to be found by the general 
 method explained in ^ 27. 
 
 Construct the following lines, and find the point of inter- 
 section : 
 
 53. 3.r — 2y + ll=0 and ?/ = 7rr. 
 
 54. 3a' + 5y-13 = and 4:i--?/-2 = 0. 
 
 55. 2a: + 3?/ = 7 and x — y = \. 
 
 56. y-f 13 = 5a; and y + 19 =1 x. 
 
 57. 3.2; + y+12 = and 2y=3a;-6. 
 
 58. 1 + 1 = 1 and 1 + 1 = 0. 
 
 59. Find the vertices of the triangle whose sides are the 
 lines 2a: + 9?/ +13 = 0, y = 7a:-38, 2y-a:=:2. 
 
42 ANALYTIC GEOMETRY. 
 
 60. Find the equation of the straight Hne passing through 
 the origin and the intersection of the lines 3a; — 2y + 4 = 
 and 3:r -|- 43/ — 5. Also find the distance between these two 
 points. 
 
 61. What is the equation of the line passing through (xi, 3/1), 
 and equally inclined to the two axes ? 
 
 62. Find the equations of the diagonals of the parallelo- 
 gram formed by the lines x = a, x = b, x — c, x^=d. 
 
 63. Show that the lines ?/ = 2:r + 3, ?/ = 3:r + 4, y = 4.r + 5 
 all pass through one point. 
 
 Find the intersection of two of the lines, and then see if its co-ordi- 
 nates will satisfy the equation of the remaining line. 
 
 64. The vertices of a triangle are (0, 0), (x^, 0), (x^, y.^. 
 Find the equations of its medians, and prove that they meet 
 in one point. 
 
 65. What must be the value of m if the line y = mx passes 
 through the point (1, 4)? 
 
 66. The line y = mx-{-2> passes through the intersection of 
 the lines y = x-{-l and y == 2:zr + 2. Determine the value of vi. 
 
 67. Find the value of h if the line y = ^x-\-h passes through 
 
 the point (2, 3). 
 
 68. What condition must be satisfied if the points (.ri, yi), 
 (^2, y-i), (^3, y-i) lie in one straight line ? 
 
 Hint. Let equation [4] represent the line through {x^, y{) and 
 (xj, 2/2) ; ^^^^ (■^3' 2/3) inust satisfy it. 
 
 69. Discuss equation [5] for the following cases : (i.) (:i'i, y^) 
 = (0, 0), (ii.)m=-0, (iii.)m = oo. 
 
 70. Discuss equation [6] for the following cases : (i.) b = 0, 
 (ii.)m — 0, (iii.) ?7i = CO, (iv.) ??^ ~ 0, and 5 = 0. 
 
 71. Discuss equation [7] for the following cases : (i.) a = b, 
 (ii.)a = 0, (iii.)a— 00, (iv.)6 = oo. 
 
the straight line. 43 
 
 General Equation of the First Degree. 
 
 42. Every equation involving x and y as variables, which 
 can be reduced by algebraic operations to the form 
 
 y^mx^ b, 
 represents a straight line having for slope the coefficient of x, 
 and for interccp)t on the axis of y the value of the constant term. 
 
 Algebraic operations never change the values of x and y 
 which will satisfy an equation ; therefore they cannot change 
 the locus represented by the equation. 
 
 The equation y = mx-\-b, from the manner in which it was 
 established, necessarily holds true for all values of m and b 
 from —00 to +00. Therefore every equation of the form 
 y^^mx-\-b must represent some particular one of the infi- 
 nite number of straight lines obtained by giving all pos- 
 sible values to the general symbols on and b ; and the values 
 of the two constants in the equation must be the values of 
 VI and b for the particular line represented by the equation. 
 
 43. Every equation which can be put in the form 
 
 a 
 represents a straight line, having for its intercepts the denomi- 
 nators of X and y, respectively. 
 
 The proof is similar to that of the preceding proposition. 
 
 44. Every equation of the first degree, with respect to the 
 variables x and y, can be put in the form 
 
 Ax + By + C = 0. [9] 
 
 where A, B, C stand for any numbers, positive or negative, 
 entire or fractional, rational or irrational. A and B cannot, 
 however, both be ; for if they could, then we should also have 
 C*=0, and the equation would vanish entirely. Equation [9] 
 is termed the General Equation of the First Degree. 
 
44 ANALYTIC GEOMETKY. 
 
 45. Every equation of the first degree in x and y represents a 
 straiyht line. 
 
 If we reduce the equation to the form Ax -}- By -f C= 0, 
 
 and solve for y, we obtain 
 
 A C 
 
 y=-B'-B 
 
 But the equation has now the form y = mx + b ; therefore 
 it represents a straight line (§ 42), and the values of the slope 
 and the intercept on the axis of y are 
 
 A . C 
 
 m-= . = — -• 
 
 B B 
 
 46. To find the intercepts and slope of a straight line, having 
 given its equation. 
 
 Method I. Find the intercepts, a and h, as in § 26. Then 
 
 (Fig. 19) m = tan7-=-tan(180°-y)=--tan^^O-=-- 
 
 a 
 
 Method II. Reduce the given equation to the form [6] ; 
 and make m equal to the coefficient of x. Reduce the equa- 
 tion to the form [7], and make a and b respectively equal to 
 the denominators of x and y. For these operations amount 
 to nothing more than substituting particular values in the 
 place of general symbols which from their nature include all 
 assignable values whatsoever. 
 
 Note. This mode of obtaining from the equation of a curve the vahies 
 of its constants is sometimes called the Method of Equating Coefficients. 
 
 Method III. Deterruine the values of ???, a, and h, in terms 
 of ^, B, and 0, by the method of equating coefficients. The 
 results are A C C 
 
 ™=__. .=--, b = --- 
 
 Nothing now remains but to reduce the given equation to the 
 general form, Ax^By + C= 0, and then substitute the par- 
 ticular values of ^, ^, and C. 
 
/ 
 
 THE STRAIGHT LINE. 45 
 
 47. Find the intercepts and slope of the straight line repre- 
 sented by the equation 3 a: — 3 = 4y + 9. 
 
 By I. If y = 0, a; = 4 ; if a; = 0, y = - 3. 
 
 Therefore ci — y, h = — Z, m = = .^ = 7- 
 
 a 4 4 
 
 By II. The equation put into the forms [6] and [7] becomes 
 
 y = |,r-3, and| + -^=l. 
 
 3 
 Therefore a = 4, ^ — — 3, m = -- 
 
 4 
 
 By III. The given equation in the form [9] is 
 
 3:r-4y-12 = 0. 
 
 Therefore A = 2>, ^ = -4, C=-12, and 
 
 -12 . ; -12 Q 3 3 
 
 a = —^ = ^, b = -—=-S, r. = - — = -. 
 
 Ex. 12. 
 
 Describe the position of the following lines by determining 
 the values of a, b, and m. 
 
 1. 1 + ^ = 1. 10. 1-1 = 1. 
 
 2. y = ^-9. 11. | + | = -l- 
 
 3. 3a: + 2 = 2y 12. 3y = rr. 
 
 4. 4y = 5a:. 13. 3rr = y. 
 
 5. 7a: + 3y = 0. 14. 5.r-4y + 20 = 0. 
 
 6. 4y = 3:^ + 24. 15. y = 6.r+12. 
 
 7. x + i/ = S. 16. y + 2-:a;-4. 
 
 8. 4y + :r + ll=0. 17. a: + V3?/-fl0 = 0. 
 
 9. 5:c-3y + 15 = 0. 18. :r— V3y-10 = 0. 
 
4G ANALYTIC GEOMETE-Y. 
 
 19. Discuss equation [9] for the following cases : 
 
 (i.) ^ = 0. (iv.) A = c^. (vii.) A=B, 0=0. 
 
 (ii.) ^ = 0. (v.) A = C=0. (viii.) A = -B, C=Q. 
 (iii.) C = 0. (vi.) A = B. 
 
 20. Eeduce equation [7] to the form of equation [6], and 
 find the value of m in terms of a and b. 
 
 21. "What value must C have in order that the line 
 4iX — 5i/-\- C may pass through the origin? Through the 
 point (2, 0) ? 
 
 22. Determine the values of ^, B, and C, so that the line 
 Ax -\- By -j- C =^ may pass through the points (3,0) and 
 (0,-12). 
 
 Since the co-ordinates of the given points must satisfy the equation, 
 we have the two relations oA -^ C = and — 12^ -f C= 0. 
 
 23. From equation [9] deduce equation [4] by the method 
 used in solving No. 22. 
 
 24. If equations [4] and [9] represent the same line, what 
 are the values A, B, C, in terms of Xi, t/^, Xo and 3/2 ? 
 
 25. In equation [4] find the values of m and h in terms of 
 
 ^X, 2/1, ^2, 2/2. 
 
 26. In equation [8] find the values of cos a, sin a, and ^, in 
 terms of the general constants A, B, C. 
 
 To reduce the form [9] to the form [S] we must multiply each term 
 by a certain quantity Ic ; to determine h, we have 
 
 lii^Ax — By -\- C) = X cos o + ysino— jo = 0, 
 whence, by equating coefficients, we obtain 
 
 cos a = kA, sin o = hB, p = — hC. 
 But cos^ a + sm^a = 1, 
 whence we have k= ± 
 
 \/A' + B' 
 By substituting the value of k, we have 
 
 A* ^. B C 
 
 y/A} + ^2 ± ^^2 ^^i ± ^j^2 ^ ^2 
 
THE STRAIGHT LINE. 
 
 47 
 
 and we should choose that sign before the radical which will make p 
 positive (^ 41), therefore the sign which is unlike that of C. 
 
 27. What are the values of cos a, sin a, and j9 for the fol- 
 lowing lines? Construct the lines. Show that they enclose a 
 parallelogram. 
 
 (a) 12a: + 52/ -26-0. (c) 12.r- Sy- 26 = 0. 
 
 (b) 12:c + 5y + 26 = 0. (d) 12:^- - 5y + 26 = 0. 
 
 Parallels and Perpendiculars. 
 
 48. If the lines represented by the equations y^7nx-\-b 
 and y = 7n'x-{- h' are parallel, then we must have, by Geom- 
 etry, 
 
 and therefore 
 
 y; 
 
 VI, 
 
 or 
 
 Fig. 21. 
 
 If the two lines are perpendicular, then, by Geometry, 
 
 y = y + 90° (see Fig. 21). 
 
 Therefore, by Trigonometry, 
 
 tan y' = — cot y, 
 
 1 
 
 m' = . 
 
 7)1 
 
 or 
 
 mm' 
 
 1. 
 
48 ANALYTIC GEOMETRY. 
 
 49. Conversely, prove that the lines represented hy the equa- 
 tions y = inx + Z> and y =^ on^x + h' are parallel if on^^m' ; 
 perpendicular, if ??^.??^' = — 1. 
 
 Hint. Use the above reasoning, taken in reverse order. 
 
 Not::. The equations m = m' and mm' = — 1 are examples of Equa- 
 tions of Condition ; they express tlve conditions which must be satisfied 
 in order that two lines may be parallel or perpendicular, respectively. 
 
 50. To find the equations of a straight line passing through 
 the point (.r^, ?/i) and (i.) parallel, (ii.) perpendicular, to the 
 line y = r}ix + h. 
 
 The slope of the required line is on in case (i.), and 
 
 in case (ii.) ; and in both cases the line passes through a given 
 point (rri, 3/i). 
 
 Therefore (§ 38) the required equations are 
 
 (1) y — yi = 'in{x — x;), 
 
 (2) 2/-yi = --(^-^i). 
 
 51. We shall now solve the problem of § 50 in another way. 
 Let the equation of a given line be 
 
 Ax + By + C-=^0. (1) 
 
 If we change C to any other value K, hut leave A and B 
 unchanged, the new equation 
 
 Ax^By^K=^ (2) 
 
 will represent a line parallel to the given line, because the 
 two lines have the same slope (§ 49). 
 
 If we change C to K, as before, and also interchange the 
 coefficients A and B, and alter the sign of one of them, the new 
 equation Bx-Ay + K=Q (3) 
 
 will represent a line perpendicular to the given line, because 
 the slopes of the two lines satisfy the condition ?7im'= — 1. 
 
THE STRAIGHT LINE. 49 
 
 By assigning different values to K, equations (1) and (2) 
 may be made to represent different parallels, and different 
 perpendiculars, respectively, to the given line ; and if we 
 regard K as entirely undetermined, then equation (2) may 
 be said to represent all parallels, and equation (3) all per- 
 pendiculars, to the given line. 
 
 But if in either case we assign a particular value to II, or 
 add a new condition which determines A'', then the equation 
 will represent one definite straight line. 
 
 Suppose we add the condition that the line must pass 
 through a given point {xi, y^ ; then these co-ordinates must 
 satisfy equations (2) and (3), and we have 
 
 Ax^-{-Bij^-^K=0, and Bx^-Ay^-^K=0. 
 Hence K=^ — {Ax^ -\- By^), and 'K=^ Ay^ — Bxi. 
 Substituting these values of K in (2) and (3), we obtain for 
 the equations of the required lines 
 
 Ax-^By = Ax,+B7/,. (4) 
 
 Bx — Ay = Bxi — Ayi. (6) 
 
 Ex. 13. 
 Find the equation of a straight line 
 
 1. Passing through (3, —7), and II to the line y=Sx — b. 
 
 2. Passing through (5, 3), and II to the' line ^y~\x = l. 
 
 3. Passing through (0, 0), and il to the line y — 4:2: = 1Q. 
 
 4. Passing through (5, 8), and || to the axis of x. 
 (^ Passing through (5, 8), and !! to the axis of y. 
 
 6. Passing through (3, —13), and JL to the line y — 4:X — 7. 
 
 7. Passing through (2, 9), and J_ to the line 7y-f 23a: — 5 = 0. 
 
 8. Passing through (0, 0), and _L to the line x-\-2y — l. 
 
 9. Perpendicular to the line bx — 'Jy -\-l = 0, and erected 
 at the point whose abscissa = 1. 
 
50 
 
 ANALYTIC GEOMETRY. 
 
 10. Perpendicular to the line y — 3a: = 2, and passing 
 through the intersection of the lines x — y = l and 2x -[• 
 3y = 7. 
 
 Angles. 
 
 52. To find the angle formed hy the lines y = mx + h, and 
 y = m^x -\-b\ 
 
 Let AB and CD (Fig. 22), represent the two lines, respec- 
 tively, meeting in the point P. 
 
 Let the angle APC=<fi, and tan <^ = ?!. By Geometry, 
 if> = y — y'. Whence, by Trigonometry, 
 
 m-m' ^^^^ 
 
 t 
 
 1 + mm' 
 
 This equation determines the value of <f>. 
 
 Fig. 22. 
 
 Fig. 23. 
 
 53. To find the equation of a straight line passing through 
 a given point (x^, 3/1), and making a given angle </> luith a given 
 line y = mx-\-b. 
 
 Let the required equation be 
 
 y -y,^m\x-x,) 
 
 where m' is not yet determined. 
 
THE STRAIGHT LINE. 61 
 
 Since the required line may lie either as PQ or PR 
 (Fig. 23), ^ve shall have ^by § b'S), 
 
 m' — on m — 7n 
 
 tan <^ = or 
 
 1 + mm^ 1 + ^Mn' 
 
 HI 7?i -4- tan <6 
 ence 7?i' = ~~ ^ , 
 
 1 ^=^?/t tan ^ 
 and the required equation is 
 
 rti ± tan <|> , . m -, 
 
 v — yi = z-— — r-^(^ — a^i» [11] 
 
 1 :f ^/i tan <j) *- -■ 
 
 and (as Fig. 23 shows) there are in general two straight lines 
 satisfying the given conditions. 
 
 Ex. 14. 
 
 1. Find the angle formed by the lines rr + 2y + l = and 
 ar-3y-4 = p. 
 
 The two slopes are — h and |. If we pvit m = — i, vi' = ^, we obtain 
 <•-— 1, ^ = 135°. If weput m = ^, m' = — I, weget i = l, = 45°. Show 
 that both these results are correct. 
 
 Find the tangent of the angle formed by the lines 
 
 2. 3a;-4y = 7 and 2:r-y = 3. 
 
 3. 2a: + 3y 4-4 = and 3a; + 4y + 5 = 0. 
 
 4. y — 01x^=1 and 2{y — l)=-nx. 
 
 Find the angle formed by the lines 
 
 5. x-\-y = \ and y = .'r-]-4. 
 
 6. y + 3 = 2a: and y-f 3;r = 2. 
 
 7. 2:r + 3y + 7 = and 5.T-2y + 4 = 0. 
 
 8. 6a: = 2y-f-3 and y-3.r = 10. 
 
 9. a; + 3 = and y — V3.r + 4 = 0. 
 
62 ANALYTIC GEOMETRY. 
 
 10. Discuss equation [11] for the cases where fp = ^° and 
 <^ = 90°. 
 
 Note, The learner should try to solve the next five exercises directly, 
 without using equation [11] ; then verify the result by means of [11]. 
 
 Find the equation of a straight line 
 
 11. Passing through the point (3, 5), and making the angle 
 45° with the line 2a;-3y + 5 = 0. 
 
 12. Passing through the point (— 2, 1), and making the 
 angle 45° with the line 2?/ = 6 — 3a;. 
 
 13. Passing through that point of the line y = 2.r — 1 for 
 wdiich X-—2, and making the angle 30° with the same line. 
 
 14. Passing through (1, 3), and making the angle 30° with 
 the line :r-2?/ + l = 0. 
 
 15. Prove that the lines represented by the equations 
 
 Axi-B2/-i-C=0, A'x-i-B'y + C' = 
 are parallel if AB' = A'B ; perpendicular, if A A' = ~BB'. 
 
 16. Given the equation 3a; + 4y+6 = 0; show that the 
 general equations representing (i.) all parallels and (ii.) all 
 perpendiculars to the given line are 
 
 (i.) 3a; + 4y + ^=0. 
 (ii.) ^x-2>y + K=0. 
 
 17. Deduce the following equations for lines passing 
 through (.Ti, 3/1) and (i.) parallel, (ii.) perpendicular, to the 
 line y = mx-\- h. 
 
 (i.) y — mx = 3/1 — r)iXx. 
 (ii.) my -\- X = m2j^-\- x^. 
 
 18. Write the equations of 3 lines parallel, and 3 lines per- 
 pendicular, to the line 2a; + 3?/-j-l = 0. 
 
THE STRAIeillT LINE. 53 
 
 10. Among the following lines select parallel lines ; per- 
 pendicular lines ; lines neither parallel nor perpendicular : 
 (i.) 2a;-f-3y-l = 0. (v.) x-y = 1. 
 
 (ii.) 3a:-2y = 20. (vi.) 5(a; + 2/) -11 = 0. 
 
 (iii.) 4.^ + 6^ = 0. (vii.) a; = 8. 
 
 (iv.) 12:r = 8y + 7. (viii.) y + 10 = 0. 
 
 20. Prove that the angle <^, between the lines 
 
 Ax-\-By-^G = ^ and ^'o; + ^'y + C" = 0, 
 
 is determined by the equation 
 
 , ^ AB'-A'B 
 ''"""^^' AA' + BB' ' 
 
 21. From the preceding equation deduce the conditions of 
 parallel lines and per^^endicular lines given in No. 15, ]d. 52. 
 
 Find the equation of a straight line 
 
 22. Parallel to 2a; + 3y + 6 = 0, and passing through (5, 7). 
 
 23. Parallel to 2a; + y — 1 = 0, and passing through the 
 intersection of 3a; + 23/ — 59 = and 5a: — 7y + 6 = 0. 
 
 24. Parallel to the line joining (—2, 7) and (~4, — 5), 
 and passing through (5, 3). 
 
 25. Parallel to y = '}nx-\-h, and at a distance d from the 
 origin. 
 
 26. Perpendicular to Ax -f By -f (7= 0, and cutting an in- 
 tercept h on the axis of y. 
 
 X 1J 
 
 27. Perpendicular to - + t = 1, and passing through (a, h). 
 
 X 1/ 
 
 28. Making the angle 45° with -+j = l, and passing 
 through {a, 0). * 
 
 29. Show that the triangle whose vertices are the points 
 (2, 1), (3, -2), (-4, -1) is a right triangle. 
 
54 ANALYTIC GEOMETRY. 
 
 30. The vertices of a triangle are (—1,-1), (—3,5), 
 (7, 11). Find the equations of its altitudes. Prove that 
 the altitudes meet in one point. 
 
 31. Find the equation of the perpendicular erected at the 
 middle point of the line joining (5, 2) to the intersection of 
 x-\-2y-ll = and 9:i-- 2y + 59 = 0. 
 
 32. Find the equations of the perpendiculars erected at the 
 middle points of the sides of the triangle whose vertices are 
 (5, —7), (1, 11), (—4, 13). Prove that these perpendiculars 
 meet in one point. 
 
 33. The equations of the sides of a triangle are 
 ,^^4-7/+l = 0, 3a: + 5y + ll=0, :i'-f2y + 4 = 0. 
 
 Find (i.) the equations of the perpendiculars erected at the 
 middle points of the sides ; (ii.) the co-ordinates of their com- 
 mon point of intersection ; (iii.) the distance of this point from 
 the vertices of the triangle. 
 
 34. Show that the straight line passing through {a, I) and 
 ((?, d) is perpendicular to the straight line passing through 
 (5, —a) and (cZ, —c). 
 
 35. What is the equation of a straight line passing through 
 (.Ti, 3/1), and making an angle <^ with the line Ax-\-By-\-C = 0'^ 
 
 Distances. 
 
 54. Find the distance from the point (—4, 1) to the line 
 3.?;-4?/ + l = 0. Ans. 3. 
 
 The required distance is the length of the perpendicular let 
 fall from the given point to the given line. The first method 
 that occurs for solving the problem is to form the equation_^f_ 
 this perpendicular, find its intersection witli, the given linej^ 
 andlhen compute the distance from this jntei^section^o the 
 given point. 
 
 Let this method be followed in solving the above problem 
 and the first five problems of Ex. 15. 
 
THE STRAIGHT LINE. 
 
 65 
 
 55. To find the distance from the point (xy, y^) to the line 
 Ax^-Bij'-\-C=Q). 
 
 Let F (Fig. 24) represent the given point {x^, y^), and AB 
 the given line Ax -\- By + C =-0, Draw PS J_ to AB, and 
 PM 1, to the axis of :r and meeting AB in a point P. Let 
 d denote the required distance PS. Then, by Trigonometry, 
 
 d=PP cos PPS = PR cos SAX= PR cos y. 
 
 Fig. 24. 
 
 Since R is in the given line, and 021= x^, 
 
 RM = 
 
 Axy-C 
 B 
 
 Therefore PR - MP - MR = ^•^t+^.Vi + ^ . 
 
 B 
 
 To find the value of cosy, we may use the relations 
 
 cos^y + sin-y = 1, and - 
 
 Eliminating sin y, we obtain 
 B 
 
 sm y 
 cosy 
 
 cos y = rh 
 
 ^A' + B' 
 Substituting these values of PR and cosy, we have 
 
 d 
 
 Ax, 4- gf/i + C^ 
 
 [12] 
 
56 ANALYTIC GEOMETRY. 
 
 So long as we are concerned with a single distance, there is 
 no occasion. for the use of both signs, and we should choose 
 that sign which will make d positive. 
 
 Hence, to find the distance from the point {x^, yj to the line 
 Ax-\-By^C = 0, we have as a practical rule : Write x^^ for 
 X and ?/i for y, and divide the value of the resulting expression 
 hy V^M^. 
 
 Solve this problem when the given point is assumed to be 
 at Q (see Fig. 24) on the same side of the line as the origin. 
 
 What is the value of d when the given j)oint is (i.) the 
 origin, (ii.) in the given line? 
 
 56. To find the distance from the point [x^, y^ to the line 
 X cos a-\-y sin a —p = 0. 
 
 Let the equation of the line through the given point, par- 
 allel to the given line, be (§ 41) 
 
 X cos a -f- y sina— p' = 0. 
 
 Since (xi, 3/1) is in this line, 
 
 r^i cos a + 3/1 sin a — ^9' = 0. 
 Therefore p' = Xi cos a + 2/i sin a. 
 
 'Now p and^:*' are the distances from the origin to the given 
 line, and its parallel, respectively ; therefore, if d denote the 
 required distance, 
 
 d= ± {p' —75) = =t [xi cos a + ?/i sin a ~ p) ; 
 and in general we should choose that sign which will make d 
 positive : the j^ositive sign, if (x-^, y^) and the origin are on 
 ojoposite sides of the given line; the negative sign, if they are 
 on the same side. 
 
 If, then, the. equation of a straight line is reduced to the 
 normal form x cos a -f y sin a — |; = 0, the distance from any 
 point (:ri, yf) to the line is found simply by substituting on the 
 left-hand side of the ecjuation x^ for x and y^ for y, and then 
 computing the value of the expression. 
 
XnE STIIAIGIIT LINE. 57 
 
 Ex. 15. 
 
 1. Find the distance from (1, 13) to the line Srr^y — 5. 
 
 2. Find the distance from (8, 4) to the line ?/ = 22: — 16. 
 
 3. Find the distance from the origin to the line 3a;-|-4?/— 20. 
 
 4. Find the distance from (2, 3) to the line 2:i- + y — 4 = 0. 
 
 5. Find the distance from (3, 3) to the line y = 4.x' — 9. 
 
 6. Prove that the distance from the point (x^, y^) to the line 
 
 y = onx 4- 5 is ^ 
 
 d= -u . Vi — ^»-'^i — ^ ^ 
 
 VT+T/? 
 
 that sign being chosen which will make d positive. Express 
 this result in the form of a rule for practice. 
 
 7. Find the distances from the line 3.'r-|-4?/ + 15 = to 
 the following points : (3, 0). (3, - 1), (3, - 2), (3, - 3), (3, - 4), 
 (3,-5), (3, -6), (3, -7), (0, 0), (-1, 0), (-2, 0), (-3, 0), 
 (-4,0), (-5,0), (-6,0). 
 
 8. Find the distances from (1, 3) to the following lines : 
 
 3.r + 4y + 15 = 0. 3:r + 4y- 5 = 0. 
 
 3:r + 4y-f 10 = 0. 3a; + 4y-10 = 0. 
 
 3.r + 4y+ 5 = 0. 3:r + 4y-15 = 0. 
 
 3.r + 4y =0. 3:c + 4y-20 = 0. 
 
 Find the following distances : 
 
 9. From the point (2, —5) to the line y — 3:r = 7. 
 
 10. From the point (4, 5) to the line 4y + 5a; = 20. 
 
 11. From the point (2, 3) to the line x -\- 1/ = 1. 
 
 12. From the point (0, 1) to the line 3.T-3y = l. 
 
 13. From the point (—1, 3) to the line 3a; + 4y+2 = 0. 
 
68 ANALYTIC GEOMETRY. 
 
 14. From the origin to the line 3:r + 2?/ — 6 = 0. 
 
 15. From the point (2, —7) to the line joining (—4, 1) and 
 (3, 2). 
 
 16. From the line y = lx to the intersection of the lines 
 y — ^x — 4: and y=bx-\-2. 
 
 17. From the origin to the line a(x — a) -\- h{x — &) = 0. 
 
 18. From the points (a, h) and (— a, — h) to fhe line 
 
 a 
 
 19. From the point (a, ^) to the line ax -\- hy = 0. 
 
 20. From the point (A, k) to the line Ax -\- By + = 1). 
 
 Find the distance between the two parallels : 
 
 21. 3a; + 4y + 15 = and 3a; + 4z/ + 5 = 0. 
 
 22. 3:^ + 4?/+15 = and 3.^ + 4?/ — 5 = 0. 
 
 23. Ax-i-By + C=0 &nd Ax + By-}-C' = 0. 
 
 24. 9^ + 3y-7=0 and 9a; + 3y-27=0. 
 
 25. y = 5x — 7 and y = bx -}-d. 
 
 26. ■^ + f = 2and^ + f=J. 
 a a A 
 
 27. Show that the locus of a point whicli is equidistant 
 from the lines 3a; + 4y — 12 = and 4:r + o?/-24 = con- 
 sists of two straight lines. Find their equations, and draw a 
 figure representing the four lines. 
 
 28. Show that the locus of a point which so moves that 
 the sum of its distances from two given straight lines is con- 
 stant is a straight line. 
 
THE STRAIGHT LINE. 59 
 
 « 
 
 Areas. 
 
 57. Find the area of the triangle whose vertices are the 
 points (2, 1), (3, - 2), (- 4, - 1). Ans. 10. 
 
 It is shown in Elementary Geometry that the area of a 
 triangle is equal to one-half the product of its base and its 
 altitude ; hence this problem may be solved by performing 
 the following operations : 
 
 (i.) Find the length of one side chosen as base, 
 (ii.) Find the equation of the altitude, 
 (iii.) Find the intersection of the base and the altitude, 
 (iv.) Find the length of the altitude, 
 (v.) Multiply this length by one-half the base. 
 
 Let the first five problems of Ex. 16 be solved in this way. 
 
 58. I'tnd the area of a triangle, having given its vertices 
 (^■i, 2/i), {x,, y,), (.!;., y,). 
 
 Solution I. If we take as base the line joining {x\ y,) 
 and {x-i, y-i), then 
 
 base -- V(y2 — yO' + {x^ — x{)\ 
 
 The altitude is the distance from (.1-3, yg) to the base. 
 Writing in equation [12] (p. 55) x^ for x^, 3/3 for 3/1, and for 
 A, B, C the values 
 
 ^^y-i — 3/1, ^=- — 0^2 — ^1), = X2yi — x,7jo, 
 obtained by equating co-eflicients in [4] and [9] (pp. 37, 43), 
 we get 
 
 altitude = (.V^ ~ .vO^s — (^2 - ^1)^3 + ^2^1 — ^1^2 
 V(?/,-yxJ^-f(x,-^)^ 
 
 The area of the triangle = i base X altitude ; therefore 
 area = J [(y., — yO^s — (^2 — oc,)y^ + x^y^ — x^y^], 
 which may be written more symmetrically thus : 
 
 area = I [x^{y., — y.) -f- x.{y-, — y,) + x,{y, — y,)\. [13] 
 
60 
 
 ANALYTIC GEOMETRY. 
 
 Solution II. Let PQR (Fig. 25) be the given triangle, 
 and let the co-ordinates of PQR be x^y^, x^y-i, x^y^, respectively. 
 Drop the perpendiculars PM, QN, EL ; then 
 
 area PQR = PQN3I+ RLNQ - PMLR. 
 By Geometry, 
 
 PQNM= ^M]V(PM+ QJST) 
 
 Similarly, 
 
 RLNQ = \{x^ - X.) (?/3 + 2/2), 
 PMLR = ^{x, - X,) {y, + 2/0. 
 Substituting these values, we have 
 
 area PQR = l[{x^ — x^) (?/, -f y^) + {x^ — x^) (3/3 + y^) 
 
 - (x, - X,) (2/3 + yO] 
 = i [•^2yi — ^12/2 + ^'33/2 — ^23/3 + ^13/3 — ^32/i] 
 = i[^i(y2 — 2/3) + ^2(^3 - 2/1) + ^3(yi - ^2)]- 
 
 
 
 Jl/ 
 
 iv 
 
 LX 
 
 Fig. 25. 
 
 Ex. 16. 
 Find the area of the triangle whose vertices are the points 
 
 1. (0, 0), (1, 2), (2, 1). 
 
 2. (3,4), (-3,-4), (0,4). 
 
 3. (2, 3), (4,-5), (-3,-6). 
 
 4. (8,3), (-2,3), (4,-5). 
 
 5. (a,0), (-a,0), (0,1). 
 
THE STRAIGHT LINE. 61 
 
 6. Compare the formula for the area of a triangle with the 
 result obtained by solving No. 68, p. 42. What, then, is the 
 geometric meaning of that result ? 
 
 Find the area of the figure having for vertices the points : 
 
 7. (3, 5), (7, 11), (9, 1). 
 
 8. (3, -2), (5, 4), (-7, 3). 
 
 9. (-1,2), (4,4), (6,-3). 
 
 10. (0, 0), (x„ y,l (t„ y,)- 
 
 11. (2,-5), (2,8), (-2,-5). 
 
 12. (10,5), (-2,5), (-5,-3), (7,-3). 
 
 13. (0, 0), (5, 0), (9, 11), (0, 3). 
 
 14. (a, 1), (0, b\ (c, 1). 
 
 15. (a, h), (b, a), (c, c). 
 
 16. (a, 5), (5, a), (c, -c). 
 
 17. Find the angles and the area of the triangle whose 
 vertices are (3, 0), (0, 3V3), (6, 3 V3). 
 
 What is the area contained by the lines 
 
 18. 07 = 0, ?/ = 0, 5a: + 4y = 20? 
 l^. x-{-y = l, x — y = 0, y = 0? 
 
 20. :r + 2y=:5, 2x-\-y = 7, y = a: + l? 
 
 21. x + y = 0, x = 9/, y=3a? 
 
 22. y = Sx, 7/ = 7x, 7/'—c? 
 
 2d. x = 0, y = 0, :r-4 = 0, y + 6^0? 
 
 24. 3a: + y + 4-0, 3:r-5y + 34 = 0, 3.r-2y + l-0? 
 
 25. ^-5y + 13=:0, 5:r + 7y + l = 0, 3:f + y-9 = 0? 
 
 26. ^' — y = 0, x-^-y = 0, x~y = a, x -\- y = b'l 
 
62 ANALYTIC GEOMETRY. 
 
 Find the area contained by the lines: 
 
 27. x = 0, y = 0, y = mx-\-b. 
 
 28. x = 0, y = 0, - + f = l. 
 
 a 
 
 29. x = 0, 7/ = 0, Ax + By + C=0. 
 
 30. y = 3p;-9, y = 2>x + b, 2y = a:-6, 23/ = a;-fl4. 
 
 31. What is the area of the triangle formed by drawing 
 straight lines from the point (2, 11) to the points in the line 
 y =^bx — 6 for which ^^^i = 4, ajj = 7 ? 
 
 Ex. 17. (Review.) 
 
 1. Deduce equation [7], p. 39, from equation [6]. 
 
 2. The equation y = mix -f ^ is not so general as the equa- 
 tion Ax -\- By -\- C '= ^ , because it cannot represent a line 
 parallel to the axis of y. Explain more fully. 
 
 Determine for the following lines the values of a, 5, y, j), and a : 
 
 3. :r=2. 6. a; + V3y = 2. 
 
 4. x^=y. 7. a; — Vo?/ = 2. 
 
 5. 3/ + l = V3(:r + 2). 8. ■\^d>x-y = 2. 
 
 9. Find the equations of the diagonals of the figure formed 
 by the lines 3:r-y + 9 = 0, 'ix = y-\, 5a; + 3y=:18, 
 5^' + 3y = 2. What kind of quadrilateral is it? Why? 
 
 10. Find the distance between the j^arallels 9a; = ?/ + l 
 and ^x = y — 7. 
 
 11. The vertices of a quadrilateral are (3, 12), (7, 9). 
 (2, — 3), (— 2, 0). Find the equations of its sides and its area. 
 
 12. The vertices of a quadrilateral are (6, —4), (4, 4), 
 (—4, 2), (—8, —6). Prove that the lines joining the middle 
 points of adjacent sides form a parallelogram. Find the area 
 of this parallelogram. 
 
THE STRAIGHT LINE. 63 
 
 Find the equation of a line passing through (3, 4), and also 
 
 13. Perpendicular to the axis of x. 
 
 14. Making the angle 45° with the axis of :r. 
 
 15. Parallel to the line 5.r -f- Gy + 8 = 0. 
 
 »1G. Intercepting on the axis of y the distance — 10. 
 
 H-'T. Passing through the point half way between (1, —4; 
 and (-5,4). 
 
 18. Perpendicular to the line joining (3, 4) and (—1, 0). 
 
 Find the equations of the following lines : 
 
 19. A line parallel to the line joining (x^, y,) and (x.^, y^), 
 and passing through (x^, y-^. 
 
 20. The lines passing through (S, 3), (4, 3), (- 5, — 2). 
 
 21. A line passing through the intersection of the lines 
 2a; + 5y + 8 = and 3:f — 4y — 7=0, and ± to the latter 
 line. 
 
 " 22. A lino J_ to the line 4.r — y = 0, and passing through 
 that point of the given line whose abscissa is 2. 
 
 23. A line li to the line 3:^: + 4y = 0, and passing through 
 the intersection of the lines :r — 2y— a = and a;-|-3y — 2a = 0. 
 
 '^ 24. A line through (4, 3), such that the given point bisects 
 the portion contained between the axes. 
 
 25. A line through (.Tj, yi), such that the given point bisects 
 the portion contained between the axes. 
 
 26. A line through (4, 3), and forming with the axes in the 
 second quadrant a triangle whose area is 8. 
 
 27. A line through (4, 3), and forming with the axes in the 
 fourth quadrant a triangle whose area is 8. 
 
 28. A line through (— 4, 3), such that the portion between 
 the axes is divided by the given point in the ratio 5 : 3. 
 
04 ANALYTIC GEOMETRY. 
 
 29. A line dividing the distance between (—3, 7) and 
 (5,-4) in the ratio 4:7, and J_ to the line joining these 
 points. 
 
 30. The two lines through (3, 5) making the angle 45° with 
 the line 2:r-3y-7-=0. 
 
 31. The two lines through (7, —5) which make the angle 
 45° with the line 6:r — 2y + 3 = 0. 
 
 32. The line making the angle 45° with the line joining 
 (7, — 1) and (—3, 5), and intercepting the distance 5 on the 
 axis of X, 
 
 •33. The two lines which pass through the origin and tri- 
 sect the portion of the line x-\-y=\ included between the 
 axes. 
 
 34. The two lines il to the line 4.r + 5?/ + 11= 0, at the 
 distance 3 from it. 
 
 35. The bisectors of the angles contained between the lines 
 y = 2a; + 4, y = %x-^^. 
 
 Hi^sTT. Every point in the bisector of an angle is equidistant from 
 the sides of the angle. 
 
 6. The bisectors of the angles contained between the lines 
 
 •o 
 
 2a;-5?/ = 0, 4:r + 3?/ = l 
 
 37. The two lines which pass through (3, 12), and whose 
 distance from (7, 2) is equal to V58. 
 
 38. The two lines which pass through (—2, 5), and are 
 each equidistant from (3, —7) and (—4, 1). 
 
 Find the angle contained between the lines: 
 
 39. y + 3 = 2a; and y+3a: = 2. 
 
 40. ?/ = 5a: — 7 and 5?/ + a;- 3 = 0. 
 
THE STRAIGUT LINE. 65 
 
 Find the distance : 
 
 41. From the intersection of the lines 3^' + 2y + 4 = 0, 
 2^- + 5y + 8=-0 to the line y = 5a: + 6. 
 
 42. From the point (h, Jc) to the line - + ^^ = 1. 
 
 43. From the origin to the line hx + ^y = <?^ 
 
 44. From the point (a, 0) to the line y = vix + — • 
 
 Find the area included between the following lines : 
 
 45. X ■= y, x-\- y ■=(), x = c. 
 
 46. x-i-y = k, 2x = y + k, 2y = x + k 
 
 47. ^ + 1 = 1. y = 2.r + 5, x = 2y + a. 
 
 48. y = 4.r + 7 and the lines which join the origin to those 
 points of the given line whose ordinates are — 1 and 19. 
 
 49. The lines joining the middle points of the sides of the 
 triangle formed by the lines x — 5y-l-ll=--0, 11. t + 6?/ — 1 = 0, 
 x + y + 4.^0. 
 
 50. Find the area of the quadrilateral whose vertices are 
 (0, 0), (0, 5), (11, 9). (7, 0). 
 
 51. AVhat point in the line 5^7 — 4y — 28 = is equidis- 
 tant from the points (1, 5) and (7, — 3) ? 
 
 52. Prove that the diagonals of a square are J. to each 
 other. 
 
 53. Prove that the line joining the middle point of two sides 
 of a triangle is parallel to the third side. 
 
 54. What is the geometric meaning of the equation 
 xy = 0? 
 
66 ANALYTIC GEOMETRY. 
 
 55. Show that the three points (3a, 0), (0, 2>h), {a, 2b)' 
 are in a straight line. 
 
 56. Show that the three lines 5x -{-Si/ — 7 =0, 3a: — 42/ 
 — 10 = 0, and ^ -f 2y = meet in a point. 
 
 57. What must be the value of a in order that the three 
 lines 3:r + 3/— 2 = 0, 2a; — y — 3 = 0, and aa: + 2y — 3 = 
 
 may meet in a point ? 
 
 What straight lines are represented by the equations : 
 
 58. x'' + (a-h)x-ab = 0? 
 
 59. xij -^-hx-^- ay + «^ = ? 
 
 60. x\j=^xif1 
 
 61. 14a;^-5.Ty — y' = 0? 
 
 In the following exercises prove that the locus of the point 
 is a straight line, and obtain its equation. 
 
 62. The locus of the vertex of a triangle having the base 
 and the area constant. 
 
 63. The locus of a j)oint equidistant from the points {x^, ?/i). 
 and (.T., y.?}. 
 
 64. The locus of a point at the distance d from the line 
 Ax^By^C = ^. 
 
 65. The locus of a point so moving that the sum of its 
 distances from the axes shall be constant and equal to Tc. 
 
 66. The locus of a jioint so moving that the sum of its 
 distances from the lines Ax^By-\-C=^, ^'a- +^'y + ^' = 
 shall be constant and equal to Iz. 
 
 67. The locus of the vertex of a triangle, having given the 
 base and the difference of the squares of the other sides. 
 
THE STRAIGUT LINE. 67 
 
 SUPPLEMENTARY PROPOSITIONS. 
 
 Lines passing through One Point. 
 
 59. If S = 0, S' —0 represent the equations of any two loci 
 with the terms all transposed to the left-hand side, and k de- 
 notes an arbitrary constant, then the locus represented hy the 
 equation /S -^ kS' = 2^(^sses throurjh every point common to 
 the tivo given loci. 
 
 For it is plain that any co-ordinates which satisfy the equa- 
 tion S=0, and also satisfy the equation jS' = 0, must like- 
 wise satisfy the equation S-\- kS' = 0. 
 
 For what values of k will the equation ^ + kS' = repre- 
 sent the lines jS==0 and /S" = 0, respectively? 
 
 60. Mnd the equation of the line joining the point (3, 4) to 
 the intersection of the lines 
 
 3:i;-2y-fl7=0 and a.- + 4y-27=0. 
 
 The method of solving this question which first occurs is 
 to find the intersection of the given lines and then apply 
 equation [4], p. 37. 
 
 Another method, almost equally obvious, is to employ equa- 
 tion [5], which gives at once 
 
 y-^ = m{x-Z), 
 and then determine m by substituting for x and y the co-ordi- 
 nates of the intersection of the given lines. 
 
 The following method, founded on the princi|)le stated in 
 § 59, is, however, sometimes preferable, on account of its 
 generality and because it saves the labor of solving the given 
 equations. According to this principle, the required equation 
 may be immediately written in the form 
 
 3a; - 2y + 17-f /<^ + 4y - 27) = 0. 
 
68 ANALYTIC GEOMETRY. 
 
 And since tlie line passes through (3, 4), we must have 
 9 - 8 + 1 7 + ^<3 + 16 - 27) = 0, 
 
 whence ^ = -• 
 
 4 
 
 Therefore 12a:-8y+68 + 9^ + 36y-243 = 0, 
 or 3:r + 4y — 25--0. 
 
 This is the equation of the required line. 
 
 61. If the equations of three straight lines are 
 
 and ive can find three constants, I, on, n, so that the relation 
 l{Ax +i??/ + C) + m(A'x+B'y+ C) + n(A"x+II''7/-\- C") = 
 is identically true, that is, true for all values of x and y, then 
 the three lines tneet in a point. 
 
 For if the co-ordinates of any point satisfy any two of the 
 equations, then the above relation shows that they will also 
 satisfy the third equation. 
 
 62. To find the equation of a straight line passing through 
 the intersection of the two lines 
 
 Ax-hI!2j + C=0, A'x + J5'iji-C' = 0, 
 and bisecting the angle between them. 
 
 There are evidently two bisectors : one bisecting the angle 
 in which the origin lies ; the other bisecting the supplementary 
 angle. 
 
 The simplest way to obtain their equations is to express 
 algebraically the fact (proved in Geometry) that any point 
 (.r, y) of the bisector is equidistant from the sides of the angle. 
 Hence from equation [12], p. 55, we immediately obtain the 
 equation ^^ _|_ ^ ^ 4. ^ ^ _^ Ax -¥ B'y + C ^ 
 
 which represents both bisectors if we use both signs on the 
 ricrht-hand side. 
 
THE STRAIGHT LINE. G9 
 
 In order to distinguish between the bisectors, it is necessary 
 to pay attention to the sign of the distance from a straight 
 line to a point. We see from equation [12] that this dis- 
 tance changes sign when the point crosses the line ; let it be 
 agreed that distances measured to points on the origin side of 
 the line shall be considered positive, and that distances meas- 
 ured to points on the side remote from the origin shall be 
 considered negative. 
 
 "Now the distance from the line Ax -^r By -\- C =^ ^ to the 
 origin itself is p 
 
 and in order that this may be always positive, we must place 
 before it the same sign as that of C. It follows that equation 
 [14] will represent the bisector of the angle in which the origin 
 lies if we choose that sign which will make the two constant 
 terms alike in sign. 
 
 If we choose the other sign, the equation of course will 
 represent the bisector of the supplementary angle. 
 
 63. To find the ^ equation of a straight line passing through 
 the intersection of the two lines 
 
 X cos a -f y sin a — ^ = 0, X cos o! -\-y sin a' — p^ — 0, 
 and bisecting the angle between them. 
 
 Taking the angle which includes the origin, and denoting 
 by {x, y) any point in the bisector, we have immediately for 
 its equation 
 
 {x cos a + y sin a — 7?) — (x cos a.' -\- y sin a — p^^ = 0. 
 
 The equation of the bisector of the supplementary angle is 
 
 {x cos a -f- 2/ sin a — 'p) + {x cos o! -\-y sin o! — pj) = 0. 
 
 It may be shown from the form of these equations that the 
 two bisectors are perpendicular- to each other. 
 
70 ANALYTIC GEOMETRY. 
 
 Ex. 18. 
 
 Find the equation of a line passing l^hrough the intersection 
 of the lines 3^^ + 2?/ + 17 = 0, :r + 4y-27--0, and 
 
 1. Passing also through the origin. 
 
 2. Parallel to the line .r + 2y + 3 = 0. 
 
 3. Perpendicular to the line 6x — 5i/ --= 0. 
 
 4. Equally inclined to the two axes. 
 
 5. Find the equation of a line parallel to the line x = y, 
 and passing through the intersection of the lines y = 2a:-l-l 
 and ?/ + 3:?: = ll. 
 
 6. Find the equation of a straight line joining (2, 3) to the 
 intersection of the lines 
 
 2.'i; + 3y + l=0 and 3a;-4y = 5. 
 
 7. Find the equation of a straight line joining (0, 0) to the 
 intersection of the lines 
 
 5a: — 2y + 3 = and lSx-}-i/ = l. 
 
 8. Find the equation of a straight line joining (P, 11) to the 
 intersection of the lines 
 
 2x-i-57/-S = and Sx—4:i/=8. 
 
 Find the equation of the straight line passing through the 
 intersection of the lines Ax -\- Bi/ -\- C = and A'x -{- J5'i/ 
 + C*' = 0, and also 
 
 9. Passing through the origin. 
 
 10. Drawn parallel to the axis of x. 
 
 11. Passing through the point (x^, y^. 
 
 12. Find the equation of a straight line passing through 
 the intersection of 5a: — 4y + 3 = and 7a: -f- lly — 1 = 0, 
 and cutting on the axis of y an intercept equal to 6. 
 
THE STRAIGHT LINE. 71 
 
 13. Find the equation of a straight line passing through 
 the intersection of ?/ = 7a:— 4 and y— - — 2:i- + 5, and forming 
 with the axis of x the angle G0°. 
 
 14. The distance of a straight line from the origin is 5 ; 
 and it passes through the intersection of the lines 3.r— 2?/ 
 + 11 = and 6 a; + 7 ?/ — 55 = 0. What is its equation ? 
 
 15. What is the equation of the straight line passing 
 through the intersection of bx -j- ay ■= ah and y = inx^ and 
 perpendicular to the former line ? 
 
 Prove that the following lines are, concurrent (or pass 
 through one point) : 
 
 16. y = 1x-\-\, y = :r + 3, ?/ = — 5.r + 15. 
 
 17. 4a:-2y-3=-0, 3:r-y + i = 0, 5a; — 2y-l = 0. 
 
 18. 2a: — y==5, 3.r-y=-6, 4a: — ?/ = 7. 
 
 19. What is the value of ??^ if the lines 
 
 -+7=1, T + - = l, y = mx 
 ah ha ^ 
 
 meet in one point ? 
 
 20. When do the straight lines y = 7nx -\-h, y -= m'o: + h\ 
 y — m^^x + Z/" pass through one point ? 
 
 21. Prove that the three altitudes of a triangle meet in one 
 point. 
 
 22. Prove that the perpendiculars erected at the middle 
 points of the sides of a triangle meet in one point. 
 
 23. Prove that the three medians of a triangle meet in one 
 point. Show also that this point is one of three points of 
 trisection for each median. 
 
 24. Prove that the bisectors of the three angles of a triangle 
 meet in one point. 
 
72 ANALYTIC GEOMETRY. 
 
 25. The vertices of a triangle are (2, 1), (3,-2), (- 4, - 1). 
 Find the lengths of its altitudes. Is the origin within or 
 without the triangle ? 
 
 26. The equations of the sides of a triangle are 
 
 3a; + 3/ + 4 = 0, 3.T-5y + 34 = 0, 3.r — 2y + l=0. 
 Find the lengths of its altitudes. 
 
 What are the equations of the lines bisecting the angles 
 between the lines 
 
 27. 3:r-4y + 7 = and 4.T-3y + 17=0? 
 
 28. 3:r-43/-9 = and 12:r + 5y-3 = 0? 
 
 29. y = 2:r-4 and 2?/ = x-\-10? 
 
 30. x-i-2/ = 2 Siud x-i/ = 0? 
 
 31. y = 7nx 4- i> and y = vi'x -}-h' ? 
 
 32. Prove that the bisectors of the two supplementary angles 
 formed by two intersecting lines are perpendicular to each 
 other. 
 
 Equations eepeesenting Straight Lines. 
 
 64. A homogeneous equation of the nth degree represents n 
 straight lines through the origin. 
 
 Let the equation be 
 
 Ax^ + Bx^'-'y + ar"-y + + Kif -= 0. 
 
 Dividing by ^y", we have 
 
 M"+|AY-+^f-^r+ + f=o. 
 
 \yj A\yJ A\gJ A 
 
 If n, I't^ ^3 ^n denote the roots of this equation, then the 
 
 equation, resolved into its factors, becomes 
 
 (i-'.)(i-)e-) e-) 
 
 0. 
 
THE STRAIGHT LINE. id 
 
 and therefore is satisfied when each one of these factors is 
 zero, and in no other cases. 
 
 Therefore the locus of the equation consists of the n straight 
 lines A ri A 
 
 65. To find the angle between the two straight lines repre- 
 sented by the equation Ax^ + Hxt/ -^-Bif = 0. 
 
 Solving the equation as a quadratic in x, we obtain 
 
 Hence the slopes of the two lines are 
 
 1A , 2A 
 
 ni 
 
 H-^H'-^AB -H+^H'-^AB 
 
 Therefore 
 
 , -JH'~^AB , A 
 
 7n7n 
 
 B B' 
 
 and (equation [10], p. 50) 
 
 ^ _ m- m' _ V//^ — ^AB 
 \-\-mm^ A+B 
 
 66. To find the condition that the general equation of the 
 second degree may represent two straight lines. 
 
 AVe may write the most general form of the equation of the 
 second degree as follows : 
 
 Ax' ■\-Hxy-\-By''-\-Dx-^Ey + C= 0. (1) 
 
 In order that this equation may represent two straight lines, 
 it must be equivalent to the product of two linear factors ; 
 that is, equivalent to an equation of the form 
 
 {Ix + my + n) {px -^qy-\-r) = 0. (2) 
 
 Equating coefficients in (1) and (2), we obtain 
 
 lp = A, mq=B, nr~0, 
 
 Iq + mp =11, lr-{-np= I), mr + nq = E. 
 
74 ANALYTIC GEOMETRY. 
 
 The product of H, D, and E is 
 
 IIDE = 2 ImniDqr + lp{n^q^ + ??i^'r^) + mq(J}r'^ + ^^i?^) 
 
 = 2^^C +^(^^-2^(7)+i?(i)^-2^C) 
 + C(//'^-2^^). 
 Hence the required condition is 
 
 ^ABC- AE' - BE' - CH' + ^BE = 0. 
 
 Ex. 19. 
 
 1. Describe the position of the two straight lines repre- 
 sented by the equation Ax'"" + ILiy -\- Bif -\-Dj: -\- Ey ^ C^ 0, 
 where (i.) A= 11^ D = 0, (ii.) B = JI--^ E= 0. 
 
 2. When will the equation aiy -j- bx ^ cij -\- d —- repre- 
 sent two straight lines ? 
 
 3. Find the conditions that the straight lines represented 
 by the equation Ax"^ -\- E'xi/ -\- B?j^ ^= may be real; imagi- 
 nary ; coincident ; perpendicular to each other. 
 
 4. Show that the two straight lines x'^ — 2xi/ sec -\- i/' = 
 make the angle with each other. 
 
 Show that the following equations represent straight lines, 
 and find their separate equations : 
 
 5. x''~2x7/-d7/ + 2x-27/ + l=0. 
 
 6. x'-4:xi/ + 57/-~6i/-{-9 = 0. 
 
 7. x'-4:Xi/-{-Si/ + (di/-~9 = 0. 
 
 8. Show that the equation x"^ -j- x?/ ~ 6 y"^ -^7 x -^ 31 7/ — 18 — 
 represents two straight lines, and find the angle between them. 
 
 Determine the values of E for which the following equa- 
 tions will represent in each case a pair of straight lines. Are 
 the lines real or imaginary? 
 
 9. 12x'-l0xi/~{-2f-}-Ux-5i/-{-K=0. 
 
THE STRAIGHT LINE. 75 
 
 10. 12.rH-A:n/ + 2y^ + lla: — 5y + 2=a 
 
 11. l:2.r^ + 36.ry + /r/ + 6:r-h6y + 3 = 0. 
 
 12. For what value of K does the equation I{'xi/-\-bx 
 + 3 y + 2 = represent two straight lines ? 
 
 Problems on Loci involving Three Variables. 
 
 67. A trapezoid is formed by drawing a line iKirallel to the 
 base of a given triangle. Find the locus of the intersection of 
 its diagonals. 
 
 If ABC be the given triangle, and we choose for axes the 
 base AB and the altitude CO, the vertices A, B, C may be 
 represented in general by {a, 0), {b, 0), (0, <?), respectively. 
 The equations of ^Cand ^Care 
 
 - + - = 1 and y + - = l. 
 a c be 
 
 Let y = m be the equation of the line parallel to the base, 
 and let it cut AC in D, BC in E ; then the co-ordinates of 
 D and E, respectively, are 
 
 fac — am \ -, fbc — bin 
 I , m and , m 
 
 (1) 
 
 \ c J \ ^ 
 
 Hence the equation of the diagonal BB is 
 y cm 
 
 X — b ac — am — be 
 and the equation of the diagonal AE is 
 
 y ^ cm .^ 
 
 x — a be — bm — ac 
 If P be the intersection of the diagonals, then the co-ordi- 
 nates X and y of the point P must satisfy both (1) and (2) ; 
 by solving these equations, therefore, we obtain for any pa,r- 
 ticular value of m the co-ordinates of the point P. But what 
 we want is the algebraic relation which is satisfied by the 
 
76 ANALYTIC GEOMETRY. 
 
 co-ordinates of P, ivJuitcver the value of m may he. To find 
 this, we have only to eliminate m from equations (1) and (2). 
 By doing this we obtain 
 
 2cx+{a + b)y = {a-^h)c, 
 
 We see from the form of this equation that the required 
 locus is the line which joins C to the middle point 'of AB. 
 
 Remakk. The above solution should be studied till it is understood. 
 In problems on loci it is often necessary to obtain relations which in- 
 volve not only the x and 2/ of a point of the locus which we are seeking, 
 but also some third variable (as m in the above example). 
 
 In such cases we must obtain two equations which involve x and y 
 and this third variable, and then eliminate the third variable ; the 
 resulting equation wall be the equation of the locus required. 
 
 Ex. 20. 
 
 1. Through a fixed point any straight line is drawn, meet- 
 ing two given parallel straight lines in P and Q ; through P 
 and Q straight lines are drawn in fixed directions, meeting 
 in P. Prove that the locus of P is a straight line, and find 
 its equation. 
 
 2. The hypotenuse of a right triangle slides between the 
 axes of X and y, its ends always touching the axes. Find the 
 locus of the vertex of the right angle. 
 
 3. Given two fixed points, A and P, one on each of the 
 axes ; if U and V are two variable points, one on each axis, 
 so taken that 0U+ OV=OA + OB, find the locus of the 
 intersection of A V and P U. 
 
 4. Find the locus of the middle points of the rectangles 
 which may be inscribed in a given triangle. 
 
 5. If PP\ QQ' nre any two parallels to the sides of a given 
 rectangle, find the locus of the intersection of FQ and P'Q'. 
 
CHAPTER III. 
 
 THE CIRCLE. 
 Equations of the Circle. 
 
 68. The Circle is the locus of a point which moves so that 
 its distance from a fixed point is constant. The fixed point 
 is the centre, and the constant distance the radius, of the circle. 
 
 Note. The word " circle," as here defined, means the same thing as 
 "circumference " in Elementary Geometry. This is the usual meaning 
 of "circle" in the higher branches of Mathematics. 
 
 69. To find the equation of a circle, Jiaving given its centre 
 {ci, b) and its radius r. 
 
 Fig. 26. 
 
 Let C (Fig. 2G) be the centre, and P any point (.r, y) of 
 the circumference. Then it is only necessary to express by 
 an equation the fact that the distance from P to (7 is constant, 
 and equal to r : the required equation evidently is (§ G) 
 
 {jc — ay + (2/ — by = r\ [15] 
 
78 ANALYTIC GEOMETRY. 
 
 If we draw Ci? li to OX, to meet the ordinate of P, then 
 we see from the figure that the legs of the rt. A CPPi, are 
 CR = x-a, PB=--y-h. 
 
 If the origin be taken at the centre, then a = 5 = 0, and the 
 equation of the circle is 
 
 This is the simplest form of the equation of a circle, and 
 the one most commonly used. 
 
 If the origin be taken on the circumference at the point A, 
 and the diameter AB be taken as the axis of x, then the centre 
 will be the point (r, 0). Writing r in place of a, and in place 
 of h in [15], and reducing, we obtain 
 
 X- H- 2/2 = 2 rx. [17] 
 
 Why is this equation without any constant term ? 
 
 70. To find the condition that the general equation of the second 
 degree, Ax- + Hxy ~^Bif ^ Dx + Ey -{-0=0, (1) 
 
 shall represent a circle. 
 
 If possible, let it be the circle whose centre is the point 
 (a, h) and whose radius is r. The equation of this circle has 
 been found to be 
 
 (^x-ay^{y-hy = r\ 
 or x'^y''-2ax-2hy + a' + h''-r'' = 0. (2) 
 
 Equating corresponding coefficients and constant terms, we 
 have A^i^ B = \ 11=0, 
 
 I) = -2a, B=-2b, C = a' + h'-r\ 
 Since, if ^ = P in equation (1), both A and B can be 
 reduced to unity by division, the two conditions necessary in 
 order that equation (1) may represent a circle are 
 
 A = B and ^=0; 
 and the general equation of a circle may be written 
 x^ -\- y' + Dx + Ey + C=0. 
 
THE CIRCLE. 79 
 
 The co-ordinates of the centre and the radius have the 
 
 values 
 
 D y E , 
 
 Ex. 21. 
 Find the equation of the circle, taking as origin 
 
 1. The point B (Fig. 26). 
 
 2. The point D (Fig. 26). 
 
 3. The point E (Fig. 26). 
 
 Write the equations of the following circles : 
 
 4. Centre (5, — 3), radius 10. 
 
 5. Centre (0, -2), radius 11. 
 
 6. Centre (5, 0), radius 5. ■ 
 
 7. Centre (— 5, 0), radius 5. 
 
 8. Centre (2, 3), diameter 10. 
 
 9. Centre (/i, h), radius VA' -f h^. 
 
 10. Determine the centre and radius of the circle 
 
 .t' + y" -10:r + 12y + 25 = 0. 
 
 In this cas e D^-IO , ^=12, C= 25 (see § 70). Therefore a = 5, 
 5 = -6, r = V25 + 3G-25 = 6. 
 
 Determine the centres and radii of the followinsf circles : 
 
 o 
 
 11. a;^ + 2/--2:r-4y = 0. 17. 6:r- — 2y(7 - 3^/) = 0. 
 
 12. 3a;2+3/-5a:-72/+l=0. 18. x'-^if=^h\ 
 
 13. x'-\-y'-'^x = ^. 19. {xAryy-^{x-yy='^h\ 
 
 14. :r2 + y2 + 8:c = 0. 20. x^ -\- if -= a^ -\- h\ 
 
 15. a;= + /-8y = 0. 21. x^ ^y"" ^Icix^lc). 
 
 16. a;' + y2a-83/ = 0. 22. x" ^ y"" = hx ■\- hj , 
 
80 ANALYTIC GEOMETRY. 
 
 23. When are the circles x" + y"" +Dx-^ Ey -\-C=0 and 
 x" + y'^-^Ux +^'?/ + <^' = concentric ? 
 
 24. What is the geometric meaning of the equation {x — df 
 -i-(y-by = 0? 
 
 25. Find the intercepts of the circles 
 
 (i.) x''}-y'-8x~8y + 1=0, 
 
 (ii.) x'-\-y'-Sx — Sy + l6 = 0, 
 
 (in.) x'''^y'-8x~8y + 20 = 0. 
 
 Putting 3/ = in each case, we have in case (i.) a;^ — 8 a; + 7=0, 
 whence x = l and 7; in case (ii.) a-'^ — 8a; + 16 = 0, whence a; = 4; in 
 case (iii.) x^ — 8 a; + 20 = 0, whence a; = ± V— 4. 
 
 Putting a; = in each case, we obtain for y values indentical with the 
 above values of x. 
 
 The geometric meaning of these results is as follows : 
 
 Circle (i.) cuts the axis of a; in the points (1, 0), (7, 0), and the axis of 
 y in the points (0, 1), (0, 7). 
 
 Circle (ii.) touches the axis of x at (4, 0), and the axis of y at (0, 4). 
 
 Circle (lii.) does riot meet the axes at all. 
 
 This is the meaning of the imaginary values of x and y in case (iii.). 
 
 If, however, we wish to make the language of Geometry conform 
 exactly to that of Algebra, then in this case we should not say that 
 the circle does not meet the axes at all, but that it meets them in imagi- 
 nary points ; just as we do not say that the equation a;^ — 8 a; + 20 = 
 has no roots, but that it has two imaginary roots. 
 
 Find the centres, radii, and intercepts on the axes of the 
 
 following circles 
 
 '• 
 
 
 
 26. 
 
 x' + f- 
 
 5x- 
 
 -72/ + 6 = 0. 
 
 
 27. 
 
 x' + y'- 
 
 -12x 
 
 -4?/ + 15 = 
 
 0. 
 
 28. x' + 7/'-4:X- 8y = 0. 
 
 29. x'-{-y'-6x + 4:y-\-4: = 0. 
 
 30. x^-\-y'-]-22x — 18y + b7=0. 
 
THE CIRCLE. 81 
 
 31. Under what conditions will the circle x' -\' if -\- Dx 
 -}-^y-]-(7=0 (i.) touch the axis of x1 (ii.) touch the axis 
 of y ? (iii.) not meet the axes at all ? 
 
 32. Show that the circle o;^ + y^lOor-lOy + 25 = 
 touches the axes and lies entirely in the second quadrant. 
 Write the equation so that it shall represent the same circle 
 touching the axes and lying in the third quadrant. 
 
 33. In what points does the straight line 3a: + y = 25 cut 
 the circle a:^ + y" = 65 ? 
 
 34. Find the points common to the loci x^ ^y^ = 25 and 
 
 y = 2a: — 4. 
 
 35. The equation of a chord of the circle o;'^ + y^ = 4 is 
 y = 2a; + ll. Find its length. 
 
 X V 
 
 36. The equation of a chord is - + V =1 ; that, of the circle 
 is x"^ -{-y"^ = r^ Find the length of the chord. 
 
 37. Find the equation of a line passing through the centre 
 oi x"^ -^ y'^ — ^ X — ^y = — 21 and perpendicular to x-\-2y= 4. 
 
 38. Find the equation of that chord of the circle x'^-\-y~=loO 
 which passes through the point for which the abscissa is 9 
 and the ordinate negative, and which is II to the straight line 
 4a:-5y — 7=0. 
 
 39. What is the equation of the chord of the circle 
 •r''^ + ?/^ = 277 which passes through (3,-5) and is bisected 
 at this point ? 
 
 40. Find the locus of the centre of a circle j'^assing through 
 the points (oTi, yi) and {x^,y^. 
 
 41. What is the locus of the centres of all the circles which 
 pass through the points (5, 3) and (—7, — 6) ? 
 
82 ANALYTIC GEOMETRY. 
 
 Find tlie equation of a circle : 
 
 42. Passing through the points (4, 0), (0, 4), (6, 4). 
 
 43. Passing through the points (0, 0), (8, 0), (0, - G). 
 
 44. Passing through the points (- G, -1), (0, 0), (0, -1). 
 
 45. Passing through the points (0, 0), (—8a, 0), (0, 6 a). 
 4G. Passing through the points (2, - 3), (3, - 4), (- 2, - 1). 
 
 47. Passing through the points (1, 2), (1, 3), (2, 5). 
 
 48. Passing through (10, 4) and (17, — 3), and radius = 13. 
 
 49. Passing through (3, G), and touching the axes. 
 
 50. Touching each axis at the distance 4 from the origin. 
 
 51. Touching each axis at the distance a from the orimn. 
 
 52. Passing through the origin, and cutting the lengths 
 a, h from the axes. 
 
 53. Passing through (5, G), and having its centre at the 
 intersection of the lines y --=^lx — 3, 4?/ — ox = 13. 
 
 54. Passing through (10, 9) and (5, 2 — 3VG), and having 
 its centre in the line 3a:— 2?/ — 17=0. 
 
 55. Passing through the origin, and cutting equal lengths 
 a from the lines :r = ?/, a; + y = 0. 
 
 5G. Circumscribincr the triano;le whose sides are the lines 
 
 X II 
 
 V = 0, y=-inx-\-h, — |-t = 1- 
 ^ -^ a b 
 
 57. Having for diameter the line joining (0, 0) and (:ri,7/,). 
 
 58. Having for diameter the line joining {x^, y^) and {x.^, ?/■.). 
 
 59. Having for diameter the line joining the points where 
 
 2 _ 9 
 
 rx. 
 
 y = mx meets x^-^-y 
 
 GO. Having for diameter the common chord of the circles 
 x^ -\-y'^ = 9'2 ^^^ ^^ _ ^y _|_ ^2 ^ ^.2^ 
 
THE CIRCLE. 
 
 83 
 
 Tangents and Normals. 
 
 71. Let QPQ (Fig. 27) represent any curve. If the secant 
 QPB be turned about the point P until the point Q approaches 
 indefinitely near to P, then the ultimate position, TT', of the 
 secant is called the Tangent to the curve at P. 
 
 The tangent TT^ is said to touch the curve at P, and the 
 point P is called the Point of Contact. 
 
 The straight line PN drawn from P, perjoendicular to 
 the tangent TT\ is called the Normal to the curve at P. 
 
 Let the curve be referred to the axes OX, Y, and let M 
 be the foot of the ordinate of the point P. Let also the tan- 
 gent and the normal at P meet the axis of x in the points 
 P, JSf, respectively. Then MT\^ called the Sabtangent for the 
 2:)oint P, and JiiYis called the Subnormal. 
 
 72. To find the equation of a tangent to the circle x^-\-y'^--r'^, 
 at the point of contact {x^, 3/1). 
 
 Let P (Fig. 28) be the point ix^, y^), and Q any other point 
 {x2, y-i) of the circle. Then the equation of the line PQ is 
 
 y-yi_y2-yi 
 
 X— Xy 
 
 (1) 
 
84 ANALYTIC GEOMETRY. 
 
 If now we make this secant become a tangent, by turning it 
 about P till P coincides with P, then x^ = x^, y-i = 2/i, and the 
 
 fraction -- — — assumes the indeterminate form 7:- 
 x-i — Xi U 
 
 But we have not yet introduced the conditions that P and 
 
 Q lie in the circle. 
 
 These conditions are 
 
 x,'+?/,' = r' 
 
 Subtracting, (a^/ - x,^) + (y/ - y,^) = 0. 
 
 Factoring, (x^ — x^ (.r, + x^ -|- (y^ - y^ (y, + y^ = 0. 
 
 Whence, by transposition and division, we have 
 
 3/2 — yi ^ _ --^2 + ^1 
 X2 — Xy 3/2 + 2/1 
 
 And by substitution in (1) the equation of the secant 
 becomes y _ y^ ^ ^.^ _|_ ^.^ 
 
 x-^i^ y-i + yi 
 
 Now let Q coincide with P, or X2 = ^1, y^ = 3/1 ; the secant 
 becomes a tangent at P, and the equation becomes 
 
 y — yi ^ _ ^^^ 
 x — x^yi 
 
 or XiX + yiy = x\^ + 3/1'. 
 
 And, since a^i^ + 3/1^ = r^ we obtain 
 
 ix^i^ + yi2/ = :>'-, [18] 
 
 an equation easily remembered from its symmetry and because 
 it may be formed from o;^ + 3/^ = r^ by merely changing x^ to 
 XiX and y^ to 3/13/. 
 
 Note. The above method of obtaining the equation of the tangent 
 to a circle is appUcable to any curve whatever. It is sometimes called 
 the secant method. 
 
THE CIRCLE. 85 
 
 73. To find the equation of the normal through {xx, y^). 
 The slope of the taiment is ^• 
 
 Therefore that of the normal will be — (§ 48). 
 
 Xi 
 
 Hence the equation of the normal is (§ 50) 
 
 0-1 
 
 which reduces to the form 
 
 y^x — x^y = O. [19] 
 
 Therefore the normal passes through the centre. 
 \Ve may also obtain the same equation by proceeding as 
 in § 51. 
 
 74. To find the equations of the tangent and normal to the 
 circle (x — af + (y — b)- = ?-^ at the point of contact {x^, ?/i). 
 
 We proceed as in § 72, only now the equations of condition 
 which place (xx, y^ and {x-i, 3/2) 011 the circle are 
 
 {x,-ay^{y,~hy^r\ 
 {x^ - aj + (2/2 - ly = r\ 
 
 After subtracting and factoring, we have 
 
 {x.i-x,) (x,+o:i-2a) + (y2-yi) (y2+yi-25) = 0, 
 
 whence . Vi yi ^2 4" ^1 ^ <^ 
 
 x^ — Xx~ y., + yx — '2.b 
 
 Hence the equation of a secant through {x^, yi) and {x^, y^) is 
 
 y-yi ^ _ ^'2 + ^1 — 2 a 
 x-xi y-i + yi — 2'5 
 
 Making X2 = o:i, and 3/2 = yi, and reducing, we obtain 
 
 (X, -a)(x- a) + (2/1 - 6) (y - 6) = r\ [20] 
 
86 ANALYTIC GEOMETRY. 
 
 Equation [20] may be immediately formed from [18] by 
 affixing — a to the x factors and — 6 to the y factors, on the 
 left-hand side.' 
 
 By proceeding as in § 73, we obtain for the equation of 
 the normal 
 
 {y^-b){x-x,)-{a^,-a){y-y,) = 0. [21] 
 
 75. To ji ad the condition that the straight line y-=inix-\-c 
 shcdl touch the circle x^ + 2/"^ = ^'^• 
 
 I. If the line touch the circle, it is evident that the perpen- 
 dicular from the origin to the line must be equal the radius r 
 
 of the circle. The length of this perpendicular is 
 
 Vl + m'^ 
 (§ 55). Therefore the required condition is expressed by the 
 
 equation 
 
 c^ = r\l + m'). 
 
 II. By eliminating y from the equations 
 y — mx + c, x^ -\-y^ = f^, 
 we obtain the quadratic in x, 
 
 (1 -}- r)t'^)x' + 2 mcx = 7-"^ — c^^ 
 the two roots of which are 
 
 — _ '^^^'^ Vr'(l-f- '^>^'0 — <?' 
 
 ^^ 1 + ni' ~ Y^ni' 
 
 If these roots are real, the line will cid the circle ; if they 
 are equal, it will touch the circle ; if they are imaginary, it 
 will not meet the circle at all. 
 
 The roots will be equal if ^r\\-\-w}) — c^ = ; that is, if 
 c- = 9-\l + m"^), a result agreeing with that previously obtained. 
 
 If in the equation y = mx + c we- substitute for c the value. 
 rVl + 'm-'*, we obtain the equation to the tangent of a circle in 
 the useful form . /r—. :t rooi 
 
THE CIRCLE. 87 
 
 This equation, if we regard ??i as an arbitrary constant, 
 represents all possible tangents to the circle x^ + if = r^ 
 
 Note 1. Method II. is applicable to any curve, and agrees with the 
 definition of a tangent given in ^ 71. 
 
 Note 2. In problems on tangents the learner should consider whether 
 the co-ordinates of the point of contact are involved. If they are, he 
 should use equation [18] ; if they are not, then in general it is. better to 
 use equation [20]. 
 
 Ex. 22. 
 
 1. Explain the meaning of the double sign in equation [22], 
 
 2. Deduce the equations of the tangent and normal to the 
 circle x^A^if = r^^ assuming that the normal passes through 
 the centre. 
 
 3. Find the equations of the tangent and the normal pass- 
 ing through the point (4, 6) of the circle .r'"^ -|- y^ = 52. Also 
 the lengths of tangtnt, normal, subtangent, subnormal, and 
 the portion of the tangent contained between the axes. 
 
 4. A straight line touches the circle x^ -\- y- ^=^ r"^ in the 
 point (a-'i, ?/,). Find the lengths of the subtangent, the sub- 
 normal, and the portion of the line contained between the 
 axes. 
 
 5. What is the equation of a tangent to the circle 
 a;2 _j_ ^2 _ o^Q ^^ ^i^g point whose abscissa is 9 and ordinate 
 negative ? 
 
 6. Find the equations of tangents to .^■^ + 2/^—10 ^t the 
 points whose common abscissa =1. 
 
 7. Tangents are drawn through the points of the circle 
 a;'-[-y2 — 25 which have abscissas numerically equal to 3. Prove 
 that these tangents enclose a rhombus, and find its area. 
 
 8. The subtangent for a certain pjoint of a circle is 5|- ; the 
 subnormal is 3. What is the equation of the circle ? 
 
88 ANALYTIC GEOMETRY. 
 
 Find the equation of a straight line 
 
 9. Touching .r^-f-y^ = 232 at the point whose abscissa =14. 
 
 10. Touching {x - 2)' + (y — 3)' = 10 at the point (5, 4). 
 
 11. Touching a:^ + y^ — 3:r — 4?/ = at the origin. 
 
 12. Touching x'^-\-y'^ — \^x — 4?/— 5 = at the point whose 
 abscissa is equal to 10. 
 
 What is the equation of a straight line touching the circle 
 x^ -\-y- ^ r^ and also 
 
 13. Passing through the point of contact (r, 0) ? 
 
 14. Parallel to the line Ax + ^y + (7= ? 
 
 15. Perpendicular to the line Ax -{-By + C= ? 
 
 16. Making the angle 45° with the axis of :r? 
 
 17. Passing through the exterior point (A, 0) ? 
 
 18. Cutting off a triangle of area F' from the axes ? 
 
 19. Find the equations of the tangents drawn from the 
 point (10, 5) to the circle x"" + y' = 100. 
 
 20. Find the equations of tangents to the circle x"^ + if 
 -\-l0x — Q>y-2 = and II to the line y = 2x-1. 
 
 21. Find the lengths of subtangent and subnormal in the 
 circle a:^ + ?/^ — 14a; — 4?/ — 5 for the point (10, 9). 
 
 22. What is the equation of the circle (centre at origin) 
 which is touched by the straight line :?; cos a -f- y sin a ^=j(:> ? 
 What are the co-ordinates of the point of contact ? 
 
 23. When will the line Ax ^ By -{-0=0 touch the circle 
 x'^y'' = r' ? the circle {x ~ a)' + (y — hf = r' ? 
 
 24. Find the equation of a straight line touching x"^ -[- y^ 
 = ax -\-hy and passing through the origin. 
 
THE CIRCLE. 89 
 
 Prove that the following circles and straight line touch, 
 and find the jiuints of contact in each case : 
 
 25. X- + / + ax -\-hij = and ax-^hy-{- a^ + Z.^ ^ 0. 
 
 26. a-- + y' — 2a.T-2Z)?/ + Z>' = and x^2a. 
 
 27. x"^ -\- if = ax -\~ hy and ax — hy-{-h'^ = 0. 
 
 28. AVhat is the equation of the circle (centre at origin) 
 "^'hich touches the line y = 3a: — 5 ? 
 
 29. AVhat must he the value of m in order that the line 
 y = mx + 10 may touch the circle x"^ -{-y"^ ^ 100 ? Show that 
 we get the same answer for the line ?/ = inx — 10, and explain 
 the reason. 
 
 30. Determine the value of cin order that the line 3a7 — 4y 
 -f- c =: may touch the circle x^-^-y"^ — ^x-^-l'iy — 44=0. 
 Explain the double answer. 
 
 31. What is the equation of the circle having for centre 
 the point (5, 3) and touching the line 3.r -{- 2?/ — 10 = ? 
 
 32. What is the equation of a circle whose radius = 10, 
 and which touches the line 4a;-{-3y — 70 = in the point 
 (10,10)? 
 
 33. About the point (5, 9) a circle touching the line 
 4:r + 3?/ + 3 = in the point (-3,3) is described. What 
 is its equation ? 
 
 34. Under what condition will the line - + ^ = 1 touch 
 the circle x^ -f y^ _ ^.2 ^ 
 
 35. What is the equation of the circle inscribed in the 
 triangle whose sides are 
 
 . = 0. y = o, -^+1=1? 
 
90 ANALYTIC GEOMETRY. 
 
 3G. Two circles touch each other when the distance between 
 their centres is equal to the sum or the difference of their 
 radii. Prove that the circles 
 
 x-^ -\-f = (T + ay, {x - af i-i/ = r' 
 
 touch each other, and find the equation of the common tangent. 
 
 37. Two circles touch each other when the length of their 
 common chord = 0. Find the length of the common chord of 
 
 (x - ay + (y - by = r', (x - by + (y - ay - r' , 
 
 and hence prove that the two circles touch each other when 
 (a-by = 2r\ 
 
 Ex. 23. (Review.) 
 
 Find the radii and centres of the following circles: 
 
 1. Sx'-6x + 32/-i-9y-V2 = 0. 
 
 2. 7x' + S7/-4:7/-{l-2xy=^0. 
 
 3. y(y~b) = x(S-x). 
 
 4. ■\/l+a;'(x'' + 7/)^2b(x + a7/). 
 
 Findxhe equation of a circle : 
 
 6. Centre (0, 0), radius = 9. 
 
 6. Centre (7, 0), radius = 3. 
 
 7. Centre (— 2, 5), radius = 10. 
 
 8. Centre (3 a, 4a), radius = 5a. 
 
 9. Centre (b -^ c, b — c), radius — c. 
 
 10. Passing through (a, 0), (0, b), (2a, 2b). 
 
 11. Passing through (0, 0), (0, 12), (5, 0). 
 
 12. Passing through (10, 9), (4,-5), (0, 5). 
 
 13. Touching each axis at the distance —7 from the origin. 
 
THE CIRCLE. 91 
 
 14. Touching both axes, and radius = r. 
 
 15. Centre (a, a), and cutting chord = h from each axis. 
 IG. Passing through (0, 0), and touching 7/ = 2x-{-S. 
 
 17. Passing through (1, — 3), and touching 2x — i/— 4 = 0. 
 
 IS. With its centre in the line 5.r — 7y— 8 = 0, and 
 touching the Hnes 2a; — ?/ = 0, x — 2i/ — 6 = 0. 
 
 19. Passing through the origin and the points common to 
 the circles ^^2 ^ ^2 _ g^ _ j^Oy - 15 = 0, 
 
 x' + f + 2x-{- 4y + 20 = 0. 
 
 20. Having its centre in the line 5x — Si/ — 7 — 0, and 
 passing through the points common to the same circles as in 
 No. 19. 
 
 21. Touching the axis of x, and passing through the points 
 common to the circles 
 
 x-'^y~-j-4:x — Ui/ - 68 = 0, 
 ^.2 ^ y2 __ Q^. _. 22^ _^ 30 == 0. 
 
 22. Find the centre and the radius of the circle which 
 passes through (9, 6), (10, 5), (3, -2). 
 
 23. What is the distance from the centre of the circle 
 passing through (2, 0), (8, 0), (5, 9) to the straight line joining 
 (0,-11) and (-16,1)? 
 
 24. What is the distance from the centre of the circle 
 x' + 7/- 4:x 4- 8y = to the line 4.i- - 3// + 30 = ? 
 
 25. W^hat portion of the line ^ = 5x-i-2 is contained within 
 the circle x' -^ f- - ISx - 4:7/ — 9 = 0? 
 
 26. Through that point of the circle x^^if = 25 for which 
 the abscissa ^= 4 and the ordinate is negative, a straight line 
 parallel to y = Zx — ^ is drawn. Find the length of the inter- 
 cepted chord. 
 
92 ANALYTIC GEOMETRY. 
 
 27. Through the point (xi,i/i), within the circle x'^ -^ 7/^ = 7'^ , 
 a chord is drawn so as to be bisected at this point. What is 
 its equation ? 
 
 28. What relation must exist among the coefficients of the 
 equation Aix' + y') + Bx +1^1/+ C = 0. 
 
 (i.) in order that the circle may touch the axis oi x? 
 (ii.) in order that the circle may touch the axis of 3/ ? 
 (iii.) in order that the circle may touch both axes ? 
 
 29. Under what condition will the straight line 2/ = 7nx -f c 
 touch the circle x' -\- y =^2 rx ? 
 
 30. What must be the value of Iz in order that the line 
 3 a: + 4?/ = Z; may touch the circle y'^ = IC-r — rr' ? 
 
 31. Find the equation of the circle which passes through 
 the origin and cuts equal lengths a from the lines ^ = y, 
 a; 4- y = 0. 
 
 32. Find the ecjuations of the four circles whose common 
 radius -= V2ct, and which cut chords from each axis equal 
 to 2 a. 
 
 33. Find the equation of the circle whose diameter is the 
 common chord of the circles x^ -\-y^ =^ r^, {x — df -\- y"^ = r"^. 
 
 Find the equation of the straight line 
 
 34. Passing through (0, 0) and the centre of the circle 
 
 ^,2 _|_ y2 ^ a{x + y). 
 
 35. Passing through the centres of the circles 
 
 a;- -j- y- = 25 and .r^ + y^ + 6 a; — Sy = 0. 
 
 36. Passing through (0, 0) and touching the circle 
 
 ^•' + y^ - 6 5; - 12y 4- 41 = 0. 
 
 37. Parallel to .r-f V3(y-12) = and touching x''^y''=lOO. 
 
THE CIRCLE. 93 
 
 38. Passing through the points common to the circles 
 
 x'-\-y-- 2x- 47/- 20-0, 
 :c- + y'-14.x"-lG?/+ 100 = 0. 
 
 39. Prove that the common chord of the circles in No. 38 
 is perpendicular to the straight line joining their centres. 
 
 40. Find the area of the triangle formed by radii of the 
 circle .r^-f2/" = 169 drawn to the points whose abscissas are 
 — 12 and +7 and ordinates positive, and the chord passing 
 through the same two points. 
 
 41. Prove that an ancrle inscribed in a semicircle is a rit^ht 
 
 o o 
 
 angle. 
 
 42. Prove that the radius of a circle drawn perpendicular 
 to a chord bisects the chord. 
 
 43. Find the inclination to the axis oi x of the line joining 
 the centres of the circles x"^ + 2a; -f y^ = 0, x^ + 2y + 2/" = 0. 
 
 44. Determine the point from which tangents drawn to the 
 circles ^^^' + v/ - 2x- 6?/+ 6 = 0, 
 
 1^ -f y2 _ ooy _ 20 ^^. + 52 = 0, 
 
 will each be equal to 4V6. 
 
 45. Find the equations of the circles which touch the 
 straight lines 6a: + 7?/ + 9 = and 72- + G?/ + 3 = 0, and 
 the latter line in the point (3, —4). 
 
 Obtain and discuss the ecpiations of the following loci : 
 
 46. Locus of the centres of a circle having the radius r and 
 passing through the point (.^i, y^). 
 
 47. Locus of the centre of a circle having the radius r' and 
 touching the circle {x — a)^ -\- (y — hy = r^ 
 
 48. Locus of all points from which tangents drawn to the 
 circle {x — a)- + (y — by = r^ have a given length t. 
 
94 ANALYTIC GEOMETRY. 
 
 49. Locus of the middle point of a chord drawn through a 
 fixed point J. of a given circle. 
 
 50. Locus of the point 31 which divides the chord AC, 
 drawn through the fixed point ^ of a given circle, in a given 
 ratio AM: MC = m : n. 
 
 5L Locus of a point whose distances from two fixed points, 
 A, B, are in a constant ratio m : n. 
 
 52. Locus of a point, the sum of the squares of whose dis- 
 tances from two fixed points, A and B, is constant, and equal 
 to Tc. 
 
 53. Locus of a point, the difference of the squares of whose 
 distances from two fixed points, ^, ^, is constant and equal 
 to Ir. 
 
 54. Locus of the middle point of a line of constant length 
 d which moves so that its ends always touch two fixed per- 
 pendicular lines. 
 
 55. Locus of the vertex of a triangle whose base is fixed 
 and of constant length, and the angle at the vertex is also 
 constant. 
 
 56. One side, AB, of a triangle is constant in length and 
 fixed in position ; another side, AC, is constant in length but 
 revolves about the point A, Find the locus of the middle 
 point of the third side, BC. 
 
 57. Find the locus of the intersections of tangents at the 
 extremities of a chord whose length is constant. 
 
 58. Given the equation of a circle x^ + lf = r"^- If its radii 
 are produced, each by a length equal to the abscissa of the 
 point where it meets the circle, find the locus of the extremi- 
 ties of the radii produced. 
 
THE CIRCLE. 
 
 95 
 
 SUPPLEMENTARY PROPOSITIONS. 
 
 76. To find the locus of the middle points of a systenn of 
 parallel chords in the circle x' -f y'^ ^= r"^. 
 
 Fig. 29. 
 
 Let the equation of any one of the chords (Fig. 29) be 
 y^=mX'\-c, and let it meet the circle in the points (xi, y^ and 
 (^'2, 2/2). 
 
 Then (§§ 37 and 72) m = - ^^^^ (1) 
 
 2/1 + ^2 
 
 Let (.r, y) be the middle point of the chord ; then 2x=Xi-]-X2, 
 ^y^^yi~^y-i (§ S)) ^^^^ % substitution we have 
 
 (2) 
 
 X 
 
 y 
 
 a relation which evidently holds true for the middle points of 
 all the chords. Therefore (2) is the equation of the locus. 
 If we write (2) in the form 
 
 we see that the locus is a straight line passing through the 
 centre, and perpendicular to the chords (§ 48). 
 
 The locus of the middle points of a system of parallel 
 chords is called a Diameter of the circle; and the chords 
 which it bisects are called the Ordinates of the diameter. 
 
96 
 
 ANALYTIC GEOMETRY. 
 
 77. Two tangents can he drawn to a ch'cle from any point ; 
 and these tangents will he real, coincident, or imaginary , ac- 
 cording as the point is outside, on, or inside the circle, respec- 
 tively. 
 
 Fig. 30. 
 
 Let the equation of the circle be 
 x' + f = r\ 
 Let {xi, 3/1) be the point of contact of a tangent, (A, h) any 
 other point in the tangent. Then {h, k) must satisfy the 
 equation of the tangent; therefore 
 
 xji + yjc = r\ (1) 
 
 Also, since (xi, y^ is on the circle, 
 
 Xi + 3/1 
 
 Eliminating 3/1, we have 
 
 ^V + 
 
 li^Y^ ,.2^ 
 
 (2) 
 
 (3) 
 
 Since equation (3) is a quadratic equation, there are two 
 points the tangents at which pass through (A, k). Solving 
 (3), we obtain _ ^.^ ^,^,.^i^ ^^_^ 
 
 and we see that the values of Xi are o^eal, coincident, or imagi- 
 nary, according as A" -|- F is greater than, eciual to, or less 
 than r^ ; that is to say, according as (A, Ic) is outside, on, or 
 inside the circle. 
 
THE CIRCLE. 97 
 
 78. Tangents arc drawn to the circle x' + y^ ~ '^'^ from any 
 point (A, I:) ; to find the equation of the straight line joining 
 the two points of contact. 
 
 Let {i\, ijx), (x.,, 7/2) be the points of contact ; then the equa- 
 tions of the tangents are (§ 72) 
 
 Since both tangents pass through (A, Tc), both these equa- 
 tions are satisfied by tlie co-ordinates A, Iz ; therefore 
 
 hx,-{-hj,=^r\ (1) 
 
 hx,-^hj, = r\ (2) 
 
 From equations (1) and (2) we see that the two points 
 G"^'!, yi)» (^'2, y-i) both satisfy the equation 
 
 hx-\-lcy = r\ (3) 
 
 which, as its form shows, represents some straight line. There- 
 fore equation (3) is the equation of the straight line passing 
 through {xx,y^ and {x^.y-^; in other words, .the equation 
 required. 
 
 The line represented by equation (3) is a real line, whether 
 (A, Ic) be outside or inside the circle. 
 
 If the point (A, Ic) be outside the circle, this line is called 
 the Chord of Contact of the two real tangents drawn from (A, U). 
 
 If the point (A, Ic) be inside the circle, the points of contact 
 and the tangents are imaginary, and we have a real line 
 joining two imaginary points. 
 
 79. The straight line passing through the points of contact 
 of the tangents (real or imaginary) which can be drawn from 
 any point to a circle is called the Polar of that point with 
 respect to the circle ; and the point is called the Pole of that 
 straight line with respect to the circle. 
 
98 ANALYTIC GEOMETRY. 
 
 80. If the polar of a 'point P pass through Q, then the 
 polar of Q will pass through JP. 
 
 Let P be the point (h, Jc), Q the point (h', Jc'), and let the 
 equation of the circle be x' + y^ = '''^ 
 
 Then the equations of the polars of P and Q are 
 
 hx + hj = 7•^ (1) 
 
 h'x + k'7/ = r\ (2) 
 
 If Q be on the polar of P, its co-ordinates must satisfy 
 equation (1) ; therefore 
 
 hh'-\-kk' = r\ 
 
 But this is also the condition that P shall be on the line 
 represented by (2) ; that is, on the polar of Q. Therefore P 
 is on the polar of Q. 
 
 81. Jf a straight line revolve about a fixed point Q, and P 
 is the pole of that line, the locus of P is the pjolar of Q. 
 
 For, since Q is on the polar of P, the point P must always 
 be on the polar of Q (§ 80). 
 
 82. If a pjoint Q move along a fixed straight line, and P is 
 the pole of that line, then the polar of Q will revolve about P. 
 
 For, by hypothesis, the polar of P passes through Q (§ 80). 
 
 83. The polar of a point luith respect to a circle is perpen- 
 dicular to the line joining the point to the centre of the circle. 
 
 Let the equation of the circle be 
 
 and let P be any point (h, k). Then the equation of the 
 polar of Pis hx-{-hj = r\ (1) 
 
 And the equation of the line joining P to the centre of 
 the circle is l:x-hy = 0. (2) 
 
THE CIRCLE. 
 
 99 
 
 The form of equations (1) and (2) shows that the lines 
 which they represent are perpendicular (§ 51). 
 
 Figs. 31 and 32 illustrate the relations of poles and polars 
 which have been established in §§ 80-83. 
 
100 
 
 ANALYTIC GEOMETRY. 
 
 84. To find a geometrical construction for the polar of a 
 point with respect to a circle. 
 
 Fig. 33. 
 
 Fig. 34. 
 
 If the notation of § 83 be retained, and OQ (Figs. 33 and 34) 
 be the perpendicular from to the pokir of P, then (§ 55) 
 
 0Q = 
 
 Also 
 
 Therefore 
 
 V/r + li' 
 OFX OQ = r\ 
 
 Hence we have the following construction : 
 Join OF, and let it cut the circle in A ; take Q in the line 
 OF, such that Qp , qj, ^ qA : OQ, 
 
 and draw through Q a line perpendicular to OF. 
 
 85. To find the length of the tangent drawn from any point 
 (h, k) to the circle (^, _ ^y -f (y — hf - r' = 0. (1) 
 
 Let F (Fig. 35) be the point {h, h), Q the point of contact, 
 Cthe centre of the circle ; then, since FQCis a right angle, 
 
 FQ' = FC'-QC\ 
 
 Now FC':=(h-ay--}-(I:-by\ QC' = r\ 
 
 Therefore 
 
 FQ' = (h - ay -i- (Ic - bf -r 
 
THE CIRCLE. 
 
 101 
 
 Hence PQ- is found, by simply substituting tlie co-ordinates 
 of F in the left-hand member of equation (1). 
 
 If for brevity we write S instead of (x — of -f (y — hy — o'"^, 
 then the equation 8— will represent the general equation 
 of the circle after division by the common coefficient of :r 
 and y^, and we may state the above result as follows : 
 
 If S^=0 be the equation of a circle, and the co-ordinates of 
 any point he substituted for x ayid y in S, the result luill he 
 equal to the square of the length of the tangent drawn from the 
 2:)oint to the circle. 
 
 If the point is inside the circle, the square is negative, and 
 the length of the tangent imaginary. 
 
 86. Two circles intersect each other ; to find the equation of 
 the straight line passing through the points of intersection. 
 
 X 
 
 Fi2. 35. 
 
 Fig. 36. 
 
 Let the equations of the circles be 
 
 (x-a)' + (y-J)'-r^=0, (1) 
 
 , {x-a'f + (,j-h'y-r" = 0. (2) 
 
 Subtract one of these equations from the other; then 
 2 (a- a')x + 2 (^- - hyj = cr - a" +b'- h" - (r' - r") = 0. 
 
 This is the equation required ; for it is the equation of 
 some straight line, and its locus passes through the intersec- 
 tions of the loci represented by (1) and (2) (§ 59). 
 
102 ANALYTIC GEOMETRY. 
 
 If we write Si and /S2 for the left-hand members of equations 
 (1) and (2) respectively, the result may be thus stated : 
 
 Jf /S^ =z 0, /S.2 = he the equations of two circles, then will 
 the equation S^ — /S^ = 0, or Si = /S^ he the equation of the 
 straigJd line through their points of intersection. 
 
 Although the two circles Si == 0, /S2 = may not cut each 
 other in real points, the straight line Si — S-^ will always be 
 real, provided the constants in it are real ; so that we have 
 a real straight line passing through imaginary points. 
 
 But another meaning may be given to the equation Si — S.^. 
 
 For if (x, y) denote any point in the line Si — /Sj, then xSi 
 is equal to the square of the tangent from {x, y) to the circle 
 Si = 0, and S-i is equal to the square of the tangent from 
 {x, ?/) to the circle S. = (§ 85). 
 
 Hence the tangents drawn to the two circles from any 
 point in the straight line Si = S^ are equal. 
 
 The straight line X — S., is called the Eadical Axis of the 
 
 O J. - 
 
 two circles /Si — 0, xS'a = 0. 
 
 It may be defined either as the straight line passing tJtrough 
 the ^ooints of intersection (real or imo,ginary) of the two circles, 
 or as the locus of the points from which tangents drawn to the 
 two circles are equal. 
 
 87. The three radical axes of three circles, tahen in pairs, 
 m.ect in a point. 
 
 Let /6'=0, Si^=0, /Sj = be the equations of the circles, 
 when the coefficient of x^ in each is unity. 
 
 Then the equations of their radical axes, taken in pairs, are 
 
 S-Si = 0, Si-S, = 0, S-S,=0. 
 
 The values of x and y that will satisfy any two of these 
 equations will also satisfy the third. Therefore the third 
 axis passes through the point of intersection of the other 
 two axes. The point of intersection of the three radical axes 
 is called the Eadical Centre of the three circles. 
 
THE CIRCLE. 103 
 
 Ex. 24. 
 
 1. What is tlie equation of the diameter of the circle 
 x'^-}-i/ = 20 which bisects chords parallel to the line 6:r + 7?/ 
 + 8 = 0? 
 
 2. What is the equation of the diameter of the circle which 
 bisects all chords whose inclination to the axis of x is 135° ? 
 
 3. Prove that the tangents at the extremities of a diameter 
 are parallel. 
 
 4. Write the equations of the chords of contact in the circle 
 ^^ + y" — ''^ foi-" tangents drawn from the following points : 
 (r,r), (2r, 3r), (a + b,a-b). 
 
 6. From the point (13, 2) tangents are drawn to the circle 
 ^' + y^ = 49 ; what is the equations of the chord of contact ? 
 
 6. What line is represented by the equation hx -{- J:y = r^ 
 w^hen (/i, Ic) is in the circle ? 
 
 7. Write the equations of the polars of the following j^oints 
 with respect to the circle o;^ + ?/^ = 4 : 
 
 (i.)(2,3). (ii.)(3,-l). (iii.) (1,-1). 
 
 8. Find the poles of the following lines with respect to the 
 circle x^ + ?/^ = 35 : 
 
 (i.) 43; + 6y=--7. (ii.) 3a;-2?/=5. (iii.) ax^ly^X. 
 
 9. Find the pole of 3.'r-f 4?/ = 7 with respect to the circle 
 X'^y'^ 14. 
 
 10. Find the pole of Ax^By^C^=^ with respect to the 
 circle x- -\- y^ = r\ 
 
 11. Find the co-ordinates of the points w^here the line x==4: 
 cuts the circle X' + v/- = 4 ; also find the equations of the tan- 
 gents at those points, and show that they intersect in the 
 point (1, 0). 
 
104 ANALYTIC GEOMETRY. 
 
 12. If the jiolars of two points P, Q meet in R, then B is 
 the pole of the line PQ. 
 
 13. If the polar of (A, h) with respect to the circle x'^-{-y''=r^ 
 touch the circle x^ -\- y^ — 'Zo-x, then k' — 2 rh = r\ 
 
 14. If the pole lie in the circle a;'-f3/^ = 4c^, then the 
 polar will touch the circle 4(:t^^ + y") = <?^- 
 
 15. Find the polar of the centre of the circle rr- + y^ = r^ 
 Trace the changes in the position of the polar as the pole is 
 supposed to move from the centre to an infinite distance. 
 
 16. AVhat is the square of the tangent dra.wn from the 
 point {h, I:) to the circle x' + y' = r' ? 
 
 17. Find the length of the tangent drawn from (2, 5) to 
 the circle x^ -{- y"^ — 2x — oy — 1= 0. 
 
 Find the radical axis of the circles 
 
 18. (x + 5y + (y + 6y = 9, (.^^_77 + (y-H)^ = 16. 
 
 19. a;^ + ?/2 + 2.T + 3?/-7-=0, x' + y- - 2x -y + 1=0. 
 
 20. X- + y"^ + bx -\- by ~ c = 0, ax"^ + «2/" + ct^^ + ^V = ^^ 
 
 21. Find the radical axis and length of the common chord 
 of the circles 
 
 .T" + y^ + ax + 5?/ + (? = 0, x"^ + yi^ + bx -\-ay-\-c = 0. 
 
 22. Find the radical centre of the three circles 
 
 .T^ + ?/2 + 4a: + 7=0, 
 
 2x^ + 27/2 + 3:i- + 5y + 9 = 0, 
 
 x'-\-f-\-y = 0. 
 
 23. Prove that the radical axis of two circles is perpen- 
 dicular to the straight line joining their centres. 
 
 24. Find a geometric construction for the radical axis of 
 two circles which do not meet each other (see § 87 and Ex. 23). 
 
CHAPTER IV. 
 
 DIFFERENT SYSTEMS OF CO-ORDINATES. 
 Oblique Co-ordinates. 
 
 88. When we define the position of a point or a line by 
 reference to some system of points or lines regarded as fixed 
 in position, we are said to employ a System of Oo-ordinates. 
 
 The system of co-ordinates which we have thus far em- 
 ployed is called the Eectangular System, because the two fixed 
 lines of reference are perj^endicular to each other. It is the 
 system to be preferred for most purposes on account of its 
 simplicity. 
 
 There are, however, two other systems in use, of such im- 
 portance that we shall briefly describe and illustrate them. 
 
 The first of these 'svstems differs from the rectangular 
 system simply in the fact that the axes of reference are not 
 perpendicular to each other. 
 
 Let OX, OF (Fig. 37) be two axes making an acute angle, 
 XO F= (u, with each other. The position of the point P is 
 
106 ANALYTIC GEOMETRY. 
 
 determined by stating its distance from each axis, measured 
 along a line parallel to the other axis. 
 
 If we draw FN II to OX, and PM II to F, then the co- 
 ordinates of F are 
 
 NF=OM=x, MF^y, 
 
 This system of co-ordinates is known as the Obhque System, 
 Eectangular and oblique co-ordinates are called Parallel 
 
 Co-ordinates ; also Cartesian Co-ordinates (from Descartes, who 
 
 first used them). 
 
 89. To find the equation of the straight line AC, referred to 
 the oblique axes OX, Y (Fig. 38), having given the intercept 
 OB = b and the angle XAC^= y. 
 
 Let F be any point {x, y) of the line. Draw BL \\ to OX, 
 meeting FM'ui D. Then, by Trigonometry, 
 
 FD sin y y — b sin y 
 . = _ L — , or ^ ■ = f- 
 
 BD sin (w — y) X sin (w — y) 
 
 If now we put m = - — - — "^ — -, we obtain as the result an 
 sin (o) — y) 
 
 equation of the same form as [6], p. 38, 
 
 y = nix -f- b. 
 
 What does the value of m become in this equation when 
 
 o> = 90° ? 
 
DIFFERENT SYSTEMS OF CO-ORDINATES. 
 
 107 
 
 90. Oblique co-ordinates are seldom used, because they 
 generally lead to more complex formulas than rectangular 
 ones. In certain cases, however, they may be employed to 
 advantage. An example of this kind is furnished by problem 
 No. 23, p. 71 : 
 
 To prove that the medians of a triangle miect in one point. 
 
 If a, h, c represent the three sides of the triangle, and we 
 take as axes the sides a and b, then the equations of the sides 
 and also of the medians may be written down with great ease, 
 as follows : 
 
 The sides, 
 The medians, 
 
 a b 
 
 y = o, 
 
 1=0. 
 
 0, 
 
 X 2y 
 a'^ b 
 
 ^ y 
 a h 
 
 1. 
 
 X y 
 
 1 = U, -~f 
 
 a 
 
 0. 
 
 On comparing the equations of the medians, we see that if 
 we subtract the second equation from the first, we obtain the 
 third ; therefore the three medians must pass through the 
 same point (§ 59). 
 
 ..-'-"^ 
 
 Fig. 39. 
 
 Polar Co-ordinates. 
 
 91. There is another system of co-ordinates, called the Polar 
 System, which is often useful. 
 
 Let (Fig. 39) be a fixed point, AOA' a fixed straight 
 line, P any point. Join OP. 
 
108 
 
 ANALYTIC GEOMETRY. 
 
 It is evident that we know the position of P, provided 
 w^e know the distance OP and the angle which OP forms 
 with OA. 
 
 Thus, if we denote the distance OP by p, and the angle 
 POA by B, the position of P is determined if p and are 
 known. 
 
 p and are called the Polar Oo-ordinates of P ; is called 
 the Pole i OA, the Polar Axis ; OP, the Kadins Vector of P. 
 
 Fig. 40. 
 
 Every point in a plane is perfectly determined by a posi- 
 tive value of p between and co, and a positive value of 
 between 0° and 360° (or and 27r, circular measure). But 
 in order to be able to represent by a single equation all the 
 points of a geometric locus, it is necessary to admit negative 
 values of p and B, and to adopt conventions suitable for this 
 purpose. 
 
 It is agreed that B shall be considered positive when it is 
 measured from the initial line, in the opposite direction to that 
 of the motion of the hands of a watch ; and negative when 
 measured in the same direction as this motion. 
 
 It is also agreed that p, or OP, shall be considered positive 
 when it forms one side of the angle 6, and negative when it 
 does not. 
 
DIFFERENT SYSTEMS OF CO-ORDINATES. 
 
 109 
 
 For example, suppose that the straight line POPi bisects 
 the first and third quadrants, and that in this line we take 
 points P, Pi, at the same distance OP — p from ; then 
 
 P is the point (p, ^tt) or (— p, ^tt) or (— p, — Itt), 
 Pi is the point (p, {tt) or (— p, ^tt) or f p, — f tt), etc. 
 
 Fig. 41. 
 
 Fig. 42. 
 
 92. To find the polo/r equation of the circle. 
 
 (i.) Let the pole be at the centre. Then, if r denote the 
 radius, the polar equation is simply p^=r. 
 
 (ii.) Let the pole be on the circumference (Fig. 41), and 
 let the diameter OB make an angle a with the initial line OA. 
 Let P be any point (r, 0) of the circle. Join BP. 
 
 Then OP = OB cos BOP, 
 
 or P = 2r cos (6 — a). [23] 
 
 If OB is taken as the initial line, the equation becomes 
 
 P = 2 r cos e. [24] 
 
 (iii.) Let the po]e be anywhere, and the centre the point 
 (/:, a). Then in the triangle OPC' (Fig. 42) 
 
 OP' - 2 OP X OCX cos POC+ OC^ - CP' = 0, 
 
 or p- — 2 p7c cos (0 — a) + k^ — r'^ = O, [25] 
 
 the "most general form of the polar equation of a circle. 
 
110 ANALYTIC GEOMETRY. 
 
 Ex. 25. 
 
 1. Find the distances from the point P in Fig. 38 to the 
 two axes. 
 
 2. Prove that the equation of a straight line, referred to 
 oblique axes in terms of its intercepts, is identical in form 
 with [7], p. 39. 
 
 3. If the straight line F.,OP^ (Fig. 39) bisects the second 
 and fourth quadrants, what are the polar co-ordinates of the 
 points Pa and Pi ? Give more than one set of values in each 
 case. 
 
 4. Construct the following points (on paper, take « = 1 in.) : 
 
 ,.0), („.^\ f„,-|\ (-a,t. Ua,-l 
 
 TT 7r\ / OTT 
 
 a, - , ::«, TT, acos-, - , a, — . oa. 
 
 o 
 
 (iy V / V 3 3y V 2 
 
 3a, ^""Y Aa, tan-^^\ Ua.i^xr^^ 
 
 4 . . 
 Note. The expression tan~^ - in higher Mathematics means "the 
 
 angle whose tangent is -•" 
 
 5. If pi, /32 denote the two values of p in equation [25], 
 p. 109, prove that pi and p^ — ^'^ — ^'^- What theorem of Ele- 
 mentary Geometry is expressed by this equation (i.) when 
 the pole is outside the circle? (ii.) when the pole is inside 
 the circle? 
 
 6. Through a fixed point Pin a circle a chord ^P is drawn, 
 and then revolved about P; find the locus of its middle point. 
 
 Note. In such problems as this there is a great advantage in using 
 polar equations. 
 
 7. If J9 denote the distance from the pole to a straight line, 
 a the angle between 'p and the polar axis, prove that the polar 
 equation of the line is p cos {B — o.)^= p. 
 
DIFFEEENT SYSTEMS OF CO-ORDINATES. 
 
 Ill 
 
 Transformation of Co-ordinates. 
 
 93. The equation of the same curve varies greatly in form 
 and simplicity, according to the system of co-ordinates adopted, 
 and the position of the fixed points and lines with respect to 
 the curve. Hence it is sometimes useful to be able to deduce 
 from the equation of a curve referred to one system of co- 
 ordinates its equation referred to another system. This pro- 
 cess is known as the Transformation of Oo-ordinates. 
 
 It consists in expressing the old co-ordinates in terms of 
 the new, and then replacing in the equation of the curve the 
 old co-ordinates by their values in terms of the new ; we thus 
 obtain a constant relation between the new co-ordinates, which 
 will represent the curve referred to the new axes. 
 
 94, To change the origin to the jjoint {h, Jc) without changing 
 the directio7i of the axes. 
 
 
 Y 
 
 
 v 
 
 f 
 
 
 
 > 
 
 
 
 \m' 
 
 
 
 a 
 
 i 
 
 
 
 
 ^ 
 
 i 
 
 M 
 
 X 
 
 Fig. 43. 
 
 Let OX, F be the old axes, O'X', 0' F' the new ; and 
 let {x, y), (.r', y') be the co-ordinates of the same point P, 
 referred to the old and new systems respectively. 
 
 Then (Fig. 43) 
 
 OA = h, AO'=k, OM-^x, MP = y, a ]\V ■= x\ 3PP = y'. 
 x= OA +AM^OA +0'jr = x'^h. 
 y = M2r -f M'P = ^0' + J/'P = y' + ^. 
 
112 
 
 ANALYTIC GEOMETRY. 
 
 These relations are equally true for rectangular and oblique 
 co-ordinates. 
 
 Hence, to find what the equation of a curve becomes when 
 the origin is transferred to a point (A, k), the new axes run- 
 ning parallel to the old, we must substitute for x and y the 
 values given above. 
 
 After the substitution, we may, of course, write x and y 
 instead of x' and y' ; so that practically the change is effected 
 by simply writing x-\-h in place of x, y-}-k in place of y. 
 
 If, however, we wish to transform a poi7it (x, y) from the 
 old to the new system, we must write x — h in place of x and 
 y~k in place of y. 
 
 95. To change the equation of a curve from one rectangular 
 system to another, the origin remaining the same. 
 
 Let {x, y) be a point P referred to the old axes OX, Y\ 
 (x', y'), the same point referred to the new axes OX', Y' 
 (Fig. 44). Then 
 
 OM=x, 3IF = y, ON=x\ NP = y\ 
 
 Let the angle XOX'=e. Draw XQ, XE ± to PM, OX, 
 respectively ; then 
 
 iVTQ = Q^^O =■■ XOR = e. 
 
DIFFERENT SYSTEMS OF CO-ORDINATES. 113 
 
 Hence OM=OFc-IiM=OE — JVQ=OJVco&e~FJVsinO. 
 
 Or X = x' cos — y' sin 6. 
 
 And PJ/- J/Q+ QP- i?A^+ QF= OjV sin + Pi\^cos 6. 
 
 Or ?/ — :r' sin 6 + y' cos ^. 
 
 Therefore, to find what the equation of a curve becomes 
 when referred to the new axes, we must write 
 
 X cos — y ^mO for x, x &m6 -{-y cos 6 for y. 
 
 96. To transform an equation frovi one rectangular system 
 to another, both the origiii and the direction of the axes being 
 changed. 
 
 First transform the equation to axes through the new origin, 
 parallel to the old axes. Then turn these axes through the 
 required angle. 
 
 If (h, k) is the new origin referred to the old axes, the 
 angle between the old and new axes of x, we obtain as the 
 values of x and y for any point P, in terms of the new co- 
 ordinates, 7 1 f a 1 ■ n 
 ' x=^ h + X cos d — y sin 6, 
 
 y -—k-{- x' sin -\- y' cos 0. 
 
 In making all these transformations, attention must be paid 
 to the sigris of h, k, and 0. 
 
 97. To transform an equation from rectangular to oblique 
 axes, the origin remaining the same. 
 
 Let a, (i be the angles formed by the positive directions of 
 the new axes 0X\ Y' (Fig. 45) with the positive direction of 
 OX. Let the old co-ordinates of a point P be x, y ; and tlie 
 new co-ordinates, x', y'. Then from the right triangles OFX, 
 FQJVwe readily obtain the formulas 
 
 x = x' cos a + ?/' cos /?, 
 y == :r' sin a -f ?/' sin /?. 
 
 Investigate the special case when yg = a + 90°. 
 
114 
 
 ANALYTIC GEOMETRY. 
 
 98. To change an equation from 'polar to rectangular co- 
 ordinates. 
 
 R X 
 
 M X 
 
 Fig. 46. 
 
 Let the co-ordinates of a point P be x, 7/ referred to the 
 rectangular system, and p, referred to the polar system. 
 
 (i.) Let the origin of rectangular co-ordinates be the pole, 
 and let the polar axis coincide with the axis of x. 
 Then (Fig. 46) 03f-^ OF cos FOJl 
 FM= OF sin FOIL 
 Or X = p cos 6, 
 
 y = p sin 6. 
 
 (ii.) If the pole is the point (h, k), we have 
 
 X = h-^ p cos 0, 
 y = k-i- p sin 0. 
 
 (iii.) If the pole coincides with the origin, but the polar 
 axis OA makes the angle a with the axes of x, we obtain 
 
 :r = p cos (^ + a), 
 y = psin (^ + a). 
 
 (iv.) If the pole is the point (h, Jc), and the polar axis 
 makes the angle a with the axis of x, 
 
 x = h-{- p cos {6 -\- a), 
 y = k-\-psm{0-{- a). 
 
DIFFERENT SYSTEMS OF CO-ORDINATES. 115 
 
 99. To change an equation from polar to rectangular co- 
 ordinates. 
 
 From the results in cases (i.) and (ii.) of § 98 (the only- 
 cases of importance), we readily obtain 
 
 In case (i.), i^ = x^ + y^ tan 6 ^ ^ -• 
 
 In case (ii.), r' = {x - hj + (y — lc)\ tan ^ -- ^• 
 
 100. The degree of an equation is not altered hy any alteration 
 of tJie axes. 
 
 For, however the axes may be changed, the new equation 
 
 is always obtained by substituting for x and y expressions of 
 
 the form i 7 i i r i zf i r 
 
 ax -\- by -{- c and a'x -\- b' y -\- c\ 
 
 These expressions are of the first degree, and therefore, if 
 they replace x and ?/ in the equation, the degree of the equa- 
 tion cannot be raised. Neither can it be lowered ; for if it 
 could be lowered, it might be raised by returning to the 
 original axes, and therefore to the original equation. 
 
 Ex. 26. 
 
 1. What does the equation y^ — 4:r-f43/ + 8 = become 
 when the origin is changed to the point (1, — 2) ? 
 
 Transform the equation of the circle (x — ay + (y — ^Y = t^^ 
 by changing the origin 
 
 2. To the centre of the circle. 
 
 3. To the left-hand end of the horizontal diameter. 
 
 4. To the upper end of the vertical diameter. 
 
 5. What does the equation x^ ■\-y'^^= r"^ become if the axes 
 are turned through the angle a ? 
 
 6. What does the equation x^ — y"^ = o^ become if the axes 
 are turned through — 45° ? 
 
116 ANALYTIC GEOMETRY. 
 
 7. The equation of a curve referred to rectangular axes is 
 X — xy — y =-0. Transform it to a new system, whose origin 
 is the point (—1, 1), and whose axes bisect the angles formed 
 by the old axes. 
 
 8. Change the following equations to polar co-ordinates, 
 taking the pole at the origin and the polar axis to coincide 
 with the axis of x : 
 
 (i.) x"^ + 2/^ = cc'^' (ii-) -'^^ — y"^ =, ci^' 
 
 9. Change the equation x'^=^^ax to polar co-ordinates, 
 (i.) taking the pole at the origin ; (ii.) taking the pole at 
 the point (a, 0). 
 
 10. Change the following equations to rectangular co-ordi- 
 nates, the origin coinciding with the pole, and the polar axis 
 
 . with the axis of x : 
 
 (i.) p = a, (ii.) p = acosO, (iii.) p"^ cos20 = a^. 
 
 Transform the following equations by changing the origin 
 to the point given as a new origin : 
 
 11. .r -f ?/ + 2 = ; the new origin (— 2, 0). 
 
 12. 2.i^-5y-10 = 0; the new origin (5, - 2). 
 
 13. Sx' + 4:xy-\-y'—5x-6y-3 = 0; new origin (|, —4). 
 
 14. x'^ + y^ — 2x — 4:y = 20; new origin (1, 2). 
 
 15. x'^ — 6xy -{- y^ — 6x -j-2y -{-1= ; new origin (0, —1). 
 
 16. Transform the equation o:^ — y^-f- 6 = by turning the 
 axes through 45°. 
 
 17. Transform the equation (x -\- y — 2ay = 4:xy by turn- 
 ing the axes through 45°. 
 
 18. Transform the equation 9 o:^ — 163/^ =144 to oblique 
 axes, such that the new axis of x makes with the old axis 
 of X a negative angle whose tangent = — | ; and the new axis 
 of y makes with the old axis of x a positive angle whose tan- 
 gent is |. 
 
DIFFERENT SYSTEMS OF CO-ORDINATES. 117 
 
 Ex. 27. (Revie^w.) 
 
 1. Find the distance from the point (—2b, h) to the origin, 
 the axes making the angle 60°. 
 
 2. The axes making the angle m, find the distance from the 
 point (1, —1) to the point (—1, 1). 
 
 3. The axes making the angle w, find the distance from the 
 point (0, 2) to the point (3, 0). 
 
 Determine the distance between the following points referred 
 to polar co-ordinates : 
 
 4. (a, 6) and (5, 0). 
 
 5. (a, 0) and {a, - 6). 
 
 6. (a, 6) and (- a, - 6). 
 
 7. (2 a, 30°) and (a, 60°). 
 
 8. Show that the polar co-ordinates (p, ^), {—p,7r-\-6), 
 (— p, O — tt) all represent the same point. 
 
 9. Transform the equation 8a;' + 8.ry+4/-f 122'-f 8y+l = 
 to the new origin {—^, — i). 
 
 10. Transform the equation 62:"-f-3?/^ — 24a:-f 6=^0 to the 
 new origin (2, 0). 
 
 X 11 
 
 11. Transform the equation - + j = l by changing the 
 origin to the point (;^'^) and turning the axes through an 
 angle <^, such that tan <^ = 
 
 12. Transform the equation 17:r' -16:r?/ -f 17?/' = 225 to 
 axes which bisect the axes of the old system. 
 
 Transform the following rectangular equations to polar 
 equations, the polar axis in each case coinciding with the 
 axis of X, and the pole being at the point whose co-ordinates 
 are given : 
 
118 ANALYTIC GEOMETRY. 
 
 13. ar-\'y' = Qax) the pole (0, 0). 
 
 14. x'-\-y'' = Qax] the pole (4«, 0). 
 
 15. y- — Gy - 5.T + 9 = ; the pole (f , 3). 
 
 16. X- - ?/- - 4.1' — 6y - 54 = ; the pole (2, - 3). 
 
 17. (.1-- + y'f = k\x' - f) ; the pole (0, 0). 
 
 Transform the following polar equations to rectangular axes, 
 the origin being at the pole and the axis of x coinciding with 
 the polar axis : 
 
 IS. p''s'm2er=2a\ 
 
 19. p = ksm20. 
 
 20. p(sin36' + cos3^) = 5^-sln^cos^. 
 
 21. Through what angle must the axes of a rectangular 
 system be turned in order that the new axis of x may pass 
 through the point (5, 7) ? 
 
 22. The equation of a straight line in rectangular axes is 
 Ax-\-JBy-{-C=0. Through what angle must the axes be 
 turned in order 
 
 (i.) that the term containing x may disappear ? 
 (ii.) that the term containing y may disappear ? 
 
 23. Deduce the following formulas for changing from one 
 oblique system to another, the origin remaining the same : 
 
 _ x^ sin (oi> — g) , y' sin (oj — /S) 
 sin w sin w 
 
 x' sin a , ?/' sin B 
 sin w sm (tj 
 
 Note. In these formulas w denotes the angle formed by the old axes, 
 a and )8 those formed by the positive directions of the new axes with the 
 positive direction of the old axis of x. 
 
 24. From the formulas of No. 23 deduce those of § 97. 
 
CHAPTER y. 
 
 THE PARABOLA. 
 
 The Equation of the Parabola. 
 
 101. A Parabola is the locus of a point whose distance from 
 a fixed point is always equal to its distance from a fixed 
 straight line. 
 
 The fixed point is called the Focus ; the fixed straight line, 
 the Directrix. 
 
 The straight line which passes through the focus, and is per- 
 pendicular to the directrix, is called the Axis of the parabola. 
 
 The intersection of the axis and the directrix is called the 
 Foot of the axis. 
 
 The point in the axis half way betw^een the focus and the 
 directrix is, from the definition, a point of the curve ; this 
 point is called the Vertex of the parabola. 
 
 The straight line joining any point of the curve to the focus 
 is called the Focal Kadius of the point. 
 
 A straight line passing through the focus and limited by 
 the curve is called a Focal Chord. 
 
 The focal chord perpendicular to the axis is called the Latus 
 Seotum or Parameter. 
 
 102. To construct a parabola, having given the focus and 
 the directrix, 
 
 I. Bij Points. Let i^(Fig. 47) be the focus, C^the direc- . 
 trix. Draw the axis FB, and bisect FB in A ; then A is the 
 vertex of the curve. At any point 3f in the axis erect a per- 
 pendicular. From F as centre, with DM as radius, cut this 
 
 -^ 
 
120 
 
 ANALYTIC GEOMETEY. 
 
 perpendicular in P and Q ; then P and Q are two points* of 
 the curve, for FP= DM= distance of P or Q from CE. In 
 the same way we can find as many points of the curve as we 
 please. After a sufficient number of points has been found, 
 we draw a smooth curve through them. 
 
 Fig. 47. 
 
 Fig. 48. 
 
 II. By Motion. Place a ruler so that one of its edges 
 shall coincide with the directi-ix DE (Fig. 48). Then place 
 a triangular ruler BCE with the edge CE against the edge 
 of the first ruler. Take a string whose length is equal to 
 BC\ fasten one end at B and the other end at F. Then 
 slide the ruler BCE along the directrix, keeping the string 
 tightly pressed against the ruler by the point of a pencil P. 
 The point P will trace a parabola ; for during the motion we 
 always have PF= PC 
 
 103. To find the equation of the 'parctbola, when its axis is 
 taken as the axis of x and its vertex as the origin. 
 
 Let F (Fig. 49) be the focus, CE the directrix, DFX the 
 axis, A the vertex and origin ; also let 2f denote the known 
 distance FD. 
 
 Let P be any point of the curve ; then its co-ordinates are 
 
 AM=x, P2I=y. 
 
THE TARABOLA. 
 
 121 
 
 •Draw FC 1. to CE \ then by the definition of the curve 
 FF =FC=-DM. 
 
 Therefore 
 
 Now 
 and 
 
 Fr^=DM\ 
 
 FF' = Fir + FM' = y'+ix-py 
 
 Whence y' = 4^x. y- ^^ ^ [26] 
 
 This is called the principal equation of a parabola. 
 
 Therefore 3/^ + (:i' — p)y = (x -j-pf. 
 
 104. Since y^ and p in equation [26] are positive, x must 
 always be positive ; therefore the curve lies wholly on the 
 positive side of the axis of y. 
 
 A further examination of equation [26] shows that the 
 curve, (i.) passes through the origin, (ii.) is symmetrical with 
 respect to the axis of .r, (iii.) extends towards the right with- 
 out limit, and (iv.) recedes from the axis of x without limit. 
 
 105. Any point (h, k) is outside, on, or inside the parabola 
 y^ ■= 4ipx, according as k'^ — 4:ph is positive, zero, or negative. 
 
 Let Q be the point (A, ^), and let its ordinate meet the 
 curve in F. 
 
122 ANALYTIC GEOMETRY. 
 
 If Ic^ — 4:ph = 0, the point (A, k) satisfies equation [2G], and 
 therefore Q coincides with P. 
 
 If ¥ — ^'ph is positive, or Ji^ > 4^A, then, since P3I ^=4:ph, 
 we have QM' > PM'\ or QiV>FM; hence Q is outside 
 the curve. 
 
 If k'^ — ^ph is negative, we may prove similarly that Q 
 must be inside the curve. 
 
 106. To find the latus rectum of a parabola. 
 
 The common abscissa of the two points where the latus 
 rectum meets the curve =^. Substituting this value for x 
 in equation [26], we have y = ±2p. Therefore the latus 
 rectum = ^p. 
 
 107. To find the points in which the straight line i/ = mx-\-c 
 meets the parabola y^ = ^px. 
 
 The co-ordinates of these points must satisfy both equations ; 
 hence, at a common point, we have the relation 
 
 \ 771 j 
 
 Since (1) is a quadratic equation, we see that every straight 
 line meets a parabola in two points. Solving, (1). we obtain 
 for the ordinates of these two points 
 
 •^ 7?l m \ rp 
 
 l^LZ^; (2) 
 
 whence it appears that the points are real, coincident, or 
 imaginary, accordmg 2C$,p — mc is positive, zero, or negative. 
 
 108. To find the equation of a parabola ivhose axis is par- 
 allel to the axis of x. 
 
 Let the vertex be the point (a, 5), and let 2p — distance 
 from focus to directrix. Then the focus will be the point 
 {a-\-p, b), and the directrix will be the line x = a —p. 
 
THE FARABOLA. 123 
 
 The distance of any point {x, y) from the focus is 
 
 and its distance from the directrix is 
 
 X — Ci^lp. 
 
 If {x, y) is a point of the parabola, these distances are equal ; 
 putting them equal, and reducing, we obtain 
 
 y2 _ 4^^ _ O^y _|_ ^2 _ 4^^ ^ Q (2) 
 
 Hence we may infer that in general an equation having 
 theform ,f j^ Ax -\- By -^ C =^^ \^^r (2) 
 
 represents a parabola having its axis parallel to the axis of x. 
 By equating coefficients in (1) and (2), we obtain 
 
 . . B'-^C -, B 
 
 whence the following results easily follow : 
 The latus rectum = —. 
 The vertex is the point 
 
 The latus rectum =^ —A. 
 
 fB'~4:C _B 
 
 [ 4:A ' 2 
 
 ine locus is the point . • 
 
 p 
 
 The axis is the line y = 
 
 2 
 
 ine directrix is the line x = ■ 
 
 4:A 
 
 If A is negative, the parabola lies to the right of the axis 
 of ?/, and may be called right-ho,nded. 
 
 If A is positive, the parabola lies to the left of the axis of ?/, 
 and may be called left-handed. 
 
124 ANALYTIC GEOMETRY. 
 
 Ex. 28. 
 
 1. Show that the distance of any point of the parabola 
 y^ = 4:2^x from the focus is ec|ual to ^) + x. 
 
 2. Find the equation of a parabola, taking as axes the 
 axis of the curve and the directrix. 
 
 3. Find the equation of a parabola, taking the axis of the 
 curve as the axis of x and the focus as the origin. 
 
 4. The distance from the focus of a parabola to the direc- 
 trix = 5. "Write its equation, 
 
 (i.) If the origin is taken at the vertex, 
 (ii.) If the origin is taken at the focus, 
 (iii.) If the axis and directrix are taken as axes. 
 
 5. The distance from the focus to the vertex of a parabola 
 is 4. Write its equations for the three cases enumerated in 
 No. 4. 
 
 6. For what point of the parabola y^^^lSx is the ordi- 
 nate equal to three times the abscissa ? 
 
 7. Find the latus rectum for the following parabolas : 
 
 Find the points common to the following parabolas and 
 straight lines : 
 
 y- = ^x, 3.r — 7y + 30 = 0. 
 
 ?/---3.r, :r- 43/ + 12 = 0. 
 
 ?/^ = 4:r, .'r = 9, 2: = 0, a; = — 2. 
 
 ?/ = 4.i', y = 6, y = — 8. 
 
 What must be the value of ^j in order that the parabola 
 
 ?/ = ^po: may pass through the point (9, —12) ? 
 
THE PARABOLA. 125 
 
 13. For what point of the parabola y- = 2>2,x is the ordi- 
 nate equal to 4 times the abscissa ? 
 
 • 14. The equation of a parabola is y' = Qx. What is the 
 equation of (i.) its axis, (ii.) its directrix, (iii.) its latus rectum, 
 (iv.) a focal chord through the point whose abscissa = 8, (v.) a 
 chord passing through the vertex and the negative end of the 
 latus rectum ? 
 
 15. The equation of a parabola is y^ = 16a;. Find the 
 equation of (i.) a chord through, the points whose abscissas 
 are 4 and 9, and ordinates positive ; (ii.) the circle passing 
 through the vertex and the ends of the latus rectum. 
 
 IG. If the distance of a point from the focus of the parabola 
 y2 — 4^;-^. is equal to the latus rectum, what is the abscissa of 
 the point ? 
 
 17. In the parabola if^^^jpx an equilateral triangle is 
 inscribed so that one vertex is at the origin. What is the 
 length of one of its sides? 
 
 18. A double ordinate of a parabola = 8p. Prove that 
 straight lines drawn from its ends to the vertex are perpen- 
 dicular to each other. 
 
 Explain how to construct a parabola, having given 
 
 19. The directrix and the vertex. 
 
 20. The focus and the vertex. 
 
 21. The axis, vertex, and latus rectum. 
 
 22. The axis, vertex, and a point of the curve. 
 
 23. The axis, focus, and latus rectum. 
 
 24. The axis, directrix, and one point. 
 
 25. The axis and two points. 
 
126 ANALYTIC GEOMETRY. 
 
 26. Determine, as regards size and position, the relations 
 of the following parabolas : 
 
 (^i.) 7/^=z4:px, (ii.) y^---=^—4:px, (iii.) a;^^4^y, (iv.) :r^ = — 4^5?/. 
 
 27. What is the locus of the equation y'^ -{- Ax ^ B// -{- C — 
 in the following special cases : 
 
 (i.) A = 0? (in.)C=0? (v.) A = C=0? 
 
 (ii.) ^ = 0? (iv.)^=^ = 0? (yI) B = C=0? 
 
 28. Show that in general the equation x^-\-Ax-\- By-^ C= 
 represents a parabola whose axis is parallel to the axis of y ; 
 and determine the latus rectum, the vertex, the focus, the 
 axis, and the directrix. 
 
 Find the latus rectum, vertex, focus, axis, and directrix of 
 the following parabolas : 
 
 29. ?/--125;-f 84-=0. . 33. :c' - 12y + 84 = 0. 
 
 30. /-12^'-84-=0. 34. ?/-- 8^^ - 8?/ + 64 = 0. 
 
 31. y^ + 12:^ + 84 = 0. 35. l+22' + 3y- = 0. 
 
 32. 2/2 + 12:^-84 = 0. 36. y = x'~x-2. 
 
 37. ?/'-4:r + 6y + l = 0. 
 
 Tangents and Normals. 
 
 109. To find the equation and the normal of the tangent to 
 the rparahola 2/^^= ^px, at the -point of contact {x^, y^). 
 
 If P, Q are the points (.Tj, ?/i), (o;^, y.^, the equation of a 
 straight line through P and Q is 
 
 If P and Q are points of the parabola y^ = 4:px, 
 y^ = 4:pxi, 
 2/2' = ^pXi. 
 
 (1) 
 
127 
 
 THE PARABOLA. 
 
 Whence ^^iZJ^^-= _i£_. 
 
 ^2-^'i 3/2 + yi 
 
 By substitution, equation (1) becomes 
 
 x-x^ y, + yi ' 
 
 whence, by clearing of fractions, and remembering that 
 yx = ^px, we obtain the equation of a secant in the form 
 
 y(yi + y-i) — yiy-i = ^px. 
 
 (2) 
 
 
 
 
 Y 
 
 
 
 
 
 
 c 
 
 
 
 '■^'.^ 
 
 .^ 
 
 
 
 \ 
 
 
 .^^ 
 
 \ 
 
 
 
 ^ 
 
 \ 
 
 E,.^-^'^y^ 
 
 7 
 
 \ 
 
 \ 
 
 
 -^"^ \ 
 
 D 
 
 A 
 
 U' 
 
 M 
 
 N 
 
 X 
 
 
 
 
 • \ 
 
 ' • 
 
 
 
 
 Fig. 50. 
 
 Now let the secant turn about F till Q coincides with F ; 
 then x\ = Xi, y2 = yi, the secant becomes the tangent at P, 
 and its equation reduces to 
 
 ViV = 2p{3c + X,). [27] 
 
 The normal passes through (x^, y^), and is perpendicular to 
 the tangent ; hence (§ 50) its equation is 
 
 'PiU — Vi) -hvA^ — oc, 
 
 O. 
 
 [28] 
 
128 ANALYTIC GEOMETEY. 
 
 110. If we make y = in equations [27] and [28], we obtain 
 
 x^ — x^ and x=^Xi-\- 2p. 
 Hence the values of the subtangent MT and the subnormal 
 
 Therefore 
 
 (i.) The subtangent is bisected at the vertex. 
 (ii.) The subnormal is constant, and equal to the distance 
 from the focus to the directrix. 
 
 111. In the triangle FFT (Fifr, 50) we have 
 
 FT= FA + A T=p + X, 
 
 FF= FC = FM=FA +AM=j) + x. 
 
 Therefore FT= FF. 
 
 Hence the angle 
 
 FFT=^ FTF= TFa or 
 
 The tangent to a parabola at any point makes equal angles 
 with the focal radius and the line passing through the point 
 parallel to the axis. 
 
 112. To find the equation of a tangent to a parabola in ten-tns 
 of its slope. 
 
 From the result obtained in § 107 we see that the straight 
 line y = mx + c touches the parabola y"^ = 4^9^, when 
 
 mc^=p, 
 
 P 
 or c = —- 
 
 tn 
 
 Hence, for all values of m, the straight line 
 
 11 = inx + — 
 ^ m 
 
 will touch the parabola y" = ^p)x. 
 
THE PARABOLA. 129 
 
 Ex. 29. 
 
 1. The normal to a parabola at any point bisects the angle 
 between the focal radius and the line drawn through the 
 point parallel to the axis. 
 
 Note. The use of parabolic reflectors depends on this property. A 
 ray of light issuing from the focus and falling on the reflector is reflected 
 in a line parallel to the axis of the reflector. 
 
 2. Explain how to draw a tangent and a normal to a given 
 parabola at a given point. 
 
 3. Prove that FC (Fig. 50) is perpendicular to PT. 
 
 4. Prove that the tangent y = mx + — touches the parabola 
 
 y- = ^px at the point f — i,' — 1* 
 
 6. Prove that the equation of a normal to the parabola 
 y"^ = 4lPX in terms of its slope is y = ^nix — '???^;(2 -{- vi^). 
 
 6. What are the equations of a tangent and a normal to 
 the parabola y^ = 5.r, passing through the jDoint whose abscissa 
 is 20 and ordinate positive ? 
 
 7. AVhat are the equations of the tangents and the normals 
 to the parabola y' = 12x, drawn through the ends of the latus 
 rectum ? Find the area of the figure which they enclose. 
 
 8. Given the parabola y' = \Ox. Through the point whose 
 abscissa is 7 and ordinate positive a tangent and a normal are 
 drawn. Find the lengths of the tangent, the normal, the 
 subtangent, and the subnormal. 
 
 9. A tangent to the parabola ?/^ = 20:r makes Avith the axis 
 of x an angle of 45°. Determine the point of contact. 
 
 10. Show that the focus i^(Fig. 50) is equidistant from the 
 points P, T, N. What easy way of drawing a tangent and a 
 normal is suggested by this theorem ? 
 
130 ANALYTIC GEOMETRY. 
 
 11. If i^is the focus of a parabola, and Q, H denote the 
 points in which a tangent cuts the directrix and the latus 
 rectum produced, prove that FQ = FB. 
 
 12. Prove that tangents drawn through the ends of the 
 latus rectum are J_ to each other. 
 
 13. Find the distances of the vertex and the focus from the 
 
 P 
 tangent y = mx -f- — • 
 
 14. The points of contact of two tangents are {x^, y^ and 
 (x2, 3/2)- Find their point of intersection. 
 
 15. A tangent to the parabola 3/^ = 4:px cuts equal inter- 
 cepts on the axes. What is its equation ? What is the point 
 of contact ? What is the value of the intercept ? 
 
 16. Through what point in the axis of x must tangents to 
 the parabola y^ = 4:px be drawn in order that they may form 
 with the tangent, through the vertex, an equilateral triangle ? 
 
 17. For what point of the parabola y'^ = ^px is the normal 
 equal to twice the subtangent? 
 
 18. For what point of the parabola y'^=^^px is the normal 
 equal to the difference between the subtangent and the sub- 
 normal ? 
 
 19. Find the equation of a tangent to the parabola ?/ = 5:r 
 parallel to the straight line 3 a; — 2?/ -1-7= 0. Also find the 
 point of contact. 
 
 20. Find the equation of the straight line which touches 
 the parabola y' = Vlx and makes an angle of 45° with the 
 line y = 3a; — 4. Also find the point of contact. 
 
 21. Find the equation of a straight line which touches the 
 parabola y^^lGo; and passes through the point (~4, 8). 
 
 22. If a normal to a parabola meet the curve again in the 
 point Q, find the length of PQ. 
 
THE PARABOLA. 131 
 
 23. Prove by the secant method that the equation of a tan- 
 gent to the parabola 'if' = 4:px — ^p"^, at the point (xi, y^ is 
 
 24. Find the equations of the tangents and normals to the 
 parabola y^ — 8:?; — 6?/ — 63 = 0, drawn through the point.s 
 whose common abscissa = — 1. 
 
 25. What are the equations of tangents to the following 
 parabolas : 
 
 (i.) 7/^- 4:2?x ? (ii.) X- = 4:2^1/ ? (iii.) x' = - 4j9y ? 
 
 Ex. 30. (He^iew.) 
 
 Note. If not otherwise specified, the axis of the parabola and the 
 tangent at the vertex are to be assumed as axes of co-ordinates. 
 
 What is the equation of a parabola, 
 
 1. If the axis and directrix are taken as axes, and the focus 
 is the point (12, 0) ? 
 
 2. If the axis and tangent at the vertex are the two axes, 
 and (25, 20) is a point on the curve ? 
 
 3. If the same axes are taken, and the focus is the point 
 (-4i,0)? 
 
 4. If the axis is taken as the axis of x, the vertex is the 
 point (5, — 3), and the latus rectum = 5|-? 
 
 5. If the axis is the line y = — 7, the abscissa of the vertex 
 = 3, and one point is (4, — 5) ? 
 
 6. If the curve passes through the points (0, 0), (3, 2), 
 (3, - 2) ? 
 
 7. If the curve passes through the points (0, 0), (3, 2), 
 (-3,2)? 
 
132 ANALYTIC GEOMETRY. 
 
 8. What is the latus rectum of the parabola 2?/^ = 3 a:? 
 What is the equation of its directrix, and of the focal chords 
 passing through the points whose abscissa = 6 ? 
 
 9. Describe the change of form which the parabola ?/^ = 4pa; 
 undergoes as we suppose ^:> to diminish without limit. 
 
 10. Find the intercepts of the parabola 7/^+4.-^; — 6?/— 16 = 0. 
 
 11. One vertex of an equilateral triangle coincides with the 
 focus, and the others lie in the parabola if = A:]px. Find the 
 length of one side. 
 
 12. The latus rectum of a parabola = 8 ; find 
 
 (i.) Equation of a tangent through its positive end. 
 (ii.) Distance from the focus to this tangent, 
 (iii.) Equation of the normal at this point. 
 
 13. What is the equation of the chord passing through the 
 two points of the parabola y'^=^x for which Xi = 2, 3/1 > 0, 
 and :i'2 = 18, y2<0? 
 
 14. Find the equation of the chord of the parabola 'f^^'ipx 
 which is bisected at a given point (:i'i, ?/i). 
 
 15. In what points does the line x-\-y — 12 meet the para- 
 bola 2/^ + 2:r-12y + 16 = 0? 
 
 16. In what points does the line ^y='2x-\-d> meet the 
 parabola ?/- — 4.'u — 8?/ + 24 = 0? 
 
 17. Find the equations of tangents from the origin to the 
 parabola {y — by = 4ip(x — a). 
 
 18. Describe the position of the parabola y^ + 2.r + 4 = 
 with respect to the axes, and determine its latus rectum, 
 vertex, focus, and directrix. 
 
 19. What is the distance from the origin to a normal drawn 
 through the end of the latus rectum of the parabola 
 
 y^ = 4 a(x — a)? 
 
THE PARABOLA. 133 
 
 Find the equation of a parabola, 
 
 20. If the equation of a tangent is 4y = 3:r — 12. 
 
 21. If a focal radius = 10, and its equation is 3?/ = 4:r — 8. 
 
 22. If for a point of the curve the focal radius = r, the 
 length of the tangent = t. 
 
 23. If for a j^oint of the curve the focal radius ■= r, the 
 length of the normal = n. 
 
 24. If for a point of the curve the length of the tangent = i, 
 the length of the normal = n. 
 
 25. If for a point of the curve the focal radius = r, the 
 subtangent = s. 
 
 26. Two parabolas have the same vertex, and the same 
 latus rectum 4^, but their axes are _L to each other. What 
 is the length of their common chord ? 
 
 27. Through the three points of the parabola 3/' = 12:r, 
 whose ordinates are 2, 3, 6, tangents are drawn. Show that 
 the circle circumscribed about the triangle formed by the 
 tangents passes through the focus. 
 
 28. A tangent to the parabola y^ = "ipx makes the angle 
 30° with the axis of x. At what point does it cut the axis ? 
 
 29. For what point of the parabola y"^ = 4j9:r is the length 
 of the tangent equal to 4 times the abscis.sa of the point of 
 contact? 
 
 30. The product of the tangent and normal is equal to 
 twice the square of the ordinate of the point of contact. Find 
 the point of contact and the inclination of the ordinate to the 
 axis of X. 
 
 31. Two tangents to a parabola are perpendicular to each 
 other. Find the product of their subtangents. 
 
134 ANALYTIC GEOMETRY. 
 
 32. Prove that the circle described on a focal radius as 
 diameter touches the tangent drawn through the vertex. 
 
 33. Prove that the circle described on a focal chord as 
 diameter touches the directrix. 
 
 Find the locus of the middle points 
 
 34. Of all the ordinates of a parabola. 
 
 35. Of all the focal radii. 
 
 36. Of all the focal chords. 
 
 37. Of all chords passing through the vertex. 
 
 38. Of all chords that meet at the foot of the axis. 
 
 Two tangents to the parabola y^ = ^'px make the angles 
 B, 6^ with the axis of x ; find the locus of their intersection 
 
 39. If cot ^ + cot ^' - Z;. 41. If tan ^ tan ^' = Z:. 
 
 40. If cot 6 - cot & = h 42. If sin sin 0^ = h 
 
 43. Find the locus of the centre of a circle which passes 
 through a given point and touches a given straight line. 
 
 SUPPLEMENTARY PROPOSITIONS. 
 
 113. Two tangents can he draivn to a j'JCirabola from any 
 point ; and they ivill he real, coincident, or imaginary, accord- 
 ing as the point is luithout, on, or within the curve. 
 
 The tangent y = mx + — will pass through the point (h, J:) 
 if ""' 
 
 Z- = mh + ^ ; 
 
 that is, if hni" -Jcm+p^O; 
 
 whence ^,, ^ >^ ± V^"'^ - 4_^. 
 
 2^; 
 
THE PARABOLA. 135 
 
 Since there are two values of m, two tangents can be drawn 
 through the point (A, k). 
 
 The values of m are real, coincident, or imaginary, accord- 
 ing as k"^ — 4 hp is positive, zero, or negative ; that is (§ 105), 
 according as (h, k) is without, on, or within the curve. 
 
 114. To find the equation of the straight line through the 
 points of contact of the two tangents drawn to the parabola 
 y' = ^px from the point (h, k). 
 
 If (xi, ?/i) and (x-i, y.J are the points of contact, the equa- 
 tions of the tangents are 
 
 y,y = 2p(x + Xi), 
 y.{y = 2p{x + x.^. 
 
 Since (A, k) is in both these lines, 
 
 hj,=^2p{x,-\-h), (1) 
 
 kij, = 2p(x, + A). (2) 
 
 But equations (1) and (2) are the conditions which make 
 the points (xi, y^) and (0:2, y.^) lie in the straight line whose 
 equation is 
 
 Hence (3) is the equation recjuired. 
 
 115. The straight line joining the points of contact of the 
 two tangents (real or imaginary) from any point P to a para- 
 bola is called the Polar of P with respect to the ^^arabola ; and 
 the point P is called the Pole of the straight line with respect 
 to the parabola. 
 
 The propositions in §§ 80-82, relating to poles and polars 
 
 with respect to a circle, also hold true for poles and polars 
 
 with respect to a parabola, and may be proved in exactly the 
 same way. 
 
136 
 
 ANALYTIC GEOMETRY. 
 
 116. To find tlic locus of the middle points of parallel chords 
 in the parabola y* = 4ipx. 
 
 Fig. 51. 
 
 Let the equation of any one of the chords FQ (Fig. 51) be 
 y = mx + c, and let it meet the curve in the points (xi, ?/i), 
 
 -- ^- (1) 
 
 Then (§ 109) 
 
 m 
 
 2/1 + 2/2 
 
 Let (x,y) be its middle point M; then 2^ = yi -{- y.^. By 
 substitution in (1) we obtain 
 
 2p 
 
 y 
 
 (2) 
 
 a relation which holds true for all the chords, because m is 
 the same for all the chords. The required locus, therefore, 
 is represented by (2), and is a straight line parallel to the 
 axis of X. 
 
 The locus of the middle points of a system of parallel chords 
 in a parabola is called a Diameter ; and the chords are called 
 the Ordinates of the diameter. 
 
THE TARABOLA. 137 
 
 Therefore every diameter of a parabola is a straight line 
 parallel to its axis. 
 
 Conversely, every straight line parallel to the axis is a 
 
 diameter ; for ??z, and therefore ■ — > may have any value what- 
 ever. 
 
 117. Let the diameter through M meet the curve at 8, and 
 conceive the straight line PQ to move parallel to itself till P 
 and Q coincide at S\ then the straight line becomes the tan- 
 gent at 8 ; therefore 
 
 The tangent drawn through the extremfiity of a diameter is 
 parallel to the ordinates of the diameter. 
 
 118. From the focus F draw FC ± to PQ, and let FC 
 meet the directrix in the point C. If 6 denote the angle 
 which the chord PQ makes with the axis of x, it easily follows 
 that DCF= ; then we have 
 
 (77) = i^D c o t (9 ^ — ^ ^' o •' ; that is, 
 m 2 ' 
 
 The perpendicular to a chord which passes through the focus 
 meets the diameter of the chord in the directrix. 
 
 119. Let the tangents drawn through P and Q meet in the 
 point T. By solving their equations, 
 
 y,y = 2p{x + a;,), 
 y.,y=r.2p{x-\-x.;), 
 
 we obtain for the value of the ordinate of T 
 
 2p(x^ — x\) 2p 3/i + ?/2 ^^ 
 y — = — = — o — . Hence 
 
 Tangents drawn through the ends of a chord meet in the 
 diameter of the chord. 
 
138 ANALYTIC GEOMETRY. 
 
 120. What is the locus of the foot of a perpendicular dropped 
 from the focus to a tangent ? 
 
 Let the equation of tlie tangent be 
 y = mx -4- ^^' 
 
 Then the equation of the perpendicular will be 
 
 X p 
 
 y = h-- 
 
 ^ m m 
 
 Since these two lines have the same intercept on the axis 
 of y, they meet in that axis ; that is, in the tangent through 
 the vertex. This tangent, therefore, is the required locus. 
 
 121. The perpendicular 
 
 X p 
 ^ rn, m 
 
 meets the directrix in the point i—p, ^-). But the ordinate 
 
 T) 
 
 of the point where the tangent y = mx + — meets the para- 
 bola y'^ = A:px is also — ; therefore 
 
 The perpendicular from the focus to a tangent, if produced, 
 meets the directrix in the diameter through the point of contact. 
 
 122. The distance of anv point (A, 1:) from the focus {p, 0) i^ 
 
 and its distance from the point ( — 7^ — ) is 
 
 >J(^+^)'+(^-? 
 
THE PARABOLA. 139 
 
 If (A, 7") is in the tangent y = mx + — » 
 
 then k — mh -f —• 
 
 Now this value of h, substituted in the two expressions for 
 distances just given, makes them equal ; therefore 
 
 Every point in the tangent is equidistant from the focus and 
 the 2^oint where the perpendicular from the focus to the tangent 
 viects the directrix. 
 
 123. What is the locus of the intersection of two perpendicular 
 tangents f 
 
 If the equation of one tangent is 
 
 p 
 
 y =:^ mx A » 
 
 ^ ' ni 
 
 then the equation of the other is found to be 
 Subtracting one equation from the other, 
 
 (a;+i5)(^m+-j = 0. 
 
 But m -j is not ; therefore 
 
 m 
 
 x+p = 0, 
 
 or x=^—p, 
 
 the equation of the directrix. 
 
 Hence the directrix is the locus required. 
 
 124. Tangents are drawn through the ends of a focal chord. 
 What is the locus of their intersection ? 
 
 Let (/i, h) be their intersection ; then we may write the 
 equation of the chord ^^ _ ^^^^ ^ ^^^ 
 
140 
 
 ANALYTIC GEOMETRY. 
 
 If the chord passes through the focus Q;, 0), we have 
 
 whence h = —p. 
 
 Therefore the required locus is the directrix. And therefore 
 combining this result with that obtained in § 123, we see that 
 
 Tangents through the ends of a focal chord are perpendicular 
 to each other. 
 
 125. To find the equation of a parabola referred to any 
 diameter and the tangent through its extremity as axes. 
 
 Transform the equation y"^ = 4:px to the diameter SJ^' 
 (Fig. 52) and the tangent through jS as new axes. Let m be 
 the slope of the tangent, the angle which the tangent makes 
 with the diameter ; then m = tan 0. 
 
 Fig. 52. 
 
 First change the origin to S without changing the direction 
 
 of either axis. o 
 
 p Zp 
 The co-ordinates of 8 are — 2'-~ (§ 116). Therefore the 
 
 nv m ^ ^ 
 
 new equation is 
 
 or 
 
 y+l-T-^/^+i?'- 
 
 my* -f ^.py = ^pmx. 
 
 (1) 
 
THE TARABOLA. 141 
 
 Now retain the axis of x, and turn the axis of y till it coin- 
 cides with the tangent at S ; then for any point JP we have 
 
 The old X = SB. The new x = SJV. 
 
 The old y = FJR. The new y = JVF. 
 
 And it is easily seen from Fig. 52 that 
 
 SB = SJV+NF cos 0, 
 Pi?--iYPsin^. 
 
 Therefore equation (1) is transferred to the new system by 
 writing aj + ycos^ in place of x, and y sin ^ in place of y. 
 Making this substitution, and reducing, we obtain 
 
 y'=i^-. (2) 
 
 an equation of the same form as y^ = 4:2?x. 
 Join /S to the focus F; then 
 
 Therefore equation (2) may be more simply written 
 
 y- = 4;y.r, (3) 
 
 where ^:»' is the distance of the origin from the focus. It is 
 easy to see that this equation includes the case where the 
 axes are the axis of the curve and the tangent at the vertex. 
 
 The quantity 4^>' is called the Parameter of the diameter 
 passing through 8. When the diameter is the axis of the 
 curve, it is called the Principal Parameter. 
 
 126. Let the equation of a parabola referred to any diam- 
 eter, arid the tangent at the end of that diameter as axes, 
 be y^ = ^jjK-c. Since the investigations in §§ 109-112 hold 
 good whether the axes are at right angles or not, it follows 
 immediately that the equation of the tangent at any point 
 (a^i, yi) is y{y = 2j)\x-^x^, and that the straight line 
 
 y = onx 4- — will touch the parabola for all values of m. 
 
142 
 
 ANALYTIC GEOMETRY. 
 
 127. To find the polar equation of a parabola, the focus 
 being the pole. 
 
 Fig. 53. 
 
 Let P (Fig. 53) be any point (p, 6) of the curve, and let 6 
 be measured from tbe vertex A of the curve in the same 
 direction as clock motion. By definition, 
 
 PN^DM. 
 
 Now 
 
 Therefore 
 
 or 
 
 FP- 
 FP = p, 
 
 D2r=DF-MF, 
 = 2^; — p cos 0. 
 p = 2p ~ p cos Oj 
 
 P = 
 
 P 
 
 1 4- cos e 
 
 [29] 
 
THE PARABOLA. 143 
 
 Ex. 31. 
 
 1. Prove that the polar of the focus is the directrix. 
 
 2. Prove that the perpendicular dropped from any point 
 of the directrix to the polar of the point passes through the 
 focus. 
 
 3. To find by construction the pole of a focal chord. 
 
 4. Prove that through any point tliree normals can be 
 drawn to a parabola. 
 
 5. Tangents are drawn through the ends of a chord. 
 Prove that the part of the corresponding diameter contained 
 between the chord and the intersection of the tangents is 
 bisected by the curve. 
 
 6. Focal radii are drawn to two points of a parabola, and 
 tangents are then drawn through these points. Prove that 
 the angle between the tangents is equal to half the angle 
 between the focal lines. 
 
 7. Prove that the locus of the intersection of two tangents 
 to the parabola y^ = ^px, which make an angle of 45°, is the 
 parabola y'^ = :r^ + %px -\-p^. 
 
 8. Explain how tangents to a parabola may be drawn from 
 an exterior point (§§ 121, 122). 
 
 9. Having given a parabola, how would you find its axis, 
 directrix, focus, and latus rectum ? 
 
 10. From the point (— 2, 5) tangents are drawn to the 
 parabola y^^^a;. What is the equation of the chord of 
 
 contact? 
 
 11. The general equation of a system of parallel chords in 
 the parabola 7?/^ = 25.r is 4:r — 7?/-f Z: = 0. What is the 
 equation of the corresponding diameter ? 
 
144 ANALYTIC GEOMETRY. 
 
 12. In the parabola i/ = \?>x, what is the equation of the 
 ordinates of the diameter ?/ +11= ? 
 
 13. In the parabola y'^= 6.t, what chord is bisected at the 
 point (4, 3) ? 
 
 14. Given the parabola y'^ = 4ipx] find the equation of the 
 chord which passes through the vertex and is bisected by the 
 diameter y = a. How can this chord be constructed ? 
 
 15. The latus rectum of a parabola = 16. What is the 
 equation of the curve if a diameter at the distance 12 from 
 the focus, and the tangent through its extremity, are taken as 
 axes? 
 
 16. Show that the equations of that chord of the parabola 
 y2 — ^pr^ which is bisected at the point (A, k) is 
 
 ]c{jj _ Jc) = 2p(x - h). 
 
 17. Prove that the parameter of any diameter is equal to 
 the double ordinate which passes through the focus. 
 
 18. Discuss the form of the parabola from its polar equation. 
 
 19. Show that if the vertex is taken as pole, the polar equa- 
 tion of a parabola is 4^5 cos^ 
 
 ^^ sin'^6> * 
 
 20. Find the locus of the foot of a perpendicular dropped 
 from the focus to the normal to a parabola. 
 
 21. Two normals to a parabola are perpendicular to each 
 other ; find the locus of their intersection. 
 
 22. Find the locus of the centre of a circle which touches a 
 given circle and also a given straight line. 
 
 23. The area and base of a triangle being given, find the 
 locus of the intersection of perpendiculars dropped from the 
 ends of the base to the opposite sides. 
 
CHAPTER VI. 
 
 THE ELLIPSE. 
 
 Simple Properties of the Ellipse. 
 
 128. The Ellipse is the locus of a point, the sum of whose 
 distances from two fixed points is constant. 
 
 The fixed points are called Foci ; and the distance from any 
 point of the curve to a focus is called a Focal Eadius. 
 
 The constant sum is denoted by 2 a, and the distance be- 
 tween the foci by 2 c. 
 
 Q 
 
 The fraction - is called the Eccentricity, and is represented 
 
 by the letter e. Therefore c = ae. 
 
 From the definition of the ellipse it is clear that if 2a < 2c, 
 or a < (?, the locus does not exist ; if a = c, the locus is simply 
 that part of the straight line joining the foci which is com- 
 prised between the foci. The ellipse is the curve obtained 
 when a'> c \ that is, when e < 1. 
 
 129. To construct an ellipse, having given the foci and the 
 co'ostant sum 2 a. 
 
 I. By Motion. Fix pins in the paper at the foci. Tie a 
 string to them, making the length of the string exactly equal 
 to 2 a. Then press a pencil against the string so as to make 
 it tense, and move the pencil, keeping the string constantly 
 stretched. The point of the pencil will trace the required 
 ellipse ; for in every position the sum of the distances from 
 the point of the pencil to the foci is equal to the length of the 
 string. 
 
146 
 
 ANALYTIC GEOMETRY. 
 
 II. By Points. Let F, F' be the foci ; then FF^ = 2c. 
 
 Bisect FF' at 0, and from lay off OA = OA^ = a. 
 
 Then AA' = 2a. 
 
 AF +AF' =- (a- c) -{- 2c -{- (a- c) = 2a. 
 AF' + A'F' = (a - c) + 2c + (a - c) = 2a. 
 
 Therefore A and A' are points of the curve. 
 
 Between A and A' mark any point J^; then describe two 
 arcs, one with i^ as centre and ^Xas radius, the other with 
 F' as centre and A'X as radius : the intersections F, Q of 
 
 these arcs are points of the curve. By merely interchanging 
 the radii, two more points, F, jS, may be found. 
 
 After a sufficient number of points has been obtained, draw 
 a continuous curve through them. 
 
 130. The line AA' is the Transverse or Major Axis, A, A' 
 the Vertices, and the Centre of the curve. 
 
 The line FB', perpendicular to the major axis at 0, is the 
 Conjugate or Minor Axis ; its length is denoted by 2 b. 
 
 Show that £ and B' are equidistant from the foci, that 
 £F= a, that BO = h, and that a' = b' + c\^ 
 
THE ELLIPSE. 147 
 
 131. To find the equation of the ellipse, having given the 
 foci and the constant sum 2a. 
 
 Take the line AA^ (Fig. 55), passing through the foci, as 
 the axis of x, and the point 0, half way between the foci, as 
 origin. Let P be any point {x,y) of the curve, and let r, r' 
 denote the focal radii of F. Then from the definition of the 
 curve, and from the right triangles FPM, F^PM, 
 
 T^' = y'^{c-{-xf (1) 
 
 ,^ = n/A^{c~xY (2) 
 
 By addition, r'- + r = 2{x' + y' + c'). (3) 
 
 By subtraction, r'" — r^ = 4:cx. (4) 
 
 Factor (4), (r' + r) (r'-r) = 4: ex. (5) 
 
 Put 2 a for r' + r, r' - r - — (6) 
 
 ex 
 
 Whence r = a ^ = a — ex. (7) 
 
 a ^ -' 
 
 ex 
 r' = a + — = a 4- ex. (8) 
 
 a' + ^^^ = 2(x' + ,/ + €')• (9) 
 
 Reduce, and substitute h"^ in place of a^ — c"^ (§ 130), 
 h'x' + ay = a'b', 
 
 a^ + iy. = ^' [30] 
 
 Substitute in (3), 
 
148 ANALYTIC GEOMETRY. 
 
 132. To trace the form of the curve from its equation. 
 
 The intercepts on the axis of x are + a and — a ; on the 
 axis of y, -\-h and — h. 
 
 Only the squares of the variables x and y appear in the 
 equation ; hence, if it is satisfied by a point {x, y), it will also 
 be satisfied by the points {x, —y), (— ^, y), (— ^, —y)- There- 
 fore we infer that 
 
 (i.) The curve is symrnetrical loith respect to the axis of x. 
 (ii.) The curve is symmetrical with respect to the axis of y. 
 (iii.) Every chord which passes through the point is bisected 
 at ; for the distance from either (:r, y) or {—x, —y) to (0, 0) 
 is ^x^-\-y'\ This explains why is called the centre. 
 
 x\ , A/^^ 
 
 Since the sum of f - J and ( 't j is 1, neither of these squares 
 
 can exceed 1 ; therefore the maximum value of x is -j- a, and 
 the minimum value — a, while the corresponding values of y 
 are -f b and — b. Therefore the curve is wholly contained 
 within the rectangle whose sides are equal to 2a and 2b, and 
 are bisected by the axes. 
 
 133. To trace the changes in the form of the ellipse when the 
 semi-axes are supposed to change. 
 
 Let a be regarded as a constant, and 5 as a variable. 
 
 (i.) Suppose b to increase. Then c decreases (since c' = a^ 
 — h^), e decreases, the foci approach the centre, and the ellipse 
 approaches the circle. 
 
 (ii.) Let b = a. Then c = 0, e = 0, the foci coincide with 
 the centre, the ellipse becomes a circle of radius a, and equa- 
 tion [301 becomes 
 
 Therefore we may regard a circle as an ellipse whose eccen- 
 tricity is equal to 0. 
 
THE ELLIPSE. 149 
 
 (iii.) Let 5 > «. The foci and major axis will now be on the 
 axis of I/, c will increase with b, e= -, and b^ = a' -j- c^. 
 
 (iv.) If we suppose b to decrease to (a remaining con- 
 stant), c will increase to a, c will increase to 1, while the 
 curve will approach, and finally coincide with, the major axis, 
 its equation at the same time becoming y = 0. 
 
 134. It follows from § 131 that a point (A, h) is on the 
 ellipse represented by equation [30], provided 
 
 - + --1 = 0. 
 d' ^b' 
 
 It may be shown by reasoning similar to that employed in 
 § 105 that the jDoint {h, Jc) is outside or inside the curve, 
 
 according as — + 7-, — 1 is positive or negative. 
 d^ b' 
 
 135. Since the constants a and b in the equation 
 
 may have any positive values, every equation reducible to the 
 form of (1) must represent an ellipse. Hence every equation 
 of the form Ax^ + Bf--=C 
 
 represents an ellipse, provided C is not zero, and A, B, and C 
 all have the same sign. Its semi-axes have the values 
 
 "=>iz *=V: 
 
 1^. 
 
 136. The chord passing through either focus perpendicular 
 to the major axis is called the Latus Eectum or Parameter. 
 To find its length, put :r = c in the equation of the ellipse. 
 
 Then ,/=^lui'-c')=-, y = d= -• 
 
 -^ a^^ ^ a^ ^ a 
 
 Therefore the latus rectum = 
 
 a 
 
150 
 
 ANALYTIC GEOMETRY. 
 
 137. The circle having for diameter the major axis of the 
 ellipse is called the Auxiliary Circle ; its equation is 
 
 x' + y' = a\ 
 
 The circle having for diameter the minor axis is called the 
 Minor Auxiliary Circle ; its equation is 
 
 x' + y'^h'. 
 
 If P (Fig. 56) is any point of an ellipse, and the ordinate 
 MP produced meets the auxiliary circle in Q, the point Q is 
 said to correspond to the point P. 
 
 The angle QOM'i^ called the Eccentric Angle of the point P, 
 and denoted by the letter <^. 
 
 138. Let y, y' represent the ordinates of points in an ellipse 
 and the auxiliary circle respectively, corresponding to the 
 same abscissa x. Then from the equations of the two curves 
 
 we have 7 
 
 y = -Va^ — ^^ ?/' = Va^ — x\ 
 
 Whence y -.y^ — h -.a, or 
 
 The ordinates of the ellipse and the auxiliary circle, corre- 
 sponding to a common abscissa, are to each other in the constant 
 ratio of the semi-axes of the ellipse. 
 
THE ELLIPSE. 151 
 
 139. Hence, if the axes 2a, 2 i of an ellipse are given, we 
 may find any number of points in the ellipse by constructing 
 the auxiliary circle, drawing ordinates at pleasure, and then 
 reducing their lengths in the ratio b : a. 
 
 In practice, it is convenient to proceed as follows : 
 Construct both the major and minor auxiliary circles; draw 
 any radius, cutting the circles in Q, B, respectively ; through 
 Q draw a line II to F, and through R draw a line li to OX: 
 the intersection P of these parallels is a point of the ellipse. 
 For from the similar triangles QOM, ORN, 
 
 ON:QM ^OR:OQ. 
 
 Now • ON^PM=y, QM=y\ 
 
 OR = b, OQ =a. 
 
 Therefore y : y' = b : a. 
 
 With the aid of the eccentric angle (f> -- QOX, the proof 
 that P is a point of the ellipse may be given as follows : 
 Let P be the point (x, y) ; then 
 
 
 X = 0M= OQ coscfi^a cos <^, 
 
 
 y = PM^- 0N= OR sin<f>^b sin 0. 
 
 Whence 
 
 we have 
 
 
 X y . 
 - = cos <f), T — sm d>. 
 a ^ b ^ 
 
 Square 
 
 X^ 7/2 
 
 -2 = cos^<^, '^-sin^t^. 
 
 Add 
 
 X^ 7/2 
 
 ^i + ^. = cos^<^ + sin2<^. 
 
 But 
 
 cos^<^ + sin^ cf> = l. 
 
 Hence 
 
 
 Therefore P is a point of the required ellipse. 
 
152 
 
 ANALYTIC GEOMETRY. 
 
 140. Another mode of constructing an ellipse from its axes 
 is shown in Fig. 57. 
 
 In the rectangle OACB, whose sides OA, OB are made 
 equal to the given semi-axes a and 6, divide the side BC 
 into any number of equal parts, and divide BO into the same 
 number of equal parts, and let 31, N denote any two corre-.. 
 spending points of division, counting from B. If we now draw 
 through the extremities A, A' of the major axis, and the points 
 M, iV, respectively, straight lines, the intersection P of the 
 lines will be a point of the required ellipse. 
 
 In order to give a general proof, let there be n equal parts, 
 and let ON and CM contain r of these parts, respectively ; 
 then 7 
 
 n n 
 
 Produce 3INio meet OB produced in Q; then 
 
 0Q:AC=0A:C3f, 
 
 OQ: h-. 
 
 nh 
 
 ra 
 a : — = n : T. 
 n 
 
 Therefore 
 
 0Q = 
 
2+ 
 
 a 
 
 ry ^ 
 nb 
 
 -1. 
 
 1+ 
 
 ny 
 
 -1. 
 
 
 = 1- 
 
 X 
 
 a 
 
 ny 
 
 -'+1- 
 
 
 --1- 
 
 
 
 = 1. 
 
 THE ELLIPSE. 153 
 
 Taking now for origin, and OA for axis of x, we have for 
 the symmetrical equations of AM and A'JY 
 
 and 
 Or 
 
 and 
 
 Multiply 
 
 that is, 
 
 This relation must hold true of the point common to the 
 lines J. J/ and A'JV; therefore this point is on the ellipse 
 whose axes are 2a and 2b. 
 
 Ex. 32. 
 
 What are a, b, c, and e in the ellipse whose equation is 
 
 ■ 25 16 
 
 2. x- + 2i/ = 2? 
 
 3. 30,- + 4/ = 12? 
 
 4. Ax'+By' = l? 
 
 5. Find the latus rectum of the ellipse Sx^ -{- 7?/^ = 18. 
 
 6. Find the eccentricity of an ellipse if its latus rectum is 
 equal to one-half its minor axis. 
 
154 ANALYTIC GEOMETRY. 
 
 What is the equation of an ellipse if 
 
 7. The axes are 12 and 8 ? 
 
 8. Major axis = 2G, distance between foci = 24 ? 
 
 9. Sum of axes = 54, distance between foci = 18 ? 
 
 10. Latus rectum = -%*-, eccentricity = | ? 
 
 11. Minor axis = 10, distance from focus to vertex = 1 ? 
 
 12. The curve j^asses through (1, 4) and (—6, 1) ? 
 
 13. Major axis = 20, minor axis = distance between foci? 
 
 14. Sum of the focal radii of a point in the curve = 3 times 
 the distance between the foci ? 
 
 15. Prove that the semi-minor axis is a mean proportional 
 between the segments of the major axis made by one of the 
 foci. 
 
 16. What is the ratio of the two axes if the centre and foci 
 divide the major axis into four equal parts ? 
 
 17. For what point of an ellipse is the abscissa equal to the 
 ordinate ? 
 
 Find the intersections of the loci 
 
 18. Sx' + 6f = ll and 7/ = x + l. 
 
 19. 2x''-\-3f = U and / = 4.r. 
 
 20. x'-\-7f = lQ and x'-{-f = lO. 
 
 21. The ordinates of the circle x'^-{-y^=r'^ are bisected; 
 find the locus of the points of bisection. 
 
 22. A straight line AB so moves that the points A and B 
 always touch two fixed perpendicular straight lines. Show 
 that any point JP in AB describes an ellipse, and find its 
 equation. 
 
THE ELLIPSE. 155 
 
 23. What is the locus of Ax^-\-By- — C when C is zero? 
 When is this locus imaginary ? 
 
 24. Prove that the abscissas of the ellipse h-oy + d'lf = a^b'' 
 
 are to the corresponding abscissas of the minor auxiliary circle, 
 X'^ -j- y2 _ ^2^ j^g ^^ . ^ 
 
 25. Construct an ellipse by the method of § 139. 
 
 26. Construct an ellipse by the method of § 140. 
 
 27. Construct the axes of an ellipse, having given the foci 
 and one point of the curve. 
 
 28. Construct the minor axis and foci, having given the 
 major axis (in magnitude and position) and one point of the 
 curve. 
 
 29. A square is inscribed in the ellipse 
 
 -4-^ = 1 
 a' ^ b' 
 
 Find the equations of the sides and the area of the square. 
 
 Tangents and Normals. 
 
 141. To find the equations of a tangent and a normal to an 
 ellipse, having given the point of contact {x^, 3/1). 
 
 Taking the equation of the ellipse, 
 Z/V + ay = a^b\ 
 and the equation of the straight line through (^1, yi) and 
 (^^'2, 3/2) , y — 7/1 _ ?/, — ?/i 
 
 and proceeding as in § 72, we obtain as the equation of a 
 chord through (xi, t/j) and (X2, y^ 
 
 x-Xi a\yi + y,) 
 
156 ANALYTIC GEOMETRY. 
 
 Now make X2 = Xi, y^ = yi ; then the chord becomes a tan- 
 
 becomes 
 
 
 which reduces to ^ + ^ = 1. [311 
 
 From the equation above it appears that the value of the 
 slope of the tangent, in terms of the co-ordinates of the point 
 of contact is 72 
 
 Xi 
 
 The normal is perpendicular to the tangent, and passes 
 through (xi, 1/1) ; therefore its equation is easily found (by the 
 method of § 51) to be 
 
 h- cC^ cc-lf- 
 
 [32] 
 
 142. To find the suhtangent and subnormal. 
 
 Making y = in [31] and [32], and then solving the equa- 
 tions for X, we obtain : 
 
 d^ 
 Intercept of tangent on axis of ^ = — > 
 
 Xi 
 
 Intercept of normal on axis of a; = — ^1 = e'Xi. 
 
 d 
 
 AVhence the values of the subtangent and the subnormal 
 (defined as in § 71) are easily found to be as follows : 
 
 Subtangent = ^ ~ ^'\ [33] 
 
 Subnormal = -~^\' [34] 
 
THE ELLIPSE. 
 
 157 
 
 143. Jf tangents to ellipses having a common major axis are 
 draicn at points having a common abscissa, they will vieet on 
 the axis of x. 
 
 For in all these ellipses the values of a and x are constant, 
 and therefore (by § 142) the tangents all cut the same inter- 
 cept from the axis of x. 
 
 Fig. 58. 
 
 144. The normal at any point of an ellipse bisects the angle 
 formed by the focal radii. 
 
 The values of the focal radii for the point P (Fig. 58) were 
 found in § 131 to be 
 
 PF= a — exi, FF' = a-\- ex^. 
 
 (§ 142) ; and therefore 
 
 NF = c — e^Xi = ae — e^Xi -= e{a — ex), 
 JVF' — c-{- e^Xi = ae-\- e^x^ -~ e(a + ex). 
 
 Therefore NF : NF^ ^ PF : PF\ 
 
 or the normal divides the side i^F' of the A PFF' into two 
 parts proportional to the other two sides. Therefore (by- 
 Geometry) FPN=FPN. 
 
 The tangent FT, being perpendicular to the normal, must 
 bisect the angle FFFi,, formed by one focal radius with the 
 other produced. 
 
158 
 
 ANALYTIC GEOMETRY. 
 
 145. To draiu a tangent ayid a normal through a given point 
 of an ellipse. 
 
 I. Let P (Fig. 59) be the given point. Describe the 
 auxiliary circle, draw the ordinate PM, produce it to meet 
 the circle in Q, draw Q 7^ tangent to the circle and meeting 
 the axis of x in T, and join PT\ then PT is a tangent 
 to the ellipse (§ 143). Draw PN J_ to PT\ PN is the 
 normal at P. 
 
 a 
 
 11. Draw the focal radii, and bisect the angles between 
 em. The bise( 
 point P (§ 144). 
 
 them. The bisectors are the tangent and the normal at the 
 
 146. To find the equation of a tangent to an ellipse, having 
 given its direction. 
 
 This problem may be solved by finding under what condi- 
 tion the straight line , ., . 
 
 will touch the ellipse Irx' + cry~ — a'^b'^. (2) 
 
 Eliminating y from (1) and (2), and then solving for x, we 
 find two values of :r : ^ 
 
 — ma-c -t ah^m'-a- -\- b'^ — c^ 
 
 vra 
 
 b' 
 
 These values will be equal if 
 
 m'a' -i-b'-c^ = 0, or c=± -Vni'd' + b' 
 
THE ELLIPSE. 159 
 
 If the two values of x are equal, the two values of y must 
 also be equal from equation (1). 
 
 Therefore the two points in which the ellipse is cut by the 
 line will coincide if d? = =b VmW + b'\ 
 
 Hence the straight line 
 
 y = mx ± {m^a"' + h"-) [35] 
 
 will touch the ellipse for all values of m. 
 
 Since either sign may be given to the radical, it follows that 
 two tangents may be drawn to an ellipse in a given direction 
 (determined by the value of m). 
 
 147. To find the locus of the intersection of two tangents to 
 an ellipse which are perpendicular to each other. 
 
 Let the equations of the tangents be 
 
 y^mx +Vm'V + ^', (1) 
 
 y = m'x + ^m'd' + b'\ (2) 
 
 The condition to be satisfied is 
 
 1 - 1 
 
 mm — — 1, or ni = 
 
 If we substitute for m' in equation (2) its value in terms of 
 m, the equations of the tangents may be written 
 
 y — 7nx = VmV + ^^, (3) 
 
 7ny -{- x = Va'^ + 7n^b'^. (4) 
 
 The co-ordinates, x and ?/, of the intersection of the tangents 
 
 satisfy both (3) and (4) ; but before we can find the constant 
 
 relation between them we must first eliminate the variable m. 
 
 This is most easily done by adding the squares of the two 
 
 equations ; the result is 
 
 (1 + m')x' + (1 + rn')y' - (1 + 7n') (a' + b'), 
 or x' + / - a' + b\ 
 
 The required locus is therefore a circle. This circle is 
 called the Director Circle of the ellipse. 
 
160 ANALYTIC GEOMETRY. 
 
 Ex. 33. 
 
 1. What are the equations of the tangent and the normal 
 to the ellipse 2a;^+3y = 35 at the points whose abscissa = 2? 
 
 2. What are the equations of the tangent and the normal to 
 the ellipse 4 :f'^ -f- 9 y' = 36 at the points whose abscissa = — f? 
 
 3. Find the equations of the tangent and the normal to the 
 ellipse x^ + 4:7/ = 20 at the point of contact (2, 2). Also find 
 the subtangent and the subnormal. 
 
 4. Show that the line 7/ ^= x -\- Vf touches the ellipse 
 2^' + 3y^ = l. 
 
 5. Required the condition which must be satisfied in order 
 that the straight line h - = 1 niay touch the ellipse 
 
 9 n 711/ 77/ 
 
 -4-^ = 1 
 
 6. In an ellipse the subtangent for the point (3, -^-^) is 16, 
 the eccentricity = 4. What is the equation of the ellipse ? 
 
 7. AVhat is the equation of a tangent to the ellipse 
 9^;' + 64?/' =--576 parallel to the line 27/ = x? 
 
 8. Find the equation of a tangent to the ellipse 3.r'-f5y'=15 
 parallel to the line 4a; — ^?/ — 1 = 0. 
 
 . ^ "h 
 
 9. In what points do the tangents which are equally inclined 
 
 to the axes touch the ellipse &V + «y = «^^^ ? 
 
 10. Through what point of the ellipse h'^x'^ + ay = n^b"^ 
 must a tangent and a normal be drawn in order that they 
 may form, with the axis of x as base, an isosceles triangle ? 
 
 11. Through a point of the ellipse 5V + «y = <^^^", and 
 the corresponding point of the auxiliary circle x- -\- if = a^ 
 normals are drawn. What is the ratio of the subnormals ? 
 
THE ELLIPSE. 161 
 
 12. For what points of the ellipse b'^x^~\- anf = c^lP' is the 
 subtangent equal numerically to the abscissa of the point of 
 contact? 
 
 13. Find the equations of tangents drawn from the point 
 (3, 4) to the ellipse 16a;'' + 25y' = 400. 
 
 14. What are the equations of the tangents drawn through 
 the extremities of the latera recta of the ellipse ^x^^^y^— 36a^ ? 
 
 15. What is the distance from the centre of an ellipse to a 
 tangent making the angle <^ with the major axis? 
 
 16. What is the area of the triangle formed by the tangent 
 in the last parabola and the axes of co-ordinates ? 
 
 17. From the point where the auxiliary circle cuts the 
 minor axis produced tangents are drawn to the ellipse. Find 
 the points of contact. 
 
 18. Prove that the tangents drawn through the ends of a 
 diameter are parallel. 
 
 19. Find the locus of the foot of a perpendicular dropped 
 from the locus to a tangent. 
 
 Ex. 34. (Revie-w.) 
 
 1. Given the ellipse Z^x" + 100/ = 3600. Find the equa- 
 tions and the lengths of focal radii drawn to the point (8, -i^). 
 
 2. Is the point (2, 1) within or without the ellipse 
 o^-2_^3^2^;L2? 
 
 Find the eccentricity of an ellipse 
 
 3. If the equation is 2x^ + 3y' = 1. 
 
 4. If the angle FBF' = 90° (see Fig. 54). 
 
 5. If LFR is the latus rectum and LOR is an equilateral 
 triangle (i^ being the focus, the centre). 
 
1G2 ANALYTIC GEOMETRY. 
 
 Find the equations of tangents to an ellipse 
 
 6. If tliey make equal intercepts on the axes. 
 
 7. If they are parallel to BF (Fig. 54). 
 
 8. Which are parallel to the line - + y = 1 (a and b being 
 the semi-axes). 
 
 9. Find the equation of a tangent in terms of the eccentric 
 angle <j!> of the point of contact. 
 
 Find the distance from the centre of an ellipse to 
 
 10. A tangent through the point of contact (xi, y^). 
 
 11. A tangent making the angle ^ with the axis of x. 
 
 12. In what ratio is the abscissa of a point divided by the 
 normal at that point ? 
 
 13. At the point (.Tj, ?/i) of an ellipse a normal is drawn. 
 What is the product of the segments into which it divides the 
 major axis? 
 
 14. Find the length of PiV^ (Fig. 58). 
 
 15. Determine the value of the eccentric angle at the end 
 of the latus rectum. 
 
 Prove that the semi-minor axis h of an ellipse is a mean 
 proportional between 
 
 16. The distances from the foci to a tangent. 
 
 17. A normal and the distance from the centre to the cor- 
 responding tangent. 
 
 Determine and describe the loci of the following points : 
 
 18. The middle point of that portion of a tangent contained 
 between the tangents drawn through the vertices. 
 
 19. The middle point of a perpendicular dropped from a 
 point of the circle (a: — of -\-y^ = r^ to the axis of y. 
 
THE ELLIPSE. 163 
 
 20. The middle point of sl chord of the ellipse iV+ay =a^Z>^ 
 drawn through the positive end of the minor axis. 
 
 21. The vertex of a triangle whose base, 2 c, and sum of 
 the other sides, 25, are given. 
 
 22. The vertex of a triangle, having given the base 2n and 
 the product k of the tangents of the angles at the base. 
 
 23. The symmetrical point of the right-hand focus of an 
 ellipse with respect to a tangent. 
 
 SUPPLEMENTARY PROPOSITIONS. 
 
 148. Two tangents can he drawn to an ellipse from any 
 point ; and they will be real, coincident, or imaginary, accord- 
 ing as the point is outside, on, or inside the curve. 
 
 If the tangent y^^7nx-\-'\/wb'c^-\-lf pass through the point 
 
 (A,/?:), then ;^ =. ^i + V^>^V+F, 
 
 or {K' - a") iw - 2 hhn + 1^ -b' = 0. 
 
 This is a quadratic equation with respect to ni, and its roots 
 give the directions of those tangents which pass through (h, k). 
 Since a quadratic equation has tivo roots, two tangents may 
 be drawn from any point (A, h) to an ellipse. 
 
 By solving the equation, w^e obtain 
 
 hk ± Vb'h:' + rr^' - a'b' . 
 
 m = > 
 
 h'-a' 
 
 and we see that the roots are real, coincident, or imaginary, 
 
 according as b'h^ -f ci^k"^ — a'b^ is positive, zero, or negative ; 
 
 7^2 k' . . ■ 
 
 that is, according as — 4-77,-1 is positive, zero, or negative ; 
 
 in other words, according as the point {h, k) is outside, on, 
 or inside the ellipse. (§ 134.) 
 
164 
 
 ANALYTIC GEOMETRY. 
 
 149. To find the equation of the straight line passing through 
 the points of contact of the tiuo tangents drawn to an ellipse from 
 the 'point (A, k). 
 
 If (^1, 2/1) and {x-i, 2/2) are the points of contact, it follows, 
 
 by reasoning similar to that employed in § 78, that the 
 
 required equation is 
 
 hx ky _ 
 
 a' ~^ b' ~ ^' 
 
 This line is always real ; but if the point (A, Jc) is within 
 the ellipse, the points (.ri, yi), {x2, 3/2) through which the line 
 passes, will be imaginary. 
 
 150. The straight line joining the points of contact of the 
 two tangents from any point P to an ellipse is called the Polar 
 of P with respect to the ellipse ; and the point P is called 
 the Pole of the straight line with respect to the ellipse. 
 
 Fig. 60. 
 
 The propositions in §§ 80-82, relating to poles and polars 
 with respect to a circle, also hold true for poles and polars 
 with respect to an ellipse, and may be proved in the same 
 way. 
 
THE ELLIPSE. 165 
 
 151. To draw a tangent to an ellipse from a given point P 
 outside the ciirve. 
 
 Suppose the 2:)roblem solved, and let the tangent meet the 
 ellipse at Q (Fig. 60). If F^Q be produced to (7, making 
 QG = QF, then A FQG is isosceles ; now Z FQF^-Z GQF 
 (§ 144) ; therefore FQ is perpendicular to FG at its middle 
 point ; therefore F is equidistant from F and G. This re- 
 duces the problem to determining the point G. 
 
 Since F'G = 2a, G lies in the circle with F' as centre and 
 2 a as radius. And G also lies in the circle with F as centre 
 and FF as radius. Hence the construction is obvious. 
 
 152. To find the locus of the middle points of the sijstcm of 
 chords represented by the equation 
 
 y — mx + Ic. (1) 
 
 Let any one of the chords meet the ellipse 
 
 in the points (:ri, y^ and (x^, t/J ; then (§ 141 (3)) 
 
 h\xy + a:,) ,oN 
 
 If {x,y) is the middle point, 2x.^^ x^-\- x^, 2y = y^~\-y^^ 
 and (3) becomes 72 
 
 m = -^, (4) 
 
 or y = ___. (5) 
 
 a^m -^ 
 
 This relation holds true for the middle points of all the 
 chords ; therefore it is the equation required. 
 
 The locus of the middle points of a system of parallel chords 
 in an ellipse is called a Diameter of the ellipse. 
 
 From the form of (5) we see that a diameter is a straight 
 line passing through the centre. 
 
166 ANALYTIC GEOMETRY. 
 
 153. Let m' denote the slope of the diameter of the chords 
 represented by the equation y = 77ix-j-k ; then from (5) of § 151 
 
 mm' = — ^* [3G] 
 
 From the symmetry of this equation we may infer at once 
 that all chords parallel to the diameter 3/ = 7n'x are bisected 
 by the diameter 1/ = vix ; hence 
 
 7/^ one diameter bisect all chords 'parcdlel to another, the 
 second diameter bisects all chords parallel to the first. 
 
 Two such diameters are called Conjugate Diameters. 
 
 154. Let a straight line cutting the case in JP and Q move 
 parallel to itself till P and Q coincide with the end of the 
 diameter bisecting PQ ; then the straight line becomes the 
 tangent at the end of the diameter. Therefore 
 
 A diameter bisects all chords parallel to the tangent at its 
 extremity. A tangent drawn through the end of a diameter 
 is parallel to the conjugate diameter (§ 153). 
 
 155. Let POP\ EOPJ (Fig. 61) be two conjugate diameters, 
 meeting the curve on the positive side of the axis of x in the 
 points P {xi, 7/1) and P {X2, y^), and making the angles a, ^ 
 respectively with the axis of x. 
 
 Let a be acute ; then it follows from equation [36] that ^ 
 must be obtuse ; whence we infer that tivo corijugate diameters 
 must lie in different quadrants. 
 
 The equation of the tangent through P is 
 
 Therefore the equation of the diameter POP', which is 
 parallel to this tangent (§ 154) and passes through 0, is 
 
 ^ + ^^ = 0. (2) 
 
THE ELLIPSE. 
 
 167 
 
 Similarly, the equation of FOF' is found to be 
 
 ^ 4_ y-'^'/ _ n 
 
 The point i? is in the locus of (2) ; therefore 
 a' ~^ b' "^• 
 
 (3) 
 
 (4) 
 
 Equation (4) is the condition which must be satisfied by 
 the co-ordinates of the extremities of every pair of conjugate 
 diameters. 
 
 Fig. 61. 
 
 156. Let the ordinates of the extremities F, R (Fig. 61) of 
 two conjugate diameters meet the auxiliary circle in Q, S 
 respectively, join QO and SO, and denote Z QOX by ^, 
 Z. SOX by <ji\ Then the values of the co-ordinates of F 
 and F are (§ 138) 
 
 Xi^= a cos ^, x.i-= a cos <^', 
 
 yi'=h sin <f>, y-i = h sin ^'. 
 
 Whence, by substitution in equation (4) of § 155, we obtain 
 
 cos (^ cos <^' + sin <^ sin <^' ^=^ 0. 
 Therefore <^'-<^ = ^7r. 
 
 That is, the difference of the eccentric cingles corresponding 
 to the ends of two conjugate diameters is equal to a right angle. 
 
168 ANALYTIC GEOMETRY. 
 
 157. Given the end {x^, y^) of a diameter, to find the end 
 {x.,, T/a) of the conjugate diameter. 
 
 From § 156 we have for one of the ends 
 
 rr^ = a cos fp' = a cos (<f> -}- ^ tt) = a sin <f), 
 
 2/2 = ^ sin </)' = Z> sin (<f> -{- ^ tt) = b cos tf). 
 
 X2 — a sin 4> a ]).i h cos ^ h 
 
 yi 5 sin ^ ~ V x^ ~ a cos ^ a 
 
 Therefore x^~ — -?/i, y.^ = -Xi. 
 
 a 
 
 Since every chord through the centre is bisected by the 
 centre, the co-ordinates of the other end of the diameter are 
 
 a 1 b 
 
 -?/i and —-x^. 
 a 
 
 158. To find the angle formed by two conjugate semi-diam- 
 eters, ivhose lengths a', b' are given. 
 
 Let the semi-diameters make the angles a, /3 respectively 
 with the axis of x, and let denote the required angle. Then 
 if (xi, 2/1) and (x.2, 3/2) are the extremities of a' and b' respect- 
 
 ively, 
 
 sm a = ^. sm p = -Tt — ^Tt' 
 a'' '^ b' ab' 
 
 Xi x\ ay^ 
 
 cos a = -,, cos p—~-r, = — TTi' 
 
 sin 6 = sin (/? — a) 
 
 = sin ^ cos tt — cos /? sin a 
 
 a6a'6' 
 a^6^ 
 
 a6a'6' 
 ab 
 
THE ELLIPSE. 169 
 
 159. The lines joining any point of an ollipse to the ends 
 of any diameter are called Supplemental Chords. 
 
 Let PQ, PQ be two supplemental chords (Fig. G2). Through 
 the centre draw OR parallel to PQ, and meeting PQ in B ; 
 also OR^ parallel to PQ, and meeting P'Q in R'. 
 
 Fig. 62. 
 
 Since is the middle point of PP\ and OR is drawn 
 parallel to PQ, and OPJ is drawn parallel to PQ, R and 
 i?' are the middle points of QP, QP respectively. Therefore 
 OR will bisect all chords parallel to QP, and OPJ will bisect 
 all chords parallel to QP. Hence OR, OP are conjugate 
 diameters. 
 
 Therefore the diameters parallel to a pair of supplemental 
 chords are conjugate diameters. 
 
 160. To find the equation of an ellipse referred to a pair of 
 conjugate diameters as axes. 
 
 The origin not being altered, we must substitute for x and y 
 expressions of the form ax -\- hy, a'x + i>'y (§ 95). 
 
 Therefore the transformed equation will have the form 
 
 Ax'-j-Cxy-]-£y' = l. 
 
170 ANALYTIC GEOMETRY. 
 
 But by hypothesis the axis of x bisects all chords parallel 
 to the axis of y ; therefore the two values of y corresponding 
 to each value of x will be equal and of opposite signs. 
 
 Hence (7= 0, and the equation 
 
 Ax"" + Cxy + ^3/2 = 1 
 
 becomes Ax"" + Bij'' = 1 . 
 
 The intercepts of the curve on the new axes are equal to 
 the semi-conjugate diameters. If we denote them by a' and 
 b', we have 
 
 ^i5 '-Vi 
 
 whence ^ = — -. B = -- » 
 
 and the equation ^^,2 1 ^^ ,2 ^ i 
 
 becomes a^ "^ 6^ ~ P'''] 
 
 This equation has the same form as the equation referred 
 to the axes of the curve ; whence it follows that formulas 
 derived from equation [30], by processes which do not pre- 
 suppose the axes of co-ordinates to be rectangular, hold true 
 ^vhen we employ as axes two conjugate diameters. 
 
 For example, the ecjuation of a tangent at the point (xi, ?/i), 
 referred to the semi-conjugate diameters a' and b', is 
 
 161. To find the conditions under ivhich an equation will 
 represent an ellipse when of the form 
 
 Ax" + By'^-^Dx + Ey-YF= 0. 
 
 If neither A nor B is zero, we mav wa-ite the equation 
 
THE ELLIPSE. 171 
 
 If we take a new origin with parallel axes at the point 
 (———-' ~qd)' ''^^^^ denote by A" the constant quantity 
 which forms the right side of the last equation, the equa- 
 tion becomes Ax^+Bj/''=K, 
 
 or ^x' + -^7/=^l; 
 
 x^ ?/^ 
 that is, H'^'k^^' 
 
 which we know (§ 135) represents an ellipse, provided K be 
 not zero and A, B, and -ST have like signs.,,, 
 
 K K * ■ 
 
 If the denominators — and — are both ncqative, it is clear 
 A B J ^ 
 
 that no real values of x and y will satisfy the equation : the 
 
 locus in this case is called an imaginary ellipse. 
 
 It is obvious that the two denominators — and — will have 
 
 A B 
 
 like signs when A and B have like signs ; and by comparing 
 the signs of the constants which enter into the value of K, it 
 appears that the common sign of the denominators will be 
 positive or negative, according as the sign of F is unlike or 
 like that of A and B. Hence an equation of the form 
 
 - Ax--^By'' + Dx-\-Ey-{-F-=0 
 
 (i.) will represent an ellipse, if A and B are neither of them 
 zero and agree in sign. 
 
 (ii.) The ellipse will he real or imaginary, according as the 
 sign of F is unlike or like that of A and B. 
 
 The axes of the ellipse are parallel to the axes of co-ordi- 
 
 nates, and the centre is the point f — — > 
 
 ^ V --^ 2i? 
 
 The major axis will be parallel to the axis of x or to the 
 axis of y, according as A is less than, or greater than, B. 
 
172 
 
 ANALYTIC GEOMETRY. 
 
 162. To find the locus of a point which 'moves so that the 
 ratio of its distances from a fixed point and a fixed straight line 
 is constant and less than unity. 
 
 Let e denote the constant ratio, 2p the distance from the 
 fixed point i^(Fig. 63) to the fixed line CE. Taking CE for 
 the axis of y, and the perpendicular to CE through F for the 
 axis of X, then from the definition of the locus 
 
 FP=exNP=ex. 
 
 Therefore we have the relation 
 
 {x — 2py-\-y'=e^x\ 
 
 or (l-e>' + y'-4;9;r + 4/ = (i (1) 
 
 Since we suppose e<\, the coefficients of x""- and y"^ are 
 both positive ; therefore the locus is an ellipse (§ 161). Com- 
 paring the coefficients of (1) with those of the first equation 
 of (§ 161), we obtain 
 
 A = l-e\ B = l, C=^-^p, i)-0, E=^p\ 
 
 -p 
 
 Therefore the centre of the locus is the point f ""-^ .^ . 
 
THE ELLIPSE. 173 
 
 Changing the origin to the centre, we obtain an equation 
 which may be written in the form 
 
 1 _ ./A 2 1 _ Ai 
 
 ) x' + -f = 1. (2) 
 
 2cp) ^{:icpr ^^ 
 
 By putting x and y successively equal to 0, we find for the 
 values of the semi-axes 
 
 2e» , 2e» 
 
 a = z. — ^,, b = , 
 
 1 - c' Vi - e^ 
 
 AVhence, by substitution in (2), ^ve get the equation in the 
 ordinary form 2 2 
 
 1 + 1 = 1. (3) 
 
 Since OD = -^ and FB ^- 2p, therefore 
 
 But the distance from the centre to the focus of the 
 ellipse is o 2 
 
 \ — &■ 
 
 Therefore the fixed point F coincides with the focus of the 
 
 ellipse. 
 
 Also -'= e, or the constant ratio e is equal to the eccen- 
 a 
 
 tricity of the ellipse as defined in § 128. 
 Whence an ellipse is often defined as 
 
 The locus of a point which moves so that the ratio of its dis- 
 tances frovi a fixed point and a fixed straight line is constant 
 and less than unity. 
 
 i^is called the Focus ; FN, the Directrix. 
 
 The symmetry of the curve with respect to the minor axis 
 shows that there is another focus and another directrix on the 
 other side of the minor axis, at distances from it equal respec- 
 tively to those of F and CE. 
 
174 
 
 ANALYTIC GEOMETRY. 
 
 163. To find the polar equation of the ellipse, the right-hand 
 focus being taken as tJte pole. 
 
 Fig. 64. 
 
 Let P be any poiDt (p, 0) of the curve. Then in the 
 triangle PFF' 
 
 FF'' ==^FF' + W + 2PF X FF^ XcosO. 
 
 But FF= p, FF' = 2c, and by the definition of the curve 
 FF' = 2a — p; therefore 
 
 (2a -py = p''-i-4:c'-t4:cp cos 6. 
 
 Reducing, and substituting aV for c-, we obtain 
 
 P = 
 
 a'' 
 
 •- _ a(l — €-} 
 a + c cos 6 ~~ 1 + e cos e 
 
 [38] 
 
 We may obtain this result more simply by using for the 
 focal radius FF the value (§ 131) 
 
 p = a — ex. 
 
 But the general value of x (0 being acute) is 
 
 X = c -\- p cos = ae -]- p cos 0. 
 
 By substituting this value of x we obtain the same polar 
 equation as before. 
 
THE ELLIPSE. 
 
 175 
 
 164. To find the area of an ellipse. 
 
 Divide the semi-major axis OA (Fig. 65) into any number 
 of equal parts, through any two adjacent points of division 
 J/, iV^ erect ordinates, and let the ordinate through M meet 
 
 the ellipse in P and the auxiliary circle in Q. Through P, Q 
 draw parallels to the axis of x, meeting the other ordinate in 
 P, jS, respectively. Then (§ 138) 
 
 area of rectangle MPP N _ MP _ b^ 
 area of rectangle MQSN MQ ~ a 
 
 And a similar proportion holds true for every correspond- 
 ing pair of rectangles. 
 
 Therefore, by the Theory of Proportions, 
 
 sum of rectangles in ellipse _ h 
 sum of rectangles in circle a 
 
 This relation holds true however great the number of rect- 
 angles. The greater their number, the nearer does the sum of 
 their areas approach the area of the elliptic quadrant in one 
 case, and the circular quadrant in the other. In other words, 
 these two quadrants are the limits of the sums of the two 
 
176 ANALYTIC GEOMETRY. 
 
 series of rectangles. Therefore, by the fundamental theorem 
 
 of limits, r ^^■ I- i i 7 
 
 ' area ot elliptic quadrant _ o 
 
 area of circular quadrant a 
 
 Multiplying both terms of the first ratio by 4, 
 
 area of the ellipse _ h 
 area of the circle a 
 
 But the area of the circle = tto^ ; therefore 
 
 area of the ellipse = trab, [39] 
 
 Ex. 35. 
 
 1. "What are the equations of the directrices (§ 162) ? 
 
 2. Prove that the polars of the foci are the directrices. 
 
 3. What is the equation of the polar of the point (5, 7) 
 with respect to the ellipse Ax^ + 9?/^= 36 ? 
 
 4. Prove that a focal chord is perpendicular to the line 
 Avhich joins its pole to the focus. In what line does the pole 
 
 lie? 
 
 6. Find the pole of the line Ax -f- Bi/ +(7=0 with respect 
 to th5 ellipse 5V -f t«Y'^ = a^^'^- 
 
 6. Each of the two tangents which can be drawn to an 
 ellipse from any point on its directrix subtends a right angle 
 at the focus. 
 
 7. The two tangents which can be drawn to an ellipse from 
 any external point subtend equal angles at the focus. 
 
 8. Find the slope vi of a diameter if the square of the 
 diameter is (i.) an arithmetic, (ii.) a geometric, (iii.) an har- 
 monic mean between the squares of the axes, 
 
 9. Given the length 2/ of a diameter, its inclination to 
 the axes, and the eccentricity ; find the major and minor axes. 
 
THE ELLirSE. 177 
 
 10. Tangents are drawn from (3, 2) to the ellipse a:^-f 4?/^~4. 
 Find the equation of the chord of contact, and of the line 
 which joins (3, 2) to the middle point of the chord. 
 
 11. Find the equation of a diameter parallel to the normal 
 at the point (xi, 3/1), the semi-axes being a and b. 
 
 12. Find the area of the rectangle whose sides are the two 
 segments into which a focal chord is divided by the focus. 
 
 13. What is the equation of a chord in the ellipse 
 13:r^ + lly^ ~ 1"13 which passes through (1,2) and is bisected 
 by the diameter 3a: — 2?/ = ? 
 
 14. In the ellipse 9 x' + SQ f = 324 find the equation of a 
 chord passing through (4, 2) and bisected at this point. 
 
 15. Write the equations of diameters conjugate to the fol- 
 lowing lines : 
 
 X — 1/ = 0, X -\- ^ = 0, ax = hi/, ay = hx. 
 
 16. Show that the lines 1x — y = 0, :r -f 3 ?/ = are con- 
 jugate diameters in the ellipse 2a;^ -f 3y''^ = 4. 
 
 17. If a', V are two semi-conjugate diameters, prove that 
 
 18. The area of the parallelogram formed by tangents 
 drawn through the ends of conjugate diameters is constant, 
 and equal to 4a6. 
 
 19. The diagonals of the parallelogram in No. 18 are also 
 conjugate diameters. 
 
 20. The angle between two semi-conjugate diameters is a 
 minimum, when they are equal. 
 
 21. The eccentric angles corresponding to equal semi-con- 
 jugate diameters are 45° and 135°. 
 
 22. The polar of a point in a diameter is parallel to the 
 conjugate diameter. 
 
178 ANALYTIC GEOMETRY. 
 
 23. Find the equations of equal conjugate diameters. 
 
 24. The length of a semi-diameter is I ; find the equation 
 of the conjugate diameter. 
 
 25. The angle between two equal conjugate diameters =: 
 60° ; find the eccentricity of the elli|)se. 
 
 26. Given a diameter, to construct the conjugate diameter. 
 
 27. To construct two conjugate diameters which shall con- 
 tain a given angle. 
 
 28. To draw a tangent to a given ellipse parallel to a given 
 straight line. 
 
 29. Given an ellipse ; to find by construction the centre, 
 foci, and axes. 
 
 30. Find the rectangular equation of the ellipse, taking the 
 origin at the left-hand vertex. 
 
 31. Find the polar equation of an ellipse, taking as pole 
 the left-hand focus. 
 
 32. Find the polar equation of the ellipse, taking the centre 
 as pole. 
 
 33. Discuss the form of the ellipse by means of its polar 
 equation. 
 
 34. If the centre of an ellipse is the point (4, 7), and the 
 major and minor axes are 14 and 8, find its equation, the axes 
 being supposed parallel to the axes of co-ordinates. 
 
 35. The equation of an ellipse, the origin being at the left- 
 hand vertex, is 2b x^ + ^\y^ — 450:^ ; find the axes. 
 
 36. If the minor axis = 12, and the latus rectum = 5, 
 what is the equation of the ellipse, the origin being taken at 
 the left-hand vertex ? 
 
THE ELLIPSE. 179 
 
 Find the centre and axes of the following ellipses : 
 
 37. 4.x'-{-i/-{-8x-2y + l = 0. 
 
 38. 9:i-"' + lGy^-3G.r-12Sy-f- 148 = 0. 
 
 39. 4.r'^ + 36y- + 36y = 0. 
 
 40. Find the eccentric angle <^ corresponding to the diam- 
 eter whose length is 2 c. 
 
 41. At the intersection of the ellipse h'x' + «y = ci^l>'' and 
 the circle x'^-j-y^ = ah tangents are drawn to both curves. 
 Find the angle between them. 
 
 42. How would you draw a normal to an ellipse from any 
 point in the minor axis ? 
 
 43.' Find the equation of a chord bisected at a point {h, k). 
 
 44. Prove that the length of a line drawn from the centre 
 to a tangent, and parallel to either focal radius of the point 
 of contact, is equal to the semi-major axis. 
 
 45. A circle described on a focal radius will touch the 
 auxiliary circle. 
 
 46. Find the locus of the intersection of tangents drawn 
 through the ends of conjugate diameters of an ellipse. 
 
 47. Find the locus of the middle point of the chord joining 
 the ends of two conjugate diameters. 
 
 48. Find the locus of the vertex of a triangle whose base 
 is the line joining the foci, and whose other sides are parallel 
 to two conjugate diameters. 
 
 49. Find the locus of the centre of a circle which passes 
 through the point (0, 3) and touches internally the circle 
 x'^y' = 2b. 
 
CHAPTER VII. 
 THE HYPERBOLA. 
 
 Simple Properties of the Hyperbola. 
 
 165. The Hyperbola is the locus of a point the difFerence of 
 whose distances from two fixed points is constant. 
 
 The fixed points are called the Foci, and a line joining any 
 point of the curve to a focus is called a Focal Eadius. 
 
 The constant difFerence is denoted by 2 a, and the distance 
 between the foci by 2(?. 
 
 The fraction - is called the Eccentricity, and is denoted by 
 a 
 
 the letter e. Therefore c = etc. 
 
 Since the difFerence of two sides of a triangle is always less 
 than the third side, we must have in the hyperbola 
 
 2a < 2(7, or a < c, or e>l. 
 
 166. To construct an hyperbola, having given the foci, and 
 the constant difference 2 a. 
 
 I. By Motion (Fig. 66). Fasten one end of a ruler to one 
 focus jP' so that it can turn freely about F'. To the other 
 end fasten a string. Make the length of the string less than 
 that of the ruler by 2 a, and fasten the free end to the focus F. 
 Press the string against the ruler by a pencil point P, and 
 turn the ruler about F'. 
 
 The point F will describe one branch of an hyperbola. 
 The other branch may be described in the same way by inter- 
 changing the fixed ends of the ruler and the string. 
 
THE IIYrEIlBOLA. 
 
 181 
 
 11. By Foinis (Fig. 67). Let F, F' be the foci ; then 
 
 FF'=2.c 
 Bisect FF' at 0, and from hiv off OA =-- OA' = a. 
 
 Then, 
 
 AA' 
 
 AF'-AF =2a + (c-a)-(c-a) = 2a. 
 A'F-A'F'=^2a-i-lc-a)-lc-a) = 2a. 
 Therefore A and A' are points of the curve. 
 
 Fig. 66. 
 
 Fig. 67. 
 
 In AA' produced mark any point F ; then describe two 
 arcs, the first with F as centre and AF as radius, the second 
 with F' as centre and A'F as radius; the intersections F, Q 
 of these arcs are points of the curve. By merely interchang- 
 ing the radii, two more points F, S may be found. 
 
 Proceed in this way till a sufficient number of points has 
 been obtained ; then draw a smooth curve through them. 
 
 Through draw FF' _L to FF' ; since the difference of 
 the distances of every point in the line FF' from the foci is 
 0, therefore the curve cannot cut the line FF'. 
 
 The locus evidently consists of two entirely distinct parts or 
 branches, symmetrically placed with respect to the line FF'. 
 
182 
 
 ANALYTIC GEOMETRY. 
 
 167. The point 0, half way between the foci, is the Centre. 
 
 The line AA' passing through the foci and limited by the 
 curve is the Transverse Axis. 
 
 The points A, A', Avhere the transverse axis meets the 
 curve, are called the Vertices. 
 
 The transverse axis is equal to the constant difference 2a, 
 and is bisected by the centre (§ IGG). 
 
 Fig. 68. 
 
 The line £B' passing through perpendicular to AA' 
 does not meet the curve (§ 166) ; but if B, B' are two points 
 whose distances from the two vertices A, A' are each equal to 
 c, then BB' is called the Conjugate Axis, and is denoted hf2b. 
 
 Since A AOB = A AOB', OB = OB'=^b; that is, the' con- 
 jugate axis is bisected by the centre. 
 
 In the triangle AOB, OA = a, OB = b, AB = c; hence 
 
 c' = a'+b\ 
 
 The chord passing through either focus perpendicular to 
 the transverse axis is called the Latus Eectum, or Parameter. 
 
 Note. Since a and h are equal to the legs of a right triangle, a 
 may be either greater or less than 6; hence the terms ''major'' and 
 ''minor'' are not appropriate in the hyperbola. 
 
THE HYPERBOLA. 183 
 
 168. To find the equation of the hyperbola, having given the 
 foci, and the constant difference 2a. 
 
 By proceeding as in the case of the ellipse (§ 131), and sub- 
 stituting h- for c- — a^, we obtain 
 
 g-|! = l- [40] 
 
 The lengths r, r' of the focal radii for any point (x, y) are 
 r = ex — a and r' = ex-j- a. 
 
 The equations of the ellipse and the hyperbola differ only 
 in the sign of b-. The equation of the hyperbola is obtained 
 from that of the ellipse by changing +^^ to —b-. In general 
 
 Any formula deduced from the equation of the ellipse is 
 changed to the corresponding formula for the hyperbola by 
 merely changing -{-V to — 5^, or b to 5 V— 1. 
 
 169. A discussion of equation [40] leads to the following 
 conclusions : 
 
 (i.) The curve cuts the axis of x at the two real points 
 (a, 0) and (-a, 0). 
 
 (ii.) The curve cuts the axis of y at the two imaginary 
 points (0, 6V^) and (0, -Z>V^). 
 
 (iii.) Xopart of the curve lies between the straight lines 
 a: = -f a and a; = — a. 
 
 (iv.) Outside these lines the curve extends without limit 
 both to the right and to the left. 
 
 (v.) The greater the abscissa, the greater the ordinate. 
 
 (vi.) The curve is symmetrical with respect to the axis of x. 
 
 (vii.) The curve is symmetrical with respect to the axis of y. 
 
 (viii.) Every chord which passes through the centre is bi- 
 sected by the centre. This explains why the point half way 
 between the foci is called the centre. 
 
 The two distinct parts of the curve are called the right-hand 
 and the left-hand branches. 
 
184 
 
 ANALYTIC GEOMETRY. 
 
 170. An hyperbola who.-o transverse and conjugate axes 
 are equal is called an Equilateral Hyperbola. Its equation is 
 
 cc- — 2/- = «"^. [41] 
 
 The equilateral hyperbola bears to the general hyperbola 
 the same relation that the auxiliary circle bears to the ellipse. 
 
 Fig. 69. 
 
 i 
 
 171. The hyperbola which has BB' for transverse axis, and 
 A A' for conjugate axis, obviously holds the same relation to 
 the axis of y that the hyperbola which has AA' for transverse 
 axis and J3£' for conjugate axis holds to the axis of .i\ 
 
 Therefore its equation is found by simply changing the 
 signs of a' and P in [40], and is 
 
 
 = 1, or 
 
 = -1. 
 
 The two hyperbolas are said to be Conjugate. 
 
THE IIYrEIlBOLA. 185 
 
 172. The straight line y = tyix, passing through the centre 
 of the hyperbola —^ — '77^ = !, meets the curve in two points, 
 the abscissas of which are 
 
 -{-ah — ah 
 
 Hence the points will be real, imaginary , or situated at 
 infinity, as h'- — ni^a'^ is positive, negative, or zero ; that is, as 
 
 in is less than, greater than, or equal to -• 
 
 The same line, y = mx, will meet the conjugate hyperbola 
 
 -^ — ^2 ~ ~ -^ i^ ^^^'^ points, whose abscissas are 
 
 — + ah _ — ah 
 
 ^rd^d^ — h'^ ^m^d^ — h'^ 
 
 Hence these points will be imaginary, real, or situated at 
 
 infinity, as mi is less than, greater than, or equal to — 
 
 a 
 
 Whence 
 
 If a straight line through the centre meet an hyperbola in 
 imaginary points, it ivill meet the conjugate hyperbola in real 
 points, and vice versa. 
 
 173. An Asymptote is a straight line which passes through 
 finite points, and meets a curve in two points at infinity. 
 We see from § 172 that the hyperbola 
 
 a' b' 
 has two real asymptotes passing through the centre of the 
 
 curve, and having for their equations y~-{--x and y~ x ; 
 
 or, expressed in one equation, 
 
 052 y2 
 
 b' 
 
 O. [42] 
 
186 ANALYTIC GEOMETRY. 
 
 Ex. 36. 
 
 What is the equation of an hyperbola, if 
 
 1. Transverse axis = 16, conjugate axis = 14 ? 
 
 2. Conjugate axis = li2, distance between foci = 13 ? 
 0. Distance between foci = twice the transverse axis ? 
 
 4. Transverse axis = 8, one point = (10, 25) ? 
 
 5. Distance between foci =^ 2(?, eccentricity = V2 ? 
 
 6. Prove that the latus rectum of an hyperbola is equal to 
 2^ 
 
 7. The equation of an hyperbola is 9^:^ — 16?/^ = 25 ; find 
 the axes, distance between the foci, eccentricity, and latus 
 rectum. 
 
 8. AVrite the equation of the hyperbola conjugate to the 
 hyperbola 9x^ — 16y^ = 25, and find its axes, distance between 
 its foci, and its latus rectum. 
 
 9. If the vertex of an hyperbola bisects the distance from 
 the centre to the focus, find the ratio of its axes. 
 
 10. Prove that the point (x, y) is without, on, or within the 
 
 hyperbola, according as —^—-.^ — 1 is negative, zero, or i^osi- 
 tive. 
 
 11. Find the eccentricity of an equilateral hyperbola. 
 
 12. The distance of any point of an equilateral hyperbola 
 from the centre is a mean proportional between its focal radii. 
 
 13. The asymptotes of an hyperbola are the diagonals of 
 the rectangle CD FF (Fig. 69, p. 184). 
 
 14. Find the foci and the asymptotes of the hyperbola 
 162;^-9y^ = 144. 
 
THE HYPERBOLA. 187 
 
 15. The asymptotes of an equilateral hyperbola are perpen- 
 dicular to each other. Hence the equilateral hyperbola is 
 also called the rectangular hyperbola. 
 
 16. An hyperbola and its conjugate have the same asymp- 
 totes. 
 
 17. Find the length of a perpendicular dropped from the 
 focus to an asymptote. 
 
 Tangents and Normals. 
 
 Note. The results stated in the following six sections are established 
 in the same way as the corresponding propositions relating to tangents 
 and normals to'afi ellipse. We shall, therefore, omit the proofs. 
 
 174. The equation of the tangent at {x^, y^ is 
 
 ^-^ = 1. [43] 
 
 175. The equation of the normal at (xi, iji) is 
 
 y^x , oc, 1/ x,y, (rt2 + &2) 
 
 ~TT "I ~ = TT-y • V^^\ 
 
 b' a a-b- ^ ^ 
 
 176. The subtan<Tent = ^, the subnormal = — 5-^- 
 
 177. The tangent and the normal at any point of an hyper- 
 bola bisect the angles formed by the focal radii of the point 
 (> 144). 
 
 178. The straight line whose equation is y = mx -. 
 is a tangent for all values of m (§ 146). 
 
 179. The equation of the director circle of an hyperbola is 
 x^j^y'^ = a'- h' (§ 147). 
 
188 ANALYTIC GEOMETRY. 
 
 Ex. 37. 
 
 1. Find the equations of tangent and normal to the hyper- 
 bola 16 x"^ — 9 y'^ — 112 at the point of contact (4,4). Also 
 find the lengths of the subtangent and the subnormal. 
 
 2. Show that in an equilateral hyperbola the subnormal is 
 equal to the abscissa of the point of contact. 
 
 3. The equations of the tangent and the normal at a point 
 of an equilateral hyperbola are 5:r — 4y = 9, 4.r -f 9?/ = 56. 
 What is the equation of the hyperbola, and what are the 
 co-ordinates of the point of contact ? 
 
 4. For what points of an hyperbola is the subtangent equal 
 to the subnormal ? 
 
 5. To draw a tangent and a normal to an hyperbola at a 
 given point of the curve. 
 
 6. If an ellipse and an hyperbola have the same foci, prove 
 that the tangents to the two curves drawn at their points of 
 intersection are perpendicular to each other. 
 
 7. Prove that the asymptotes of an hyperbola are tangent 
 to it at infinity. 
 
 8. Prove that the length of a normal in an equilateral 
 hyperbola is equal to the distance of the j)oint of contact from 
 the centre. 
 
 9. Find the distance from the origin to the tangent through 
 the end of the latus rectum of the equilateral hyperbola 
 x^ — 'if' = a^. 
 
 10. What condition must be satisfied in order that the 
 
 X II ?/'" ?/'" 
 
 straight line h - = 1 niay touch the hvperbola ^ — Vr, =1? 
 
 ^ m n ^ ' ^ a b' 
 
 11. When is the director circle of an hyperbola imaginary ? 
 
 12. Find the locus of the foot of a perpendicular dropped 
 from the focus of an hyperbola to a tangent. 
 
THE HYPERBOLA. 189 
 
 Ex. 38. (Review.) 
 
 1. The ordinate through the focus of an hyperbola, pro- 
 duced, cuts the asymptotes in P and Q. Find PQ and the 
 distances of P and Q from the centre. 
 
 2. In the hyperbola 9:c^ — IGy^ = 144 what are the focal 
 radii of the points whose common abscissa is 8 ? What other 
 points have equal focal radii ? 
 
 3. What relation exists between the surti of the focal radii 
 of a point of an hyperbola and the abscissa of the point ? 
 
 4. Prove that in the equilateral hyperbola every ordinate 
 is a mean proportional between the distances of its foot from 
 the vertices of the curve. Hence find a method of construct- 
 ing an equilateral hyperbola when the axes are given. 
 
 5. In the equilateral hyperbola the distance of a point from 
 the centre is a mean proportional between its focal radii. 
 
 6. In the equilateral hyperbola the bisectors of the angles 
 formed by lines draw^n from the vertices to any point of the 
 curve are parallel to the asymptotes. 
 
 7. If e, e' are the eccentricities of two conjugate hyperbolas, 
 
 8. Through the positive vertex of an hyperbola a tangent is 
 drawn. In what points does it cut the conjugate hyperbola ? 
 
 9. The sum of the reciprocals of two focal chords perpen- 
 dicular to each other is constant. 
 
 10. Through the foot of the ordinate of a point in an equi- 
 lateral hyperbola a tangent is drawn to the circle described 
 upon the transverse axis as diameter. What relation exists 
 between the lengths of this tangent and the ordinate of the 
 point ? 
 
190 ANALYTIC GEOMETRY. 
 
 11. In an equilateral hyperbola find the equations of tan- 
 gents drawn from the positive end of the conjugate axis. 
 
 12. From what point in the conjugate axis of an hyperbola 
 must tangents be drawn in order that they may be perpen- 
 dicular to each other? 
 
 13. What condition must be satisfied that a square may be 
 constructed whose sides shall be parallel to the axes of an 
 hyperbola and whose vertices shall lie in the curve? 
 
 14. Find the equation of the chord of the hyperbola 
 16a:' — 9?/' = 144 which is bisected at the point (12,3). 
 
 15. Find the equation of a tangent to the hyperbola 
 16a;' — 9?/' = 144 parallel to the line ?/ = 4.r — 3. 
 
 16. Determine the points in an hyperbola for ^vhich the 
 length of the tangent is twice that of the normal. 
 
 17. A chord of an hyperbola which touches the conjugate 
 hyperbola is bisected at the point of contact. 
 
 SUPPLEMENTARY PROPOSITIONS. 
 
 Note. Many of the following propositions are closely analogous to 
 propositions already established for the ellipse ; hence the proofs are 
 omitted, and references given to the chapter on the ellipse. 
 
 180. Two tangents can he drawn to an hyperbola from any 
 
 point (A, h) ; and they will be reed, coincident, or imaginary, 
 
 as the point is without, on, or within the curve (^ 148). 
 
 A' B 
 
 The two tancrents will be real if — - — 1 is negative. 
 
 a' y^ 
 
 Likew^ise two real tangents can be drawn from (A, Ic) to the 
 conjugate hyperbola if — 2~72"^-^ ^^ negative. 
 
THE HYPERBOLA. 191 
 
 Hence it follows that if — — -, lias any value between 
 a- 0^ 
 
 and — 2, a pair of real tangents can be drawn from (A, Ic) to 
 each hyperbola, 
 
 181. The straight line passing through the points of contact 
 of the two tangents drawn to an hyperbola from any point P 
 is called the Polar of P with respect to the hyperbola ; and P 
 is called the Pole of this straight line. 
 
 The polars of the foci are called the Directrices. 
 The equation of the polar of the point (h, k) is 
 
 ^-^ = 1. (§149) 
 
 The propositions in §§ 80-82 hold true for poles and polars 
 with respect to an hyperbola, and may be proved in the same 
 way. 
 
 182. The locus of the middle points of chords parallel to 
 
 the line ?/ = 7?ix is 72 
 
 l/ = ~ (§152) 
 
 a'm 
 
 This locus is called a Diameter of the hyperbola. 
 Every diameter passes through the centre. 
 The chords bisected by a diameter are called the Ordinates 
 of the diameter. 
 
 183. If m' is the slope of the diameter, bisecting chords 
 parallel to the line y = mx, then 
 
 in?n=^J [45] 
 
 and from the symmetry of this equation we infer that 
 
 If one diameter bisects all chords parcdlel to another, the 
 second diameter will bisect all chords parallel to the first. 
 
 Two diameters drawn so that each bisects all chords par- 
 allel to the other are called Conjugate Diameters (§ 153). 
 
192 ANALYTIC GEOMETRY. 
 
 184. Since the product of the slopes of two conjugate diam- 
 eters is ip- 
 
 it follows that vi and m' must agree in sign ; therefore 
 Two conjugate diameters lie in the savie quadrant. 
 
 Also, if m in absolute magnitude is less than -, then m' 
 
 h ^^ 
 
 must be greater than — But the slope of the asymptotes is 
 
 h ^^ 
 
 equal to ± - Therefore 
 a 
 
 Two conjugate diameters lie on ojyposite sides of the asymp- 
 tote in the same quadrant ; and of two conjugate diam,eters, one 
 Tneets the curve in real points and the other in imaginary points 
 (§ 172). 
 
 185. The length of a diameter which meets the hyperbola 
 in real points is the length of the chord between these points. 
 
 If a diameter meets the hyperbola in imaginary points, 
 that is, does not meet it at all, it will meet the conjugate 
 hyperbola in real points (§ 172) ; and its length is the length 
 of the chord between these points. 
 
 A comparison of the equations of two conjugate hyper- 
 bolas will show that if a diameter meet one of the hyper- 
 bolas in the imaginary point (AV— 1, Z*V— 1), it will meet 
 the other in the real point (A, h) ; hence the length of the 
 semi-diameter will be VA^ -f- U\ 
 
 186. The equations of an hyperbola and its conjugate differ 
 only in the signs of a?- and h^. But this interchange of signs 
 does not effect the equation 
 
 m??i' = — . Therefore 
 d^ 
 
 If two diameters are conjugate with respect to one of two con- 
 
 jugate hxjperholas, they loill he conjugate with respect to the other. 
 
THE IIYrERBOLA. 
 
 193 
 
 Thus, let POP and QOQ' (Fig. 70) be two conjugate di- 
 ameters. Then FOP bisects all chords parallel to QOQ' that 
 lie luiihin the branches of the original hyperbola and between 
 the branches of the conjugate hyperbola ; and QOQ' bisects all 
 chords parallel to POP' that lie within the branches of the 
 conjugate hyperbola and between the branches of the original 
 hyperbola. 
 
 % 
 
 1C/ 
 
 By 
 
 'B 
 
 M 
 
 P? 
 
 [B' 
 
 Fig. 70. 
 
 From the above theorem it follows immediately that 
 
 If a straight line nncet each of two conjugate hyperbolas in 
 two real points, the two portions of the line contained between 
 the liyperbolas are equal {thus, BD = B'D', Fig. 70). 
 
 187. TJie tangent drawn through the end of a diameter is 
 parallel to the conjugate diameter (§ 154). 
 
194 ANALYTIC GEOMETRY. 
 
 188. Having given tlie end {xi, y{) of a diameter, to find the 
 end {x., 1/2) of ike conjugate diameter. 
 
 If (.rj, ?/i) is a point of the hyperbola 
 
 1-f!=i. (1) 
 
 a^ 
 
 we know (§ 172) that (5:2, 3/2) will be a point of the conjugate 
 
 hyperbola 2 ,,2 
 
 --•^ = -1. (2) 
 
 If the equation of the diameter through {x^, 3/1) be 
 y = mx, 
 
 then m = — 
 
 Xi 
 
 If the equation of the diameter through (x^, y^ be 
 y == m^x, 
 
 then nn^ = -^ (§ 182). 
 
 am 
 
 Hence the equation y = 7n'x may be written 
 
 y-^- (3) 
 
 The values of x^ and 3/2 are now found by solving equations 
 (2) and (3), and are 
 
 a h 
 
 -2/1, y-i = ± -^*i. 
 
 b^ 
 
 a 
 
 The positive signs belong to one end, and the negative signs 
 to the other end, of the conjugate diameter. 
 
 189. If' deyiote the angle formed by two conjugate scini- 
 diameters, a' and b', their lengths, then sin 6 = — — (§ 15S). 
 
THE IIYPEEBOLA. 195 
 
 190. To fuid the equation of an hyperbola referred to any 
 pair of conjugate diameters as axes of eo-ordinates. 
 
 If a', b' denote the two semi-diameters, the required equa- 
 tion is ,2 -i 
 
 ^ ^ =1. (1) 
 
 a" b'' ^^ 
 
 The method of solving the problem is the same as that used 
 in § IGO ; but since the intercept of the curve on the axis of y 
 is imaginary, § 167, the sign of ^ in § 160 will be negative. 
 
 Since the form of equation (1) is the same as that of the 
 equation referred to the axes of the curve, it follows that all 
 formulas which have been obtained without assuming the axes 
 of co-ordinates to be at right angles to each other hold good 
 when the axes of co-ordinates are any two conjugate diam- 
 eters. For example, the equation of the asymptotes of the 
 hyperbola re^Dresented by equation (1) is 
 
 ^--^--=0. 
 a'' ¥' 
 
 191. The tangents through the ends of two conjugate diam- 
 eters meet in the asymptotes. 
 
 Let the ends of the diameters be the points {x^, y^ and 
 
 fe, 2/2) ; i^Q"^ the equations of tangents through {x^, y^ and 
 
 (^2, 2/2) will be x^ _y{y__ .,^ 
 
 a' ~ b' ' ^ ^ 
 
 and ^_^=:_1. (2) 
 
 a b^ ^ ^ 
 
 The asymptote y^=-'x meets the tangent represented by 
 (1) in the point / a:b ah" 
 
 })x^— ay^^bx^ — ay,^ 
 and the tangent represented by (2) in the point 
 / ccb ab"- 
 
 \ay2 — bx2 ay-i — bx.^, 
 
196 ANALYTIC GEOMETRY. 
 
 But from § 188, hx^ = ay^ and aij2 = bxi. 
 
 Therefore ay2 — hx.j, = hx^ — ayy. 
 
 Hence we see that the two jDoints of meeting coincide. 
 
 192, To find under what condition an equation will repre- 
 sent an hyperbola ivhen of the form 
 
 Ax' + ^y + Dx -\-Ey+F=0. 
 
 If neither A nor B is zero, the equation may be w^ritten 
 
 By proceeding as in § 161, we find that the equation repre- 
 sents in general an hyperbola if A and B are neither of them 
 zero, and have unlike signs. 
 
 The axes of the hyperbola are parallel to the axes of x and y, 
 
 f T) y 
 
 and the centre is the point f 7' - 
 
 193. If a straight line cut an hyperbola and its asymptotes 
 the p)ortions of the line intercepted between the curve and its 
 asymptotes are equal. 
 
 Let CC^ (Fig. 71) be the line meeting the asymptotes in C, C" 
 and the curve in B, B\ and let the equation of the line be 
 
 y = ')nx-\-c. (1) 
 
 Let il/be the middle point of the chord BB' ; then (§ 182) 
 the equation of the diameter through m is 
 
 y = ^- (2) 
 
 am 
 
 By combining equation (1) with the equations of the asymp- 
 totes, we obtain the co-ordinates of the points (7 and C" ; taking 
 
THE IIYPERLOLA. 
 
 197 
 
 the half-sum of these values, we get for the co-ordinates of the 
 point halfway between (7 and C" the values 
 
 Dj 
 
 M 
 
 X 
 
 P? 
 
 KB' 
 
 Q' 
 
 Fig. 71. 
 
 These values satisfy equation (2) ; therefore the point half 
 way betw^een Cand C coincides with M; therefore MC^MC. 
 And since 3IB = MB\ therefore BC=B'C\ 
 
 Let CC^ be moved parallel to itself till it becomes a tangent 
 at P, meeting the asymptotes m R, 8\ then the points B, B' 
 coincide at P, and we have PB = P8. Hence 
 
 The portion of a tangent intercepted by the asymptotes is 
 bisected by the point of contact. 
 
198 
 
 ANALYTIC GEOMETRY. 
 
 194. To find the equation of an lujijerhola referred to the 
 asymptotes as axes of co-ordinates. 
 
 Fig. 72. 
 
 Let the lines OB, OC (Fig. 72) be the asymptotes, A the 
 vertex of the curve, and let the angle AOC=' a. 
 
 Let the co-ordinates of any point P of the curve be x, y 
 when referred to the axes of the curve, and x\ y' when referred 
 to OB, OC 9.S axes of co-ordinates. 
 
 Draw F3I± to OA, FNW to OC; then 
 
 x = 03f, y = MP, x'^ON, y' = NP. 
 x = ON cos a -j- NP cos a = (.r' + y') <^os a, 
 y = NP sin a — ON sin a = (y' — x^) sin a. 
 
 Hence, by substitution [40], we obtain 
 
 {x' -f y^y cos'^ a (?/' — a-')- sin^ a 
 
 But from the relation tan a 
 
 1. 
 
 we have 
 
 sm a 
 
 cos* a 
 
 a' + ¥ a' -f b' 
 
 Substituting these values, and dropping accents, we have 
 
 4x?/ = a- + 6-. [46] 
 
THE HYPERBOLA. 199 
 
 195. The following method of showing that an hyperbola 
 has asymptotes, and finding their equations, is more general 
 than the method given in §§ 172, 173. 
 
 The abscissas of the points where the straight line y = mx-\-c 
 meets an hyperbola are found by solving the equation 
 
 a;" {mx + <^)^ _ i 
 
 Ir — 7?i^a^ 2 2 inc 1/ + c^ 
 
 x^ : — X— ' 
 
 (1) 
 
 ieb' b' b' 
 
 This equation is identical with the equation 
 
 Ax' + 2Bx-^C=0 (2) 
 
 c A b^' — ni^a? T) r>ic ^ b"^ '\- c^ 
 
 it Ji = ; > 1j — — -> O = ; 
 
 d'h' b' b' 
 
 If Xi, x-i are the roots of (2), 
 
 ^ _ -B + ^B'-AC C 
 
 ^ -B--^~B'-AC 
 
 -B ~-^B'- AC C_ 
 
 ^ -B + VB'-AC 
 
 
 c 
 
 If ^ = and also ^ = 0, we have ^^i = .Tg = — = co ; and 
 
 the line y ^^ mx -{- c w^ll meet the curve in two points at 
 infinity. 
 
 1( A = 0, m — zt — li B = 0, c = 0. Therefore there are 
 
 , " ■ * , b 
 two asymptotes ; their equations are ?/ = ~x and y = x. 
 
 If only A = 0, then ni = rb -< the straight line is 2'>arallel 
 
 to an asymptote, and 
 
 ar, = — — =■■ ^^^— J X2^= oo. Therefore 
 
 2B 2mc 
 
 A straight line parallel to an asymptote meets the curve in 
 one finite point and in one point at infinity. 
 
200 
 
 ANALYTIC GEOMETRY. 
 
 196. To find the locus of a point which moves so that its 
 distance from a fixed point bears to its distance from a fixed 
 straight line a constant ratio greater than unity. 
 
 Fig. 73. 
 
 Let e = the constant ratio, and 2p = the distance from the 
 
 fixed point F (Fig. 73) to the fixed line DN. Choosing for 
 
 axes the line drawn through i^ perpendicular to DN, and the 
 
 fixed line DN, we obtain the same equation as that found in 
 
 § 162, 
 
 ^ ' (l-eV + y2-4;?:r+4/-0. (1) 
 
 But since e is greater than 1, the equation now will repre- 
 sent an hyperbola (§ 195), the centre of which is at the 
 
 2p 
 
 (2) 
 
 point [ T " J 
 
 Transferring the origin to the centre 0, we get 
 
 2ep) ^{2epy 
 
 .r 
 
 1. 
 
THE IIYl'ERBOLA. 201 
 
 Putting a = -,^-^2' h = ^^?-- (2) becomes 
 -'- ^ we — 1 
 
 the equation of an hyperbola in the ordinary form. 
 
 It may be shown, as in § 162, that the fixed point F coin- 
 cides with the focus of the curve, and that the constant ratio 
 e is equal to the eccentricity. 
 
 This locus is often taken as the definition of an hyperbola, 
 F being called the Focus, and the fixed line DN the Directrix. 
 
 The symmetry of the curve with respect to the conjugate 
 axis shows that there are two foci and two directrices sym- 
 metrically placed with respect to the conjugate axis. 
 
 197. To find the polar equation of an hyperhola, the left- 
 hand focus being taken as pole. 
 
 For a point in the right-hand branch, the distance to the 
 pole, which is the remote focus, is (§ 168) 
 
 p = ex -}- a, 
 
 the distance x being reckoned from the centre. 
 
 Now X = p cos 6 — c = p cos — ae. 
 
 Whence, by substitution and reduction. 
 
 e cos — 1 
 
 [47] 
 
 Ex. 39. 
 
 1. "What is the polar of the point (— 9, 7) with respect to 
 the hyperbola 7^* - 12?/^ = 112 ? 
 
 2. Find the equations of the directrices of an hyperbola. 
 
 3. Find the angle formed by a focal chord and the line 
 which joins its pole to the focus. 
 
^\'^2 Vxlh n. 
 
 202 ANALYTIC GEOMETRY. 
 
 4. Find the pole of the line Ax-\-By-\- C with respect to 
 an hyperbola. 
 
 5. Find the polar of the right-hand vertex of an hyperbola 
 with respect to the conjugate hyperbola. 
 
 6. Find the distance from the centre of an hyperbola to 
 the point where the directrix cuts the asymptote. 
 
 7. If (.Tx, ?/i) and (^^2, 1/2) are the ends of two conjugate 
 diameters, then 
 
 8. The equation of a diameter in the hyperbola 25a'^— 16y^ 
 = 400 is Zy = x. Find the equation of the conjugate diam- 
 eter. 
 
 9. In the hyperbola 49 rr^ — 4y- = 196, find the equation of 
 that chord which is bisected at the point (5, 3). 
 
 10. Find the length of the semi-diameter conjugate to the 
 diameter y = 3.r in the hyperbola ^x^ — 4?/^ = 36. 
 
 11. The area of the parallelogram formed by drawing tan- 
 gents through the ends of two conjugate diameters is constant, 
 and equal to 4aZ>. 
 
 12. If a\ h' are the lengths of two conjugate semi-diameters, 
 then ^,r2_5r2^^2_ j2 
 
 13. Prove that PQ (Fig. 70) is parallel to one asymptote 
 and bisected by the other. 
 
 14. An asymptote is its own conjugate diameter. 
 
 15. The conjugate diameters of an equilateral hyperbola 
 are equal. 
 
 16. Having given two conjugate diameters in length and 
 position, to find by construction the asymptotes and the axes. 
 
 17. To draw a tangent to an hyperbola from a given point. 
 
THE HYPERBOLA. 203 
 
 18. Find the equation of a conjugate hyperbola referred to 
 its asymptotes as axes. 
 
 19. Find the equation of a tangent at any point (x^, 3/1) of 
 the hyperbola 4 .ry = a^ + ^^ 
 
 20. Find the equation of an hyperbola, taking as the axis 
 
 ^ (i.) the tangent through the left-hand vertex ; 
 (ii.) the tangent through the right-hand vertex. 
 
 21. Trace the form of an hyperbola by means of the polar 
 equation, p. 201. 
 
 22. Find the polar equation of an hyperbola, taking the 
 right-hand focus as pole. 
 
 23. Find the polar equation of an hyperbola, taking the ■ 
 centre as pole. 
 
 24. Show that the equation 
 
 a:--?/'-2p;-4y + l = 
 represents an hyperbola. Find its centre and axes, and con- 
 struct roughly the curve. 
 
 25. The distance from a fixed point to a fixed straight line 
 is 10. Find the locus of a point which moves so that its dis- 
 tance from the fixed point is always twice its distance from 
 the fixed line. 
 
 26. Through the point (— 4, 7) a straight line is drawn to 
 meet the axes of co-ordinates, and then revolved about this 
 point. Find the locus of its middle point. 
 
 27. A straight line has its ends in two fixed perpendicular 
 lines, and forms with them a triangle of constant area «^ 
 Find the locus of the middle point of the line (see § 000). 
 
 28. The base a of a triangle is fixed in length and position, 
 and the vertex so moves that one of the base angles is always 
 double the other. Find the locus of the vertex. 
 
CHAPTER VIII. 
 LOCI OF THE SECOND ORDER. 
 
 198. The locus represented by an equation of the second 
 degree is called a Locus of the Second Order. 
 
 We have seen, in the preceding chapters, that the circle, 
 parabola, ellipse, and hyperbola are loci of the second order. 
 We now propose to inquire whether there are other loci of 
 the second order besides the four curves just named; in other 
 words, to find what loci may be represented by equations of 
 the second degree. 
 
 For this purpose we shall write the general equation of the 
 second degree in the form 
 
 Ax' + ^/ + Ox2j + Dx+i:i/-\-F=0, (1) 
 
 and shall assume that the axes of co-ordinates are rectangular. 
 This assumption will in nowise diminish the generality of 
 our conclusions ; for if the axes be supposed oblique, we can 
 change them to rectangular axes, and this change will not 
 alter the degree of the equation or the nature of the locus 
 which it represents (§ 100). 
 
 199. If we suppose the coefficients of equation (1) to be 
 susceptible of all values including zero, we see that (1) in- 
 cludes the general equation of the first degree as a special 
 case when A = B = C= 0. 
 
 But even if no one of the coefficients is zero, they may 
 stand in such a relation to one another that the equation can 
 be resolved into two linear factors, and therefore represents 
 straight lines, real or imaginary. 
 
LOCI OF THE SECOND ORDER. 205 
 
 In order to find what this relation is, let us solve (1) with 
 resj^ect to one of the variables. Choosing y for this purpose, 
 we obtain n ^ \p \ 
 
 y = --4^±2l^VSM^l^Ti^ (2) 
 
 where 
 
 L = C''-^AB, M=1{CE--2BB), N=E''-^BF. 
 
 If Lx^ A^Mx-\-N be the square of a binomial of the form 
 &x + T, then the value of y may be written 
 
 Cx^E Sx + T 
 
 y = — ■ — ■■ ± — . 
 
 -^ 2B 2B 
 
 and the locus of (1) will in general be a pair of straight lines. 
 Xow, from Algebra, we know that the condition that 
 Bx'^ + Mx + iV should be a perfect square is 
 
 ]IP-4:BjV=0; 
 or, substituting the values of B, M, and iV, 
 
 (CE - 2BBy - (C - ^AB) (E' - 4.BF) = 0, 
 
 or F(C'~4:AB)+AE'+BB'-CBE=0. (3) 
 
 The quantity on the left-hand side of equation (3) is usually 
 denoted by A, and is called the Discriminant of equation (1). 
 And we may conclude that (1) represents straight lines (real 
 or imaginary) whenever A — 0. 
 
 200. In order to simplify the form of equation (1), let us 
 change the origin to the point (A, k), and then so choose the 
 values of h and k that the terms involving the first powers of 
 X and y will vanish. Making the change by substituting in 
 (1) x-^ h for X, and y-f k for ?/, we find that the coefficients 
 A, B, and C remain unaltered, and we may write the trans- 
 formed equation 
 
 Ax' -f By' 4- Cxy + B'x i-E'y = B (4) 
 
206 ANALYTIC GEOMETRY. 
 
 where D' = 2^A -f- a + A 
 
 E =-[Ah^ + BU'-^Chh + Dh + Eh-\-F]. 
 
 The values of h and Ic which will make i>' and E vanish 
 are evidently found by solving the equations 
 
 2Ah+Ck+n = 0, 
 2EIc-\-Ch + E = 0, 
 
 4AE-C" 4:A£-C' 
 
 The value of E can now be reduced to a form very easily 
 remembered : 
 
 E = - [Ah' + EIc' -f Chk + Eh+ Ek + E] 
 
 = - ^[{2.Ah+Ck+D)h-\-{2Bk-^Ch+E)k^nh^Ek+2E'] 
 = - ^{D'h •+ E'k + Eh + Ek + 2E) 
 ■=-i(Eh + Ek + 2E) 
 
 ^ , 2BE' - CEE+ 2AE'-CEE+ 2E(C'-4:AB) 
 ' C'-4:AB 
 
 2' 
 
 where ^ = 4:AB~C\ 
 
 Equation (4) may now be written 
 
 Ax'+By' + Cx7/=E. (5) 
 
 From the form of (5) we see that if (x, y) be a point of the 
 locus, so also is {—x, —y) ; that is, the new origin is a point 
 so placed that it bisects every chord passing through it. A 
 point having this property is called a Centre of a locus of 
 the second order. 
 
 The values of h and k are evidently single ; hence a locus 
 of the second order cannot have more than one centre. 
 
 The values of h and k are finite, provided 2 or 4:AB — C 
 is not zero. If, however, ^ = 0, the values of h and k become 
 
LOCI OF THE SECOND ORDER. 207 
 
 infinite or indeterminate. In this case a change of origin to 
 the centre is obviously impossible, and a different method of 
 reduction must be found. 
 
 Hence it will be convenient to divide loci of the second 
 order into two classes : those which have a finite centre, and 
 those which do not. The ellipse and the hyperbola belong 
 to the first class ; the parabola, to the second class. 
 
 The class to which the locus of a given equation belongs 
 is ascertained by seeing whether the value of 2, namely, 
 AAB — C'\ is or is not equal to zero; on this account 2 
 may be called the Criterion of the general equation of the 
 second degree. 
 
 CLASS I. :S NOT ZERO. 
 
 201. Equation (5) is the general equation of loci of the 
 second order wdiich have a centre, referred to the centre as 
 the origin of co-ordinates. 
 
 If in equation (5) we place a: = 0, we obtain two values of 
 y equal in magnitude and opposite in sign. Since the axis of 
 1/ is not limited as to direction, we infer that cveri/ chord pass- 
 ing through the centre is bisected at the centre. Hence a chord 
 passing through the centre is called a Diameter. 
 
 We can get rid of the term involving xg by another change 
 of axes. For this purpose we must change the direction of 
 the axes through an angle 6, keeping the origin unaltered, 
 and then determine the value of $ by putting the new term 
 which involves xg equal to zero. 
 
 The change is made by substituting for x and g, in equation 
 (5), the respective values (§ 95), 
 
 X cosO — g sin 6, 
 5; sin 0-}-g cos ; 
 
 and equation (5) now becomes 
 
 Fx' + Qg' + C'xg = B, 
 
208 ANALYTIC GEOMETRY. 
 
 where F = A co.s" +.11 siir O+CainO cos 0, (6) 
 
 Q =Asm'0+J3co^'0-Csme cos 6, (7) 
 
 C' = 2(B-A)sinecoiiO-i-C(coii'0-sm'0). (8) 
 
 Putting C = 0, we obtain 
 
 (A -B) sh)^J — Cco^,e = 0, (9) 
 
 or tan 2^ = -—^, (10) 
 
 A—B 
 
 a relation which will always give real values for 6, since the 
 tangent of an angle may have any value, positive or negative. 
 
 Since the angles which correspond to a given value of a 
 tangent differ by 180°, the values of 6 obtained from (10) will 
 differ by 90° ; hence the new axes determined by (10) are 
 limited to a single definite pair of perpendicular lines, passing 
 through the centre. 
 
 The coefficients P, Q, A, B, C are connected by simple 
 relations, which are independent of the value of 6, and which 
 may be found as follows. 
 
 From (6) and (7), by addition and subtraction, 
 
 Bi-Q = A+B, (11) 
 
 F~Q = (A-B) co^^O + Csirfe. (12) 
 
 Equation (9) may be written 
 
 = (A-B) siTi\e - Cco^'^0. (13) 
 
 Adding the squares of (12) and (13), we have 
 
 (P _ Qf = {A- By + C\ (14) 
 
 F-Q =±^{A^W+^'' (IS) 
 
 Whence, from (11) and (15), 
 
 F=^^[A + B±^{A-By + C'l (16) 
 
 Q = i[A+B=fV{A-By + C-^]. (17) 
 
 These values of F and Q are always real. 
 
LOCI OF THE SECOND ORDER. 209 
 
 Finally, by squaring (11) and subtracting (14), we obtain 
 
 4:FQ = 4:AIl-C'^l. (18) 
 
 In applying these formulas the question arises which sign 
 should be chosen before the radical in equation (15). If we 
 take for 20 the smallest positive angle which corresponds to 
 the value of tan 2^ in (10), then the value of 2^ must lie 
 between 0° and 180°, and sin 20 must be positive. If, now, 
 in equation (12) we substitute for cos 2^, its value, obtained 
 from (9), equation (12) will become 
 
 
 
 The form of this equation shows that JP— Q must have the 
 savie sign as that of C. 
 
 Equation (5) is now reduced to the simple form 
 
 Px'-\-Qu'=E. (19) 
 
 We see from (19) that the axes of co-ordinates are now 
 so placed that each axis bisects all chords parallel to the 
 other axis. Two lines so drawn through the centre of a curve 
 of the second order that they have this property are called 
 Conjugate Diameters. 
 
 From what proceeds we may infer that in those loci of the 
 second order which have a centre there exists one, and only 
 one, pair of rectangular conjugate diameters. 
 
 These two diameters are called the Axes of the curve. Hence 
 equation (19) is the general equation of curves of the second 
 order, referred to the centre as origin and to their axes as the 
 axes of co-ordinates. 
 
 202. The nature of the locus represented by equation (19) 
 depends upon the signs of P, Q, and R. There are two 
 groups of cases, according as 5 is positive or negative, and 
 three cases in each group. 
 
210 ANALYTIC GEOMETRY. 
 
 Group 1. 2 Positive. 
 
 Then equation (18) shows that P and Q must agree in sign. 
 {A and B must also agree in sign.) 
 
 Case 1. li R agree in sign with P and Q, then (§ 135) 
 the locus is an ellipse, having for semiaxes 
 
 Case 2. If i? differ from P and Q in sign, no real values 
 of X and y will satisfy (19), so that no real locus exists. But 
 it is usual to say in this case that the locus is an imaginary 
 ellipse. 
 
 Case 3. If ^ = 0, the locus is a single point, namely, the 
 origin. 
 
 Group 2. % Negative. 
 
 Then (18) shows that P and Q must have unlike signs. 
 
 Case 1. If i? agrees in sign with P, we may, by division 
 (and changing the signs of all the terms if necessary), put 
 equation (19) into the form of equation [40], page 183. 
 Therefore the locus is an hyperbola, with its transverse axis 
 on the axis of x, and having for semiaxes 
 
 Case 2. If R agrees in sign with Q, we may, by division 
 (and change of signs if necessary), put equation (19) into the 
 form of equation (2), page 184. Therefore the locus is an 
 hyperbola, with its transverse axis on the axis of y. 
 
 Case 3. If P = 0, the locus consists of two straight lines, 
 intersecting at the origin, and having for their equations 
 
 y-^-'l^ 
 
 X. 
 
LOCI OF THE SECOND ORDER. 211 
 
 CLASS II. 2 = 0. 
 
 203. Let us now suppose that, in the general equation, 2, 
 or 4:AB — (7^ is equal to zero. When this is the case, A and 
 B must have like signs, and we shall assume that A and B 
 are positive ; if they happen to be negative, we may make 
 them positive by multiplying the equation by the factor —1. 
 
 The existence of a curve of this class is immediately shown 
 by the form of the equation ; for the condition 4:AB — C^ = 
 is also the condition that the first three terms, Ax'-'rBi/'^-^-Cxy, 
 form a complete square. 
 
 Throughout this section we shall also assume that C is 
 not zero ; this being assumed, it follows from the relation 
 ^AB — (7^ = that neither A nor B can be zero. 
 
 When 2 = 0, the values of the co-ordinates of the centre, 
 
 J ^ CE-2BD ^ _ CB-'IAE 
 ^ 4:AB-C''' 4:AB-C'' 
 
 become indeterminate or infinite, according as the numerators 
 of the values are, or are not, equal to zero... In both cases a 
 change of origin to the centre (as in § 200) is impossible. 
 
 In the case where h and h are indeterminate, however, the 
 nature of the locus can be determined without any transfor- 
 mation of co-ordinates whatever. 
 For from the relation 
 
 ^AB-C' = (20) 
 
 it follows that, in case the condition 
 
 CE~2BB = (21) 
 
 is satisfied, then the condition 
 
 CD-2AE=0 (22) 
 
 must also be satisfied {C being supposed not equal to zero). 
 That is, the numerators of h and h must vanish together ; 
 whence it also follows from equation (3) that in this case the 
 condition A — 
 
212 ANALYTIC GEOMETRY. 
 
 is satisfied. Therefore tlie locus must consist of straight 
 lines. 
 
 Tlie equations of these lines are given immediately by equa- 
 tion (2), which may now be written 
 
 2% + Cx^E±,-jE''-^BF= 0. (23) 
 
 We see that the locus consists of two parallel straight 
 lines, which are real, imaginary, or coincident according as 
 E'^ — 4 BF is positive, negative, or zero. 
 
 This result may also be obtained by solving the general 
 equation with respect to x, and then introducing the condi- 
 tions of (20), (21), and (22). We thereby obtain the equation 
 
 2Ax+Cij + D±: ^~D'-^AF= 0. (24) 
 
 This equation, by means of (20)-(22), may be shown to 
 be identical with (23). 
 
 When, in addition to the conditions given in (20)-(22), 
 E^ = ^BF, it follows that i)' = 4^i^ and the locus is the 
 single straight line represented by either of the equations 
 
 2Ax-\-Cij-\-D = 0, 
 2By+Cx+E = 0. 
 
 If the numerators of h and 1c are not zero, the centre is at 
 
 infinity. Although we cannot now transform the origin to 
 
 the centre, we can make the term involving xy disappear by 
 
 proceeding exactly as in § 201 ; that is, by turning the axes 
 
 through an angle 6, the value of which is determined by the 
 
 equation n 
 
 ^ tan 2 6) = —- (25) 
 
 A-B ^ ^ 
 
 If P, Q, U, V represent the new coefficients of :r^ ?/', x, y, 
 respectively, P and Q will have values identical with those 
 of F and Q given in § 201, and 
 
 U= F cose + E sine, (26) 
 
 V=-FsmO + Ecose. (27) 
 
LOCI OF THE SECOND ORDER. 213 
 
 The relations of T, Q, A, B, and C\ found in § 201, also 
 hold true, namely, 
 
 F+Q^A+B, 
 
 P-B^±z V(yl - By + C\ 
 
 where the sign before the radical should be the same as that 
 of C. But since now C^ = ^AB, the value oi P—Q become? 
 
 P-Q = ±{A+B)- 
 
 whence, if C is negative, 
 
 P=0, Q^A-\-B. 
 
 But if C is positive, 
 
 P = A+B, Q = 0. 
 
 Suppose that C is negative. Then the general equation 
 
 becomes Qif + Ux + Vy -{-F=0. (28) 
 
 Divide by Q, and we have 
 
 •^ Q Q'^ Q 
 
 01" y' + y.y + TTT^ + ^P^+T^- 
 
 or [y -\ = { x-\ 
 
 If we now take as a new origin the point 
 ^Q F-V V_ 
 
 equation (28) becomes 
 
 U 
 
 which represents a parabola whose axis coincides with the 
 axis of X, and is situated on the positive or the negative side of 
 
214 ANALYTIC GEOMETRY. 
 
 the new origin, according as U and Q are unlike or like in 
 sign (§ 103). 
 
 The vertex of the parabola is the new origin, and the 
 parameter is equal to the coefficient of x in the erpiation of 
 the curve. 
 
 Suppose that C is positive. Then the general equation 
 
 ^^^^^^^^ Px' + Ux + Yy + F= 0. (29) 
 
 And this, by changing the origin to the point 
 
 4.PF-U' _U_ 
 4FP ' 2P 
 
 becomes x"- = — -y. 
 
 This represents ^parabola having the axis of y for its axis, 
 and placed on the positive or the negative side of the new 
 origin, according as "p^and P are unlike or like in sign. 
 
 We have already found that the value of P or Q, when not 
 zero, is ^ + -^• 
 
 We may obtain the values of U and V in terms of the 
 original coefficients, as follows : 
 
 From (25) we find, by Trigonometry, 
 
 Introducing the condition ^AB = C^, we obtain 
 tan 6 = — ^^' if C is negative ; 
 
 o p 
 
 = ^^-^, if C is positive ; 
 G 
 
 whence, if Cis negative, 
 
 sin^ = ^^ cos^= ~^ 
 
 V4^^+(7'^ V4^^ + C 
 
 '>2 
 / 
 
LOCI OF THE SECOND ORDER. 215 
 
 And if C is positive, 
 
 . . 2B . C 
 
 sin 6 — ) cos 
 
 By substitution we obtain from (26) and (27) 
 
 ^^2AE-CD^ (30) 
 
 V= ^^^^zM^, (31) 
 
 the positive sign being placed before both radicals. 
 
 204. Special Cases. When certain of the coefhcients in 
 equation (1) are equal to zero, the preceding investigations 
 may require some modification as to details, not as to results. 
 
 Only three special cases recjuire any notice. 
 
 1. Suppose that B = 0. In this case we cannot solve the 
 general equation with respect to y, and hence find the value 
 of A, as has been done in § 199. If, however, we solve the 
 equation with respect to x, we obtain 
 
 ^=-^^±^ViJ¥+W7+n; (32) 
 
 where 
 X' = (7^-4ylA J\r = 2{CD-2AE), N'=B'-4:AF] 
 
 and by proceeding as in § 199, we obtain exactly the same 
 value of A as before. 
 
 In this case 2 cannot be zero ; so that the locus always 
 belongs to the first class of curves. 
 
 2. Suppose that A =^ B ^=^0. The general equation now 
 becomes Cxy + Dx + Ey + F=^0. (33) 
 
 In this case the value of A cannot be found directly by the 
 method of § 199. If, however, (33) represents straight lines, 
 
216 • ANALYTIC GEOMETRY. 
 
 the equation may be written as the product of two linear 
 ^'^^^^'^^^ {Px^3I){Qy-\-N) = 0. (34) 
 
 Equating coefficients in (33) and (34), we obtain 
 
 PQ = C, PN=D, QM=E, 3IN=F, 
 whence CF=DE. 
 
 When this condition is satisfied, (33) represents two straight 
 lines, and their equations are 
 
 Cx+E = 0, 
 Cy-\-D = 0. 
 
 The two lines are parallel to the axes, and therefore per- 
 pendicular to each other. 
 
 These results may also be obtained by putting A = and 
 ^ = in the results reached by the general investigation. 
 
 In general, since 2 is negative, ecjuation (33) represents an 
 hyperbola. By changing the origin to the centre, the equa- 
 tion takes the form , , 
 ory = a constant, 
 
 which we know (§ 194) represents an hyperbola, referred to 
 its asymptotes as axes. Therefore, in general, equation (33) 
 represents an equilateral hyperbola. 
 
 3. Suppose :S = 0, and also C=0. Then either ^ or ^ 
 must also be zero. A and B cannot both be zero, for in this 
 case equation (1) would cease to be an equation of the second 
 degree. 
 
 If ^ = 0, equation (1) becomes 
 
 Bif + Dx-\-Ey-\-F=0, 
 
 an equation of the same form as (28). Therefore the locus is 
 a parabola. 
 
 If B=0, equation (1) becomes 
 
 Ax''-\-Dx-\-Ey-\-Fr^O, 
 
 which has the same form as (29), and the locus is a parabola, 
 with its axis on the axis of y. 
 
LOCI OF THE SECOND ORDER. 
 
 205. The main results of the investigation are given in the 
 followincT Table : 
 
 Loci kepresexted by tue General Equation of the 
 
 Second Degree, 
 
 .Ir-' + Bxf + Cxy + Dx +Ey +F=^0. 
 
 CLASS. 1 CONDITIONS. 
 
 LOCUS. 
 
 I. 
 
 Loci 
 
 having a 
 
 centre. 
 
 2 positive, A not zero. 
 2 positive, A = 0. 
 2 negative, A not zero. 
 2 negative, A = 0. 
 
 Ellipse (real or imaginary). 
 
 Point. 
 
 Hyperbola. 
 
 Two intersecting straight lines. 
 
 II. 
 Loci not 
 having a 
 
 centre. 
 
 2 = 0, A not zero. 
 2 = 0, A = 0. 
 
 Parabola, 
 
 Two parallel straight lines. 
 
 Thus it appears that there are no loci of the second order 
 besides those whose properties have been studied in the pre- 
 ceding chapters. 
 
 206. Examples. 1. Determine the nature of the locus 
 represented by the equation 
 
 Ix" -llxy + 6y' + 23a: - 2y + 20 = 0. 
 
 Here we have S = — 121, A = 0. 
 
 Therefore the equation represents two intersecting straight 
 lines. By substitution in the values of h and h (§ 200), we 
 find that the lines intersect at the point (2, 3). 
 
 If we change the origin to the centre, the equation of the 
 lines become 
 
218 ANALYTIC GEOMETRY. 
 
 2. Determine the nature of the locus represented by the 
 equation ^x^ J^bf -\-2xy -12.x -I2y = 0, 
 
 and reduce the equation to its simplest form. 
 
 Here 2=96, A = 8G4, A = 1, lc = l, 
 
 ^ E = 12, F=6, Q = 4. 
 
 Therefore the locus is a real ellipse, the centre is the point 
 (1, 1), and the equation, in its simplest form, is 
 
 The value of 0, found from equation (10), is 45°. 
 Therefore the equations of the new axes of x and y referred 
 to the original axes of co-ordinates, are respectively 
 x-y =0, 
 
 The form of the reduced equation shows that the major 
 axis of the ellipse is situated on the axis of y. 
 
 3. Determine the nature of the locus of the equation 
 
 x-" + y' - 5xy + Sx - 20y + 15 = 0, 
 and reduce the equation to its simplest form. 
 In this case 
 
 S=-21, A =-21, h=-4, k^O, 
 i^=l, P = -i Q = h ^ = 45°. 
 
 Therefore the locus is an hyperbola ; and since H agrees in 
 sign with Q, the transverse axis is situated on the new axis 
 
 of y- 
 
 The equation of the curve, in its simplest form, is 
 1y'-Sx' = 2. 
 
 And the equations of the axes of the curve referred to the 
 original axes of co-ordinates are 
 
 x-y + 4: = 0, 
 :r -f 2/ + -i = 0. 
 
LOCI OF THE SECOND ORDER. 219 
 
 4. Determine the nature of the locus of the equation 
 
 x" ^ if - 2x}j + 2x — y -l-= 0, 
 
 and reduce the equation to its simplest form. 
 
 Here :S = 0, A-1, P=0, Q = 2. ^ = 45°. 
 
 Therefore the locus is a parabola, the axis of which coin- 
 cides with the new axis of x. 
 
 V2 
 From equation (30) we have U—-—-' 
 
 A 
 
 Hence the equation of the curve, in its simplest form, is 
 
 Since C/^and Q agree in sign, the parabola is situated on 
 the negative side of the new origin. 
 
 After the original axes of co-ordinates have been turned 
 through the angle -q _ 4^0 
 
 the vertex of the parabola is the point 
 
 / 25 3V2\ 
 V8V2' 8 -/ 
 and the equation of its axis is 
 
 3V2 
 
 If the axes of co-ordinates are turned through the anf^le 
 
 to their original position, the vertex becomes the point 
 /19 31\ 
 
 and the equation of the axis becomes 
 
 4:r-4y-f 3 = 0. 
 
220 ANALYTIC GEOMETRY. 
 
 207. The locus of a point which so movos that its distance 
 from a fixed point bears a constant ratio to its distance from a 
 fixed straight line, is called a Conic. 
 
 The fixed point is called the Pocus ; the fixed straight line, 
 the Directrix ; the constant ratio, the Eccentricity. 
 
 If we denote the distance from the focus to the directrix by 
 d, the eccentricity by e, and take for the axis of x the perpen- 
 dicular from the focus to the directrix, and for the axis of y 
 the directiix, the equation of a conic is easily found to be 
 
 f-\-{x-dy = e'x\ 
 or (1- e>^ + 2/' - '2.xd-{-d' = 0. 
 
 This is an equation of the second degree ; hence a conic is 
 always a locus of the second order. 
 
 If e — 0, the curve is a circle (§ 70). 
 If c < 1, the curve is an ellipse (§ 162). 
 If e = 1, the conic is a parabola (§ 101). 
 If e > 1, the conic is an hyperbola (§ 195). 
 
 In many treatises the properties of these curves are deduced 
 from the definition above given. 
 
 208. The term " conic " is an abbreviated form of " conic 
 section." The four curves, circle, parabola, ellipse, and hy- 
 perbola, were originally called conic sections, because it was 
 discovered that they could all be obtained by making a plane 
 cut the surface of a cone of revolution in different ways. 
 
 If the plane be perpendicular to the axis of the cone, the 
 section made by the plane will be a circle. 
 
 If the plane be parallel to an element of the surface, the 
 section will be o, parabola. 
 
 If the plane make a greater angle with the axis than the 
 elements make, the section will be an ellipse. 
 
 If the plane make a less angle with the axis than the ele- 
 ments make, the section will be an hyperbola. 
 
LOCI OF THE SECOND ORDER. 221 
 
 Still further : 
 
 If the plane pass through the axis, the section will consist 
 of two intersecting straight lines. 
 
 If the plane pass through the vertex perpendicular to the 
 axis, the section will be s, point. 
 
 If the plane be parallel to an element, and we conceive the 
 vertex removed to an infinite distance, the cone will become 
 a cylinder, and the section will consist of two parallel lines. 
 
 Hence conic sections, in the most general sense of the term, 
 embrace all loci of the second order. 
 
 Ex. 40. 
 
 Determine the nature of the following loci, and reduce each 
 equation to its simplest form. 
 
 1. 3:r' + 2?/'-2:i: + y-l=0. 
 
 2. l+2:r + 3?/^ = 0. 
 
 3. y" -1xy-\-x^-%x^\<6 = ^. 
 
 4. 3.z-^ + 2r<;7/+3y^-16y + 23 = 0. 
 
 5. a;^-10xy + 3/' + 5; + y + l = 0. 
 
 6. rr^-2.ry + y'-6a;-63/ + 9 = 0. 
 
 7. a;2 + .r?/+y' + ^' + y-5 = 0. 
 
 8. f-x'-y^^. 
 
 9. mx' + 24.7:7/ + 29y^ -72:r + I263/ + 81= 0. 
 
 10. ?/2-2.r-8y + 10-=0. 
 
 11. 4a;' + 9y^ + 8.r + 36?/ + 4 = 0. 
 
 12. 52;r'H-72:ry + 73y' = 0. 
 
 13. 9y2-4:r'-8.r + 18y + 41 = 0. 
 
 14. y^ — .ry — S.'T-f 5y = 0. 
 
 15. 16:^' - 24.17/ + 9y' - Ihx - lOOy = 0. 
 
 16. 4.T^ + 4:ry + y' + 8.T + 4y-5--0. 
 
) < 
 
ANSWERS. 
 
 Ex. 3. Page 7. 
 
 1. Let Xj = — 2, 2/i = 5, x^ = — 8, 2/2 = — 3. Substituting in [1], we 
 have 
 
 d = V^- ()/ + (- 8)"^ = vTuo = 10. 
 
 In Fig. 3 the points Panel Q are plotted to represent this case. If 
 we choose to solve the question without the aid of [1], we may neglect 
 algebraic signs, and we have 
 
 QE = NO -MO = S- 2= 6 ; 
 FB = PM + i/i2 = 5 + 3 = 8 ; 
 .-. PQ2 = QRi + PE^ = 3G + 64 = 100, and PQ = 10. 
 
 2. 13. 8. 5, 5, G. 
 
 3. '5. 9. a, i, Va^Ti^. 
 
 4. 10. 10. V29^, 2VI0, 4V5; 
 
 5. 2\/a2TP. ,2Vl0, 9^/2. 
 
 6. 25, 29, 20V2.*^^^ 11. 8 or -16. 
 
 7. 2\/r7, 5\/2, VlOe. 12. (x- - Tf + (y - 2)^ = 121. 
 
 13. {x-%^ +{y- 3)2 = {x- 4)2 + (3/ - 5)2, which reduces to .r 4- 3/ = 7. 
 
 Ex. 4. PageN9. 
 
 1. (6,6). 3. (2,-2). ^ 5. (7,1). 
 
 2. (-1, 0). 4. ^(3, -1), (*,"-V-), (- J, -f). 6. (a, -6). 
 7. Take the origin of co-ordinates al the intersection of the two legs, 
 
 and the axes of x and y in the directions of the legs. Then, if a and b 
 denote the lengths of the legs, the co-ordinates of the three vertices will 
 be (0, 0), (a, 0), and (0, h). 
 
 10. Observe that now the distances RB and BQ will be x — x^ and 
 2^-2^^^ ^ 12. ill). 14. (7i-31|). 
 
 11. A 2). 13. (4,8). 15. (13,-1). 
 
ANALYTIC GEOMETIIY. 
 
 
 
 
 Ex. 7. 
 
 ■ Page 23. 
 
 1. 
 
 12, 16. 
 
 
 
 
 14. 
 
 Locus does not cut the axes. 
 
 2. 
 
 -10,6. 
 
 
 
 
 15. 
 
 (5, 7). 
 
 3. 
 
 ± 4, ± 4. 
 
 
 
 
 16. 
 
 (2, 1). 
 
 4. 
 
 ±!.±2. 
 
 
 
 
 17. 
 
 (3, -4) and (-4, 3). 
 
 5. 
 
 ± I, imaginary. 
 
 
 
 18. 
 
 (3, 4). 
 
 6. 
 
 ±f,-4. 
 
 
 
 
 19. 
 
 (5, 3) and (3, 5). 
 
 7. 
 
 ± 6, ± a. 
 
 
 
 
 20. 
 
 (0, 0) and (2, 4). 
 
 8. 
 
 3 on OX. 
 
 
 
 
 21. 
 
 (5, -3), (6, 4), (-1,-4). 
 
 9. 
 
 + 3 on OX 
 
 
 
 
 22. 
 
 \/6l, V265, VlO^L^. ^ 
 
 10. 
 
 Locus passes 
 
 throu 
 
 g^ 
 
 origin. 
 
 23. 
 
 3^4;^^ 
 
 11. 
 
 Locus passes 
 
 throu 
 
 gb 
 
 origin. 
 
 24. 
 
 (a, h){-a, h),{-a,-h){a,- 
 
 12. 
 
 ( On OX, 8, 
 .On OF, 2: 
 
 and - 
 
 -4. 
 
 
 25. 
 
 No. 
 
 4:\/48 
 
 
 
 26. 
 
 10. 
 
 13. 
 
 Locus passes 
 
 throu 
 
 gl^ 
 
 origin. 
 
 
 
 h). 
 
 Ex. 9, Page 31. 
 
 1. Let X and y denote the variable co-ordinates of the moving point. 
 Then it is evident that for all positions of the point x==Sy. There- 
 fore the required equation is x = Sy or x — 3y = 0. Does the locus of 
 this equation pass through the origin ? 
 
 2. rr-6 = 0, a: + 6 = 0, x = 0. 
 
 3. 7/ -4 = 0, y + l = 0, y = 0. 
 
 4. The line x = 3 is the line AB (Fig. 74); how is this line drawn? 
 The locus of the variable point consists of the two parallels to AB, drawn 
 at the distance 2 from AB. Let CD, EF, be these parallels, and {x, y) 
 denote in general the variable point, then for all points in CD x = 2> 
 + 2 = 5, and for all points in EF cc = 3 — 2 = 1. Therefore the equation 
 of the line CD is x -5 = 0, and that of the line ^i^ is a;-l = 0. The 
 product of these two equations is the equation (a; — 5) (a; — 1) = 0. This 
 equation is evidently satisfied by every point in each of the lines CD 
 and EF, and by no other points. Therefore the required equation 
 is (x — 5) (a; -1) = 0, or a;'^ — 6a; + 5 = 0. Verify that this equation is 
 satisfied by points taken at random in the lines CD and EF. 
 
ANSWERS. 
 
 3 
 
 5. 3/^ — 11 ?/ + 2i = 0, two parallel lines. 
 
 6. x^ + 8x — d = 0, two parallel lines. 7. a: + 3=0, y-2 = 0. 
 8. It is proved in elementary geometry that all points equidistant 
 
 from two given points lie in the perpendicular erected at the middle 
 point of the line joining the two given points. This perpendicular is 
 the locus required, and its equation evidently is x = 3. 
 
 Y 
 
 
 F 
 
 B 
 
 D 
 
 
 
 
 E 
 
 A 
 
 X 
 
 C 
 
 Y 
 
 P 
 
 7) A X 
 
 Fig. 74. 
 
 Fig. 75. 
 
 Let us now solve this problem by the analytic method. Let 
 (Fig. 75) be the origin, A the point (6, 0), and let F represent any posi- 
 tion of a point equidistant from and A, x and y its two co-ordinates. 
 Then from the given condition 
 
 PO = PA, 
 
 Therefore x'^+y'^ = {x- Qf + {y- Of, 
 
 or a;2 + 3/2 = a;2-12a;-h36+y2. 
 
 whence rr = 3, 
 
 the equation of the locus required. 
 9. a; -1=0. 10. 7/- 2-0. 
 
 y- 
 
 11. x-y-l^O. 12. x-y = 0. 
 
 13. a;2 + y2 ^ ]^00, a circle with the origin for centre and 10 for radius. 
 
 14. Express by an equation the fact that the distance from the point 
 {x, y) to the point (4, —3) is equal to 5. The equation is (.r — 4)^ 
 -f(2/ + 3)^ = 25. 
 
 15. (x + 4)2 + (7/ + 7)2 = 64. 16. x2-i-y2 = Sl. 
 
 17. Draw ^0 ± to 5C(Fig. 76). Take AO for the axis of x, and 
 BC for the axis of y ; then A is the point (3, 0). 
 
 Let P represent any position of the vessel, x and y its co-ordinates 
 
4 ANALYTIC GEOMETRY. 
 
 OM and PM. Join PA, and draw PQ ± EC, and meeting it in Q. 
 Then from the given condition 
 
 PA =PQ=^OM. 
 
 Therefore PA^ = 0M\ 
 
 Now PA^ = AM'' + PIP = (x - 3)2 + 3/2, and OilP = a,.2. Substituting, 
 we have 
 
 {x-Zf-Vy-^ 
 
 whence 
 
 
 2/2 = 6a; -9. 
 
 
 
 B 
 
 P^ 
 
 ** 
 
 (i 
 
 
 
 
 
 
 1 A M 
 
 X 
 
 c 
 
 
 "^ 
 
 
 Fig. 76. 
 
 The locus is the curve called the parabola. "We leave the discussion 
 of the equation as an exercise for the learner. 
 
 18. If BCis taken for the axis oft/, and the perpendicular from A to 
 5Cas the axis of x, the required equation is 3/2 = 12cc — 36. 
 
 19. x^ — 37/2 = 0, two straight lines. 
 
 20. x2 + 1/2 = JT^ — a2, a circle. 
 
 21. 4aa; ± ^2 ^ 0^ two straight lines. 
 
 Ex. 10, 
 
 4. cZ = \/] 
 
 + .71 
 
 Page 33. 
 
 6. re + 3/ = 7. 
 
 7. (¥.!); fV2. 
 
 5. (x - 4)2 -1- (3/ - 6) = G4. 
 
 8. Take two sides of the rectangle for the axes, and let a and b rep- 
 resent their lengths ; then the vertices of the rectangle will be the points 
 (0, 0), (a, 0), (a, h), (a, b). 
 
 9. Take one vertex as the origin, and one side, a, as the axis of x; 
 then (0, 0) and (a, 0) will be two vertices. Let {b, c) be a third vertex ; 
 then (a -i- J, c) will represent the fourth. 
 
ANSWERS. 5 
 
 10. (11,2). 11. (5,-2),(^,-i).(|,-V)- 
 
 12. (1, - 1). 13. V^. 14. (I, I). 16. (6, 23). 
 
 1^- 1^ 4 ' 4 V ^ 2 ' 2 J' V 4 ' 4 j' 
 
 21. 3 or ~ 23. 23. (4, 8) and (4, - 8). 
 
 1 3 and 2 on OX. 24. (2 a, a) and (- 2 a, a). 
 
 16 and 1 on or. 25. (a, 0) and (- a, 0). 
 
 26. 10, VIuI, V52. 
 
 27. Taking the fixed lines for axes, the equation is 3/ = 6x or a = 6?/. 
 
 28. Taking A for origin, and AB for the axis of x, the equation is 
 X - V3y = 0. 
 
 29. Taking the fixed line and the perpendicular to it from the fixed 
 point as the axes of x and y respectively, the required equation is 
 a? — ^y^ = [y — af. 
 
 Ex- 11. Page 40. 
 
 1. x-y+l = 0. 20. 2/ + 3 = 0. 
 
 2. 2x-y-3 = 0. 21. a;-2 = 0. 
 
 3. a; + y - 1 = 0. 22. a- - y + 2 = 0. 
 
 4. a; - y = 0. 23. a: - y + 5 = 0. 
 
 5. 3a; + 2y-12 = 0. 24. a;-y-4 = 0. 
 
 6. 2.r-3y + 6 = 0. 25. a; - VSy - 4V3 = 0. 
 
 7. ^tc + y - 7 = 0. 26. y + 4 = 0. 
 
 8. 4a:-3y = 0. 27. V3a:-y-4 = 0. 
 
 9. y=0. 28. a; = 0. 
 
 10. y = 4. 29. \/3 a; + y + 4 = 0. 
 
 30. a: + y + 4 = 0. 
 
 31. a; + V'3y + 4\/3 = 0. 
 3 = 0. 32. a: - y - 4 = 0. 
 
 14. VSx-y +7-2\/3 = 0. 33. 3a; + 4y-12 = 0. 
 
 15. a; - y + 14 = 0. 34. x- - 3y - 6 = 0. 
 
 16. VSx + Sy +12-13^3 = 0, 35. aj + y + 3 = 0. 
 
 17. x-v/3y-^^0. 36. 3a; -5y -15 = 0. 
 
 18. a; + y-3=^.^ 37. a;-2y+10 = 0. 
 
 19. V3a;+y=0. 38. a--y-l=0. 
 
6 ANALYTIC GEOMETRY. 
 
 39. x-y-n = 0. 52. x + y -\- C^y/i 
 
 40. •l,T + 3/-4n = 0. 53. (1, 7). 
 
 41. .r + y-5V2 = 0. 54. (1,2). 
 
 42. X - y V3 + iO = 0. 55. (2, 1). 
 
 43. ,r+y\/3+ 10 = 0. 56. (3, 2). 
 
 44. x-yy/2> -10 = 0. 57. (-2,-6). 
 
 45. 
 46. 
 
 a;+73/ + ll = 0, a;-32/+l=0, 58. (-4,6). 
 
 3x + y-7 = 0. 59. (5, - 3), (6, 4), (- 4, -1). 
 
 a; -7 3/ = 39, 9a;-5y = 3, 60. 9a' + 2?/ = 0, \V^b. 
 
 4x + 2/ = ll. 61. y — X = yy^ — x-^. 
 
 (llx-'dy = 25, 7a: + 9?/ r (t?-c)a;-(6-a)3/ = atZ~6c, 
 
 1 =-17, 5.T-6y-21 = * \ {d-c)x + {h-a)y = hd-ac. 
 
 ■ 5a; - 3/ = 0, 5 a; + 6?/ - 35 = 0, f 23/2^7: + (a-^ -2x^)y - x^y^ = 0, 
 
 48. ■ 3a;-2/ = 21, 9a; + 4y = 0, 64. \ y^x + {2xi- x.;)y - Xiy2 = 0, 
 
 ^y = 0,Ux + 3y = 2d. [ y^x - {x^ + x^)y= 0. 
 
 49. a;-y\/3-7*=-0. 65. m = 4. 
 
 50. y = a: + 3. 66. m = 3. 
 
 51. a; + y-6\/2 = 0. 67. 6 = - 9. 
 
 6^- !^ = '?-^^^ ^^-^ ^i(y2-3/3) + ^2(y3-3/i) + ^3(yi-3/2)=o. 
 
 3 1 2 1 
 
 Ex. 12. Page 45. 
 
 1. o = 4, 6 = 7, m = - |. 12. a = 6 = 0, m =^ i 
 
 2. a = 27, 6 = - 9, m = i 13. a = & = 0, m = 3. 
 
 3. a = -f, 6 = 1, m = |. 14. a = - 4, 6 = 5, m=|. 
 
 4. a = J = 0, m = f. 15. a = -2, 6 = 12, 7n = 6. 
 
 5. a = 6 = 0, m = -|. 16. a = 6, 6 = - 6, m = 1. 
 
 6. a = 8, 6 = -6, m = f. 17. a=-10, 6=-J/V3, w=-^\/3. 
 
 7. a = 6 = 3, m = -l. 18. a = 10, 6 = i^ V3, m = i\/3. 
 
 8. a = -11, 6 = --V,m = -i. ^ 
 
 20. m = -^. 
 
 9. a = -3, 6 = 5, m = -|. a 
 
 10. a = 2, 6 = -3, m = f. 21. ; - 8. 
 
 11. a=-2, 6 = -3, m = -|. 22. ^ = - 4, ^ = - 1. 
 
ANSWERS. 
 24. I ^ ^ (3/2 - yiX ^ = - (-^2 - .Ti), 25 ^ _ 3/2 -.Vi J _ ^2^1 - a^i3/2 
 
 ^ ^^ = ^2yi - •1^2- ■ •'•i-^'l' ^-2-2^1 
 
 27. (a) cos a = j^, sin a = ^V P = 2. 
 
 (6) cos a = — jf , sin a = — /j, p = 2. 
 
 (c) cos a = II, sin a = — y\, p = 2. 
 
 (c?) cos a = — |-|, sin a = -j^, p = 2. 
 
 Ex. 13. Page 49. 
 
 1. 3 a; -y- 16 = 0. 6. .t + 43/ + 49 = 0. 
 
 2. 3a; -43/ -3=0. 7. 7.T-232/ + 193 = 0. 
 
 3. 4a; -3/ = 0. 8. y = 2x.^ 
 
 4. 3/- 8 = 0. 9. 7x + 5y-ll = 0. 
 
 5. a;- 5 = 0. 10. a; + 83/ - 5 = 0. 
 
 n 
 
 Ex. 14. Page 51. 
 
 2. tan d> = J. 3. tan (p = J^ • 4. tan A 
 
 V 2 r 18 ^ w2 + 2 
 
 5. 90°. 6. 45°. 7. 90°. 8. 0°. 9. 30°. 
 
 11 (y = '^-'^-10, j3 |y-3 = m'(a;-2), 
 
 ■ la; + 53/ = 28. '1 and m' = - (5 ± 3\/3). 
 
 12. 
 
 3/ = 5a; + 11, / 3/ — 3 = m^(a; — 1), 
 
 l. + 53/-3 = 0. 14.^ andm-^*^^- 
 
 22. 2a; + 3y-31 = 0. '^ H 
 
 23. 62a; + 31 V -1115 = 0. f ^-3y+26 = 0, 
 
 30. < 5x + 3y+ 8 = 0, 
 
 24. v = 6a; -27. 0,00 a' 
 ^ ^ 2x + 3y— 9 = 0. 
 
 25. y = mx±d\/l + m\ ^^ a:_6 = 
 
 26. Bx = Aiy-b). ' 3.- '93/+ 12 = 0. 
 
 27. ax-hy^f-h\ 32. J 10a;- 43/+ 63 = 0. 
 
 28. (a±%-(aT6)(a;-a) = 0. U8a;-40y + 111 = 0. 
 
 {a;— y— 6 = -j meeting in the point 
 2x~ y- 2 = I (-4, -10). 
 
 5a;_3y_10 = oJ Distance = V85. 
 
 OK -7l±5tand>,^ ^N 
 
 35. y — y, = : (x — x,). 
 
 -^ -^^ ^i^tand) ^ '^ 
 
ANALYTIC GEOMETRY. 
 
 Ex. 15. Page 57. 
 
 |V5. 3. 4. 4. yVVl3. 5. 0. 
 
 The learner should construct the given line, 
 and observe how the sign of the required dis- 
 tance changes when we cross the line. 
 
 8. 6, 5, 4, 3, 2, 1, 0, —1. The learner should construct the lines, and 
 observe the change of sign of the distance, as in No. 7. A study of 
 Nos. 7 and 8 will make it evident that, in equation [12], if Ai\ +Bij^ + C 
 has the same sign as C, then the point {x^, y^) lies on the same side of the 
 line as the orio-in, and vice versa. 
 
 1. ^ViO. 
 
 
 2. 
 
 7. ¥-,¥. 
 
 ¥> 
 
 ¥, 
 
 1. 1. 
 
 0, 
 
 -I 
 
 ¥.¥. 
 
 I 
 
 f. 
 
 1, 0, 
 
 -I 
 
 
 9. 
 
 fViO. 
 
 17. 
 
 Va' + h\ 
 
 22. 
 
 4. 
 
 10. 
 
 IfVil. 
 
 18. 
 
 ah 3 ah 
 
 23. 
 
 , C-C' 
 
 11. 
 
 2V2. 
 
 
 y/d' + h'^ Va"^ + 6- 
 
 
 y/A' + B' 
 
 12. 
 
 |VT8. 
 
 -V- 
 
 19. 
 
 a2-62 
 
 24. 
 25. 
 
 fVlO. 
 
 13. 
 
 Va^ + 62 
 
 T3^^^- 
 
 14. 
 
 -AV13. 
 
 -V-V2. 
 
 4V2. 
 
 20. 
 21. 
 
 Ah+Bk + C-D 
 
 26. 
 
 
 15. 
 
 y/A' + ^2 
 2. 
 
 Zah 
 
 16. 
 
 2y/d' -r 62 
 
 Ex. 16. Page 60. 
 
 1. IJ. 4. 40. 
 
 8. 35. 
 
 11. 26. 
 
 2. 12. 5. ah. 
 
 9. 19.^. 
 
 12. 96. 
 
 3. 29. 7. 26. 
 
 10. Kx^y2-x^/{). 
 
 13. 61. 
 
 14. ^(a-c)(6-l). 
 
 15. l(a-h){2c-a-h). 
 
 16. Uo^-a^). 
 
 21. da'. 
 
 22. 2^. 
 21 
 
 27.^. 
 27n 
 
 28. },ah. 
 
 17. 60°, 60°, 60°; 9 V3. 
 
 18. 10. 
 
 23. 24. 
 
 24. 36. 
 
 29. ^^ 
 
 2AB 
 
 19. }. 
 
 25. 16. 
 
 30. 56. 
 
 20. H. 26. ^ah. 31. 10^. 
 
ANSWERS. 
 
 Ex. 17. Page 62. 
 3. 2, 00, 90°, 2, 0°. 4. 0, 0, 45°, 0, 135°. 
 
 2>/3-l 
 
 
 5. ^V3-2, 
 
 2V3- 
 
 -1, 60 
 
 ^ '^-^' ^^^ • 
 
 6. I 
 
 , §V3, 150°, 1, GO 
 
 0_ 
 
 26. 
 
 4y = .r + 8. 
 
 7. 1 
 
 I, -5v/3, 30°, 1, - 
 
 60°. 
 
 27. 
 
 4y = 9a;-21. 
 
 8. I 
 9. 
 
 >/o, -2, G0°, 1, - 
 a; + 23 7/-18 = 0, 
 49x+ 7y-82 = 0. 
 
 30°. 
 
 28. 
 29. 
 
 •9a;-203/ + 9G = 0, 
 .5x- 4y + 32 = 0. 
 88x-12l3/ + 371 = 0. 
 
 10., 
 11. 
 
 -\>/82. 
 
 • 3a; + 43/-57 = 0, 
 
 3a; + 4y+ 6 = 0, 
 
 12^-52/ -39 = 0, 
 
 
 30. 
 31. 
 
 5x- y-10 = 0, 
 
 . x + by -2^ = 0. 
 
 2x + y- 9 = 0, 
 
 , a;-2y-17 = 0. 
 
 
 12a; -5y + 24 = 0. 
 
 
 32. 
 
 a; ± 2/ - 5 = 0. 
 
 
 area = 63. 
 
 
 33. 
 
 2x = 3/, 2y = r.. 
 
 12. 4 
 
 3. 
 
 
 34. 
 
 4a; + 5y + 11 ± 3 vTl 
 
 13. a 
 
 ; = 3. 
 
 
 35. 
 
 2/ = (l±V2)(,r + 2). 
 
 14. 1 
 
 x - ?/ + 1 = 0, 
 
 
 36. 
 
 13.y-12 23 + 5V29 
 
 20. 
 
 (x + y-l = 0. 13a;-30 14 
 
 15. 5x + 6y-39 = 0. 37 f 7a;~32/ + 15 = 0, 
 
 16. 14a;-3y-30 = 0. * I 3a; + 72/- 19 = 0. 
 
 17 4a:-5v + 8 = 38. | 8.' + 72/-19 = 0, 
 Lt. ^x 02/ + 5 u. Il6x' + 3y + 17 = 0. 
 
 18. a: + 2/ -7-0. 39^ 450^ 
 
 19. y-y^ - v^-yx . 4o. 90°. 
 
 41. 
 
 31n/26 
 
 7/ = 3, 13.y = 5x-l, ^- 143 
 
 "51. 92a' + 692/ + 102 = 0. ' \/d' 
 
 22. a; + 4?/ = 34. 
 
 43. 
 
 23. 3a; + 42/- 5a = 0. Vli^ + k^ 
 
 24.- 3a; + 42/ = 24. 
 
 44. -Vl + m2. 
 
 25. 3/-2/i = --|;(^-^'^i). 45, 
 
 m 
 
/t^ 
 
 54. 
 
 b 
 
 
 2a2 + 5a6 + 262 
 
 57. 
 
 6 
 
 58. 
 
 172^ 
 
 59. 
 
 59. 
 60. 
 
 no. 5R 
 
 61. 
 
 10 ANALYTIC GEOMETRY. 
 
 46. _. 54. xy represents the two axes. 
 
 a = 5. 
 
 X + a = 0, X — h = 0. 
 
 X + a=-0, y + b = 0. 
 KQ rg o^' The axes and x = y. 
 
 51. (10, 5|). 61. 2a; -2/ = 0, 7a;+2/ = 0. 
 
 62. If h denotes the altitude of the triangle, and the base is taken as 
 the axis of a;, the locus is the straight line y = h. 
 
 63. The equation of the locus is 
 
 {x - x,f + {y - y,f = {x- x,f + (y - y,)^ 
 This is the equation of the straight line bisecting the line joining (x^, 3/^) 
 and {x.^, 7/2), and ± to it. 
 
 64. The two parallel lines represented by 
 
 Ax-\-By + C± c^VZM-"^ = 0. 
 
 65. x^y = h 66. A^±By±C _^ A^x + B^y + C^ ^j^^ 
 
 y/A^-\-B' y/A'HB'-' 
 
 67. Let h denote the base, k"^ the constant difference of the squares of 
 the other two sides. Taking the base as axis of x, and middle point 
 of the base as origin, the equation of the locus is 2bx= ¥■. 
 
 Ex. 18. Page 70. 
 
 1. 7a? + 7/ = 0. 
 
 2. a; + 2.y + 13 = 0. 
 
 3. 5a; + 6?/ -37 = 0, 
 
 9. {AC'-A'C)x^i^BC'-B'C)y = ^. 
 
 10 {BA' -AB')y ^CA'~AC^ = ^. 
 
 11 ^.^ + By + C ____ A'x + B'y + C' 
 ^,Ti + %i + (7 A'x^^B'y^^& 
 
 12. 472a; -291/ +174 = 0. 13. 3/ =xV3 + 3 - VS. 
 
 j4 I 4a; + 32/ -25 = 0, y x _ m'b-a 
 
 ' 1 3a;-4y+ 25 = 0. "a h ma + b 
 
 16-18. Generally the easiest way to solve such exercises as these is 
 to find the intersection of two of the lines, and then substitute its co- 
 ordinates in the equation of the third line. 
 
 4 (a;-y + 8=0, 
 \ x + y-6 = 0. 
 
 6. 
 
 64 a; -233/ = 59. 
 
 7. 
 
 Ux+y = 0. 
 
 5. y = x + 3. 
 
 8. 
 
 5a; +2/ -16 = 0. 
 
ANSWERS. 11 
 
 to 1 OA -JIM vi^^—m 1/^—h 
 19. m = l. 20. uheii — - = 
 
 21. If we choose as axes one side of the triangle and the correspond- 
 ing altitude, we may represent the three vertices by (a, 0), (— c, 0), (0, /;). 
 
 22. Choosing as axes one side and the perpendicular erected at its 
 middle point, the vertices may be represented by (a, 0), (—a, 0), (&, c). 
 
 23. It is well here to choose the same axes as in No. 21. 
 
 24. Choosing the origin anywhere within the triangle, it is evident 
 that the equations of the bisectors in the normal form may be written as 
 follows : 
 
 {x cos a-\- y sin a—p) — {x cos a^ + y sin a^ — p^) = 0, 
 
 (x cos a^ + y sin a^ —p^) — (x cos a^^ + y sin a'^ — i>^0 = 0. 
 (x cos a^^ + y sin a^^—p^^) — (x cos a -f y sin o —p) = 0. 
 Now, by adding any two of these equations, we obtain the third ; 
 therefore the three bisectors must pass through one point. 
 
 25. 
 
 2V2, VIU, 2VW. 
 . Origin within the A. 
 
 29. 
 
 J a: + y - 2 = 0, 
 
 26. 
 27. 
 
 .uvio, ifVsi, i|Vl3. 
 
 r rr + 3/ + 79 = 0, 
 I 7x-1y + 23 = 0. 
 
 30. 
 
 .r - 1 = 0, 
 ly-l = 0. 
 
 ^8 
 
 7x-9y + S4: = 0, 
 Ox + 72/ -12 = 0. 
 
 31. 
 
 2/ — mx — b y — m^x — b^ 
 
 
 
 Vl + w^ Vl + m'2 
 
 Ex. 19. Page 74. 
 
 1. (i.) Parallel to the axis of x, (ii.) parallel to the axis of y. 
 
 2. When ad = be. 
 
 3. The two lines are real, imaginary, or coincident, according as 
 H'^ — 4AB is j^ositive, negative, or zero. The two lines are ± to each 
 other when A + B = 0. 
 
 5. a; + y + 1 = 0, and x — 3 y + 1 = 0. 
 
 6. .T-2y ±(y-3) V^1 = 0. 
 
 7. .r-y-3 = 0, and x-Sy + 3 = 0. 8.45°. 9. ir= 2. 
 10. Jr=-10 or --V-. 11. ^=28. 12. /r= V- 
 
12 ANALYTIC GEOMETRY. 
 
 Ex. 20. Page 76. 
 
 1. Take the point as origin, and the axis of 3/ parallel to the given 
 lines. If the equations of the given lines are x = a, x = b, and if the 
 slopes of the lines drawn in the two fixed directions are denoted by 
 m^, m^^, the equation of the locus is 
 
 (b — a)y = m^b{x — a) — m^^a{x — b). 
 
 2. If a and b are the segments of the hypoten,use made by a perpen- 
 dicular dropped from the vertex of the right angle, the equation of the 
 locus is y^xJi 
 
 3. Let OA = a, OB-=h. Then the equation of the locus is a- + y = a + 5. 
 
 4. Take as axes the base and the altitude of the triangle. Let b 
 denote the base, a one segment of the base, h the altitude. *Then the 
 equation of the locus is Ox 2?/ 
 
 b — a h 
 This is a straight line joining the middle points of the base and the 
 altitude. 
 
 5. Take as axes the sides of the rectangle, and let a, b denote their 
 lengths. The equation of the locus is 
 
 bx — ay = 0. 
 Hence the locus is a diagonal of the rectangle. 
 
 
 Ex. 21. 
 
 Page 79. 
 
 1. 
 
 x'' + y^ = -2rx. 
 
 13. 
 
 (4. 0), 4. 
 
 2 
 
 xUy^=2ry. 
 
 14. 
 
 (-4,0), 4. 
 
 3. 
 
 x^ +y^=-2ry. 
 
 15. 
 
 (0, 4), 4. 
 
 4. 
 
 (a; _ 5)2 + {y + 3)2 = 100. 
 
 16. 
 
 (0,-4), 4. 
 
 5. 
 
 x' + (y + 2y^l2h^ 
 
 17. 
 
 (0, 1), I 
 
 6. 
 
 ^x - 5)2 +y^ = 25. 
 
 18. 
 
 (0,0), Sk. 
 
 7. 
 
 {x + 5)2 + 2/2 = 25. 
 
 {x - 2)2 + {y- 3)2 =. 25. 
 
 19. 
 20. 
 
 (0, 0), 2 k. 
 
 8. 
 
 (0, 0), Vd' + 62. 
 
 9. 
 
 x^+y^-2hx-2Jcy = 0. 
 
 21. 
 
 (|0)JV5. 
 
 1 1 
 
 (1, 2), Vs. 
 
 (i 1). w^. 
 
 22. 
 
 Ll. 
 
 L2. 
 
 fh k\ Vh^ + k'' 
 
ANSWERS. 13 
 
 23. When D^T)' and E=E^ \ in other words, wlion the two equa- 
 tions differ only in their constant terms. 
 
 24. In this case r = 0. Hence the equation represents simply the 
 point (o, h). We may also say that it is the equation of an infinitely 
 small circle, having this point for centre. 
 
 
 r{i.) C+and=i)2^ 
 
 31. 
 
 \ (ii.) C+and =E\ 
 
 
 I (ill.) C + and > D^ and > E'^. 
 
 32. 
 
 a;^ + y2_i0a; -102/ + 25 = 0. 
 
 33. 
 
 (7, 4) and (8, 1). 
 
 34. 
 
 (2. 0) and (f. -f). 
 
 35. 
 
 |V5. 
 
 36. 
 
 V d' + b-'l 
 
 37. 
 
 2a; -2/ -2 = 0. 
 
 38. 
 
 4:x-5y +71 = 0. 
 
 39. 
 
 3x-5y-34: = 0. 
 
 26. ^ On OX, 3 and 2 ; 
 I On OY, 6 and 1. 
 
 r(0, 2),5; 
 
 27. I On OX, fi ± V2I ; 
 I On Y imaginary points. 
 .(2, 4),<;/5; 
 
 28. ] On OX, and 4 ; 
 ( On Y, and 8. 
 
 (-(3,-2). 3; 
 
 29. ^ On OX, 7and-l; 
 i On r, - 2 ± VT3. 
 /-(-ll, 0), VRT); 
 
 30. ] On OX, - 3 and - 19 ; 
 iOn or, 9±2v/(i. 
 
 40. Let (.r, y) be any point in the required locus ; then the distance 
 of (x, y) from {xi, y^ must always be equal to its distance from (0^2, y^) ; 
 therefore {x - x^f + (3/ - y^f = {x - x^f + (y - 3/2)2 . 
 whence 2x{x^- 'x.,) + 2y(y^- y.^) = {x^^- + yi^ - x,;' - y.^^). 
 
 Show that this represents a straight line ± to the line joining [x^, yj 
 and {x^, y^). 
 
 41. 8a: + 6y +17=0. 
 
 42. First Method. Substitute successively the co-ordinates of the 
 given points in the general equation of the circle ; this gives three equa- 
 tions of condition, and by solving them we find the values of a, 6, r. 
 
 Second Method. Join (4, 0) to (0, 4) and also to (6, 4) by straight 
 lines, then erect perpendiculars at the middle points of these two lines ; 
 their intersection will be the centre of the circle, and the distance frojtn ^u 
 the centre to either one of the given points will be the radius. /J^^;^lt»t* r -^ 
 
 43. a;2 + y2 - 8 ;c 
 
 44. x^ + y2 + G X 
 
 
 Ans. a;2 + y2 - 6 a; - By + 8 = 0. 
 
 6y = 0. 
 
 45. a;2 + y^ + 8 a.-i; - 6 ay = 0. 
 
 y-0. 
 
 46. .T2+y2 + 8a- + 20y + 31 = 0. 
 
14 ANALYTIC GEOMETRY. 
 
 47. x-2-i-3/2_9a'-53/ + 14 = 0. 51. x'-vy'-2ax 
 
 48 10^- 5)2 + (y + 8)2 = 169, 52. x'' +y^ =.ax + hy. 
 
 ■ I (.T - 22)2 ^ty_ 9)2 = 169. 
 
 ^ ^ vy ; g3 (a,_i)2^/ 4^2_20. 
 
 • la;2+2/2- 6(a; + 2/)+ 9 = 0. 54. rt^ +3/2_14a;-4y-5 = 0. 
 
 50. a'2 + 2/2-8a;-8y + 16 = 0. 55. x-^ + y2 + 20?/ = 0. 
 
 56. ???. (x^ + ?/2) 4- aZ> = (??ia — h)x + (mi — a)y. bl . x^ + w^ = x^.^ + y^y. 
 
 58. (.r - .r,) {x - x^) + (y - y,) {y - y,) = 0. g^ ^,, _ ^^, ^ ^., ^ ^,, _ a^ 
 
 59. (1 H- ?«-) (x2 + 2/2) - 2 r (x + my) = 0. ^ 
 
 Ex. 22. Page 87. 
 
 1. The double sign corresponds to the geometric fact that two tangents 
 having the same direction may always be drawn to a given circle. 
 3. 2a; + 3y = 26, Sx-2y=^0, VlTY, V;12, 9, 4, -^S-VlS. 
 
 4 L, J. 
 
 5. 9x-13?/ = 250. 
 
 6. x±?jy^lO. \/ 
 
 7. 104i 
 
 8. ic2 + ^2_25. 
 
 9. 14x±62/ = 232. 
 
 10. 3x + ?/ = 19. 
 
 11. 3x-42/ = 0. 
 12 f3.T + 72/ = 93, 
 
 l3;y_7y = 65. 
 
 13. x = r. 
 
 14. ^x + % ± r\/^2+^2 _ 0. 31. (x- _ 5)2 + (y - 3)2 = -L2JL. 
 
 15. i?.K-^2/±rVJ.2+^2_o. gg |(a^- 2)2 + (3/ - 4)2 = 100, 
 
 22. 
 
 1 (p COS a, p 
 
 2 
 
 sin 0). 
 
 23. 
 
 ■ When (72 = 
 When r = 
 
 ^.r + By + C 
 V.-i2 + ^2 
 
 24. 
 
 ax + hy = 0. 
 
 
 25. 
 
 (- a, - h). 
 
 
 26. 
 
 (2 a, 6). 
 
 
 27. 
 
 (0, h). 
 
 
 28. 
 
 .t2 + 2/2 = |. 
 
 
 29. 
 
 m = ±f. 
 
 
 30. 
 
 c = 28 or - 
 
 52. 
 
 16. x-3/±rV2 = 0. I (x - 18)2 + (7/ - 16)2 = 100. 
 
 17. The equation of the two tan- 33. (,r - 1)2 + (y _ 0)2 = 25. 
 
 gents is {h^—r^)y'^=r^{x — Tif. Ill 
 
 18. x + y = r V2. ' r2 ~ a2 ^ 62' 
 
 jg {x = 10, 35. (a2+62)(x2 + 2/2)-2a6Va2-i-62 
 1 3 a- + 43/ = oO. X {x + 2/) + a^6^ = 0. 
 
 20, y = 2 X + 13 ± GV5. 36. x = a + r. 
 
 21. 21, 3«-. 37. [4r2-2(a-6)2]5. 
 
ANSWERS. 15 
 
 X. Ex. 23. Page 90. 
 
 1- (1, -^). 2^29- 5- •'■-' +/ = 81. 
 
 2. K2 ± V7), i Vn. 6. (.r - 7)2 +3/2 = 9. 
 
 3. (i, I), V-^. 7. {X + 2)2 + (y - 5)2 = 100. 
 
 b ah 
 
 8. ,r2+y2_2a(3a; + 4y) = 0. 
 
 VVl + a- Vl + a2 
 9. a;2 + y2 + 262 + c2 = 2[(& + ^^^^ + (j _ c),/]. 
 
 10. 3 ah{x^ + 2/^) + 2 a6(a2 + h"") = (5 o2 + o ^2)60; + (5 62 -^ 2 a2)ay. 
 
 11. .T2^2/2-5x-12y = 0. 13. x2 + 2/2 + 14a; +U2/ + 49 = 0. 
 
 12. a;2 + 3/2 - 14a; - 43/ - 5 = 0. 14. a- + 7/2 _ 2ra; - 2 ?'?/ + r2 = 0. 
 
 15. a;2 + 3/2 _ 2a.r -2ay + 3a2-- = 0. 
 
 16. a.'2 + 3/2 = |. 17 . 5(a'2 -\- 1/) - 1 .t + 30 3/ + 49 = 0. 
 
 [ (a; _ -y-)2 + (y _ Y)2 = ^. 20. a;2 + y2 + 50 a; + 88 y + 230 = 0. 
 
 21 f ^•'+ y2_3(3^^_ 46y + 324 = 0, 
 ( 25 a;2 + 25 y2 - 80 a- - 494 y + 64 = 0. 
 
 22. (6,2), 5. 25. iV234. 
 
 23. 15. 28. VIO. 
 
 24. 10. 27. x^x + y^y = x^^ + yi2. 
 
 28. (i.)Z)2=4^(7^ (ii.)^2_4^C'_ (m.)Z)2=^2_4^(7_ 
 
 29. r2 = 2rmc + c2. 35. 6.6- - 8y - 25 = 0. 
 
 30. ;>; = 40. 36. 2(37 ± 3 V41)x- + 25y = 0. 
 
 31. a2+y2 = ay\/2. 37. a; + V3y± 20 = 0. 
 
 32. a;2 + y2 ± 2a{x ± y) = 0. 38. a; + y -10 = 0. 
 
 33. 2(a;2 - ax + y2 - ?-2) + a2 = 0. 40. .](35 + 24 V30). 
 
 34. x-y = 0. 43. 135°. 
 
 |- ix 4- 4)2 + (y + 10)2 _ 85. 44. (7, - 5) and (- 63^ 92-6). 
 
 45. \ I oUy , / , 670\2 85 
 
 U'^ 169; ^(,^"^169 j 1692' 
 
 46. The circle {x — x\f + (y — yi)2 = r^. 
 
 47. The circle {x - a)2 + (y - 6)2 = (r + r'f. 
 
 48. The circle (a; - aj- + (y ~ hf = r^ + t^. 
 
16 ANALYTIC GEOMETRY. 
 
 49. Take A as origin, and let the radius of the circle = r ; then the 
 locus is tlie circle x^ + y"^ = rx. 
 
 50. Take A as origin, and let the radius of the circle = r ; then the 
 
 2,mrx 
 locus is the circle sf -\- y^ = — ; — 
 
 51. Take A as origin, AB as axis of x, and let AB = a ■ then the 
 locus is the circle (m^ + 7^"') (x^ + y"^) -2am?x + aHi^ = 0. 
 
 52. Take AB as the axis of x, the middle point of AB as origin, and 
 let AB = 2a ; then the locus is the circle 2{f + y"^) = Ic^ -2a}. 
 
 53. Using the same notation as in No. 52, the locus is the straight line 
 4 ax = P. 
 
 54. Taking the fixed lines as axes, the locus is the circle "^{x"^ ^- y"^) = d^ . 
 
 55. Take the base as axes of x, its middle point as origin, and let the 
 length of the base = 2a, and the constant angle at the vertex = d. Then 
 the locus is the circle x^ + ?/2 — 2a cot ^ 3/ = a-. 
 
 56. Take A as origin, AB as axis of x, and let AB = a, AC=h. 
 
 52 
 Then the locus is the circle (x - a)^ + 2/^ = — • 
 
 57. The circle x^ + v^ = _ll where I is the length of the chord. 
 
 4 ^2 _ ^2 
 
 58. The locus consists of the two circles x^ + y^ ± rx = 0. 
 
 Ex. 24. Page 103. 
 
 1. 7a;_63/ = 0. 2. 2x-2y + 9 = 0. 
 
 4. x+2/ = r, 2x + oy = r, {a + h)x + {a - b)y = r-^. 
 
 5. 13x + 22/ = 49. 6. The tangent at (A, ^). 
 
 7. (i.) 2x + 3 3/ = 4, (ii.) 3x-3/ = 4, (iii.) x - 3/ = 4. 
 
 8. (i.) (20, 30), (ii.) (21, -14), (iii.) (35 a, 356). 
 
 9. (6, 8). 18. 12x + 17y-51 = 0. 
 
 10. (-'^''l -^\ 19- x + y-2 = 0. 
 
 ^ , , 20. (a2-a6)x-(a6-62)y + ac = 0. 
 
 11. (4, ±V-12), 4x4:V-122/ = 4. ^ ' ______ 
 
 16 A2 + F-r2. 2^- a:-2/ = 0, Vata+6/^-4c. 
 
 17. 3. ' * 22. (-2,-1). 
 
ANSWERS. 17 
 
 Ex. 25. Page 115. 
 
 1. Writing a; + i for x, und y — 2 for y, ami reducing, we have y"^ = 4a;. 
 
 2. x' + 2/' = r\ 6. x' + y-' = r\ 
 
 3. a;* + .V^ = 2 rx. 6. 2 xy = a?. 
 
 4. a;2+y2__2ry. 7. a;2_3/2 = 2. 
 
 8. (i.) p = a, (ii.)/)2cos20-a2. 
 
 9. (i.) p = 4acot^ COS0, (ii.) /!)(1 — cos^) = 2a. 
 
 10. (i.) x^ + 2/^ = a^, (ii-) a;"^ + y^ = «a;. (iii-) -^'^ — y"^ = o?- 
 
 11. re + y = 0. 15. a;^ — G ary + 3/^ = 0. 
 
 12. 2a;-5y = 0. 16. a-?/ = 3. 
 
 13. 12a;2 + 16 a-y + 43/2 = 1. 17. y^ = 2a(a;V2 - a). 
 
 14. a;2+y2 = 25. 18. 4ary = 25. 
 
 
 Ex, 
 
 . 26. 
 
 Page 117. 
 
 1. 
 
 6V3. 
 
 
 12. 
 
 dx^ + 2dy^ = 225. 
 
 2. 
 
 4 sin J ft). 
 
 
 13. 
 
 p = Sacosd. 
 
 3. 
 
 \/l3-12cosft,. 
 
 
 14 
 15. 
 
 p = ia. 
 
 4. 
 
 Va"^ + b'^-2abcoii{0- 
 
 7). 
 
 p = f csc^^e. 
 
 5. 
 
 2 a sin 6. 
 
 
 16. 
 
 p = 49 sec 2^. 
 
 6. 
 
 2 a cose. 
 
 
 17. 
 18. 
 
 p = k^ cos 26. 
 
 7. 
 
 aVo - 2 V3. 
 
 xy = a^. 
 
 9. 
 
 2a;2 + 2a;y +y^ = l. 
 
 
 19. 
 
 {x^+y^)^ = 2kxy. 
 
 10. 
 
 2a;2 + y2 = 6. 
 
 
 20. 
 
 ar' + y3 - 5 ^a;y = 0. 
 
 11 
 
 y = 0. 
 
 
 21. 
 
 tan-^ }. 
 
 
 22. (i.)tan-i('- 
 
 -i\ 
 
 (ii.)tan-^4 
 
 Ex. 28. Page 124. 
 
 2. y2 = ipx - 4p2. 3. y2 = ipx + \f. 
 
 4. (i.)y2 = 10a;, (ii.)y2 = 10a; + 25, (iii.) y^ = lOx - 25. 
 
 5. (i.)y2=16a;, (ii.) y^ = 16a; + 64, (iii.) y^ = 16 a: - 64. 
 
18 ANALYTIC GEOMETRY. 
 
 6. (2, ±G). 8. (4,6) and (25,15). 
 
 7. 6, 15, ^. 9. (12,6). 
 
 
 
 10. The line x = 9 meets the parabola in (9, 6) and (9, - 6). The 
 line x = passes through the vertex. The line x = — 2 meets the para- 
 bola in the imaginary points (— 2, ± V— 8). 
 
 11. The line 3/ = 6 meets the parabola in (9, 6). The line y = — 8 
 meets the parabola in (16, — 8). 
 
 12. p = 4. 13. The point (2, 8). 
 
 14. (i.)3/ = 0, (ii.)a; = -2, (iii.) a; - 2, (iv.) x±y- 2 = 0, {v.)y = 2x. 
 
 15. (i.) 4a; - 57/ + 24 = 0, (ii.) a:^ + y^ = 20.r. 16. 3;^. 17. S^/3p. 
 
 26. The common latus rectum = 4j3. The common vertex is at the 
 origin. The axis of :r is the axis of (i.) and (ii.) ; that of y is the axis of 
 (iii.) and (iv.). Parabola (i.) lies wholly to the right of the origin, 
 (ii.) wholly to the left, (iii.) wholly above, (iv.) wholly below. AVe may 
 name them as follows : — 
 
 (i.) is a right-handed X-parabola. 
 (ii.) is a left-handed X-parabola. 
 (iii.) is an upward I^'-parabola. 
 (iv.) is a downward F-parabola. 
 
 27. (i.) If A = 0, the equation becomes y^ + By + 0= 0, a quadratic 
 in y, and representing the two straight lines, parallel to the axis of x, 
 
 T) 
 
 whose equations are y±~± ^VB^ — 4 AC=- 0. (ii.) If 5 = 9, the origin 
 
 A 
 is in the axis of the curve, (iii.) If C= 0, the origin is in the curve. 
 (iv.) If J. = .S = 0, the locus consists of the two parallels y = ± V— C. 
 (v.) \i A= (7=0, the locus consists of the straight lines y = 0, y + B = 0. 
 (vi.) If B= C= 0, the vertex is at the origin. 
 
 28. Latus rectum = — B. 
 
 A A^-^C 
 
 The vertex is the point 
 
 The focus is the point ( 
 
 The axis is the line x = - 
 
 2 
 
 The directrix is the line y = 
 
 2 45 
 
 A A^_-±C_B 
 
 2' 45 4 
 A 
 
 A^ + B'--iC 
 
 4:B 
 
 If B is negative, the curve is above the axis of x. 
 If B is positive, the curve is below the axis of x. 
 
ANSWERS. 
 
 19 
 
 
 Latus 
 Rectum. 
 
 Vertex. 
 
 Focus. 
 
 Axis. 
 
 Directrix. 
 
 29. 
 
 12 
 
 (7,0) 
 
 (10, 0) 
 
 2/ = 
 
 x = i 
 
 30. 
 
 12 
 
 (- 7, 0) 
 
 (-4,0) 
 
 y = o 
 
 a; = -10 
 
 31. 
 
 -12 
 
 (- 7, 0) 
 
 (- 10, 0) 
 
 y = o 
 
 .T ■-= - 4 
 
 32. 
 
 -12 
 
 (7,0) 
 
 (4,0) 
 
 y = o 
 
 a- = 10 
 
 33. 
 
 12 
 
 (0,7) 
 
 (0, 10) 
 
 a; = 
 
 y = 4 
 
 34. 
 
 8 
 
 (6,4) 
 
 (8,4) 
 
 .V = 4 
 
 .r = 4 
 
 35. 
 
 -f 
 
 (-^0) 
 
 (-iO) 
 
 2/ = 
 
 x = -i 
 
 36. 
 
 1 
 
 (^.-f) 
 
 (^. - 2) 
 
 x = h 
 
 2/ = -f 
 
 37. 
 
 4 
 
 (-2,-3) 
 
 (-1.-3) 
 
 y = -3 
 
 a; = -3 
 
 Ex. 29. Page 129. 
 
 2. On the axis of x lay off from the vertex to the left a length equal 
 to the abscissa of the point of contact ; this determines a second point in 
 the tangent. Lay off from the foot of the ordinate of the given point, 
 towards the right, a length equal to 2p ; this determines a second point 
 in the normal. 
 
 6. x-4y + 20 = 0, 4a;+y-90 = 0. 
 
 7. Tangents I ^-3/ + 3 = 0, normals | '-^ + ^ " ^ == ^' 
 
 ^ [a; + .y + 3 = 0; \x-y-d = 0. 
 
 These lines enclose a square whose area = 72. 
 
 8. Tangent = V266, normal = V95, subtangent = 14, subnormal = 5. 
 P 
 
 9. (5, 10). 
 
 13. 
 
 14. 
 
 x,x. 
 
 r'2. 
 
 mv 1 + m^ 
 X — y + p = 0, point of contact {p, 2p), intercept = p. 
 
 2 (yi + 2/2)]- 
 
 15. 
 
 16. Equation of the tangents yVS = ±x±3p, required point (— 3p, 0). 
 
 17. For the two points whose co-ordinates are 
 
 = ^(1 
 8^ 
 
 Vl7), 
 
 y 
 
 ^p\ 
 
 I + VfT 
 
 18. For the points (0, 0) and {3p, ± 2pV3). 
 
 19. 9.r-6y + 5 = 0, (|, |). 
 
 20. a; -22/ +12 = 0, (12, 12). 
 
20 ANALYTIC GEOMETRY. 
 
 21. 3/ = a;(±>/2-l) + 4(±V2-l). 22. — ^-^ -- 
 
 24. One of the points of contact is (—1, 11). The vertex of the para- 
 bola is the point (— 9, 3) ; therefore the distance from the vertex to the 
 intersection of the axis and the ordinate of the point of contact is 
 equal to 8; therefore the subtangent = 10; therefore a second point of 
 the tangent is (—17, 3) ; therefore the equation of the tangent through 
 (-1, 11) is a;-2y + 23 = 0. 
 
 25. j (ii.) a-ix= 2p{y + y{), 
 l(iii.) x^x = -2p{y +y^). 
 
 Ex. 30. Page 131. 
 
 1. 
 
 3/2 = 24 a; -144. 
 
 2. 
 
 f 
 
 = 16x. 
 
 3. 3/2 = _17rc. 
 
 4. 
 
 f 22/2-ll,r + 12,y + 7 
 l2a;2+lLr + 12y-3 
 
 ■3 = 
 
 -0; 
 -0. 
 
 or 
 
 5. (y+7y = 4:{x-3). 
 
 6. 3y2 = 4x. 
 
 7. 
 
 3a'2 = 4y. 
 
 8. 
 
 h 
 
 8x + 3 
 
 i = 0, 8x±15y-3 = 0. 
 
 10 
 
 ■ 4 on OX; 
 
 8 and -2 on OY. 
 4(2 + \/3)p. 
 ^-(1.) y=-x + 2, 
 \{n.) 2V2, 
 l(iii.) a: + y-6 = 0. 
 
 
 
 20. 
 
 3/2 = 9a:. 
 
 J.V. 
 
 11. 
 
 12. 
 
 
 
 21. 
 22. 
 
 23. 
 
 3/2 = 8. T. 
 
 3/2 = 4^^-^. 
 
 3/2 = -.. 
 
 13. 
 
 x + y-Q = 0. 
 2p 
 
 y-yi = -^i^-^i)- 
 
 
 
 24. 
 
 V/l2 + P 
 
 14. 
 
 
 
 25 
 
 3/2=2(2r-s)a;. 
 
 15. 
 16. 
 
 (8, 4), (2, 10). 
 (2, 4), (11, 10). 
 
 
 
 26. 
 
 27. 
 
 28. 
 
 4pV2. 
 
 The equation of the circle is 
 
 (.T- 3)2 + (3/ --1)2 =2^. 
 
 (-3i>,0). 
 
 17. 
 
 h±\/h-^-iap 
 
 18. 
 
 f A left-handed A'-parab( 
 j Latus rectum = —.2. 
 i Vortex, (-2, 0). 
 Focus, (- I, 0). 
 
 3la. 
 
 29. 
 30. 
 
 j(i'. ±2p); 
 1 45° and 135°. 
 
 
 [ Directrix, x= — |. 
 
 
 
 31. 
 
 4p2. 
 
 19. 
 
 4 a. 
 
 
 
 34 
 
 The parabola y'^ = px. 
 
35. 
 
 y' = 2px-p\ 
 
 36. 
 
 f = 2px-2p\ 
 
 37. 
 
 y- = -p^- 
 
 38. 
 
 y2 = 2px + 2f. 
 
 43. 
 
 Take the given line as 
 
 ANSWERS. ;21 
 
 The loci in exercises 35-38 are parabolas, the latus rectum in each 
 being half that of the given parabola. If the given parabola is y^ = ipx, 
 the equations of the loci are : 
 
 39. The straight line y =pk. 
 
 40. The parabola y* — 4pa; =2>2^^ 
 
 41. The straight line kx=p. 
 
 42. The parabola {^—pY+y'^—ri- 
 
 the axis of y, and a perpendicular 
 through the given point as the axis of x, and let the distance from the 
 point to the line = a. The locus is the parabola y^ = 2al x ]• 
 
 Ex. 31. Page 143. 
 10. 3x-5y-6 = 0. ^2ay==ipx. 
 
 11 8v— '^5 = 14 ] The chord is parallel to the 
 
 tangent at the end of the 
 
 12. 13a: + 22y + ^ = 0. I diameter. 
 
 13. x-y-l=0. 15. 3/2 = 52a\ 
 
 .20. If the equation of the given parabola is y^ = 4ipx, the locus is 
 the parabola y'^ = p{x —p). 
 
 21. If the equation of the given parabola is y^ =, 4^^.^ the locus is 
 the parabola y'^ = p{x — 3p). 
 
 22. Take the given line as the axis of y, and a perpendicular through 
 the centre of the given circle as the axis of x. Let the radius of the 
 circle = r ; distance from the centre to the given line = a. There are 
 two cases to consider, since the circles may touch the given circle either 
 externally or internally. The two loci are the parabolas 
 
 2/2 = 2(o + r)a;+r2-a2, 
 3/2 = 2(a-r).T + r2_a2. 
 
 23. Let 2a be the given base, ab the given area; take the base as 
 axis of X, its middle point as origin ; then the locus is the parabola 
 
 x^ + by = a^. 
 
 Ex. 32. Page 153. 
 1. 5, 4, 3, f. 2. V2, ], 1, VI. 3. 2, VS, Vl, iVl. 
 
 4. J-. J-. .'AE^. ^V^^- 
 
13. 
 
 a;2 + 2y2 = l00. 
 
 14. 
 
 8a;2 + 9y2=8a2. 
 
 15. 
 
 2:V3. 
 
 16 
 
 ah 
 X — V — -\ 
 
 
 Va'^ + h' 
 
 18. 
 
 a.f),(-f,-§)- 
 
 22 ANALYTIC GEOMETRY. 
 
 5. «\/6. 
 
 6. c = ^\/3. 
 
 7. 4x2 + 97/2 = 144. 
 
 8. 250-2+169^2 = 4225. 
 
 9. 144x2 + 2251/2 = 32,400. 
 
 10. 16x2 + 252/2 = 1600. 
 
 11. 25x2 + 1692/2 = 4225. 19. (1, 2), (1, - 2). 
 
 12. 115x2 + 252y2 = 4140. 20. (3, 1), (3, -1), (_3, 1), (-3, - 
 
 21. See No. 11. The equation of the locus is x2 + 2?/2 = r-. 
 
 22. Taking as axes the two fixed lines, and putting AP==a, BP-- 
 the acute angle between AB and the axis of x = cp, we find that 
 
 X = a cos 0, y = & sin (}>. 
 Therefore P describes an ellipse whose equation is 
 
 e! + ^ = i. 
 
 a* 
 
 A 
 23. The two straight lines y = ±— x. The locus is imaginary when 
 
 the values of y are imaginary ; that is, when A and B have unlike .«igns. 
 29. The equations of the sides are 
 
 ah ah 
 
 X = ± — -————, y = -— — - , 
 
 Va2 + 62 Va2 + 62 
 
 4a262 
 
 area 
 
 62 
 
 Ex. 33. Page 160. 
 
 4x±9y = 35, 
 9 X ± 4 y = 6. 
 2x±3yV3-12 = 0,_ 
 6xV3±4y+9V3+2V6 = 0. 
 4x+ y = 10, 
 x-4y + 6 = 0; 
 
 5. :i-+^ = l 
 
 m 
 
 2 ,,2 
 
 6. 9x2 + 25 y' 
 
 7. 
 
 2y = x±10. 
 
 8. 
 
 4x-3y ± Vl07 = 0. 
 
 9. 
 
 «2 62 
 
 x = ± , ?/ = + 
 
 
 Va2+62 Va2+62 
 
 10. 
 
 Same answers as No. 9. 
 
 11. 
 
 62 : a\ 
 
 12. 
 
 x = ±-«, y = ±A. 
 V2 V2 
 
 13. 
 
 y = 4, 3x + 34y=17. 
 
ANSWERS. 23 
 
 14. Tlio equation ± Vox ± 3?/ = 9a represents the four tangents. 
 
 15. a Vl — e^ cos"'' <p. 16. ^ (a^ tan <^ + 6"^ cot <(>). 
 17. The extremities of the latera recta. 
 
 19. The method of solving this question is similar to that employed 
 in \ 147. The required locus is the auxiliary circle x^ -Vy^ = a?. 
 
 1. 
 
 2. 
 
 x = 8, 40y = 
 
 1 -¥-. ¥• 
 
 Within. 
 1 
 
 Va' 
 
 1 
 
 V2' 
 
 = 9i 
 
 Ex. 
 
 C + 72: 
 
 = 0. 
 
 34. 
 
 Pag( 
 8. 
 9. 
 
 10. 
 
 11. 
 12. 
 
 13. 
 
 14. 
 15. 
 
 s 161. 
 
 hx + ay = ab-\/~l. 
 
 X y . 
 
 - cosd) + y sin d) = 1. 
 
 a ^ b ^ 
 
 ah 
 
 3. 
 
 Va^-eV 
 
 4. 
 
 oVl — e^ cos'^'c^. 
 e^ : h\ 
 
 5. 
 
 Vl3±l 
 
 2V3 
 
 e%y^ 
 aW 
 
 6. 
 
 a; ± 2/ ± Va"'' + 
 hx + c_y = 6c -y 
 
 6-'': 
 
 V(l-e2)(a2-e'^Xi2). 
 
 7. 
 
 /2q 
 
 ,2 , Z.-i 
 
 tan0 V^-^ 
 
 18. The locus is the minor axis produced. 
 
 19. The ellipse [ x — -\ + 2^ = — ; centre is ( -, ) ; semiaxes are - 
 , V 2/ 4 4 V2 / 2 
 
 and r. ^ ' ^ '' 
 
 20. The ellipse a^{y \-\-h'^x^ = ; centre is [ 0, -] ; semiaxes 
 
 a 1 6 \ -'/ V ^/ 
 
 are - and ~ 
 2 2 
 
 In 21-23 take the base of the triangle as the axis of re, and the origin 
 
 at its middle point. 
 
 21. The ellipse (s^ - e-) x'- + sh/ = s'is'' - c'). 
 
 22. The ellipse kx"^ + 2/^ = ^C^- 
 
 23. The circle {x + cf + y^ = 4a2. 
 
 Ex. 35. Page 176. 
 1. a; = ±- = ±-- 3. 20a; + 63y- 36 = 0. 
 
 . / Aa" Bh^\ Q /•^ 2 &' /••^ 2 * /•••x 2 1 
 
24- ANALYTIC GEOMETRY. 
 
 W44f 
 
 9. a^l\ '-' '"'"^ ^ b = lVT 
 
 10. 3 .r + 8 y = 4, 2 a; - 3 y = 0. 11. a^yja; = Rrj?/. 
 
 12. Area = — (7?2 + n), m and 7i being the two segments (use th 
 
 2a 
 polar equation). 
 
 13. 26a; + 33?/ -92 = 0. 14. .r + 2y = 8. 
 15. h^x + a^?/ = 0, b\x - ahj = 0, cv^y + Z^li- = 0. 
 
 16. 
 
 Z>,T + ay = 
 
 0. 
 
 
 
 26. 
 
 See I 156. 
 
 17. 
 
 See I 157. 
 
 
 
 
 27. 
 
 See I 159. 
 
 18. 
 23. 
 
 See g 158. 
 a b 
 
 
 
 
 30. 
 31. 
 
 1 f cc Q 
 
 24. 
 
 y^d' - P 
 
 xVP- 
 
 -b\ 
 
 b'- 
 
 
 e = ^\/ti. 
 
 
 
 
 32. 
 
 25. 
 
 ^ 1 - e2 cos2 d 
 
 34. 16a;2 + 49^/2 _ 128a: - 6863/ + 1473 = 0. 
 
 35. 2a = 18, 2i = 10. 36. --^ + y'^ = bx. 
 
 144 "^ 
 
 37. Centre is (-1, 1), 2a = 2, 26 = 4. 
 . 38. Centre is (2, 4), 2a = 8, 26 = 6. 
 39. Centre is (0, - J), 2a = 3, 26 = 1. 
 
 40. cos (/) = a/4— r!- 41. tan = — 
 
 » /-/2 — /i2 
 
 6 
 
 42. Find the ratio of y^ to the intercept on the axis of 3/ 
 
 43. b^hx + a^ky = bVi"" + a''k\ .- t^, ,,. a;^ y^ , 
 
 '^ 47. The elapse ^ -^ = J 
 
 a , 
 pse - + 
 
 46. The ellipse i!i! + ^ = 2 
 
 62 48. The ellipse 62.1-2 + a2y2 = b^c 
 
 49. The ellipse 25a;2 + I63/2 - 48y = 64. 
 
 Ex. 36. Page 186. 
 
 yi .,2 
 
 1 i^_2C==i 3. 8.x2_3/2_Sa2. 
 
 64 49 
 2 2 4. 625a:2-84y2 = 10,000. 
 
 25 144 5. 2.T2_2y2 = c2. 
 
 7. a = 4, 6 = 3, c=»5, c = f. latus rectum == |. 
 
ANSWERS. 25 
 
 8. IGy"^ — dx^^ 25, transverse axis = 6, conjugate axis = 8, distance 
 between focus = 10, latus rectum = y. 
 
 9. a : 6 = 1 : V3. 11. e = VI 
 
 14. Foci, (5, 0), (—5, 0); asymptotes, y = ±^x. 17. b. 
 
 Ex. 37. Page 188. 
 
 1. 16 a: - 97/ = 28, 9 .r + I63/ = 100, f , -%\ 3. x" -2/^ = 9, (5, 4). 
 
 4. The four points represented by 
 
 ± a' ± &2 
 
 X = ■! 7/ = • 
 
 Va-' - ^2 Va'' - h' 
 
 If J'^ > a^ the points are imaginary. If the hyperbola is equilateral, 
 the points are at an infinite distance. 
 
 9. ^. 10. ^--^' = 1. 
 
 V3 
 
 m' n 
 
 11. "When a is less than h. 12. The circle a;^ + y^ = ^2^ 
 
 Ex. 38. Page 189. 
 
 1. 2he, ae^. 11. y = ±x's/2 + a. 
 
 2. U and 6. 12. (0, ± Va^ _ 6'^). 
 
 3. The sum = 2 car. 13. 6^ > ^2. 
 
 8. (a, 6 \/2), (a, - 6 \/2). 14. 64 a; - 9 y + 741 = 0. 
 
 10. They are equal. 15. ?/ = 4 a; ± 8 \/2. 
 ,2 52 
 
 16. a; = ± 
 
 Va2 - 4 i-^ Va2-462 
 
 Ex. 39. Page 201. 
 1. 9a; + 12y + 16 = 0. 6. a. 
 
 2. X 
 
 a 
 
 8. 75.r-167/=0. 
 
 <^ 9. 245.^-123/ -1189 = 0. 
 
 3. ^. 10. |V3. 
 
 ^ 17. See § 150. 
 
 4. (-^^ f \ 18. 4.ry = -(a^ + 6^). 
 
 •^ X y 
 
 5. a- + a = 0. 19- F, + y-,=-- 
 
26 ANALYTIC GEOMETRY. 
 
 a) ^_^_^' = o 22. p=«^(^l::ill 
 
 (ii.) ^-^'+?5 = o. 23. -^ ^'' 
 
 a- b'^ a ' *^ e^co^^d-l 
 
 24. The equation represents an equilateral hyperbola, with it? trans- 
 verse (real) axis parallel to the axis of y. The centre is (1, — 2) ; the 
 semiaxes are each equal to 2. 
 
 25. The hyperbola Zx^-y"^ + 20 x- 100 = 0. The centre is the point 
 (— J3O-, 0). Changing the origin to the centre, we obtain 9a;2 — 33/2.= 400. 
 
 26. The locus is the curve 2a;y — 7 a; + 4?/ = 0. If we change the 
 origin .0 the point (A, h), we c*an so choose the values of h and k as to 
 get rid of the terms containing x and y. Making the change, we obtain 
 
 2xy + (2k-7)x + {2h + A)y-7h + ik + hk = 0. 
 If we choose h and k so that 2 A + 4 = 0, and 2 A; — 7=0, that is, if 
 we take A = — 2, ^ = |, the terms containing x and y vani^, and the 
 equation becomes 2xy = 7. Hence we see (§ 194) that the locus is an 
 equilateral hyperbola, and that the axes of co-ordinates are now the 
 asymptotes. 
 
 27. The equilateral hyperbola 2xy = a^. 
 
 28. Taking the base as axis of x, and the vertex of the smaller angle 
 as origin, the loci are the axis of x and the hyperbola 3x^ — y^ — 2ax = 0. 
 
 Ex. 40. Page 221. 
 
 1. The ellipse 72 a:' + 48 y^ = 35. 9. The ellipse 4.^2 + 9y^= 36. 
 
 2. The parabola 2/^^ = — ^.^. 10. The parabola 3/^ = 2a;. 
 
 3. The parabola y^ = 2.rV2. 11. The ellipse 4a'2 + 9^/2 = 36. 
 
 4. The ellipse 4^2 + 2/ = 1. 12. The ellipse 4a;2 4- y'^ = 100. 
 
 5. The hyperbola 32^2-48/ = 9. 13. The hyperbola 4 a;^- 9 7/^= 36. 
 
 6. The parabola _y 2 = 3,^.^0 -^^ The straight lines 3/=.r,y= — 5. 
 
 7. The ellipse 9x2 + 3 7/2 = 32. 15. The parabola 7/2 = 5 a;. 
 
 8. The hyperbola 4x2-4y2^1_Q jg The parabola 25a;2+2j/V5 = 0. 
 
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