QA !>11 W4/W 1«^H ■*■» UNIVERSITY Of CALIFORNIA SAN Di^^^^^^ 3 1822 01062 3031 r- LIBRARY UNIVERSITY OP CALIFORNIA SAN DI£SQ tl4 511 W478 18<)« ; 3 1822 01062 3031 ^A - SSE MATHEMATICAL TEXT-BOOKS BY GEORGE A. WENTWORTH Mental Arithmetic Elementary Arithmetic Practical Arithmetic Primary Arithmetic (Wentworth and Reed) Grammar School Arithmetic Advanced Arithmetic First Steps in Algebra New School Algebra School Algebra Elements of Algebra (Revised Edition) Shorter Course in Algebra Complete Algebra Higher Algebra College Algebra (Revised Edition) First Steps in Geometry (Wentworth and Hill) Plane Geometry (Revised) Solid Geometry (Revised) Plane and Solid Geometry (Revised) Syllabus of Geometry Geometrical Exercises Analytic Geometry Plane and Solid Geometry and Plane Trigonometry (Second Revised Edition i Plane Trigonometry (Second Revised Edition) Plane Trigonometry and Tables (Second Revised Edition) Plane and Spherical Trigonometry (Second Revised Edition) Plane and Spherical Trigonometry and Tables (Second Revised Edition) Plane and Spherical Trigonometry, Surveying, and Navigation (Second Revised Edition) Surveying and Tables (Second Revised Edition) Plane Trigonometry, Surveying, and Tables (Second Revised Editioni Plane and Spherical Trigonometry, Surveying, and Tables (Second Revised Edition) Logarithms, Metric Measures, etc. ELEMENTS Analytic Geometry G. A. WENTWUKTH, A.M. ACTHOE OF A SKKIKS OK TEXT-BOOKS IN MATHEMATICS GINN & COMPANY BOSTON ■ NEW YORK • CHICAGO • LONDON Copyright, 1886, by GEORGE A. WENTWORTH ALL RIGHTS RESERVED G8.10 Che gtftengum ^re gg GINN & COMPANY • PRO- FRIETORS • BOSTON • U.S.A. NOTE TO THE SECOND EDITION. In this edition such changes have been made as actual experi- ence in the class-room has shown to be desirable. A chapter on Higher Plane Curves, and four chapters on Solid Geometry have been added, making the work sufficiently extensive for our best schools and colleges. An effort has been made to have this edition free from errors. It is not likely, however, that this effort has been entirely success- ful, and the author will be very grateful to any reader who will notify him of any needed corrections. NOTE TO THE EDITION OF 1898. The old plates have become so worn that it is necessary to have new plates. A few verbal changes liave been made. No changes, however, have been made that will prevent the using of old and new books together. March, 1898. G. A WENTWORTH. PREFACE. o>Ko rpmS book is intended for beginners. As beginners -*~ generally find great diiticulty in comprehending the connection between a locus and its equation, the opening chapter is devoted maiidy to an attempt, by means of easy illustrations and exam[)les, to niaki; this connection clear. Each chapter abounds in exercises ; for it is only by solving })roblems which require some degree of original thought that any real mastery of the study can be gained. The more difficult propositions have been })ut 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. -ill. and Chapters V.-VII., with Review Exercises and Supplementary Propositions left out. Between this course and the entire work the teacher can exercise 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 tlie English treatise of Charles Smith. Special acknowledgment is due to ii. A. Hill, A.M., of Cambridge, Mass., and to Prof. J. M. Taylor, Colgate University, Hamilton, New York, for assistance in the preparation of the book. Corrections and suggestions will be thankfully received. G. A. WENTWORTH. Exeter, N.H., January, 1888. O O IsT T E N T S. ol» 1. From the point (—2, 5) to the 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, — b). Find the lengths of the sides of a triangle 6. If the vertices are the points (15, — 4), (—9, 3) (11, 24). V 7. If the verticesare 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, —b). "J 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. V 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 coordinates of the point (x, 7/) satisfy if 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 square of a length, it is immaterial whether the length is considered positive or negative. Why ? Division of a Line. 8. To bisect the line joining two given 2^oints. Let P and Q (Fig. 4) be the given points (x^, i/i) and (^2) 3/2)- Let X and y be the coordinates of H, the mid- point of FQ. The meaning of the problem is to find the values of x and y in terms of .Tj, 2/1, and x^, y^. M b Fig. 4. N Draw P3I, ES, QJVW to OY; also draw PA, EB || to OX. Then rt. A PEA = rt. A EQB (hypotenuse and one acute angle equal). Therefore, PA = EB, and AE = BQ; also, MS=SN. By substitution, ic — a-^ = .To — x, and y—y\^yi — y\ whence, ^^S^.^^mpi., ,-2] LOCI AND THEIR EQUATIONS. M 9. To divide the line joining two given points into two parts having a given ratio m : n. Let P and Q (Fig. 5) be the given points (x^ y-^ and {x^y^. Let R be the required point, such that PR : RQ = m : n, and let x and y denote the coordinates of R. Complete the figure by drawing lines as in Fig. 4. The rt. A PR A and RQB, being mutually equiangular, are similar ; therefore PA PR m , AR PR m -= — , and RB RQ n' BQ RQ n Substituting for the lines their values, we have X — Xi m , 1/ — 7/, m =— , and ■ ■ — 1^ Vi—V *i Solving these equations for x and ?/, we obtain _ ^nx 5 the values of y are imaginary ; this last result means that there is no point with an abscissa greater than 5 whose coordinates will satisfy the given equation. 18 ANALYTIC GEOMETRY. By assigning values of x differing by unity, we obtain tlie following sets of values of x and // ; and by plotting the points, and then drawing through them a continuous curve, we obtain the curve shown in Fig. 11. alues of X. Val aes of y. . . ±5. 1 . . . ±4.9. 2 . . ±4.6. 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 foiind as follows : Let P (Fig. 11) be any point so placed that its coordinates, x=OM, i/ = MP, satisfy the equation x^ -\-y-^^25. Join OP; then x'-{- f^'oF; therefore, OF = 5. Hence, if P is any point in the circumference described with as centre and 5 for radius, its coordinates will satisfy the given equation ; and if P is not in this circumference, its coor- dinates will not satisfy the equation. This circumference, then, is the locus of the equation. 23. The points whose coordinates satisfy the equation y^ = 4x lie neither in a straight line nor in a circumference. .Nevertheless, they do all lie in a certain line, which is, therefore, completely determined by the equation. To con- struct this line, we first find a number of points that satisfy the equation (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. LOCI AND THEIR EQUATIONS. 19 The coordinates of a number of such points are given in the table below. It is evident that for each positive value of X there are two values of ?/, equal numerically but un- like in sign. For a negative value of x, the value of >/ is imaginary ; this means that there are no points to the left of the axis of y that will satisfy the given equation. Values of x. Values of y. • 0. 1 ±2. 2 ±2.83. 3 ±3.46. 4 ±4. 5 ±4.47. 6 ±4.90. 7 ±5.29. 8 ±5.66. 9 ±6. — 1 imaginary. 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 wholly on the right of the axis of f/, 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 foundation of the science of Analytic Geometry: I. Every algebraic equation involving x and i/ is satis- fied by an unlimited number of sets of values of x and i/; in other words, x and // may be treated as variaMes, or quantities varying continuously, yet always so related that their values constantly satisfy the equation. Fig. 12. 20 ANALYTIC GEOMETRY. II. The letters x and ?/ may also be regarded as repre- senting the coordinates 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 continuous, the point may be conceived to move con- tinuously, and will therefore describe a definite line, or group of lines. The line, or group of lines, described by a point moving so that its coordinates always satisfy the equation is called the Locus of the Equation; conversely, the equation satis- fied by the coordinates of every point in a certain line is called the Equation of the Line. An equation, therefore, containing the variables x and y is the algebraic representation of a line. In Analytic Geometry the loci considered are represented by their equations, and the investigation of their properties is carried on by means of these equations. Exercise 6. Determine and construct the loci of the following equa- tions (the locus in each case being either a straight line or a circumference of a circle) : 1. a; — 6 = 0. 9. 9x- — 25 = 0. 2. a; + 5 = 0. 10. ^x-—y'^ — ^. 3. y= — l. 11. a;2 — 16^2 = 0. 4. x = 0. 12. a;2_f_y2^36_ 5. ?/ = 0. 13. x''-\-y'' — l=Q. 6. x-\-y = 0. 14. x(?/H-5)=0. 7. x — 2y = 0. 15. (x — 2)(x — 3) = G. 8. 2x + 3?/+10 = 0. 16. (y-4)(3/+l)=0. LOCI AND TIIEIK EQUATIONS. 21 17. What is the geometric meaning of the equation 5a;2 — 17x — 12 = 0? Hint. Resolve the equation into two binomial factors. 18. What is the geometric meaning of the equation 19. What two lines form the locus of the equation xy-\-4:X^0? 20. Is the point (2, — 5) situated in the locus of the equa- tion 4ic — 3// - 22 = ? Hint. See if the coordinates of the point satisfy the equation. 21. Is the point (4, — 6) in the locus of the equation 22. Is the point ( — 1, — 1) in the locus of the equation 16a;2 + V + 15cc — 6?/— 18 = ? 23. Does the locus of the equation a---l-y" = 100 pass through the point ( — 6, 8)? 24. Which of the loci represented by the following equations pass through the origin ? (l)3x-[-2 = 0. (5)3x = 2i/. (2)3x-lUj-{-7 = 0. {e)3x-n>/ = 0. (3) a;2 - 16/ -10 = 0. (7) a- — 1 Oy- = 0. (4) aa; + % + c = 0. (8) ax-\-b>/ = 0. 25. The abscissa of a point in the locus of the equation Sx — 4?/ — 7^0 is 9 ; wliat is the value of the ordinate ? Ans. 5. 26. Determine that point in the locus of // — 4a- ^0 for which the ordinate =: — 6. A]is. The point (0, - 6). 27. Determine the point where the line represented by the equation lx-\- y — 14 = cuts tfie axis of x. Ans. The point (2, 0). 22 analytic geomexky. 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 to the points where the curve cuts the axes. 26. To find the intercepts 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 inter- cept, put v/ = in the given equation of the curve, and then solve the equation for x ; the resulting real values of X will be the intercepts required. If the equation is of a higher degree than the first, there will in general be more than one real value of x; and the curve will intersect the axis of x in as many points as there are real values of x. To an imaginary value of x there corresponds no inter- cept ; but it is sometimes convenient to speak of such a value as an imaginary intercept. 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 coordinates must satisfy both equations. Tlierefore, to find their coordinates, solve the two equations, regarding the variables x and y as unknown quantities. If the equations are both of the first degree, there will LOCI AND THEIR EQUATIONS. 23 be only one pair of values of x and y, and one point of intersection. 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 2>asses throxujh the oriyin, its equation, re- duced to its slmjdest form, cannot have a constant term; that is, cannot have a term free from both x and y. Since in this case the point (0, 0) is a point of the curve, its equation must be satisfied by the values cc = 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 constant term. Therefore the equation cannot have a constant term. 29. If an equation has no constant term, its locus must pass through the origm. For, the values x = 0, y = must evidently satisfy the equation, and therefore the point (0, 0) must be a point of the locus. Exercise 7. Find the intercepts of the following curves : 1. 4a- + 3// — 48 = 0. 8. 0- — 3 = 0. 2. 5y — ox — 30^0. 9. a- — 9=0. 3. ic2 + 2/2 = 16. 10. a-'- — //- = 0. 4. 9a- + 4/ = 16. 11. / = 4a-. 5. 9.r2 — 4/ = 16. 12. .T- + // — 4.r — 8// = 32. 6. 9x^ — 4// = 1 6. 13. X- + // — 4.r — 8// = 0. 7. a:'x' -\- by = a'b\ 14. (x — 5)' -\- (y — ())' = 20. 24 ANALYTIC GEOMETKY. Find the points of intersection of the following curver; 15. Sx — 4:1/ -{-13 = 0, lla;+7^— 104=0. 16. 2a; + 3^ = 7, x—i/ = l. 17. a;-7// + 25 = 0, x'-\-i/^ = 25. 18. 3x-j-ii/ = 2o, x'-\-i/ = 25. 19. x-\-y = 8, x--{-f = 34.. 20. 2x=i/, x^ + i/^ — 10x = 0. 21. The equations of the sides of a triangle are 2x-{- 9y + 17=r0, 7a; — ^—38 = 0, a; — 2^-^2 = 0. Find the coordinates of its three vertices. 22. The equations of tlie sides of a triangle are 5a; + 6^ = 12, 3a; — 4^ = 30, a; + 5y = 10. Find the lengths of its sides. 23. Find the lengths of the sides of a triangle if the equations of the sides are a; = 0, //^O, and 4a; + 3^ = 12. 24. What are the vertices of the quadrilateral enclosed by the straight lines a; — a = 0, a; + a = 0, y — b = 0, y-\-b = 0? What kind of a quadrilateral is it? 25. Does the straight line 5a; + 4y = 20 cut the circle 26. Find the length of that part of the straight line 3a; — 4y = which is contained within the circle x^-\-if = 25. 27. Which of the following curves pass through the origin of coordinates? (1) 7a; — 2y+4 = 0. (4) ax-\-by = 0. (2) 7a; — 2// = 0. (5) ax + % + c = 0. (3) tf — x' = 4:y. (6) x'^ — y-\-a = a-\-xy. 28. Change the equation 4a'H-2// — 7^0 so that its locus shall pass through the origin. LOCI AND THEIK EQUATIONS, 25 CONSTKUCTION 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 necessary 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 circumference, and can find its centre and its radius, the entire locus can then be described with the aid of a pair of compasses. It will appear later that the form of the given equation enables us at once to tell whether its locus is a straight line or a circumference. If the locus of an equation is neither a straight line nor a circumference, 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, diifering not mucli 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 coordinates 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 constructing the curve, and as a means of detecting nu- merical errors made in computing the coordinates 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 metliod. But the nearer tlie points are to one another, tlie nearer the curve will approacli 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 tlie size of the paper compared with the greatest length to be laid off upon it. Paper sold under the name of " coordinate paper," ruled in small squares, j^^ of an inch on a side, will be found very convenient in practice. 32. Construct the locus of the equation 9^2 + 4/ -576 = 0. If we solve for both x and y, we obtain the following values : '=:±aV64-; ; = ±§Vl44 — ?/2 (1) (2) 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 y, 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 x Values of y. . . ± 12. ±1 ± 11.91. ±2 ±11.62. ±3 ±11.13. ±4 ± 10.39. ±5 ± 9.36. ±6 ± 7.93. ±7 ± 5.80. ±8 ± 0. ±9 ± imaginary LOCI AND THEIR EQUATIONS. 27 Discussion. From equations (1) and (2) we see that if X =0, ij = ±i 12, and if y = 0, x = ±:8; therefore, the intercepts of the curve on the axis of x are +8 and — 8, and those on the axis of y are + 12 and — 12. These inter- cepts are the lengths OA, OA', and OB, OB', in Eig. 13. If we assign to a? a numerical value greater than 8, posi- tive or negative, we find by substitution in equation (1) that the corresponding value of ij 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 +12 and — 12. The greater the numerical value of x, between the limits and 4" 8 or and — 8, the less the corresponding value of y numerically ; why ? From equation (1) we see that for each value of x, between the limits and i 8, there are two real values of y, equal numerically and unlike in sign. Hence, for each value of X between and ± 8 there are two points of the curve placed equally distant from the axis of x. Therefore, the curve is symmetrical with respect to the axis of x ; iu other words, if the portion of the curve above the axis of x is 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 4x — /+16 = 0. Solving for both x and y/, we obtain y=±2^ir^, (1) -^-^- (2) 28 ANALYTIC GEOMETRY. We may either assign values to x, and then compute values of y by means of (1), or assign values to y, and com- pute values 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 J'ig 14. Values of y. ±0 Fig. 14. ±1 ±2 ±3 ±4 ±5 ±6 t9 10 Values of x. ■ -4. — 3.7o. -3. -1.75. 0. 2.25. 5. 8.25. 12. 16.25. 21. Discussiox. 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 6»C= — 4. If we draw through A the line AD _\_ 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 res})ect 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 AX its Axis. 34. Construct the locus of the equation y = sin X. If we assume for x the values 0°, 10°, 20°, 30°, etc., the corresponding values of >/ are the natural sines of these angles, and are as follows : Values of X. Values of y. Values of x. Values of y 0° ... 0. 50° . . . . 0.77. 10° . . . 0.17. 60° . . . . 0.87 20° . . . 0.34. 70° . . . . 0.94. 30° . . . 0.50. 80° . . . . 0.98. 40° . . . 0.64. 90° . . . . 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 a; = 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. 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 G0° = 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 x 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. Exercise 8. Construct the loci of the following equations : 1. 3x — y — 2 == 0. 13. r-i=o. 2. y = 2x. 14. y = x\ 3. x'^^y\ 15. xy = 12. 4. ^2+,/ =100. 16. X ■= sin y. 5. ^2-^=^ = 25. 17. y =^ 2 sin x. 6. 4a;2-/r=0. 18. y =^ sin 2x. 7. 4a;2+V = 144. 19. 2/ = cos X. 8. y — 16x = 0. 20. y = tan x. 9. y/2_|_ 1(5^=0. 21. y = cot x. 10. .T^— 2.T — lOy — 5==0. 22. y = sec x. 11. y--2y-10a- = 0. 23. y =^ CSC X. 12. (.«-3)^+(/y-2y^=:25. 24. y = sin X + cos x. loci and theik equatioxs. 31 Equation of a Curve. 35. From what precedes, we may conclude that every equation 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 statement in algebraic language of a geometric con- dition which must always be satisfied by a point (x, y), as we imagine it to move in the plane of the axes. For ex- ample, 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^ -[-1/^=4: states the condition that the point must so move that the sum of the squares of its coordinates 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 coordinates 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 be satisfied by a point, to find the equation of its locus. The importance of this problem is that in the practical applications of Analytic Geometry the law of a moving point is commonly the one thing known, so that the first step must consist in finding the equation of its locus. Exercise 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. What is the equation of its locus? 2. "What is the equation of the locus of a point that moves so that its abscissa is always equal to + 6? — 6? 0? 32 ANALYTIC GEOMETRY. 3. What is the e(;[uation of the locus of a point that moves so that its ordinate is always equal to -f- 4 ? — 1? 0? 4. A point so moves that its distance from the straight line X == 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 = 5 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 lina X + 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 x == — 6 ? (2) from the parallels y=^l 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 construct 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 coordinates as centre, with aradius equal to 5, a circle is described. A point outside this circle so moves that its distance from the circumference of the cir- cle 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 BC. 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 2k^; find the equation of its locus. 21. A point 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. Exercise 10. (Review^.) 1. If we plot all possible points for which x = — 5, how will they be situated ? 2. Construct the point (x, y) if a- = 2 and (l)7/ = 4x-3, (2)3x-2y-8. 3. The vertices of a rectangle are the points (a, b), ( — a, b), ( — a,—b), and (a, — b). Find the lengths of its sides, the lengths of its diagonals, and show tliat the vertices are equidistant 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 ANALVTIO GEOMETRY. 5. Express by an equation that the distance of the point (x, if) from the point (4, 6) is equal to 8. 6. Express by an equation tliat the point (x, y) is equi- distant from the i)oints (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 coordinates of three vertices of a parallelogram are known : (5, 3), (7, 10), (13, 9). What are the coordi- nates of the remaining vertex ? 11. The coordinates of the vertices of a triangle are (3,5), (7,-9), (2,-4). Find the coordinates 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 triseetion nearer tliat side. Find the centre of gravity of the triangle whose vertices are the points (2, 3), (4,-5), (-3,-6). 13. Tlie vertices of a triangle are (5, — 3), (7, 9), (—9, 6). Find the distance from its centre of gravity to tlie origin. 14. If tlie vertices of a quadrilateral are (0, 0), (5, 0), (9, 11), (0,3), what are the coordinates of the intersection of the two straight lines that join the middle points of the opposite sides ? 15. Prove that the two straight lines which join the middle points of tlie 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 coordinates of the other end? LOCI AND THEIR EQUATIONS. 35 17. The line juiiiiiig the jnjiuts (xi. y^) and (x^, y^) is divided into four equal parts. Find the coordinates of the points of division. 18. Explain and illustrate the relation tliat exists between an equation and its locus. 19. Construct the two lines that form the locus of the equation x" — 7a" =; 0. 20. Is the point (2, ~5) in the locus of the equation 4a:=^ — V = 36? 21. Tlie ordinate of a certain point in the locus of the equation .r- + if -\- 20x' — 70 =; is 1. What is the abscissa of this point? 22. Find the intercepts of the curve a:^ + y— 5..-7y + 6 = 0. Find the points common to the curves : 23. x'-\-y^ = 100, and y- — — = ^- 24. a;^ + //^ =^ 5«^, and ic^ = 4«7/. 25. U^x" + ay = tt^W, and a;^ + / = a\ 26. Find the lengths of the sides of a triangle, if its vertices 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 Hue. Find the equation of its locus. CHAPTER II. 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 axis of x and the line, m = tan y. p = the perpendicular from the origin to the line. a = the angle between the axis of x and p. These six quantities are general constants; a, b, and 7n may have any values from — ao to + cc ; ^, any value from to -)- oc ; y, any value from 0° to 180° ; a, any value 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 conditions must be given. 37. To find the equation of a straight line passing through two given points (x-^, y^ and (x2, y^. Let A (Fig. 16) be the point (xj, y-^, B the point (^2, 2/2) ; and let P be any point of tlie line drawn through A and B, x and ?/ its coordinates. Draw AC, BD, PM \\ to OY, and AEF \\ to OX. THE STRAIGHT LIXE. 37 The triangles APF, ABE are similar; therefore, PF _BE AF~ AE A^iow, FF— 1/ — j/i, AF=x — Xi, BE=i/o— >/i, AE=X2 — Xi. Therefore, ^^-^^^ ^^^^^y W This is the equation required. Y Fig. 16. Fig. 17. It is evident that the angle FAP = y. Therefore, each side of equation [4] is equal to tan y or m. The first side contains the two variables x and ?/, and the equation tells us that they must vary in such a way that the fraction ' — remains constant in value and equals 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 lead to equation [4] whatever be the positions of ,these points. In Fig. 17 the points are in different quadrants. The triangles APF, ABE are to be constructed as shown in the figure. They are similar ; and by taking proper account of the algebraic signs of the quantities, we arrive at equation [4], as before. The learner should study this case with care, and should study other cases devised by himself, till he is convinced that equation [4] is perfectly general. 38 ANALYTIC (JEOMETRV. 38. To find the equation of a straujld line, given one point (xy, ?/i) in the line and the slope ni. Let the figure be constructed like Fig. 16, omitting the point B and the line BED. Then it is evident that whence, PF y—yx AF ic — aji' y — y\ = tn(3c — oci). Y ir I Y c y () J I X Y [5] Fig. 18. M A\ X Fig. 19. 39. To find the equation of a strai/jht line, gicen the intercept h and, the angle y. Let the line cut the axes in the points A and B (Fig. 18). Let P be any point (x, y) in tlie line. Draw PM^ to OY, and7?C|| to OX. Then OB = h, rBC = y, BC = x, PC = y — b; therefore, m = ' : X whence, y = utx + h. [6] 40. To find the equation of a straight line, given its inter- cepts a and h. Let the line cut the axes in tlie points A and B (Fig. 19), and let P be any point (x, y) in the line. Then OA = a, THE STKAIGHT LINK. 39 OB = h. Drawi^J/Xtu OX. The triangles F MA, BOA are similar ; therefore, PM MA OA — OM J OA OA y - b a — X ^ X J- 5 a a a b whence, 7; + a ~ -'^ [^] This is called the Symmetrical Equation of the straight line. 41. To find the e(ivat\oii of a straight line, given its dis- tance p from the origin, and tlce angle a. i />' vS r/ / '*' \r \^i .1/ \.Y Fig. 20. Let AB (Fig. 20) be the line, P any jioint in it. Draw OS L to AB, meeting AB in ,S; 7^J/ _L to OA'; Mii || to AB, meeting OS in U\ ami /'^ J_ to All. Tlien p=OS= Oil + QB, a = XOS = PMQ. Now, OR — OM cos a = .^' cos a, and QP = PM sin a — y sin a. Therefore, OR + (>F =;> = .r cos a + v/ sin a, or -z^ cos a + // sin a = 7>. [8] This is called the Normal Equation of the straight line. The coefficients cos a and sin a determine the dii-ection of the line, and p its distance from the origin. 40 AXALYTIC GEOMETRY. Note. Observe that all the equations of the straight line that have been obtained are of the first degree. Their differences in form are due to the constants which enter them. The form of each, and the signification of its constants, should be thoroughly fixed in mind. Exercise 11. Find the equation of the straight line passing through the two points : 1. (2, 3) and (4, 5). 7. (2, 5) and (0, 7). 2. (4, 5) and (7, 11). 8. (3, 4) and (0, 0). 3. (— 1, 2) and (3, — 2). 9. (3, 0) and (0, 0). 4. (—2,-2) and (-3, -3). 10. (3, 4) and (—2, 4). 5. (4, 0) and (2, 3). 11. (2, 5) and (-2,-5). 6. (0, 2) and (— 3, 0). 12. (m, 7i) and (— m, — n). Find the equation of a straight line, given : 13. (4, 1) andy = 45^ 2*9. b = —4,y=120°. 14. (2, 7) and y = 60°. 30. 6 = -4, y = 135°. 15. (— 3, 11) andy = 45°. 31. b = — 4=, y = 150°. 16. (13,-4) and y=:.150°. 32. b = — 4:, y = 180°. 17. (3, 0) and y = 30°. 33. a = 4:,b = S. 18. (0, 3) andy = 135°. 34. a^ — 6,b = 2. 19. (0, 0) andy = 120°. 35. a = — 3,b = — 3. 20. (2, — 3) and y == 0°. 36. a = 5,b = — o. 21. (2, — 3) and y = 90°. 37. a = — 10,b = 5. 22. b = 2, y = 4:0°. 38.a = l,b = — l. 23. b = o, y = 4o°. 39. a = n, b = — n. / 24. 6 = — 4, y = 45°. 40. a = n,b = 'in. 25. & = — 4, y = 30°. 41. 79 = 5, a = 45°. 26. ^- = -4, y = 0°. 42. ^J = 5, a =120°. 27. i = - 4, y = 60°. 43. 2i = 5,a = 240°. 28. ^; = — 4, y = 90°. 44. J^ = 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). 46. If its vertices are the points (2, 3), (4, — 5), ( — 3, — 6). 47. Form the equations of the medians of the triangle described in example 46. 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 = 7^, y = 30°. '51. j!> = 6, y = 45°. 50. « = — 3, (xi, ?/i) is (2, 5). 52. p = 6, y = 135°. Reduce the following equations to the symmetrical form, and construct each by its intercepts : 53. 3x — 2^ + 11 = and ?/=7x + l. 54. 3x + 5^ — 13 = and4a; — y — 2 = 0. 55. Reduce Ax-{- Bi/=C to the symmetrical form; also y = mx-{-h. What are the values of a and b in terras of A, B, C, and m? Reduce the following equations to the form i/ = mx + b, and construct each by its slope, and intercept on the axis of y: -56. y + 13 =5x and ?/ + 19 = 7x. 57. 3:c + ?/ + 2 = and2y = 3a' + 6. 58. Reduce Ax -\- Bi/= C to the form // = tnx + b ; also --}-- = l. What are the values of 7?i and b in terms of a b A, B, C, and a ? 59. Find the vertices of the triangle whose sides are the lines 2.r + 9y + 17 = 0. // = 7.r - 38, 2i/-x= 2. 42 ANALYTIC GEOMETRY. ' • 60. Find the oqiuitioii of the str;iiglit line passing tlirongh the origin and the intersection of tlie lines '3x — 1*//-[-4 = and o.r-|-4y = 5. Also find the distance between these two points. 61. AVhat is the equation of the line passing through (•^'i; ^i)) ^^^^ equally inclined to the two axes ? 62. Find the equations of the diagonals of the parallelo- gram formed by the lines x = a, x = h, y = c, y^^d. 63. Show that the lines ^=2x + 3, ?/ = 'ox + 4, y = 4.^ + 5 all pass tlirough one point. Hint. Find the intersection of two of tlie lines, and then see if its coordinates will satisfy the equation of the remaining line. 64. The vertices of a triangle are (0, 0), (xi, 0), (.rg, y^). Find the equations of its medians, and prove tliat they meet in one point. 7 65. What must be the value of in if the line y = mx passes through the point (1. 4)? r 66. The line y = mx + 3 passes through the intersection of the lines y =^ x -\- 1 and y = 2x- -|- 2. Determine the value of /«. 67. Find the value of h if the line v/ = Ox + ^ passes through the point (2, 3). 68. What condition must be satisfied if the points (•^1) yi); (^2) Z/2) (^3? ys) lie in one straight line? Hint. Let equation [4] represent the line through (xi, ?/i) and (-'^2, 2/2); then (X3, 2/3) must satisfy it. 69. Discuss equation [5] for the following cases : (i) (xi, yi) is (0, 0), (ii) m = 0, (iii) ni = cc. 70. Discuss equation [6] for the following cases: (i) h = 0, (ii) m = 0, (iii) 7n = 00, (iv) m = 0, and b^=0. 71. Discuss equation [7] for the following cases: (i) a^^h, (ii) rt^O, (iii) a = x, (iv) b = co. the stkaioht link. 43 General Equation of tijk First Dkguee. 42. If any -point P(xi, 7/1) is connected with the origin ; then -^ = cos XOP, i^~ = sin XOF, and OP = \/x{' + yf . OF OF i I ji Hence, if an// two real quantities are each divided by the square root of the sum, of their squares, the quotients are the cosine and sine of some angle. 43. The locus of every equation of the first degree in x and y is a strciight line. Any simple equation in ,r and // can be reduced to the form Aoc+By=C, [9] in which C is positive or zero. Dividing both members of [9] by yj A'-\- r>'-^, we obtain ^ ^ ^' /1^ V^'+5' V^P+7>''' V^-'+/^' Now, by § 42, the coefficients of x and y in (1) are a set of values of cos a and sin a, and the second member, being posi- tive, is some value of ^; (§41). Hence (1) is in the normal form, and its locus is some straight line. Whence the proposition. Cor. 1. To reduce any simple equation to the normal form, put it in the form of [9], and divide both members by the square root of the sum of the squares of the coefficients of X and y. CoK. 2. To construct (1), locate the point '{A, B), con- nect it with the origin, and on this line lay off OS equal to the second member of (1) ; the perpendicular to OiS through Sis the locus of (1). or [9]. 44. The locus of an cquntidu of the first degree in .t and y is called a Locus of the First Order. 44 ANALYTIC GEOMETRY. Exercise 12. Reduce the following equations to the normal form, and thus determine p, or the distance of eacli locus from the origin : 1. 3x- — 2^+11 = 0. 5. y-\-VS = 5x. 2. 3^3 + 5^ — 13 = 0. 6. y+ 19 = 70:. 3. 4x— y— 2 = 0. 7. e.r + ry + ?i = 0. 4. 2x-{-o//=7. 8. 7i//-\-cx — r^^O. Eeduce the following equations to one of the forms [6], [7], [8], and determine by the signs of the constants which of the four quadrants each lodus crosses : 9. y = ^x — 9. 14. 5a;4-4y — 20 = 0. 10. 3x-+2 = 2y. 15. i/ = 6x-\- 12. 11. 4.y = o.r — 1. -^16. )/-\-2 = x—4:. 12. 4?/ = 3a: + 24. 17. a; + V% + 10 = 0. 13. 5x + 3?/ + 15 = 0. 18. a;— VSy — 10 = 0. 19. Discuss equation [9] for the following cases : (i) ^ = 0. (iv) A = cc. (vii) A = B,C=0. (ii) i? = 0. (v) A=C = 0. (viii) A = -B,C = 0. (iii) C = 0. (vi) A = B. 20. Reduce equation [7] to the form of equation [6], and find the value of m in terms of a and b. 21. What must be the value of C that the line 4x — 5y = C may pass through the origin ? through (2, 0) ? 22. Determine the values of A, B, and C, that the line Ax-\- By=C may pass through (3, 0) and (0, — 12). Hint. Since the coordinates of the given points must satisfy the equation, we have the two relations ZA = C and — 12B = C. 23. From [9] deduce [4] by the method used in No. 22. THE STRAIGHT LINE. 45 24. If equations [4] and [9] represent the same line, what are the values of A, B, C, in terms of Xi, i/i, x^, y-i"^ 25. In equation [4] find the values of vi and b in terms of Xi, iji, X2, 1/2- Angles. 45. To find the angle formed by the lines y ^ mx -\- b, and y = m'x -\- b'. Let AB and CD (Fig. 21) represent the two lines respec- tively, meeting in the point P. Let the angle AFC^^cf) ; then, by Geometry, <^ = y — y'. Whence, by Trigonometry, 7n — in' tan = -, , 1 + mm This equation determines the value of <^ Y [10] Fig. 21. Fig. 22. Cor. 1. If the lines are parallel, tan <^= ; hence, 7n= m'. Conversely, if m^=m', <^ = 0, and the lines are parallel. CoR. 2. If the lines are perpendicular, tan <^=x ; hence, 1 4- m )ii' = 0, or ni' — . Conversely, if 1 + m m' = 0, m <^:=90°, and the lines are perpendicular. 46 ANALYTIC GEOMETRY. 46. To find the equatioti of a strnujlit line jjcissiiif/ throiujk the ijoint (a'l, y/i) and (i) 'parallel, (ii) ijerjiendicMlat', to the line y = mx -f- i>- The slope of the required line is m in case (i), and ■ in case (ii) ; and in both cases the line passes through a given point (xi, y^). Therefore (§ 38), the required equation is (i) !/ — I/i = '>^K^—'-^i), i3xercise 13. Find the equation of tlie straight line : 1. Passing through (3, — 7), and || to the line y = 8x — 5. 2. Passing through (5, 3), and || to the line ^y — ^x=l. 3. Passing through (0, 0), and || to the line // — 4x^10. 4. Passing through (5, 8), and || to the axis of x. 5. Passing througli (5, 8), and || to the axis of y. 6. Passing through (3, —13), and _L to the line y=4:X — 7. 7. Passing through (2, 9), and _L to the line 7y-\-23x -5 = 0. 8. Passing through (0, 0), and _L to the line x-\-2y=^l. 9. Perpendicular to the line 5x — 7y + 1 = 0, and erected at the point whose abscissa = 1. 47. To find the eqiiation of a straight line passing through a given point (xi, y^, and malting a given angle ^ with a given line y = mx -\- b. Let the required equation be y — !Ji = '":^'{^ — ^i)j where in' is not yet determined. THE STRAIGHT LINK. 47 Since the required line may lie eiciier as PQ or I'R (Fig. 22), we shall liave (§45), m' — m VI — m' tan <^ = -T— ^ or 1 + fti^^' 1 + tnm' , m ± tan 1 ip ni tan ^ and the required equation is tn ± tail = — 1,

= 0. 4. y — nx = 1 and 2 (// — 1) ^ "•^• Find the angle formed by the lines : 5 . .r -f- ?/ = 1 and ?/ = ,T + 4. 6 . .V + 3 = 2x and ?/ + 3,r = 2. 7. 2x + 3// + 7 = and 3.r — 2// + 4=0. 8. G.r = 2// + r> and // — 3.r = 1 0. 9. X + 3 = and y — V3.r + 4 = 0. 48 ANALYTIC GEOMETRY. 10. Discuss equation [11] for the cases when <^ = 0* and <^ = 90°. Note. The learner should solve the next four 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 maliing the angle 45° with the line 2x — 3y + 5 = 0. 12. Passing through the point ( — 2, 1), and making the angle 45° with the line 2// = 6 — 3x. 13. Passing through that point of the line y^=2x — l for which x = 2, and making the angle 30° with the same line. 14. Passing through (1, 3), and making the angle 30° with the line x — 2// + 1 = 0- • 15. Prove that tlie lines represented by the equations Ax + Bi/-{-C = 0, A'x + B'i/-\-C' = are parallel if AB'=^A'B ; i:>erpendicular, if AA'^ — BB'. y 16. Given the equation 3aj + 4y-|- 6 = 0; show that the general equations representing (i) all parallels and (ii) all perpendiculars to the given line are (i) 3.r + 4y + 7i = 0. (ii) 4x — 3y + /i = 0. 17. Deduce the following equations for lines passing through (xi, ?/i) and (i) parallel, (ii) perpendicular, to the line y = vix + b. (i) y — mx^ ?/i — mxi. (ii) my + x = my^ + x^. 18. Write the equations of three lines parallel, and three lines perpendicular, to the line 2a; + 3^ + 1=^0. THE STRAIGHT LINE. 49 y 19. Among the following lines select parallel lines ; per- pendicular lines ; lines neither parallel nor perpendicular : (i) 2a; + 3?/ — 1 = 0. (v)a;-?/ = 2. (ii) ?>x — 2ii = 20. (vi) 5 (a; + ?/) - 1 1 = 0. (iii) ^x-\-^y = 0. (vii) cc = 8. (iv) 12x = 8^ -I- 7. (viii) 7/ + 10 = 0. p^ 20. Prove that the angle <^, between the lines Ja; + % + C' = and A'x + BUj + C'= 0, is determined by the equation , A'B — AB' 21. From the preceding equation deduce the conditions of parallel lines and perpendicular lines given in Xo. 15. Find the equation of the straight line : V22. Parallel to 2x -f-3^ + 6 =0, and passing through (5, 7). ^23. Parallel to 2x-{- 1/ — 1 = 0, and passing through the intersection of 3x + 2^ — 59 ^ and 5x — 7v/ + 6 = 0. y^ 24. Parallel to the line joining ( — 2, 7) and (—4, — 5), and passing through (5, 3). 25. Parallel to y = mx-\-b, and at a distance d from the origin. 26. Perpendicular to Ax-\- Bi/-{- C=^0, and cutting an intercept b on the axis of i/. X ■?/ 27. Perpendicular to - + j- = 1, and passing through (a, b). 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. 50 ANALYTIC GEOMKTRY. 30. The vertices of a triangle are ( — 1, — 1), ( — 3, 5), (7, 11). Find the equations of its altitudes. ^*rove that the altitudes meet in one point. y^ 31. Find the equation of the perpendicular erected at the middle point of the line joining (5, 2) to the intersec- tion of a- + 2y — 11 = and 9a; — 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), (— -1, 13). Prove that these perpen- diculars meet in one point. -^3. The equations of the sides of a triangle are x-\-i/-\-l = 0, 3.r + 5^ + ll = 0, cr + 2^ + 4 = 0. Find (i) the equations of the perpendiculars erected at the middle points of the sides ; (ii) the coordinates of their common point of intersection ; (iii) the distance of this point from the vertices of the triangle. 34. Show that the straight line passing through («, h) and (c, d) is perpendicular to the straight line passing through (b, — a) and ((1, — c). 35. What is the equation of the straight line passing through (iPi, ?/i), and making an angle ^ witli the line Ax + Bij+C^O? Distances. 48. Find the distance from the point ( — 4, 1) to the line 3a; - 4y + 1 = 0. . Ans. 3. The required distance iS the length of the perpendicular from the given point to the given line. The method that first suggests itself is to form the equation of this perpen- dicular, find its intersection with the given line, and com- pute the distance from this intersection to the given point. Let this metliod be followed in solving the above problem and the first iiv(; pro1)Ienis of Exercise 15. THE STKAIGHT LINE. 51 49. To find tlte distance from the point (rt\, //i) to tJia Hue X cos a 4".'/ siu a^=j). Let the line a cos a + // sin a=:y/, (1) which is evidently parallel to the given line, pass tlirough the given point (oi\, y/j); then we have Xi cos a + i/y sin a =y>'. Therefore, x^ cos a + .'/i sin a — p^p — p- But p'' — p equals numerically the required distance. Tliereforc, the distance from the point (a-,, ij-^ to tlie line X cos a-\- ij sin a =7> is obtained by substituting Xx for x and 2/1 for y in the expression x cos u + // sin a — p. CoR. 1. The distance as obtained from the formula will evidently be positice or negatwe according as the point and origin are on o})posite sides of the line, or on the same side. CoR. 2. If the equation of the line is xix -\- Bij = C, and d denotes the distance from (.r,, ij{) to the line ; then, evidently, Axx -t B\)x — € p..,-. Hence, to find the distance from the point (x^, y^) to the line Ax-\-B}j— C, write x^ for x, and iji for y i n the e x- pression Ax-\- By— C, and divide the residt by ^A--\- />'-. For example, let (—1, 3) be the point, and 2x-\-4: = 'Sy the equation of the line. Putting this equation in tlie form of [9], we have -2x-j-3y = 4. -2 (-1)4-3x3-4 , ,_ Whence, d = ^pf^l^^^ = + /rr Vl3. Hence, the point and origin are on opposite sides of the line. If only the length of d is sought, its sign nuiy be neglected. 52 ANALYTIC GEOMETRY. Exercise 15. - 1. Find the distance from (1, 13) to the line Sx=^y — 5. 2. Find the distance from (8, 4) to the line y^=1x — 16. 3. Find the distance from the origin to the line 3x- + 4y = 20. 4 . Find the distance from (2, 3) to the line 2x-\-y — A. =0, 5. Find the distance from (3, 3) to the line y=^^x — 9. 6. Prove that the distance from the point (x^, y^) to the line y = mx -(- & is v/i — mxi — h d^^'- , — > VI +m^ the upper or lower sign being used according as b is posi- tive or negative. Express this result in the form of a rule for practice. 7. Find the distances from the line 3a: -f 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: 3a;4-4y + 15 = 0. 3a; + 4?/— 5 = 0. 3a; + 4//+ 10 = 0. 3a; + 4^—10 = 0. 3a; + 4y+ 5 = 0. Sx-\- ^y — 15 = 0. 3x^Ay =0. 3x + 4?/ — 20 = 0. Find the following distances: 9. From the point (2, — 5) to the line y — 3x = 7. 10. From the point (4, 5) to the line 4:y-\-5x = 20. 11. From the point (2, 3) to the line x-\-y^l. .12. From the point (0, 1) to the line Sx — 3y = l. 13. From the point (—1, 3) to the line Sx -{-4y-\-2 = 0. THE STRAIGHT LINE. 53 -/ 14. From the origin to the line ?>x-\-2i/ — G^O. 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 = 3x — 4 and y=^ox-\-2. 17. From the origin to the line a(x — a) + b()/ — ^) = 0. 18. From the points (a, h) and ( — a, —b) to the line ^+1=1. a b 19. From the point {a, b) to the line ax -\-bij=^0. 20. From the point (Ji, k) to the line Ax-\- Bf/-\- C = D. Find the distance between the two parallels : 21. 3x' + 4^ + lo = and 3a: + 4y + 5 = 0. ^22. 3a; + 4y + lo = and 3.>; + 4y — 5 = 0. 23. Ax-\-Bt/=C and Ax-\-B/j=C'. 24. Ax + Bi/=Caud—Ax — Bi/=C'. 25. y = 5x — 7 and y = 5x -\- 3. 26. - + f = 2 and - + 7 = 0- a b a b Z 27. Show that the locus of a point which is equidistant from the lines Sx + 4y — 12 =0 and 4.r + 3// — 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 constant is a straight line. 54 ANALYTIC GEOMETRY. r / \ / / \ > ^r L I v" ~1 IX Areas. 50. To find fhe area of a triangle, having given its vertices. Solution I. Let PQR (Fig. 23) be the given triangle, and let the vertices of PQR be (x^, y{), (x^, y^, {x^, y^, re- spectively. Drop the perpendiculars PM, QN, RL; then area PQR = PQNM+ RLNQ — PMLR. By Geometry, PQNM= ^ NM{MP -\- NQ) ^^(a-i — 3-2)(//i + y2). Similarly, RL]S'Q = ^{x^-x,)iy, + y.^. PMLR = ^{x,-x,)(y, + y,). Substituting these values, we have Fig 23. ° area PQR = ^ \_{j\ — x^) {y. + ?/i) + (.^2 — x^){yz + .^2) — (^1 — ^?)(!h + //i)] = i [— -'"i.'/i + ^i//2 ~ -"^zlh + -^'"J/i ~ •'•1//3 + ■'•;;//!]• .-. area = i [ici (2/2 - */a) + x-i (j/3 -!/i)+ .^s (yi-2/2)]. [13] Solution II. Since the area of a triangle is equal to one half the product of its base and its altitude, this prob- lem may be solved as follows : (i) Find the length of any side as base, (ii) Find the equation of the base. (iii) Find the distance of the base from the opposite vertex, (iv) Multii)ly this distance by one half the base. Exercise 16. Find the area of the triangle whose vertices are the points : 1. (0,0), (1,2), (2,1). 3. (2,3),(4,-5),(-3,-G). 2. (3, 4), (-3, -4), (0,4). 4. (8, 3), (-2, 3), (4, -5). 5. (a,0), (-«,0), (0,^.). THE STKAIGIIT LINE. 6. Compare the ionuula 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,,!/,), (x„,!/,). 11. (2,-r,), (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, b), {h, a), (c, c). 16. {a,h), {h,a), {c, — <'). 17. Find tlie angles and the area of the triangle whose vertices are (3, 0), (0, 3 V3), (G, 3 V3). What is the area contained by the lines : 18. a: = 0, y = 0, 5j-1-4// = 20? 19. x-\-l/=l, .r — //=(), v/=0? 20. a; + 2// = 5, 2.r + // = 7, y = x-\-l? 21. a' + y = 0, .r = //, // = 3ft ? 22. // = 3a-, If = 7.r, // ^ r- ? 23. a- = 0, >j = 0, j--4 = 0, // + (;=0? 24. 3x + // 4- 4 = 0, 3.r - 5// + 34 = 0, 3.i- - 2// +1=0? 25. X- 5// + 13 = 0, 5.r + 7// + 1 = 0. 3.r + //-9 = 0? 26. X — I/ — 0, X + // = 0, X— i/ — (f, X -\- // = b ? 50 ANALYTIC GEOMETRY. Find the area contained by the lines: 27. x = 0, y = 0, y^:^mx-\-b. 28. a; = 0, tj^Q, - + '{ = 1. 29. x = 0, v/ = 0, Ax-{-Bi/-\-C = 0. 30. 7/== 3a; — 9, y — Sx-'rS, 2i/ = x — 6, 2?/ = a; + 14. 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= 5x — 6 for which x = 4:, a: = 7 ? Exercise 17. (Revie\Ar.) 1. Deduce equation [7], p. 39, from equation [6]. 2. The equation y = mx -\-b \^ not so general as the equation Ax-\- By-\- C=0, because it cannot represent a line parallel to the axis of y. Explain more fully. Determine for the following lines the values of a, b, y, p, and a : 3. a; = 2. 6. x+\/3y = 2. 4. x = 7/. 7. x — \/3y = 2. 5. 7/ + l = V3(.x + 2). 8. V3x — ^ = 2. 9. Find the equations of the diagonals of the figure formed by the lines 3a- — y + 9 = 0, 3x = y—l, 5a; + 3// ^18, 5x + 3y = 2. What kind of quadrilateral is it ? Why ? 10. Find the distance between the parallels 9x=^y-\-l and 9a; = ?/ — 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. 57 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 x. 15. Parallel to the line ox + 6^ -}- 8 == 0. 16. Intercepting on the axis of y the distance — 10. " 17. Passing through the point halfway 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 (xi, y^) and (a-j, ya)? and passing through (ccj, y^. 20. The lines passing through (8, 3), (4,3), (—5,-2). 21. A line passing through the intersection of the lines 2ic + 5y + 8 = and 2>x — 4v/ — 7 = 0, and X to the latter line. 22. A line JL to the line 4x — ?/ =; 0, and passing through that point of the given line whose abscissa is 2. 23. A line || to the line 3a:-|-4y = 0, and passing through the intersection of the lines x — 2y — a = and x-\-'6y — 2a = ^. 24. A line through (4, 3), such that the given point bi- sects the portion contained between the axes. 25. A line through (a-j, ?/i), 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), sucli that the jwrtion be- tween the axes is divided by the given point in the ratio 5:3. 58 ANALYTIC GEOMETRY. 29. A line dividing the distance between ( — 3, 7) and (5, — 4) in the ratio 4:7, and JL to the line joining these points. 30. The two lines through (3, 5) making the angle 45° with the line 2x — 3i/ — 7=0. 31. The two lines through (7, — 5) that make the angle 45° with the line ijx — 2i/-\-o = 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 that pass through the origin and trisect the portion of the line x-\- i/ = l included between tlie axes. 34. The two lines || to the line 4^- -1-5^ + 11 = 0, at the distance 3 from it. 35. The bisectors of the angles contained between the lines // = 2x-{-4: and — ]/^3x -\-6. Hint. Every point in the bisector of an angle is equidistant from tlie sides of the angle. 36. The bisectors of the angles contained between the lines 2x — 5y == and 4a- + 3y = 12. 37. The two lines that pass through (3, 12), and whose distance from (7, 2) is equal to V58. 38. The two lines that pass through ( — 2, 5), and are each equidistant from (3, — 7) and (—4, 1). Find the angle contained between the lines: 39. // + 3 = 2x and // + 3.r = 2. 40. >/ = 5x — 7 and 5y -{- Ji — 3 = 0. THE STRAIGHT JASK. 59 Find the distance : 41. From the intersection of the lines 3x-\-2i/-\- 4 = 0, 2x + 5y + 8 = to the line y = 5x + 6. X II 42. From the point (A, Ic) to the line -4--= 1. ah 43. From the origin to the line hx -\- hy = c'^. 44. From the point (a, 0) to the line y=:??u-|- -• m Find the area included by the following lines: 45. x = //, X -\- y =^ 0, X = r. 46. x-\-i/ = k, 2x = y + /.-, 2// — X -\- k. 47. - + '^ = 1, y = 2x-\-h, x=2ii-\-a. a h ^ ' 48. ?/=4.r + 7 and the lines that 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 tlie triangle formed by the lines a: — 5^+1 1 =0, 11a- -|-C// — 1 =<>. a; + ?/ + 4 = 0. 50. Find the area of the quadrilateral whose vertices are ((VO), (0,5), (11, 9), (7,0). 51. What point in the line 5.)- — 4// — 28 = is equidis- tant from the points (1, 5) and (7, — 3) ? 52. Prove that the diagonals of a square are perpen- dicular to each other. 53. Prove that the line joining the middle points of two sides of a triangle is parallel to the third side. 54. What is the geometric meaning of the equation xy = 0? 60 ANALYTIC GEOMETKY. 55. Show that the three points (3a, 0), (0, 'SO), (a, 2U) are in a straight line. 56. Show that the three lines 5a; + 3//— 7 = 0, 3a; — Aij — 10 = 0, and x-\-2i/^0 meet in a point. 57. What must be the value of a in order that tlie three lines 3x + ?/ — 2 = 0, 2x—y — S=0, and aa; + 2y — 3 = may meet in a point ? What straight lines are represented by the equations : 58. x^-\-{a — b)x — ah=^0? 59. xy-[-hx-\-(.uj-\-ah—^'^ 60. x~ij = xif ? 61. 14a;^ — 5a-^ — y- = ? 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 point equidistant from the points (a;i, yi) and {x^, 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 k. 66. The locus of a point so moving that the sura of its distances from the lines Ax-\-By-\- C=0, A'x-\-B'y-\- C"=0 shall be constant and equal to k. 67. The locus of tlie vertex of a triangle, having given the base and the difference of the squares of the other sides. THE STRAIGHT LINE. 61 SUPPLEMENTARY PPtOPOSITIONS. Lines passing through One Point. 51. If S=0, S' = represent the equations of any tivo loci with the terms all transposed to the left-hand side, and k denotes an arbitrary constant, then the locus represented by the equation S-\- kS' = pxisses throuc/h every point common to the two given loci. For if any coordinates satisfy the equation *S'=0, and also satisfy the equation *S' = 0, they must likewise satisfy the equation S + kS^ = 0. For what values of k will the equation ^'-|- A-,S'' = rep- resent the lines *S' = and >S" = 0, respectively ? 52. Find the equation of the line joining the point (3, 4) to the intersection of the lines 3a? — 2^+17=0 and a; + 4?/ — 27 = 0. The method of solving this question that first suggests itself is to find the intersection of the given lines and then apply equation [4], p. 37. Another method, almost equally obvious, is to employ equation [5], which gives at once y — 4 = m(x — 3), and then determine m by substituting for x and y the coor- dinates of the intersection of the given lines. The following method, founded on the principle stated in § 51, 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 3x - 2y + 1 7 + A-(.r -f Ay - 27 ) - 0. 62 ANALYTIC OEOMETKY. And since the line passes through (3, 4j, we must have 9 - 8 + 17 + /c(3 + 16 - 27) = 0, whence, k = ' • Therefore, 12^; — 8// + 08 -\-9x-\- 36i/ — 243 = 0, or ox + 4// — 25 = 0. This is the equation of the required line 53. . If the equations of three straight lines are Ax +/.'// -f-C =0, A'x ^ />"// + 6" =0, A"x +/>'"// + C"' ^0, and we can find three constants, I, m, n, so that the relation l{Ax-\- By-[- C)^m{A'x+ir ij^C)-\-n{A"x-]- B" i/-^ C")^0 is identically trne, that is, true for all values of x and y, then the three lines 'meet in a 'point. For if the coordinates of any point satisfy any two of the equations, then the above relation shows that they will also satisfy the third equation. 54. To find the equation of the bisector of the angle between the two lines X cos a + // sin a =: p, and X cos a' -\- y sin a' =^jj'. There are evidently two bisectors : one bisecting the angle in which the origin lies ; the other bisecting the supplementary angle. Now, every point in either bisector is equally distant from the sides of the angle. Let (x, y) be any point in the bisector of the angle that includes the origin ; then (§ 49) X cos a+ //sin a— 7^:= a- cos a' 4-y sin a' — p\ (1) Since (x, y) is any point in this bisector, (1) is its equation. THE STRAUJIIT LINE. 63 The equation of the other bisector is xcosa-{- 1/ sin a — 2^ — — (;r cos a' + y sin a* — //). (2) To distinguish equations (1) and (2) we note that in tlie first the constant terms in the two members have like signs ; while in the second the constant terms have unlike signs. CoR. 1. If the equations of the lines are in the form Ax +Bi/ = C, A'x + B'l/ = C, the equations of the bisectors are evidently Ax + Btj- C A'jr, + By - C V^2+^^ V^'2+"j5'i 1[4] Equation [14] represents the bisector of the angle in which the origin lies, or of its supplementary angle, accord- ing as we take the upper or lower sign. For example, let the equations of the lines be 2x = 4?/ -f- 9, and 5{/^=ox — 7. Putting these equations in the form of [9], we have 2x — 4// = 9, and 3x — 5// = 7. Hence, the equations of the bisectors of tlieir included angles are ^ . ^ ., ^ ^ 2x — 4y — 9 ox — oy — / V20 ~ V34 in which the upper sign gives the equation of the bisector of the angle in which tlie origin lies. CoK. 2. If S=0 and *S"=:0 represent two simple equa- tions in tlie normal form, with the terms all transposed to the first members, then the equations of the bisectors of their included angles may be written ^=±^', or .S'^^' = 0. 64 AJTALYTIC GEOMETRY. Exercise 18. Find the equation of the line passing through the inter- section of the lines 3x — 2y + 1^ = 0, x + 4y — 27 = 0, and : 1. Passing also through the origin. ' 2. Parallel to the line x + 2i/-\-3 = 0. 3. Perpendicular to the line 6x — 5?/ = 0. 4. Equally inclined to the two axes. 5. Find the equation of the line parallel to the line x = y, and passing through the intersection of the lines y = 2x-\-l and i/-\-3x^= 11. 6. Find the equation of the straight line joining (2, 3) to the intersection of the lines 2ic + 3^+l=0 ■dnd3x — 4i/ = 5. 7. Find the equation of the straight line joining (0, 0) to the intersection of the lines 5x — 2i/-\-3=0 and 13x-\-2j = l. 8. Find the equation of the straight line joining (1, 11) to the intersection of the lines 2x-{-5i/ — 8 = 2ind3x — 4>/ = 8. Find the equation of the straight line passing through the intersection of the lines Ax -\- Bi/ -\- C^O and A'x-{- B't/-{-C' = 0, and also: 9. Passing through the origin. 10. Drawn parallel to the axis of x. 11. Passing through the point (xi, t/i). 12. Find the equation of the straight line passing through the intersection of 5ar — 4?/ + 3^0 and 7a;-[- lly— 1^0, and cutting on the axis of y an intercept equal to 6. THE STRAIGHT LINE. 65 13. Find the equation of the straight line passing through the intersection of y^=-lx — 4 and y = — 2a3 + 5, and form- ing with the axis of "a; the angle 60°. 14. The distance of a straight line from the origin is 5 ; and it passes through the intersection of the lines 3x — 1y + 11 = and 6x + 7^ — 55 = 0. What is its equation ? 15. What is the equation of the straight line passing through the intersection of bx-{-ay=^ab and y=mx, and perpendicular to the former line ? Prove that the following lines are concurrent (or pas?' through one point) : 16. ?/ = 2a- + l, yz=x-{-3, ?/=: — 5x + 15. 17. Ax—2y — ^ = 0, 3x-7/ + ^ = 0, 5x — 2y — l = 0. 18. 2x — y = 5, 3cc — y = G, 4x — y = 7- 19. What is the value of 77i if the lines a a meet in one point ? 20. When do the straight lines y = mx -\-b,y=^ vi'x -\~ h', y^m"x-\-b" 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 the two points of trisection for each median. 24. Prove that the bisectors of the three angles of a tri- angle meet in one point. 06 ANALYTIC GEOMETKY. 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 3j^ + /y + 4=0, 3.x- -5^ + 34=^0, 3a; — 2^ + l:=0. Find the lengtlis of its altitudes. What are the equations of the lines bisecting the angles between the lines : 27. 3x — 4// + 7 = and 4.r — 3y + 17 = ? 28. 'Sx + 4// — 9 = and 12x + 5.7 — 3 = 0? 29. i/ = 2.r — 4:and2t/ = x-\-10? 30. x-\-f/ = 2 Sindx—>/ = 0? 31. y = mx + b and y = in'x -\- //? 32. Prove that the bisectors of the two supplementary angles formed by two intersecting lines are perpendicular to each other. Equations representing Straight Lines. 55. A homogeneous eqvatwn of the nth degree rejjresents n straight lines through the origin. Let the equation be j.r" + i;x"-'n + 6'.r" y + + Ay = 0. Dividing by .///", we have If ^'i5 >'2> '*3) '>'n denote the roots of this equation, then the equation, resolved into its factors, becomes THE STRAIGHT LINE. 67 and therefore is satisfied when any one of these factors is zero, and in no other cases. Therefore, the locus of the equation consists of the n straight lines x — i\y = 0, x — r.,i/ = 0, , x — r„y = 0; and these lines evidently all pass through the origin. 56. To find the angle between the two stniiyht lines rep- resented hxj the equation Ax^ + Cxy -\- Bif =■ 0. Solving the equation as a quadratic in x, we obtain 2 Ax -I- ((7 ± ^C"-4AB) t/ = 0. Hence, the slopes of the two lines are 2A . 2A — C- VC2 - 4:AB ■ — C + VC'2 - 4:AB Therefore, , VC'2— 4^^ , A VI — m ^ , mm = — ; B B and (equation [10], p. 45), m—m' ^C^—AAB tan (b = ~-, ,= — • 57. To find the condition that the general emtation of the second degree may represent two straight lines. We may write the most general form of tlie equation of the second degree as follows : Ax' + By- + Cxy + Bx + By + /<^ = 0. (1 ) That this equation may represent two straight lines, its first member must be the product of two linear factors in X and y ; that is, the equation can be written in the form (lx-]-7ny-\-n) (px -\- qy-\-i') = 0. (2) Equating coefficients in (1) and (2), we obtain Ij) = A, niq = />, n r = F. Iq -f- nip = 6', //• + np = Z>, nir -\- nq = E. 68 AXALYTIC GEOMETRY. The product of C, D, and E is CDE = 'llmvpqr -\- lp{ii?(f -\- m^i^) + mq(Pi^ + n^p^y = 2ABF-\- A {E^ — 2BF) + B(I>'- 2AF) + F{C^ — 2AB). Hence, the required condition is F{C^ — AAB) + AE' + BD- - CDE= 0. (3) Exercise 19. 1. Describe the position of the two straight lines repre- sented by the equation Ax^-{-Cxi/-\- Bi/'-^-\-Dx-{-Ei/-\-F=0, when (i) A=C = I) = (), (ii) B= C = E=0. 2. When will the equation ax>/-\-bx-\-ci/ -j-cl^O repre- sent two straight lines ? 3. Find the conditions that the straight lines represented by the equation Ax--\- Cxy-\- Bt/^^0 may be real ; imagi- nary ; coincident ; perpendicular to each other. 4. Show that the two straight lines x^ — 2xy sec 6+ ^^ = make the angle 6 with each other. Show that the following equations represent straight lines, and find their separate equations : 6. x'' — 2xy — ^if-\-2x — 2y-]-l=0. 7. a;2 — 4a!y + 3/ + 6?/— 9 = 0. 8. Show that the equation x^-\-xy — 6^^ -\-lx-\- Sly — 18 = represents two straight lines, and find the angle between them. Determine the values of K for which the following equa- tions will re})resent in each case a pair of straight lines. Are the lines real or imaginary ? 9. 12x^ — 10xy-\-2y^-{-llx — 5y + K=0. THE STRAIGHT LIXE. 69 10. 12x^ + Kx>/-{-2if-\-llx — 5i/-\-2 = 0. 11. 12a;2 + 36a;?/ + Ay + 0./;+6y+3 = 0. 12. For what value of if does the equation Kxy-\-bx + 3y+2 = represent two straight lines? Problems on Loci involving Three Variables. 58. A trapezoid is formed by drawing a line parallel to the base of a given triangle. Find the locus of the intersec- tion of its diagonals. If ABC is 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), (h, 0), (0, c), respectively. The equations oi AC and BC are, respectively, - + ^ = land ? + ^ = l. a c be Let y^m be the equation of the line parallel to the base, and let it cut ^C in D, ^C in ^; then the coordinates of D and E, respectively, are f ac — am \ ^ (be — biyi \ ( J m 1 and ( > m I • (1) X — a be — bm — ac ^ ' If P is the intersection of the diagonals, then the coordi- nates X and y of the point P must satisfy both (1) and (2) ; by solving these equations, therefore, we obtain for any particular value of m the coordinates of the point P. But what we want is the algebraic relation that is satisfied Hence, the equation of the diagonal BD is y cm X — b ac — am — be and the equation of the diagonal AE is y cm 70 ANALYTIC GKOMETllV. by the coordinates of P, whatever the mine of m may be. To find this, we have only to eliminate m from equations (1) and (2). By doing this we obtain 2cx + {a + h)y ={a-\- b)c, X , >/ . i(« + ^) c We see from the form of this equation that the required locus is the line that joins C to the middle point of AB. Remark. The above solution should be studied till it is understood. In problems on loci it is often necessary to obtain relations that involve not only the x and y of a point of the locus which we are seek- ing, but also some third variable (as m in the above example). In such cases we must obtain two equations that involve x and y and this third variable, and then eliminate the third variable ; tlie resulting equation will be the equation of the locus required. Exercise 20. 1. Through a fixed point any straight line is drawn, meeting two given parallel straight lines in P and Q ; through P and Q straight lines are drawn in fixed direc- tions, meeting in R. Prove that the locus of i2 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 B, one on each of the axes ; if Z7and Fare two variable points, one on each axis, so taken that 0U-{- 0F= OA-^OB, find tlie locus of the intersection of A V and B U. 4. Find the locus of the middle points of the rectangles that may be inscribed in a given triangle. 5. If PP\ QQ' are any two parallels to the sides of a given rectangle, find the locus of the intersection of P'Q and PQ'. CHAPTER III. THE CIRCLE. Equations of the Circle. 59. Tlie Circle is the locus of a point which moves so tliat its distance from a fixed point is constant. Tlie fixed point is the centre, and the constant distance the radius, of the circle. Note. The word "circle," us here detined, means the same thing as "circumference" in Elementary Geometry. This is the usual meaning of "circle" in the higher branches of Mathematics. 60. To find the equation of a circle, having given its centre («, b) and its radius r. E Fig. 24. Let C (Fig. 24) be the centre, and P any point (x, y) of the circumference. Then it is only necessary to express by an equation the fact that the distance from P to C is con- stant, and equal to r : the required equation evidently is (§ 6) (X - aY +(!/- W = /•■-. [15] 72 ANALYTIC GEOMETRY. If we draw CR || to OX, to meet the ordinate of P, then we see from the figure that the legs of the rt. A CPR are CR = x — a, PR = 1/ — b. If the origin is taken at the centre, then a = b:=0, and the equation of the circle is ic2 + 2/2 = r2. [16] This is the simplest form of the equation of a circle, and the one most commonly used. If the origin is taken on the circle at the point A, and the diameter AB is 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 b in [15], and reducing, we obtain x"+ y^ = 2rx. [17] Why is this equation without any constant term? 61. The locus of any equation of the second degree in x and y in which the term in xy is wanting and the coefficients of x^ and y^ are equal is a circle. Any such equation can evidently be reduced to the form x''-^if-\-2Dx-\-2Ey-{-F=0. (1) Therefore, {x^ + 2Dx + 2)2) + (^fJ^2Ey + U^) = D" -\- E^ — F, or (x-\-Dy+{y-\-Ey=(iy-\-E^ — F). (2) Now, from [15] it follows that the locus of (2) is a circle whose centre is ( — i>, — E), and whose radius is \/l)-^ + E^ — F. Cor. If Z)2 + ^2 > F, the radius is real and the circle is readily constructed. If D^ -\- E^=^F, the radius is zero, and the locus is the single point ( — D, — E). If D^-\-E'^ < F, the radius is imaginary, and the equation represents no real locus. THE CIRCLE. 73 '62. Ajit/ point (h, k) is without, on, or within the circle x^-{- y^:= r^, according as h^ -\- k- >, ==, or < r^. For a point is without, on, or within a circle, according as its distance from the centre >, =, or < the radius. Exercise 21. Find the equation of the circle, taking as origin : 1. The point B (Fig. 24) ; and BA as axis of x. 2. The point D (Fig. 24) ; and DE as axis of y. 3. The point E (Fig. 24) ; and ED as axis of y. 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 (Ji, k), radius V/i^ + H 10. Determine the centre and radius of the circle ic^ + Z — 10^+12^+25 = 0. Here (x - 5)2 + (y + 6)2 = 36. . •. a = 5, 6 = - 6, r = 6. Determine the centres and radii of the following circles: 11. x''-\-f/ — 2x — 4.y = 0. 17. 6x2 — 2y(7 — 3y)=0. 12. 3a;2+3/— 5x— 7?/+l=0. 18. x-+if = 9k\ 13. x''-\-y^ — Sx = 0. 19. (x + yy- + (x — yy-=8k-. 14. cc2 + /+8.r = 0. 20. x--\-y- = a- + b^ 15. x^-{-y--8y = 0. 21. x^ + y' = k{x -\- k). 16. x^-\-f-j-Sy = 0. 22. x'-\-y- = hx-\-ky. 74 ANALYTIC GEOMETRY. 23. When are the circles x- -\- if + />./• -\- Eij -[- 6' = and ^•- + //' + J^>'-'^ + ^^'// + C' = concentric ? 24. Wliat is the geometric meaning of the equation {x-af-\-{ij-l>f = i'^ 25. Find the intercepts of the circles (i) x2 + /-8.«-8y+ 7 = 0, (ii) a;- + / — 8a; — 8y+16 = 0, (iii) x' + }j- — 8ic — 8y + 20 = 0. Putting 2/ = in each case, we Iiave in case (i) x- — 8x + 7 = 0, whence x = 1 and 7; in case (ii) x'^ — 8x + 16= 0, whence x = 4; in case (iii) x- — 8x + 20 = 0, whence x = ± V— 4. Putting X = in each case, we obtain for y values identical with the above values of x. The geometric meaning of these results is as follows: Circle (i) cuts the axis of x 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 (iii) does not 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 tlie language of Geometry conform more exactly to that of Algebra, then in this case we should say that the circle meets the axes in imaginary points. This form of statement, however, must be understood as simply another way of saying that the circle does not meet the axes. Find the centres, radii, and intercepts on the axes of the following circles: 26. a-2 4-;//2_5;y — 7y4-6 = 0. 27 . x" + / — 12x — 4?/ + 15 = 0. 28. a-- + if — 4 J- — 8// = 0. 29. x' + // - (;./• + 4// + 4 = 0. 30. x' + y- + Tlx — 1 8y + 57 = 0. THE CIRCLE. 75 31. Under what conditions will the circle y^ -\- if -\- Dx -\- Eif-\- C = (i) touch the axis of j! ? (ii) touch the axis of 1/ ? (iii) not meet the axes at all ? 32. Show that the circle x'^-{-ir -\-10x — 10 i/i- 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 3x-\- u = 25 cut the circle x- -\- if = 65 ? 34. Find the points common to the loci x'^-\-if=^^ and y = 2x — 4. 35. The equation of a chord of the circle a--+ ?/- = 25 is y^2x-\- 11. Find the length of the chord. ~ X 1/ 36. The equation of a chord is - -f- 'r ^= 1 j that of the circle is x^-\- )f = 't'^. Find the length of the chord. 37. Find the equation of a line passing through the centre of a;^ + ^^ — 6a; — ^y = — 21 and perpendicular to a; + 2// = 4. 38. Find the equation of that chord of the circle x--\- if- = 130 that passes through the point for wliich the abscissa is 9 and the ordinate negative, and that is parallel to the straight line 4.x — 5// — 7 = 0. 39. What is the equation of the chord of the circle a;2_|_^-'_-277 tliat passes through (3, — 5) and is bisected at this point ? 40. Find the locus of the centre of a circle passing through the points (xi, y^) and (j-.^, //o). 41. What is the locus of the centres of all the circles that pass through the points (5, 3) and ( — 7, — 6) ? 76 ANALYTIC GEOMETRY. Find the equation of tlie circle : 42. Passing through the points (4, 0), (0, 4), ((5, 4). 43. Passing through the points (0, 0), (8, 0), (0,-6). 44. Passing through the points (— 6, — 1), (0, 0), (0, — 1). 45. Passing through the points (0, 0), (—8a, 0), (0, 6a). 46. Passingthroughtliepoints(2, — 3),(3, — 4),(— 2, — 1). •y 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, 6), 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 origin. 52. Passing through the origin, and cutting the lengths a, h from the axes. 53. Passing through (5, 6), and having its centre at the intersection of the lines y = lx — 3, 4^/ — 3a; = 13. : 54. Passing through (10, 9) and (5, 2 — 3V6), and hav- ing its centre in the line 3x — 2y — 17 = 0. 55. Passing through the origin, and cutting equal lengths a from the lines x = y, x-\-y = 0. 56. Circumscribing the triangle whose sides are the lines y = 0, y = mx-\-b, - + ^=1. 57. Having for diameter the line joining (0, 0) and (x^, y^). 58. Having for diameter the line joining (x^, ?/i)and(iC2, y^- 59. Having for diameter the line joining the points where y ^= nnx meets ic^ -|- y/^ = Irx. 60. Having for diameter the common chord of the circles x^ -\-y^^^ r^ and (x — a)^ -\-y^ = r\ THE CIRCLE. 77 Tangents and jSTormals. 63. Let QPQ' (Fig. 25) represent any curve. If the secant QPR is turned about the point P until the point Q approaches indefinitely near to P, then the limiting position, TT, of the secant is called tlie Tangent to the curve at P. Fig. 25. Fig. 26. 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, perpendicular to the tangent TT', is called the Normal to the curve at P. Let the curve be referred to the axes OX, OY, and let M be the foot of the ordinate of the point P. Let also the tangent and the normal at P meet the axis of x in the points T, N, respectively. Then TM is called tlio Sub- tangent for the point P, and MN is called the SubnormaL 64. To find the equation of the tangent to the circle x^-\-y'^=^ r"^, at the point of contact (xi, y^. Let P (Fig. 26) be the point (x^, ?/i), and ()any other point (^2j 2/2) of the circle. Then the equation of the secant PQ is y — lh^ lh — yi 0) 78 ANALYTIC GEOMETRY. Now, since (xi, ij^) and (jv, y.^ are on this circle, we have Subtracting, (x^^ — oc^^) + (jh" — y^) = 0. _ Factoring, (a^ — Xi) (.r., + x,) + (^, — y ,) (//, + ?/i) = 0, Whence, by transposition and division, we have y^ — yi _ _ a-2 + a'i a-2 — a-i v/2 + y\ By substituting in (1), the equation of the secant becomes y ~ yi _ •'^2 + a^i Now let Q coincide with P, or a-g^^u y^^^Uu ^^^^ secant becomes a tangent at P, and the equation becomes y — yi ^ a-1 ^ — ^1 Z/i' or ^ia; + z/iZ/ = •'^1^ + 2/1". And, since x^ -\- yi^ = r"^, we obtain Xi(ic+yii/ = r% [18] which is the equation required. NoTK. If we had put Xn = xi, 1/2 = 2/1, in (1) before we introduced the condition that (xi, 2/1) and (X2, 2/2) were on the circle, the slope of the tangent would have assumed the indeterminate form -• The above method of obtaining the equation of the tangent to a circle is applicable to any curve whatever. It is sometimes called the secant method. Equation [18] is easily remembered from its sym- metry, and because it may be formed from x^ + y- — r^ by merely changing x- to XiX, and y"^ to yiy. THE CIRCLE. 79 65. To find tlie P(p((tf'tmi nf the normal throxujh (xj, ?/i). The sloi)e of the tangent is — — • Therefore, that of the normal is — (§ 45, Cor. 2). Hence, the equation of the normal is (§ 38) which reduces to the form yix - XiU = O. [19] Therefore, the normal passes through the centre. 66. To find the equations of the tangent and normal to the circle (x — aY -\- (// — Ij)- = r at the point of contact (x^, ij{). We proceed as in § 64, only now the equations of con- dition which place (.rj, //j) and (.j-o, ?/,) on the circle are (.ri-«)'+ 0/1 -*)' = '•". After subtracting and factoring, \\'e have {x-x,) {xA-x-2a)+{!,-y,) (^,+^,-2i)=0, , Vi — III 3*2 + 3"i — 2a whence, ^ = — — • X2 — X1 2/2 + 2/1— 2^> Hence, the equation of a secant tlirough (x^.?/{) and (xn, ?/.,) is 1/ — 2/1 .Tg -{-Xi — 2« .-»■ — '^-1 y-i + //i — 2^^ Making .ro^.i,, and //2 = //i, and reducing, we obtain (a?i - a) {x-a) + {iji -h){y-h) = r\ [20] 80 ANALYTIC GEOMETKY. Equation [20] may be immediately formed from [18] by affixing — a to the x factors and — b to tlie t/ factors, on the left-hand side. By proceeding as in § 65, we obtain for the equation of the normal {yi - b) {X - aci) - (oci - a) (y - yi) = O. [21] 67, To find the condition that the straight line y = mx + c shall touch the circle x"-\-if=^ r^. I. If the line touches the circle, it is evident that the per- pendicular from the origin to the line must be equal to the radius r of the circle. The length of this perpendicular is — :^^z=: (§ 49). Therefore, the required condition is Vl -1" nv^ c- = r^(l + «r). II. By eliminating y from the equations y = mx -\- c, x^ + ^^ = ^, we obtain the quadratic in x, (1 "1- m^a;^ -|- 2mca; = v^ — c^, the two roots of which are mc Vy^(l -f- m^) — c' 1 -\- vi^ ~~ 1 + m^ If these roots are real and unequal, the line will cut 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 Vr^(l + w^) — c^:=0; that is, if c^= r^(l -|-m^), a result agreeing with that previously obtained. If in the equation y^^mx-{-c we substitute for c the value 7'Vl -\-m^, we obtain the equation of the tangent of a circle in the useful form y = mx ± »■ Vl + m^. [22] THE CIRCLE. 81 This equation, if we regard m as an arbitrary constant, represents all possible tangents to the circle x'^-\-]f=^r^. Note 1. Method II is applicable to any curve, and agi-ees with the definition of a tangent given in § 63. Note 2. In i^roblems on tangents the learner should consider whether the cooi'dinates 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 [22]. Exercise 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'^-\-i/ = t^, assuming that the normal passes through the centre. 3. Find the equations of the tangent and the normal to a;2-[-y2_.52 at the point (4, 6). Find, also, the lengths of the tangent, normal, subtangent, subnormal, and the portion of the tangent contained between the axes. 4. A straight line touches the circle x^-\-if=r- in the point (xi, t/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 the tangent to the circle x^ -\-y^:= 250 at the point whose abscissa is 9 and ordinate negative ? 6. Find the equations of tangents to x^-{- ij^ = 10 at the points whose common abscissa = 1. 7. Tangents are drawn through the points of the circle x^-{-y^ = 25 that have abscissas numerically equal to 3. Prove that these tangents enclose a rhombus, and find its area. 8. The subtangent for a certain point of a circle is 5^ ; the subnormal is 3. What is the equation of the circle ? 82 ANALYTIC GEOMETRY. Find the equation of the straight line : 9. Touching x~-\-;/-=232 at the point whose abscissa=14. 10. Touching (« — 2)- + (// — 3)' = 10 at the point (5, 4). 11. Touching x^ -\- if — 3x — 4?/ ^ at the origin. 12. Touching x'^-^if — 14x — 4y/ — 5=0 at the point whose abscissa is equal to 10. What is the equation of a straight line toucliing the circle x^-\-y'^= i^, and also : 13. Passing through the point of contact (r, 0)? 14. Parallel to the line Ax-\-Bi/~\- (7=0? 15. Perpendicular to the line Ax -)- /:>//+ C= ? 16. Making the angle 45° with the axis of x? .17. Passing through tlie exterior point (Ji, 0) ? 18. Forming with the axes a triangle of area r^? 19. Find the equations of tlie tangents drawn from the point (10,5) to the circle .t- -f" //" = 100. 20. Find the equations of tangents to the circle x^-\-if + 10a; — Qnj — 2 = and parallel to the line y = 2x — 7. 21. Find the lengths of the subtangent and subnormal in the cii'cle x^-\-if — l^x^ — 4// = 5 for the point (10,9). 22. What is the equation of the circle (centre at origin) that is touched by the straight line x cos a-(- .V sina:=/>? What are the coordinates of the point of contact? 23. When Avill tlie line Ax -\- B//— C = touch the circle x^ + ?/ = v-^ ? the circle (x — a.y + (y — f')' == r^ ? 24. Find the equation of the straight line toucliing .t^+// = ax-\-hy and passing through the origin. THE CIRCLE. 83 Prove that the following circles and straight lines touch, and tind tlie point of contact in each case : 25. x^ -{- //'■^ -\- ax -\-b//^^0 and ax + iy + «'■' -\-b'^ = 0. 26. x^+i/- — 2ax — 2b y -\-h^ = {) and x = 2a. 27. iK^H- if = ax -\- bij and ax — bij -\- 1!^ = 0. 28. What is the equation of the circle (centre at origin) that touches the Hue i/ = 3x — 5 ? 29. What must be the value of 7)i in order that the line t/ = mx -\- 10 may touch the circle x^-\-y^ = 100 ? Show that we get the same answer for the line y=^mx — 10, and explain the reason. y 30. Determine the value of c in order that the line ox — 4//-l-c = may touch the circle a;^+^^ — 8a- + 12y — 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 3x-\-2y — 10 = ? 32. What is the equation of a circle whose radius =: 10, and which touches the line 4a; -\-oy — 70 = in the point (10, 10) ? 33. A circle touching the line ■ix-\-oy-\-o=^0 in the point ( — 3, 3) passes through the point (o, 9). What is its equation ? yT 34. Under what condition will the line --f-^^l touch the circle x^-\-y'^^^ r' ? 35. What is the equation of the circle inscribed in the triangle whose sides are x = 0, y=0, - + -^=1? a 84 ANALYTIC GEOMETRY. 36. 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 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 + (l/-by = r', (x-by + (>/-ay = ^^, and hence prove that the two circles touch each other when (a — by =^2)^. * Exercise 23. (Review.) Find the radii and centres of the following circles : 1. 3x' — 6x-\-3i/'-\-9i/ — 12 = 0. 2. 7x^- + 3>/ — 4u—(l — 2xy^0. 3- y (l/ — 5) =x{3 — x). 4. Vl + a' (x' + /) = 2b(x + uij). Find the equation of the circle : 5. Centre (0, 0), radius = 9. 6. Centre (7, 0), radius =3. 7. Centre (—2, 5), radius = 10. 8. Centre (3a, 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). 1 3. Touching each axis at the distance — 7 from the origin. THE CIRCLE. 85 14. Touching both axes, and radius =r. 15. Centre (a, a), and cutting chord = ^ from each axis. 16. Having the centre (0, 0), and touching ?/ = 2a; -f- 3. 17. Having the centre (1, — 3), and touching 2x — y = 4. 18. With its centre in the line 5x — 7y — 8=0, and touching the lines 2x — 1/ = 0, x — 2y — 6 = 0. 19. Passing through the origin and the points common to the circles a^' + / - 6x — 10?/ — 15 = 0, x''-\-u''-\-2x-\- 4?/-20 = 0. 20. Having its centre in the line ox — 3)/ — T = 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^-^7f -\- Ax- U/j- 68 = 0, x^ + f — 6x — 22i/+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^-\-y^ — 'ix-\-8>/=0to the line ix — 3y -f 30= ? 25. What portion of the line ?/ = o./- + 2 is contained within the circle x^-\-y- — 13a; — 4y — 9^0? 26. Through that point of the circle a;^ + ^=25 for which the abscissa = 4 and the ordinate is negative, a straight line parallel to i/^ox — 5 is drawn. Find the length of the intercepted chord. 86 ANALYTIC GEOMETRY. 27. Through the point (;/•,, y^), within tlie circle x--\-y" = r% a chord is drawn so as to l)o bisected at this \)oint. What is its equation ? 28. What rehation must exist among the coefficients of the equation J (x- -\- ;/) -\- D.r -\- Ey + 6' = 0, (i) in order that the circle may touch the axis of x ? (ii) in order that the circle may touch thc^ axis ot y ? (iii) in order that the circle may touch both axes ? 29. Under what condition will the straight line ?/ = wa; -{- e touch the circle a?- + y- = 2rx ? 30. What must be the value of k in order that the line 3.x + 4y = k may touch the circle ?/ = IOj; — x'- ? 31. Find the equation of the circle that passes through the origin and cuts equal lengths a from the lines x=^y, x-\ry = 0. 32. Find the equations of the four circles whose com- mon radius =a\/'2, and which cut chords from each axis equal to 2a. 33. Fin'd the equation of the circle whose diameter is the common chord of the circles x"^ -{- y''^ =^ r"^, (x. — , =, or ,--{- ki/=r"-; (3) hence, (3) is the eqnation of tlie recpiirc^d locus. Since (3) is of the first degree, the locus is a straight line. 92 ANALYTIC GEOMETKT. The line hx + kij = r- is called the Polar of the point (h, k) with regard to the circle x^-{-7/= 7^, and the point (h, k) is called the Pole of the line. The pole (A, k) may be with- out, on, or within the curve. In Fig. 29 it is within, while in Fig. 30 it is without the circle. Fig. 29. THE CIRCLE. 93 Cor. 1. If the point (h, k) is on the circle, (3) is evi- dently the equation of the tangent at (A, k) ; hence, The polar of any point on the circle is identical with the tangent at that 2^0 int. Cor. 2. If (Ji, k) is an external point, by § 71, (3) is the equation of the chord of contact of tangents from (h, k) to the circle ; hence, The polar of any external point is the same line as the chord of contact of tangents draivn from that point. Thus, in Fig. 30, iOf is the polar of P, or the chord of contact of tangents drawn from P. 73. The polar and pole of a circle may be defined as follows : If a chord of a circle is turned round a fixed point (Ji, k), the locus of the intersection of the two tan- gents at its extremities is the polar of the pole (h, k) with regard to that curve. 74. If the polar of a point F passes through P', then the polar of P' loill pass through P. Let P be the point (A, k~), P' the point (A', k'), and let the equation of the circle be x- -{- y'^=^ 1^. Then the equations of the polars of P and P' are hx-\-ky=i^ (1) h'x + k'y = i^. (2) If P' is on the polar of P, its coordinates must satisfy equation (1) ; therefore, hh'-\-kk' = i^. But this is also the condition that P shall be on the line represented by (2) ; that is, on the polar of P'. Therefore, P is on the polar of P'. This relation of poles and polars is illustrated in the Figs. 29 and 30. 94 ANALYTIC GEOMKTUY. 75. To find a (jeometvical conslvnction for the polar of a point with respect to a circle. Fig. 32. The equation of the line through any point P (Ji, k) and the centre of the circle, or the origin, is hx—hi/ = 0. (1) Now, the equation of tlie polar of P is /'.« + %=>^. (2) But the loci of (l)and (U) are perpendicular (§ 45, Cor. 2). Hence, if PC is the polar of P, OP is perpendicular to BC, and 00 = ^h^+Jc" (§41) Also, OP^-sJh^'+k-. Therefore, OPXOQ = A Hence, to construct the polar of P : Join OP, and let it cut the circle in A ; take Q in the line OP, so that OP : 0A = OA : OQ. Tlie line through Q per^x'ndicular to OP is tlie ]iolar of P. To locate the pole of BC, draw OQ perpendicular to BC, andtakePsoth.t oQ:OA=OA:OP. TIIK CIRCLE. 95 76. To find the length of tJie tangent drawn from any point (h, k) to the circle (x — ay-\-(g — Of — >'^' = 0, (1) Let F (Fig. 33) be the point (h, k), Q the [)oiiit of contact, C the centre of the circle; then, since FQC is a riglit angle, Fq'=fc^—qc\ Now, FC' = (h - a)- + (/.■ - by, and 'QC' = i\ Therefore, FQ- = (h — a)"- + (k - hf — i^. Hence FQ^ is fonnd by simply substituting the coordinates of F for X and xj in the expression {x — 0)''^+ {ij — hf — v"^. P Fig. 33. Fig. 34. If for brevity we write 'y — r'^=0. (2) 96 ANALYTIC GEOMETRY. Then, if the tangents drawn from P (x, y) to the circles (1) and (2) are eq^ual, we have {x — ay-\-{y-hy- r2 ={x — a'f -{-{y-b'f- r'% (3) which is the equation of the required locus. Cor. 1. Performing the indicated operations in (3), and transposing, we have 2 (« — a') ic + 2 {b — b') y = a' — a'^^b^ — b'^ — ',-^r''', (4) which shows that the locus is a straight line. This locus is called the Radical Axis of the two circles. Hence, if S-i = 0, 82,=^ are the equations of two circles, then Si = 025 or Si — *S'2 =^^ 0, 7vill be the equation of their radical axis. Cor. 2. When the circles *S'i = and *S'2=:0 intersect, the locus of Si = S2 passes through their common points. Hence, ivhen two circles intersect or are tangent, their radical axis is their common chord or common tangent. CoR. 3. The slope of (4) is the negative reciprocal of the slope of the line joining the centres of (1) and (2). Hence, the radical axis of two circles is ji&i'petidicular to the line joining their centres. 78. Let >S'=0, *S'i = 0, *S^2 = be the equations of three circles, in each of which the coefficient of x^ is unity. Then the equations of their radical axes, taken in pairs, are S — Si = U, Si — 02 ■"" ^t s — S2 ^= 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 common to the other two. Hence, The three radical axes of three circles, taken in pairs, vieet in a point. This point is called the Radical Centre of the three circles. THE CIRCLE, 97 Exercise 24. 1. What is the equation of the diameter of the circle x^ -{- 7j- = 20 that bisects chords parallel to the line 2. What is the equation of the diameter of the circle that 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 x^ -\- y^ = 7'^ fov tangents drawn from the following points: (r, r), {2r, 3?'), {a-\-h, a — b). 5. From the point (13, 2) tangents are drawn to the circle cc- + y^ = 49 ; what is the equation of the chord of contact ? 6. What line is represented by the equation hx + ky=^ ?'^ when (Ji, k) is on the circle ? 7. Write the equations of the polars of the following points with respect to the circle a:^ -f ?/- ^ 4 : (i)(2,3). (ii)(3,-l). (iii)(l,-l). 8. Find the poles of the following lines with respect to the circle x^-{-y^=^?>5: (i) 4a; + 6?/ == 7. (ii) 3a; — 2y = o. (iii) ax + h)j = 1 . 9. Find the pole of 3x-\-4:y = 7 with respect to the circle x--\- y-= 14. 10. Find the pole of Ax-\-By-\-C^O with respect to the circle x^-\- y- = r. 11. Find the coordinates of the points in which the line a; = 4 cuts the circle x^ + y = 2o; also find the equations of the tangents at those points, and show that they inter- sect in the point i^^-, 0). 98 anai.vtk; gkomktrv. 12. If tlio polavs of two points /', Q meet in B, then R is the pole of the line FQ. 13. If the polar of (A, k) witli respect to tlie circle 3.2_j_y2_.j.2 touches the circle x^-\->f = 2rx, then k'^-\-'2rh 14. If the polar of (A, k) with respect to the circle a:^ -\- f/^ = c^ touches the circle 4(^x^-\-y^)^c'^, then the pole (h, k) will lie on the circle a;^+?/- = 4cl 15. Find the polar of the centre of tlie circle x'^ -\- //- = r^. Trace the changes in the position of the polar as tlie pole is supposed to move from the centre to an infinite distance. 16. Wliat is the square of the length of the tangent drawn from tlie point (Ji, k) to the circle .t" + //' ^ ?•- ? 17. Find the length of the tangent drawn from (2, 5) to the circle x^- + f — 2x — ,% — 1 = 0. Find the radical axis of the circles : 18. (.r + 5)^ + (y + 6)~o, (x-7y-\-(>/-ny-=i6. 1 9. .7-2 + ff -f 2x + 3// — 7 = 0, X- + //•-' — 2x — 7/4-1=0. 20. X- + f + hx + />// — c = 0, ax"^ + air + ""-> + />'// = ^^■ 21. Find the radical axis and length of tlie common choiwl of tlie circles .r," 4- y' -\- (IX -\- Jnj -|- *" = 0, :i^ -\- //- -|- hx -f- (nj 4- c = 0. 22. Find the radical centre of the three circles a;2 + yH 4x + 7 = 0, 2.^2+ LY- + o.r + 5/y + 9 = 0, a;^-|- ?/"-(- // = 0. CHAPTER IV. DIFFERENT SYSTEMS OF COORDINATES. Rectilinear System. 79. When we define the position of a point, with refer- ence to any fixed lines or points, we are said to nse a System of Coordinates. In the Rectilinear System, already described, we liave thus far employed only rectangular axes, or coordinates, which are to be preferred for most purposes, on account of their greater simplicity. When the axes of reference intersect at oblique angles, the axes and coordinates are called Oblique. ,Y Let OX, OF (Fig. 35) be two axes making an acute angle, A'OF=w, with each other. If we draw PN^ to OX, and PJ/|| to OY, then the coordinates of P are NP = 0M= X, MP = //. Since oblique and rectangular coordinates differ only in the angle included between the axes, any of the previously deduced formulas that do not depend on any property of the right angle are apjdicable wlien the axes are oblicjue. Thus, formulas [2], [3], [4], [7J liold for oblique axes as 100 ANALYTIC GEOMETRY. well as for rectangular, and therefore are general formulas for the Rectilinear System. When the axes are oblique, instead of [1], we evidently have (Fig. 35) PQ = \IpW + UQ' — 2PR XBQ cos PBQ, ••• d= '^ (^2 — xiY + (yz — yif -\- ''^ix2 — ^i) (yz — yi)cosa>, which reduces to [1] when w=:90°. The Rectilinear System is sometimes called the Cartesian System, from Descartes, who first used it. 80. To find the equation of the straight line AC, referred to the oblique axes OX, OF (Fig. 36), having given the inter- cept OB=^b and the angle XAC=y. Let P be any point (x, y) of the line. Draw BD || to OX, meeting PM in D. Then, by Trigonometry, PD sin y y — h sin y BD sin (a» — y)' X sin (w — y) If now we put m = —. — ; — - — r , we obtain as the result an sin ((!> — y) equation of the same form as [6], p. 38, y = mx -\- b. Here 771 = the ratio of the sines of the angles which the line AC makes with the axes ; that is, m = sin JT^P-^ sin PBY, which equals tan X^P^tan y when w = 90°. DIFFERENT SYSTEMS OF COORDINATES. 101 81. Oblique coordinates are seldom used, because they generally lead to more complex formulas than rectangular coordinates. In many cases, however, they may be em- ployed to advantage. An example of this kind is furnished by problem No. 23, p. 65 : To prove that the medians of a triangle meet in one point. If a, b, 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 with great ease, as follows : The sides, !/ = 0, x = 0, - + {^ = 1. ^0, '-+■''■■ a b The medians, a a b 1=0, a b 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 (§ 53). Polar System of Coordinates. 82. Next to the rectilinear, tlie system of coordinates most frequently used is the Polar System. Fig. 37 Let (Fig. 37) be a fixed point, OA a fixed straight line, P any point. Join OP. 102 ANALYTIC GEOMETRY. It is evident tliat we know the position of P, provided we know the distance OP and the angle wliieh OT I'oi'uis with OA. Thus, if we denote the distance OP by p, and tlie angle AOP by 6, the position of P is determined if p and 6 are known. The fixed point O is called the Pole, anut, in order to represent by a single equation all the points of a geometric locus, we often employ negative valnes of p and 6, and adopt the following laws of signs : (i) 6 is positive when measured from right to left, and negative when measured in tlie opposite direction. (ii) p is positive or negative according as it extends in the direction of the terminal side of 6 or in the opposite direction. Thus, any given point may be determined in four different ways. DIFFERENT SYSTEMS OF COOKDINATES. 103 Fur example, suppose tluit the straight line r()l\ bisects the first and third quadrants, and that in this line we take points F, Pi, at the same distance OF = a from O ; then F is the point(//, i.7r)or( — a,|7r)or(— «, — ^7r)or( a, — In); Fi is the point (^A, ^Tr)o\-{—a,\Tr) or( a, — ^7r)()r(— «, — ;j7r). Fig. 39. Fig. 40. 'Y 83. To Jiml the jJolar equation of a circle. (i) Let the pole be at the centre (Fig. ,38). Then, if r denotes the radius, the polar equation is simply p=^r. (ii) Let the pole be on the circumference (Fig. o*.)), and let the diameter OF make an angle a with the initial line OA. Let F be any point (p, 6) of the circle. Join BF. Then, OF=^OD cos EOF, or P = 2»' cos (9 - a). [23] If OB is taken as the initial line, the equation becomes P = 2r cos 9. [24] (iii) Let the pole be any point, and the centre the point (p', 6'). Then in the triangle OCF (Fig. 40), 0F~ — 20FX OCXcos COF-]- OC^— aF- = Q, or 9- - 2pp' cos (8 - 6') + p'-2 - r'^ = O, [2r)] the most general form of tlie polar equation of a circle. 104 ANALYTIC GEOMETRY. Exercise 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 P2O-P3 (Fig. 38) bisects the second and fourth quadrants, what are the polar coordinates of the points P2 and P3 ? Give more than one set of values in each case. 4. Construct the following points (on paper, take a ^ 1 in.) : «, V [a, IT ), [t»m^ (x, y) from the new to the old system, we must write x — It in place of x and y — k in place of y. 86. To chanye the reference of a curve from one set of rectanyidar axes to another, the orhjin remainintj the same. Let (a:, y) be a point P referred to the old axes OX, Y; (x', y') the same point referred to the new axes OA"', OY (Fig. 42). Then 03f=x, MP = y, ON=x', Nr = y'. Let the angle XOX' == 6. Draw NQ, NR i. to PM, OX, respectively; tlien NPQ= QXO = EON= e. DIKKKIJENT SYSTEMS OF COOliDIN A'I'KS. 107 Hence, 0M= OR — RM= OR — NQ = ON cos 6 — FN si ii 6. Or x^=x' COS 6 — y' sin 6. And PJ/= MQ + QF = KN-\- QF = ON sin Q + 7'.V cos ^. Or y^=x' sin ^ + y' cos 6. Therefore, to find what the equation of a curve becomes when referred to the new axes, we must write X cos 6 — 1/ sin 6 for x, x sin. 6 -\- // cos 6 for y. "TT"^ 87. To change the reference of a curve from one set of rectangular axes to another, both the origin 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 (Ji, 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 coordinates, a* = A 4" x' cos 6 — y' sin 6, y^k-{-x' sin 6 -\- y' cos 6. In making all these transformations, attention must be paid to the signs of h, k, and 6. 88. To change the reference of a curve from, rectangular to oblique axes, the origin remaining the same. Let a, ^ be the angles formed bj'' the 2iositive directions of the new axes OX', OY' (Fig. 43) with the positive direction of OX Let tlie old coordinates of a point P be x, y; and the new coordinates, x\ //'. Then from the right triangles ORN, PQN we readily obtain tlie formulas X = x' cos a + y' cos (3, y = x' sin a-\-y' sin (3. Investigate the special case when fS=^a-\-[)0°. 108 ANALYTIC GEOMKTltV. 89. To deduce the formulas for finding the polar equation of a curoe from its rectangular equation. Let cc, y be the rectangular coordinates of any point P, and p, 6 its polar coordinates. Y 'ix-^ \ vr ^^ M R X Y /P / ^^-^"^^^^ M X Fig. 43. Fig. 44. (i) Let the origin of rectangular coordinates be the pole, and let the polar axis coincide with the axis of x. Then (Fig. 44) 031= OP cos 3I0P, PM= OP sin 3£0P. Or cc = /o cos 6, y=^p sin 6. (ii) If the pole is the point (h, k), we have x = h-{- p cos $, y = k-\-p sin 0. (iii) If the pole coincides with the origin, but the polar axis OA makes the angle a with the axis of x, we obtain x = p cos (^ + a), y = p sin (^ + a). (iv) If the pole is the point (h, k), and the polar axis makes the angle a with the axis of x, x = h-]r p cos (6-\-a), y = k-\-p sin (<9 + a). DIFFERENT SYSTEMS OF COORDINATES. 109 90. To deduce the fonnulas for finding the rectangular equation of a curve from its polar equation. From the results in cases (i) and (ii) of § 89 (the only cases of importance), we readily obtain In case (i), p^ = x--\- i/, tan 6 = -- In case (ii), p^ = (x — hy-\-{y — k),^ tan ^ = X y—k X — h 91. The degree of an equation is not altered by passing from one set of axes to another. For, however the axes may be changed, the new equation is always obtained by substituting for x and y expressions of the form ax-\-by-\rc and a'x -\- />'// -\- c'. These expressions are of the first degree, and, therefore, if they replace x and y in the equation, the degree of the equa- tion cannot be raised. Neither can it be lowered ; for if it could be lowered, it would be raised by returning to the original axes, and therefore to the original equation. Exercise 26. 1. What does the equation y"^ — 4x + 4?/ + 8 = become when the origin is changed to the point (1, — 2) ? Transform the equation of the circle (x — «)" -f (y — by = j"^ by changing the origin : 2. To the centre of the circle. 3. To the left-hand end of the horizontal diameter. 4. To the npper end of tlie 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^ = a^ become if the axes are turned through — 45° ? 110 ANALYTIC GEOMETRY. 7. The equation of a curve referred to rectangular axes is X — xy — y^= 0. Transform it to new axes, whose origin is the point ( — 1, 1), the new axis of y bisecting two of the angles formed by the old axes. 8. Change the following equations to polar coordinates, taking the pole at the origin and the polar axis to coincide with the axis of x : (i) a;^ -|- y^ = a^. (ii) x^ — )f = a^. 9. Change the equation x^=:4a?/ to polar coordinates, (i) taking the pole at the origin ; (ii) taking the pole at the point (a, 0). 10. Change the following equations to rectangular coordi- nates, the origin coinciding with the pole, and the polar axis with the axis of x : (i) p ^= a, (ii) p = a cos 6, (iii) p^ cos 26 == al Transform the following equations by changing the origin to the point given as a new origin : 11. x-\-i/-\- 2 = 0; the new origin ( — 2, 0). 12. 2x — 5y — 10 = ; the new origin (5, — 2). 13. ^x"^ -\- -ixij -\- if — 5x — Qij — = 0; new origin (J, — 4). 14. x^ -\-y'^ — 2x — 4y = 20 ; new origin (1, 2). 1 5 . a;^ — Q)xy -\- if — Cj:' + 2 // -|- 1 = ; new origin (0, — 1) . 16. Transform the equation x' — v/2-|-6 = by turning the axes through 45°. 17. Transform the equation (x-\- y — 2rt)^=: 4a:-^ by turn- ing the axes through 45°. 18. Transform the equation 9^^ — 16//= 144 to oblique axes, such that the new axis of x makes witli the old axis 3 of x the negative angle tan~' — - ; and the new axis of y . . '-> makes with the old axis of x the i)0sitive angle tan ~^-. 4 DIFFKKKNT SYSTEMS OF C06ui>INATFS. Ill Exercise 27. (Review^.) 1. Find the distance from the point ( — 2b, b) to the origin, the axes making the angle 60°. 2. The axes making the angle w, 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 given by polar coordinates : 4. {a, 6) and {b, ). 5. {a, 6) and (a, — 6). 6. {a, 6) and (— a, — 0). 7. (2a, 30°) and (a, 60°). 8. Show that the polar coordinates (p, 6), ( — ,o, 7r + ^), ( — p, 6 — tt) all represent the same point. 9. Transform the equation H.r--\-^.ri/-^4//--\-l2x-\-S}/ + 1 = to the new origin ( — ^, — ^). 10. Transform the equation 6x^+3^" — 24.r + 6 = to the new origin (2, 0). 11. Transform the equation --|-^ = 1 by changing the origin to the point ( -, - ) and turning the axes through an angle cb, such that tan d> = "^ a 12. Transform the equation 17.r- — 16.r// -f- 1 7;/- = 225 to axes that bisect the axes of the old system. Transform the following rectangular equations to polar equations, the polar axis in each case coinciding with, or being parallel to, the axis of x, and the pole being at the point whose coordinates are given : 112 ANALYTIC GEOMETRY. 13. x^-\-i/=Sax ; the pole (0, 0). 14. X- -\- ij-^'6ax; the pole (4a, 0). 15. 3/- — 6y — 5.x + 9 = ; the pole (| , 3). 16. x^—if — Ax-Q>j — U = 0; the pole (2, — 3). 17. ix--{-y-y = k\x''—ij') ; the pole (0, 0). Transform the following polar equations to rectangular axes, the origin being at the pole and the axis of x coincid- ing with the polar axis : 18. p^sm2e = 2a^ 19. p = k sin 2$. 20. p(sin 3^ + cos 3^) = 5k sin $ cos d. 21. Through what angle must a set of rectangular axes be turned in order that the new axis of x may pass through the point (5, 7) ? 22. The rectangular equation of a straight line is Ax + By ■j- C=0. Through what angle must the axes be turned in order that (i) the term containing x may disappear ? (ii) the term containing y may disappear ? 23. Deduce the following formulas for changing from one set of oblique axes to another, the origin remaining the same : x' sin (o> — a) j^ y' sin (w — /3) sin 0) sin w x' sin a , y' sin B y=' — ■■ r' — '■ — sm 0} sm w Note. In these formulas w denotes the angle formed by the old axes, a and ^ 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 I^o. 23 deduce those of § 88, CHAPTEE V. THE PARABOLA. The Equation of the Parabola. 92. 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 through the focus perpendicular 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 halfway between the focus and the directrix is, from the definition, a point of tlie 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 Radius 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 Rectum or Parameter. 93. To construct a parabola, having given the focus and the directrix. I. Bij Points. Let F (Fig. 45) be the focus, CE the directrix. Draw the axis FD, and bisect FD in A ; tlien A is the vertex of the curve. At any point M in the axis erect a perpendicular. From F as centre, with DM as 114 ANALYTIC (iKOMKTUV. radius, cut this perpendicular in 1* and (J ; then /* and Q are two points of the curve, for 7*7^ = X'iJ/= distance of F 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 continuous curve through them. Fig. 45. Fig. 46. II. Bij Motion. Place a ruler so that one of its edges shall coincide with the directrix DE (Fig. 46). 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 tlie point of a pencil F. The point F will trace a parabola ; for during the motion we always have FF=FC. 94. To find the rectangular equation oftheparahola, when its axis is taken as the axis of x and its vertex as the origin. Let F (Fig. 45) be the focus, CE the directrix, DF.Y the axis, A the vertex and origin; also let 2^) denote the known distance FF. Let F be any point of the curve ; then its coordinates are A3I=x, MF = y. THK I'AKAHOLA. 115 Draw PC_Lto CE\ then by the detiuition of the curve FF^PC = 1)M. Therefore, FF' = ljJr. Now FF''^JIF'+l^'=i/--\-(x—jjy, and lJJl'=^(x-\-py. Therefore, y^ + (•*' ~ P) ' =^ (•'■f + p)"- Whence, y^= 4.pjc. [2G] This is called the principal equation of a parabola. 95. Since ?/ and jj in equation [2G] are positive, x must always be positive; therefore, the curve lies wholly on the positive side of the axis of y. An examination of equation [26] shows that the curve, (i) passes through the origin, (ii) is symmetrical with respect to the axis of x, (iii) extends towards the riglit without limit, (iv) recedes from the axis of x without limit. 96. Any point (h, />-) is outside, on, or inside the jiorahola jf=^A:px, arrordin;/ FJI', or QM>FM; hence, Q is outside the curve. If k'^—Aph is negative, we may prove similarly that Q must be inside the curve. 97. If X =-p, y = ± 2/?. But these two values of y make up the latus rectum. Hence, the latus rectum = 4:2j. Cor. From the equation y^ = 4px, it follows that x:y = y:4p; that is, the latus rectum is a third proportional to any abscissa and its corresponding ordinate. 98. If (xi, yi) and (x^, 2/2) s^re any two points on the parabola, we have yj^ = 4:pxi, y^^ = ipx2. Hence, yi^ ■ y2^ = ^i- ^2', that is, the squares of the ordinates of any two points on the parabola are to each other as their abscissas. 99. To find the points in which the straight line y=^mx + G meets the parabola y"^ = 4px. Eegarding these equations as simultaneous, and eliminat- ing X, we have . Whence, y=^±^_^ j p-mc ^ ^2) -^ 7n 7n \ p ^ ^ From (2) it follows that y = vix-\-c has two distinct, two coincident, or no points in common with g-^^ipx, according as ^ — mc >, =, or < 0. CoR. If pj — mc = 0, or c^=p-^ni, y = mx-\-c will be a tangent; that is, y = i7ix-\-— (3) is a tangent to y'^^Apx in terras of its slope. THE PAKAUOLA. 117 Exercise 28. 1. Show that the distance of any point of the parabola y^ =: ^px, from the focus is equal to p -\- 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 directrix = 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 parab- ola is 4. Write its equations for the three cases enumerated in No. 4. 6. For what point of the parabola ?/-:=18x is the ordi- nate equal to three times the abscissa? 7. Find the latus rectum of tlie following parabolas : y'^=^Qx, f/'^=lox, hif^ax. Find the points common to the following ])arabolas and straight lines : 8. if = ^x, 3.C — 7y -f oO = 0. ^ 9. f = ?>x, ic — 4//+12 = 0. 10. if = '^.x, a.- = 9, a=0, a- = — 2. 11. ?/- = 4x, ?/ = 6, y = — 8. 12. What must be the value of ^; in order that the parab- ola ij'^ = 'ipx may pass through the point (9, — 12) ? 118 ANALYTIC GEOMETRY. 13. For what point of the pavabohx i/'' = o2x is the ordi- nate equal to 4 times the abscissa ? 14. The equation of a pavaboLa is i/- = Sx. What is the equation of (i) its axis, (ii) its directrix, (iii) its hitus rectum, (iv) a focal chord tli rough the point whose ab- scissa =8, (t) a chord })assing through the vertex and the negative end of the latus rectum ? 15. The equation of a parabola is ?/^ = 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, 16. If the distance of a point from the focus of the parabola if = 4:j)x is equal to the latus rectum, what is the abscissa of the point ? 17. In the parabola y^ = A2jx 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 = 8/). Prove that straight lines drawn from its ends to the vertex are jperpen- 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. Determine, as regards size and position, the relations of tlie following parabolas : (i) y^z=4jjx, (ii) f = — 4:px, (iii) x- = 4:j)i/, (iv) x^ = — 4rp//. the pakabola, 119 Tangents and Nokmals. 1 00. To find the equation of the tangent and of the normal to tJte parabola y^='^2^x at any j^oint (a*!, ?/i). Let (xi, yi), (x.2, ?/o) be any two points on the parabola; then the equation of the secant through them is Ju «/ 1 i/^2 ~~^ OC-t Since (.Ti, ?/i) and (.r^, //o) are on tlie cvirve y- = ^jjx, we have 2/2 — !h _ k^ . a-2 — a-i V/2 + Vi By substituting in (1), the et^uation of the secant becomes rzjh^_J^. (2) x — x^ y^ + yi Now, if (xo, jfo) is made to coincide witli (x^, //i), (2) be- comes the equation of tlie tangent ix\>(xi,y^. Putting //a:=yi, clearing of fractions, and remembering that y^^-^px^, Ave obtain as the equation of the tangent at (xi,y^, yr!/=22){.r + x,). [27] The normal passes through (x^, y^, and is perpendicular to the tangent; hence, its equation is, by [27] and §415, 101. If in [27] we put // = 0, we obtain .<• = — .Ti, or TA = AM(Y\g. 48). Therefore, tlie snbtangent Is bisected at the vertex. If in [28] we put y=^0, we obtain x = x^ + 2p, or X — xi [= J/xV] = 2/; (Fig. 48). Hence, the stibnoi-inal is constant and e 0, and x = 18, y < ? 14. Find the equation of the chord of the parabola fz=4,px that is bisected at a given point (x^, y^). 15. In what points does the line x-]ry ^=12 meet the parabola ^/^ + 2x — 12?/ + 16 = ? 16. In what points does the line 3?/^ 2a; + 8 meet the parabola y'^ — 4cX~Sy -\-24: = ? 17. Find the equations of tangents from the origin to the parabola (y — by=^ A'p(x — a). 18. Describe the position of the parabola y^ + 2a:; + 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. 125 Find the equation of the parabola : 20. If the equation of a tangent is 4y = 3x — 12. 21. If a focal radius = 10, and its equation is 3^ = 4a; — 8. 22. If for a point of the curve the focal radius = r, and the length of the tangent = ^. 23. If for a point of the curve the focal radius = r, and the length of the normal = n. 24. If for a point of tlie curve tlie length of the tangent = t, and the length of the normal = n. 25. If for a point of the curve the focal radius = r, and the subtangent =: s. 26. Two parabolas have the same vertex, and the same latus rectum 4p, but their axes are J_ to each other. What is the length of their common chord ? 27. Through the three points of the parabola 7/^ = 12a', 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 7f=^px makes the angle 30° with the axis of x. At what point does it cut the axis? 29. For what point of tlie parabola if^Apx is the length of the tangent equal to 4 times the abscissa 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 tangent to the axis of x. 31. Two tangents to a parabola are perpendicular to each other. Find the product of their subtangents. 126 ANALYTIC GEOMETKY. 32. Prove that the circle described on a focal radius as diameter touches the tangent drawn through the vertex. 33. Trove that the circle described on a focal chord as diameter tpuches tiie directrix. Find the locus of the middle points : 34. Of all the ordinatos 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 if=^^px make the angles 6, 6' with the axis of x ; find the locus of their intersection : 39. It cot 6 -\- cot 6' = k. 41. If tan ^ tan ^' = 7c. 40. If cot e — cot e' = k. 42. If sin 6 sin e'^k. 43. Find the locus of the centre of a circle that passes through a given point and touches a given straight line. SUPPLEMENTARY PPtOPOSITIONS. 1 03. Two distmct, two coincident, or no real tunyents can he drawn to a. parabola from any point (Ji, k), accordiny as the jjoint is without, on, or icit/iin the curve. Let the tangent y = mx + — pass through the point (h, k) P then, k = mh -\- — , m or h7n'- — km -|- ^v = 0. Whence, m = ^ — • THK I'AKAI'.OT.A. 127 These values of m arc real ami unequal, real and equal, or imaginary, according as /.-"•^ — 4yy //>,=, or <0; that is, according as (Ji, k) is without, on, or within the parabola ; hence the proposition (§ 90). 104. To find the equation of the chord of contact of two tangents drawn from any external point {h,k) to the parabola Let (xi. iji) and {x^, yi) be the points of contact ; then the equations of the tangents are yiij = 'lp{x-\-x^. Since (/i, k) is in both these lines, we have ky^ = ^p{x,^h), (1) ky, = 2p{x^^h). (2) From equations (1) and (2) we see that both tlie points (xi, 2/1) and (j-2, y^ lie in the straight line whose equation is ky = 2p{x-^h). (3) Hence, (3) is the equation required. 105. To find the equation ofth e pola r of the pole (h, k) u'ith regard to the parabola y'^ = 4yw;. Let P be the fixed point (A, k), PQE one position of the revolv- ing chord, and let the tangents at Q and R intersect in Piixi, y^) ; it is required to find the locus of Pi, as the chord turns about P. Since PH is the chord of con- tact of tangents drawn from the point Pi(xi, yi), its equation is (§ 104) yiy=^j'(->- + ^'i)- Fig. 49. (1) 128 ANALYTIC GEOMETRY. Since (1) passes through (Ji, Ic) we have y,k = 2p{h + x,). (2) But (xi, ?/i) is any point on the required locus, and by (2) its coordinates satisfy the equation ky = 2p(x + h). (3) Hence, (3) is the required equation, and the polar is a straight line. Cor. When the pole (Ji, k) is on the curve, the polar is evidently a tangent at (A, k) ; when the pole (A, k) is with- out the curve, the polar is the chord of contact of tangents from (A, k). Thus the tangent and chord of contact are par- ticular cases of the polar. The Proposition of § 74 may be proved for poles and polars with respect to a parabola. 106. To find the locus of the middle points of parallel chords in the jiarahola if' = 4/>x. NT.,-^^^ ^ >-7 > \ / ; 'A Fig. 50. Let any one of the chords PQ (Fig. 50) be y^=mx-\-c, and let it meet the curve in the points (x^, i/i), {x^, y^)- THE PARABOLA. 129 Then (§ 100), vi= -^' • (1) Let M (x, y) be the middle point of PQ ; tlien 2^ = 2/1 + 2/2- By substitution in (1) we obtain 7H — — or y = -^' (2) y ^ m ^ ^ a relation that holds true for all the parallel chords, because m is the same for all the chords. The required locus, there- fore, is represented by (2), and is a straight line- parallel to the axis of x, and called a diameter of the parabola. Hence, Every diameter of a parabola is a straight line parallel to its axis. Every straight Ihie parallel to the axis is a diameter ; for 2» m, and, therefore -^, may have any value whatever. m 107. Let the diameter through Jf meet the curve at 8, and conceive the straight line P^ to move parallel to itself till P and Q coincide at 8 ; then the straight line becomes the tangent at S ; therefore. The tangent draivn through the extremity of a diameter is parallel to the chords of that diameter. 108. From the focus i^draw FC ± to PQ, and let EC meet the directrix in the point C. If 6 denotes the angle which the chord PQ makes with the axis of x, it easily follows that DCF^d ; then we have CD = FD cot = ^-^ m hence, by (2) § 106, The perpendicular from the focus to a chord meets the diameter of the chord in the directrix. Moreover, since DS (Fig. 50) is parallel to QP, the per- pendicular from the focus to a tangent and the diameter through the point of contact meet in the directrix. 130 ANALYTIC GEOMETRY. 109. Let the tangents drawn throngh /' and Q meet in the point T. Regarding their equations, y^ij = 2p {x + .T2), as simultaneous, we obtain for tlie value of the ordinate of T y — -J-LJ h> _ _L. Hence, y 2 — 1/1 I'l Tangents draum throufjh the ends of a chord meet m the diameter of the chord. 110. To find the locus of the foot of a per2yendicular from tlie focus to a tangent. Let the equation of a tangent be ni Then the equation of the perpendicular will be X p y = 1 — ni VI Since these two lines have the same intercept on the axis of y, they meet in that axis ; hence, the tangent tlirough tlie vertex is the required locus. 111. Since FP = PC (Fig. 48) and angle EPC = EPF, therefore the tangent P T is perpendicular to FC at its middle point, and every point in it is equally distant from i^and C. 112. Tangents at right angles intersect in the directrix. Let the equation of one tangent be y = inx -\ (1) Then the equation of the other is y = — --7np. (2) Subtracting (2) from (1) we obtain for tlieir common point THE PARABOLA. 131 (x-\-p)(vi-\- -)=0. But m -\ cannot be zero ; lience, x -\-p = 0, or x = — p, which is the equation of the directrix. 113. The polar of the focus (p, 0) is = 2p (x -{-p), or X = —p. Hence, the polar of the focus is the directrix, or tangents at the ends of a focal chord intersect in the directrix. CoR. From this result and § 112 it follows that tanyents throufjh the ends of a focal chord intersect at right angles. 114. To find the equation of a parabola referred to any diameter and the tangent thnnigh its extremitij as axes. Transform the equation ]/ = 4:px to the diameter SX' (Fig. 51) and the tangent through S as new axes. Let m be the slope of the tangent, 9 the angle which the tangent makes with the diameter ; then m = tan 6. First transform to new parallel axes through S. Now, by § 106, BS=2p^m; hence, from y" = Apx we obtain AB=^p-^vr. Therefore, the now equation is ..■+4 III' or my- -\- 4py = 4j>iii.v. (1) 132 ANALYTIC GKOMETRT. 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 P we have The old X = SR. The new x = SN. The old ?/ =: RP. The new y = NP. Now it is easily seen from Fig. 52 that >S'^ = SN + NP cos e, RP = NP sin 6. Therefore, equation (1) is transformed to the new system by writing x -\r y cos 6 in place of x, and y sin 6 in place of y. Making this substitution, remembering that m = tan 6, and reducing, we obtain an equation of the same form as y'^=^^px. Join S to the focus F, and denote FS by p' ; then ^ ^ ?>i'' m^ sin^ ^ Therefore, equation (2) may be more simply written y- = ^p'x, (3) where p' is the distance of the origin from the. focus. It is easy to see that this equation includes the case when 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 >S^. When the diameter is tlie axis of the curve, Ap' is called the Principal Parameter. THE PARABOLA. 133 Cor. Let the equation of a parabola referred to any diameter, and the tangent at the end of that diameter as axes, be y^=-\p'x. Since the investigations in §§ 99, 100 hold good whether the axes are at right angles or not, it r/ follows immediately that the straight line y=^mx-\- — will touch the parabola for all values of m, and that the equation of the tangent at any point (xi, y^ is yxy=^'2,xj'{x-\-x^. 115. To find the 2)olar equation of the jmrabola, the focus being the j^ole. Let P be any point (p, 6) of the curve ; then p = FF = XP = DM= 2p + FM ■=2p-\- p cos Q. . - 2y> ' ' ^ 1 — cose Discussion of [29] [29] Fig. 53. Since cos Q cannot exceed + 1, p is positive for all values of ^. If ^^0, cos ^==1, and p=^v:. This shows that the axis of the parabola does not cut the curve to the right of the focus. If ^r=|,r, cos ^^ If ^ = TT, cos ^ = — 1 If ^=3^, cos ^= If ^ = 27r, cos ^= 1 p = 2p = semi-latus rectum. P= p = FA. p = 2p = FR'. p = X. As 6 increases from zero to tt, p decreases from x to p. As 6 increases from n to 2ir, p increases from p to x. 134 ANALYTIC GEOMETRY. Exercise 31. 1. Given a parabola, to draw its axis (§106). 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 tJiree 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 con- tained 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 poiuts. Prove that the angle between the tangents is equal to half the angle between the focal lines. 7. Show that if the vertex is tnken as pole, the polar equation of a parabola is A}) cos 6 ^~ s\n-B 8. Explain how tangents to a ]>arabola may be drawn from an exterior point (§ 102). 9. Having given a ])arabola, how would you fiud its axis, directrix, focus, and latus rectum? 10. From the point ( — 2, 5) tangents are drawn to the parabola //- = 6,/-. What is the equation of the chord of contact? 11. The general equation of a system of parallel chords in the parabola lif = 25a- is 4.r — 7// + ^•' = 0. What is the equation of the corresponding diameter? THE PARABOLA. 135 12. lu the parabola if = Vox, \vh;t,t is the equation of the ordinates of the diameter y -f~ H = ? 13. In the parabola / = (jj', wliat chord is Ijisccted at the point (4, 3) ? 14. Given the parabola // = 4/?:r ; find the equation of the chord that passes through tlie vertex and is bisected by the diameter y = a. How can this chord be constructed ? 15. The latus rectum of a parabola = 16. Y/hat 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 equation of that chord of the i)arab- ola if^^-ipx which is bisected at the point (//, k) is l-{l,-k)=^2p{x-h). 17. Prove that the parameter of any diameter is equal to the focal chord of that diameter. 18. Prove that the locus of ij- — 8// — G.'- + -8=:0 is a parabola wliose axis is parallel to the axis of x ; and deter- mine the latus rectum, the vertex, the focus, the axis, and the directrix. 19. Prove that in general the locus of ?/-+-''•'■ + ^'// + C=0 is a parabola Avhose axis is parallel to the axis of ./• ; and determine its latus rectum, vertex, and axis. 20. Prove that in general the \oqx\^ oi x^ -\- Ax -\- B i/ -\- C = Q is a parabola whose axis is parallel to the axis of //; and determine its latus rectum, vertex, and axis. 21. Find the locus of the centres of circles that touch a given circle and also a given straight line. 22. 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. 116. 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 Radius. The constant sum is denoted by 2a, and the distance between the foci by 2c. The fraction - is called the Eccentricity, and is repre- sented by the letter e. Therefore, c = ae. In the ellipse a > c ; that is, e / Motion. Fix pins in the paper at the foci. Tie a string to them, making the length of the string exactly equal to 2a. 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 lensrth of the strinsr. THE ELLIPSE. 137 II. Bij Points. Let F, F' be the loci; then FF' = 2c. Bisect FF' at 0, and from lay off OA=OA' = a. Then A' A = 2a, F'A' = FA, A'F-\-A'F' = AF-{-FA =2a, AF'+AF =AF'-\-F'A' = 2a. Therefore, A and A' are points of the curve. Between i^and F' mark any point X; then describe two arcs, one with F as centre and AX as radius, the other with F' as centre and AX as radius ; the intersections P, Q of these arcs are points of the curve. By merely interchang- ing the radii, two more points, E, S, may be found. After a sufficient number of points has been obtained, draw a continuous curve through them. 118. The line AA' is the Transverse or Major Axis, A, A' the Vertices, and the Centre of the curve. The line BB\ perpendicular to the major axis at 0, is the Conjugate or Minor Axis; its length is denoted by 2b. Show that B and B' are equidistant from the foci, tliat BF=a, that BO = h, and that a- = Ir -\- r ^ Ir -[- a-e\ 138 ANALYTIC GKOMETRY. 119. To find the equation of the ell'qjse, having given, the foci and the constant sum 2a. B Take the line AA' (Fig. 55), passing through the foci, as the axis of x, and the point 0, halfway between the foci, as origin. Let F be any point (x, y) of the curve, and let r, r' denote the focal radii of P. Then from the definition of the curve, and from the right triangles F'PM, FPM, r"=>/+{c + xy, (1) ?•-= if-\- (c — xy. (2) By addition, r''J^v^=^2(x' + y^ + c'). (3) By subtraction. r'- — r = 4:cx. (4) But r'-^r = 2a. (5) By division, 2ex a (6) By subtraction, ex r^=a = \a — ex\. a '- -^ (7) By addition, r' = a + -- = [a + ex~\. (8) Substitute in (3) remembering that h'=^a~ — c' (§ 118). Then U'x''-\-a-if ^a'U', or ^! + ^ = 1. [30] THE ELLIPSE. 139 Cor. If the transverse axis is on the axis of y, and tlie conjugate on the axis of x, the equation of the elli})se is ^+-A=i. (10) 120. To trace the form of ike ellipse from its equxtion. The intercepts on the axis of x are + a and — a\ on tlie axis oi y, -\-b and — h. Only the squares of tlie 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), ( — x, y), (— x, — //). Therefore, we infer that (i) The curve is symmetrical vnfh respect to the axis of x. (ii) Tlie curve is symmetrical with respect to the axis of y. (iii) The curve is symmetrical with resperf to the centre O, which bisects every chord passing/ through it. This explains why is called the centre. \ 2 / \ 2 Since the sum of I - | and [ t\ is 1, neither of these \aj \bj squares can exceed 1 ; therefore, the maximum value of .r is + a, and the minimum value — o, while the corresponding values of y are -f- b and — b. Therefore, the curve is wholly con- tained within the rectangle whose sides are x^±a, y = ±b. 121, To trace the chanyes in the form of the ellijjse wlicu the semi-axes are supposed to change. Let a be regarded as a constant, and b as a variable. (i) Sujipose b to increase. Then c decreases (since c'^ = a^ — i'^), 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 (/, and equation [30] becomes the equation of the circle, 140 ANALYTIC GEOMETRY. (iii) If we suppose b to decrease to (a remaining con- stant), c will increase to a, e will increase to 1, while the curve will approach, and linally coincide with, the major axis, its equation at the same time becoming i/ = 0. 122. Let {Xi,i/i) and (^'2,2/2) be any two points on the ellipse i^x^ + a-^^ = a-(5''^; then we have Dividing and factoring, we have 2/1^ : 2/2^^ : : (a — Xj) {a + x^) : (a — x^) {a + a^g). This is, the squares of any two ordinates of the ellipse are to each other as the products of the segments into which they divide the major axis. 123. It follows from § 119 that a point (Ji, k) is on the ellipse represented by the equation [30], provided -. + 7-2-1=0. a^ V It may be shown by reasoning similar to that employed in § 96 that the point (A, li) is outside or inside the curve, h\ k"" , . according as — 2 ryi — 1 is positive or negative. 1 24. If A, B, C all have the same sign, every equation of the form ^^2+^^2_C' (1) may be reduced to the form x" . y . X , y ^ a^ W W a? Hence, every equation of the form of (1) represents an ellipse whose semi-axes are a/— and -y „* The transverse axis lies on the axis of x or the axis of y, according as A is less than or greater than B. THE ELLIPSE. 141 125. The chord passing through either focus perpendicu- lar to the majoraxis is called the Latus Rectum or Parameter. To find its length, put a; = c in the. equation of the ellipse. b"" , „ „. b* b'' Then, r=^(s^-<^) = ^ Therefore, the latus rectum 21/ _ r4i/ a |_2a Forming a proportion from this equation, we have 2a : 2b : : 2b : latus rectum ; that is, the lathis rectum is a third li^'ojiortional to the major and minor axes. 126. The circle having for diameter the major axis of the ellipse is called the Auxiliary Circle ; its equation is cc^ -f- y" = al The circle having for diameter the minor axis is called the Minor Auxiliary Circle ; its equation is a- + u = ti-. 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 tlie point J^. The angle QOM is called the Eccentric Angle of the point P, and is denoted by the letter (^. 142 ANALYTIC GKOMETRY. 127. Lft ij, ij represent the orcliuates of points in an ellipse and the auxiliary circle respectively, corresponding to the same abscissa x. Tlieu from the equations of the two curves we have b .—^ j , , i—^ :j- y = rb - V « — X , y —± 'si a — X-. Whence, y:y'^b:a, or, the ordinates of the eUijise and the auxiliary circle, corre- sponding to a common abscissa, are to each other in the con- stant ratio of the semi-minor a7id semi-major axes of the ellipse. 128. The principle of § 127 furnishes the following easy method of constructing an ellipse by points when its axes are given: Construct both the major and minor auxiliary circles (Fig. 57); draw any radius cutting the circles in B and Q; through Q draw a line parallel to BO, and through R draw a line parallel to OA; the intersection P of these lines is a point on the ellipse. For we have MP:MQ = OB:OQ, or MP \%J — b\a. From this proportion and that in § 127, we have MP = y; hence, P is a point on the ellipse. In like manner any num- ber of points may be found. THE ELLIPSE. 143 (1) Cor. From Fig. 57, we have ic= 0M= OQ cos , y=MP= 0N= OR sill , and 7 = sin -}- sin- '//' = C when C is zero ? When is this locus imaginary ? 24. Prove that the al)scissas of the ellipse Ir^x"^ -\-a'^if = 11%"^ are to the corresponding abscissas of the minor auxiliary circle, x^-\-if^=U^, SiS a : b. 25. Construct an ellipse by the method of § 128. 26. Construct an ellipse, having given c and b. 27. Construct tlie 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 ellipse. 29. A square is inscribed in the ellipse a^^ b' Find the equations of the sides and the area of the square. Tangents and Normals. 130. To find the equations of a tangent and of .r — .7'i .Tg — .r 1 and proceeding, as in § G4, we obtain as the equation of a secant through (a^i, y^ and (xj, v/o) y — V\ ^ _ l>\^x + a-o) a-— a-i «^(/yi + Z/2) THE ELLIPSE. 147 Now make x^^x^, ij„=zy^-^ then the chord becomes a tan- gent, and x—xj^ (r{!/i + 7/2) y ~ Vi ^^^^'i becomes '— = — > X — Xi c^ y\ which reduces to ^+'Jj!/_ = 1, ri^n From the equation above it appears tliat the value of the slope of the tangent, in terms of the coordinates of the point of contact, is b'^Xi a% The normal is perpendicular to the tangent, and passes through (xi, ?/i) ; therefore, its equation is easily found (by the method of § 46) to be i/-!/i = |!^;(^-^i). [33] 131. To find the suhtangent and subnoDnal. Making y = in [32] and [33], and tlien solving the equations for x, we obtain : Intercept of tangent on axis of a- = — ('"^ Intercept of normal on axis of x =: — .r, = e^x^. Whence, the values of the subtangent and the subnormal (defined as in § 63) are easily found to be as follows : Siibtang-eiit = [34] Subiiorinal = a- [:ib^ 148 ANALYTIC GEOMETRY. 132. If tangents to ellipses having a common major axis are drawn at points having a common abscissa, they will meet on the axis of x. For in all these ellipses the values of a -and. x are con- stant, and therefore (by § 131) the tangents all cut the same intercept from the axis of x. Y Q Fig. 59. 133. 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. 59) were found, in § 119 to be PF=a — ex^, PF'=a-\- exy. If the normal through P meets the axis of x in N, ON = e'^Xi (§ 131) ; and, therefore, NF =c — e^Xi = ae — e^.r^ = e(a — ex). NF' ^c-\- e^Xi = ae -\- e-x^ = e(a + ex). Therefore, NF : NF' = PF : PF', or the normal divides the side FF' of the A PFF' into two parts proportional to the other two sides. Therefore (by Geometry), Z FPN= Z FPN. The tangent PT, being perpendicular to the normal, must bisect the angle FPG, formed by one focal radius with the otlier produced. THE ELLIPSE. 149 134. To draw a taiujent and a "lornud tliroiujh a (jlven point of an ellipse. I. Let P (Fig. 60) be the given point. Describe the aux- iliary circle, draw the ordinate MP, produce it to meet the circle in Q, draw QT tangent to the circle and meeting the axis of X in T, and join PT; then PT is a tangent to tlie ellipse (§ 132). Draw PNl, to PT; PNis the normal at P. C II. Draw the focal radii, and bisect the angles between them. The bisectors are the tangent and the normal at the point P (§ 133), 135. To find the equation of a tangent to an ellipse in terms of its slope. This problem may be solved by finding under what con- dition the straight line y=^mx-]-c (1) will touch the ellipse b-x- + iV + i^ pass through the point (A, k); then k = mh -\- Vm'^a- + ^^ or (Ji" — a^) vi" — 2hkin + k^ — V = 0. m- hk± \IVK'^a^k^ — a^lP' h:' — a^ ' <^^) Hence, there will be two distinct, two coincident, or no tangents through {h, k), according as h^h^-\- aVc^ — ci-h^^, =, or<0; that is, according as (h, k) is without, on, or within the ellipse. 138. To fijid the equation of the chord of contact of the two tangents drawn from an cjcternal point (Jt, k) to the ellipse, a^~^ V Let the student prove, by a course of reasoning similar to that employed in §§71 and 104, that the required equation is hx kg . THE ELLII'SK. ir>r, 139. To find the equatioii of the iiohir of the pole (h, /:), with regard to the ellipse. Let the student prove, by a course of reasoning similar to that employed in §§ 72 and 105, that the required equation is hx ky a^ 0- CoR. The tangent and cliord of contact are particular cases of the polar. The proposition of § 74 holds true for poles and polars with regard to the ellipse. 140. To draw a tangent to an ellipse from a given jjoint P outside the curve. Fig. CI. Suppose the problem solved, and let the tangent touch the ellipse at Q (Fig. (11). If F'Q is produced to G, making QG=QF, then A FQG is isosceles; now ZFQP = Z GQP (§ 1.33) ; therefore, PQ is per])ondicnlar to FG at its middle point ; therefore, P is equidistant from /♦" and G. This reduces the problem to determining the point G. Since F'G=^2a, (7 lies on the circle with F' as centre and 2a as radius. And G also lies on the circle with 7' as centre and PF as radius. Hence the constrnction is obvious. 156 ANALYTIC GEOMETRY. 141. To Jind the locus of thi; middla points of anij system of jmrallel chords in the ellipse. Let any one of the parallel chords y = mx + c meet the ellipse b'x'^ -\- (Cif = a-y^ in the points (xi, yi) and {x«. y^) ; then, by § 130, /v-(a-i+a-2) If (a*, y) is the middle point, 2a; = iCi + X2, 2y = yi-{- y^, and (1) becomes Px or y^ — —— (2) This relation holds true for the middle points of all the chords ; therefore, it is the equation of the locus required. From (2) we see that any straight line passing through the centre of an ellipse is a diameter. 142. Let m! denote the slope of the diameter of the chords whose slope is m; then from (2) of § 141 m = i' or mm' = r. \ ot \ ma^ a- ^ -■ Thus [37] is the equation of condition that the diameter yz=m'x bisects all chords parallel to the diameter y = mx', but [37] is evidently also the equation of condition that y =■ mx bisects all chords parallel to ?/ = vi'x ; hence, If one diameter bisects all chords parallel to another, the second diameter bisects all chords parallel to the first. Two such diameters are called Conjugate Diameters. CoR. From [37] the slopes of two conjugate diameters must have opposite signs ; hence, two conjugate diameters of an ellijyse lie on opposite sides of the miiior axis. THE ELLIPSE. 157 143. Let a straight line cutting the ellipse in P and Q move parallel to itself till P and Q coincide with the end of the diameter bisecting PQ; then the straight line be- comes the tangent at the end of the diameter. Therefore, The tangents at the extremities of any diameter are parallel to the chords of that diameter , and also to its conjugate diameter. 144. Let POP' and HOP' (Fig. 62) be two conjugate diameters meeting the ellipse in the points P (xi, y^ and R (^2, 3/2)- The slope of the tangent through P is ^; a yi hence, the equation of the diameter ^0^', which is parallel to this tangent (§ 143), is h\r^ Xix . yiv „ y = ^x, or — - + •— - = 0. ayi a^ ¥ (1) Fig. G2. Now R {x2, 2/2) is on (1), and also on the ellipse ; hence we have 3. 3. , , , , -J— 2 J_ •ZLZj! = (0\ and 1. Solving (2) and (3) for x^ and y^, we obtain a b o a (3) 158 ANALYTIC GKOMICTRY. The up]ter signs give the coordinates of li, and the lower those of E' in terms of a\ and //i. Equation (2) is the condition that must be satisfied by the coordinates of the extremities of every pair of conju- gate diameters. 145. Denoting the semi-conjugate diameters OP and OB (Fig. G2) by a' and //, respectively, we have a'2 — xf + //i" = .rj- + (((- — xA = ^^ + "^^V^ -= H' + ^'^.\ (1) a and h"' = jv + 1/2' = J, Z/r + ^^vTi'^ (§ 144) = a' - j-;' + K x,^ = a' — eW. (2) Adding (1) and (2), we have a'2 + ^/2 = a- + i'^. That is, the sum of the stianres of any pair of semi^cov jugate diameters is equal to the sum of the squares of the semi-axes. Equations (1) and (2) ex])ress the lengths of tlie semi- conjugate diameters a' and // in terms of a, b, and x^ (the abscissa of the extremity of a'). 146. Let the ordinates of the extremities P, R (Fig. 62) of two conjugate diameters meet the auxiliary circle in Q, S, respectively, join QO and SO, and denote Z. QOX by <^, Z^ SOX by <^'. Then the values of the coordinates of P and R are (§ 127), Xi = a cos , Xo = a cos (f>', l/i^b sin ^, yi'=b -sin <^'. Whence, by substitution in etpiation (2) of § 144, we obtain cos <^ cos ^'-|-sin <^ sin <^'=z{). Therefore, cos (^' — ^) = 0, or <^' — <^ = ^tt. THK ELLirSE. 159 That is, tlie difference of the eccentric avrjles correspond i>ir/ to the ends of two conjugate diameters is equal to a right angle. CoK. The angle FOR (Fig. 62) is obtuse, siiu^e QOS=l_'Tr. 147. To find the angle formed by two conjugate semi-diam- eters, whose lengths a', b' are gioen. Let the semi-diameters make the angles a, (3, respectively, with the axis of .r, and let denote the required angle. Then if (a-j, ?/i) and (x^, i/o) are the extremities of a' and b\ respectively, sin a 7/ ''''f^=b'=7b'- cos a = —5 COS (3 or., aj/i «■ ■ 'b'^~Jb'' sin 6^ sin (/S — a) = sin )8 cos a — cos /8 sin a b^x-^-\-a^Ui^ aHr aba'b' aba'b' ah a'l^ (1) CoR. 1. Clearing (1) of fractions, we liave a'V sin 6=^ ah, which shows that the area of the parallelogram HEKE is equal to the rectangle L21QN. (O ClfJiS=a'b' sin 6.) R N ^ jy w Fig. G3. That is, the j>arallelogram formed by tangents at the extremities of any pair of conjugate diameters is equal to the rectangle on the axes. IGO ANALYTIC GEOMETRY. CoK. 2. If CT (Fig. G3) is perpendicular to tlie tangent KK, then, CT^ CD sin CDR = a' sin $ = j,- 148. The lines joining any point of an ellipse to the ends of any diameter are called Supplemental Chords. Let FQ, F'Q be two supplemental chords (Fig. 64). Through the centre draw OR parallel to P'Q, and meeting PQ'm R; also OR' parallel to FQ, and meeting F'Q in R'. Q Fig. 64. Since is the middle point of FF', and OR is drawn parallel to F'Q, and OR' is drawn parallel to FQ, R and R' are the middle points of QF, QF', respectively. Therefore, OR will bisect all chords parallel to QF, and OR' will bisect all chords parallel to QF'. Hence, OR, OR' are conjugate diameters. Therefore, the diameters jxirallel to a pair of supplemental chords are conjugate diameters. CoR. 1. This principle affords the following easy method of drawing a pair of conjugate diameters which shall in- clude a given angle. On the transverse axis AA' describe a segment of a circle which shall include the sriven angle. Let the arc of this THE ELLirSE. 161 segment cut the ellipse in Q and *S'; then the diameters parallel to QA and QA', or SA and SA', are conjugate and include the required angle. CoR. 2. If B is the upper vertex of the conjugate axis, the conjugate diameters parallel to BA and BA' will evi- dently be equal, and will lie on the diagonals of the rec- tangle on the axes of the ellipse. 149. To find the equation of an ellipse referred to a pair of conjugate diameters as axes. Since each of two conjugate diameters of the ellipse bisects the chords parallel to the other, the curve is (obliquely) symmetrical with respect to each of the new axes; hence, as the required equation is of the second degree, it contains only the squares of x and y, and is of the form Ax^ + Bi/=C. (1) The intercepts of the curve on the new axes are the semi- conjugate diameters. Denoting them by a' and b', we have Substituting these values in (1), we obtain which is the required equation in terms of the semi-conjugate diameters. 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 that do not pre- suppose the axes of coordinates to be rectangular, hold true when we employ as axes two conjugate diameters. For example, the equation of a tangent at the point(a:i,2/i), referred to the semi-conjugate diameters a' and b', is a'2 "•" b''' 1G2 ANALYTIC OKOMKTRY. 150. To consfritrt the, polar of a fucus Since the po focus (ae, 0) is Since the polar of (h, k) is -7 + ^ = 1, tlie polar of the aex = a; or x Hence, ae : a ^= a -. x. Therefore, if OD (Fig. 65) is taken so that OF: 0A=^ OA : OD, and DC is drawn perpendicular to OD, DC will be the polar of the focus F. The polar of a focus is called a Directrix of the ellipse. Hence, DC is the directrix corresponding to the focus F. In like manner we may constnict F'C, or the directrix corresponding to the focus F'. Cor. Let Q (x, ?/) bo any ])oint on the ellipse; then, QS=OD-OM^'^-x = '^ = ^. Hence, e = FQ-^QS. That is, tJie distances of ant/ poivt on the ellipse from a focus, and the corresj)ondi)if/ directrix, bear the constant ratio e. THE ELLIPSE. 163 Whence, the ellipse is often defined as : Tlte locus of a point ivliicli moves so that its distances from a fixed 2)0 int and a fixed stvahjlit line hear a constant ratio less than unity. 151. To find tlie polar equation of the ellipse, the left- hand focus being taken as the jjole. Fig. 66. Let F be any point (p, 0) of tlie ellipse ; then, from equa- tion (8) of § 119, we have p = a^ ex. (1) Now X = 0M= F'M — F'0 = pcoQe — ae. Substituting tins value of x in (1), we have p = « -|- ep cos 9 — ae^. Whence, p 1 — ecos6 [39] Cob. Since e < 1, and cos 6 cannot exceed unity, p is always positive. ), p^ -^ = ./ + ae = F'A. If If 1 = 0. ; Itt, p=:a(l — e'-) = F'Ji =■ senii-latus rectum. 164 ANALYTIC GEOMETRY. If e = ^, p = ''-^^^ = a-ae = F'A'. ^ l-\-e If = ^-,r, p = «(1 — e^) = semi-latus rectum. If 6 = 2ir,p = a + ae = F'A. While 6 increases from zero to tt, p decreases from a-\-ae to a — ae; and while 6 increases from tt to 27r, p increases from a — ae to a-\- ae. If F is taken as the pole, the polar equation is a(l — e}) " 1 + e cos 6 Exercise 35. 1. Find the area of the ellipse a:;^ + 4y^ = 16. 2. Find the distances of the directrices from the centre in No. 1. 3. What is the equation of the polar of the point (5, 7) with respect to the ellipse 4a;^ + %^ = 36 ? 4. Prove that a focal chord is perpendicular to the line that joins its pole to the focus. In what line does the pole lie ? 5. Find the pole of the line Ax-\- By -\- C = with respect to the ellipse b'^x^ -\- a? if = a%^. 6. Each of the two tangents that can be drawn to an ellipse from any point on its directrix subtends a right angle at the focus. 7. The two tangents that can be drawn to an ellipse from any external point subtend equal angles at the focus. 8. Find the slope mi 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 21 of a diameter, its inclination 6 to the axis, and the eccentricity; find the major and minor axes. THE ELLIPSE. 165 10. Tangents at the extremities of any chord intersect on the diameter which bisects that chord. 11. Tangents are drawn from (3, 2) to the ellipse x^ + 4//- = 4. Find the equation of the chox-d of contact, and of the line that joins (3, 2) to the middle point of the chord. 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^- + 11^^143 that passes through (1, 2) and is bisected by the diameter ^x — 2?/ = ? 14. In the ellipse 9x^ + 36 ?/^ = 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 following lines : X — y^O, a" -|- y = 0, ax = hy, ay = bx. 16. Show that the lines 2x — y^=0, x-\-Sy=^0 are con- jugate diameters in the ellipse 2x^ -\- 3y^ ^= 4. 17. Find the equation of a diameter parallel to tlie normal at the point (xi, y^), the semi-axes being a and b. . 18. The rectangle of the focal perpendiculars upon any tangent is constant and equal to the square of the semi-minor axis. 19. The diagonals of the parallelogram in Fig. 63, § 147, are also conjugate diameters. 20. The angle between two semi-conjugate diameters is a maximum when they are equal. 21. The eccentric angles corresponding to equal semi- conjugate diameters are 45° and 135°. 22. The polar of a point in a diameter is parallel to tlie conjugate diameter. 166 ANALYTIC GEOMKTKV. 23. Find the ecjuations of equal conjugate diameters. 24. Tlie length of a semi-diameter is I; find the equation of the conjugate diameter. 25. The angl(! between two equal conjugate diameters is 120° ; find the (;c(u'ntricity of the ellipse. 26. Given a diameter, to construct the conjugate diameter. 27. To draw a tangent to a given ellipse parallel to a given straight line. 28. Given an ellipse, to find by construction the centre, foci, and axes. 29. Find the rectangular equation of the ellipse, taking the origin at the right-hand vertex. 30. Find the polar equation of an ellipse, taking as i)ole tlie right-hand focus. 31. Find the polar equation of the ellipse, taking the centre as pole. 32. 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 C()()rdinates. • 33. The equation of an ellipse, the origin being at tlie left-hand vertex, is 25.r--l- 81^ ;=450.r ; find the axes. 34. 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 ? 35. Find the eccentric angle <^ corresponding to the diameter whose length is 2c. 36. At the intersection of the ellipse Ji^x^ -\- c/^ = aHi'^ and the circle x--\-/f^(ih tangents are drawn to both curves. Find the ansrle between them. THE ELLIPSE. 167 37. How would you draw a normal to an ellipse from auy point in the minor axis ? 38. Find the equation of a chord that is bisected at the point (Ji, k). 39. 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. 40. A circle described on a focal radius will touch the auxiliary circle. 41. Find the locus of the intersection of tangents drawn through the ends of conjugate diameters of an ellipse. 42. Find the locus of the middle jtoint of the chord joining the ends of two conjugate diameters. 43. Find the locus of the vertex of a triangle Avhose base is the line joining the foci, and whose other sides are parallel to two conjugated iameters. 44. Show that ^iX^ + y^ + 8a? — 2y + 1 = represents an ellipse ; find its centre and axes. 45. If A and B have like signs, sliow that tlie locus of Ax- + 11 y- -\- iKr -f- Jiy + i''= is in general an ellipse whose axes are parallel to the coordiuate axes ; and determine its semi-axes. 46. Find the locus of the centre of a circle that passes through the point (0, 3) and touches internally the circle x^ + 1/' ^ 25. CHAPTER VII. THE HYPERBOLA. Simple Properties of the Hyperbola. 152. 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 Radius. The constant difference is denoted by 2a, and the dis- tance between the foci by 2c. The fraction - is called the Eccentricity, and is denoted a by the letter e. Therefore, c-=ae. Since the difference of two sides of a triangle is always less than the third side, we must have in the hyperbola 2a 1. 153. To construct an hyperhola, having r/iven the foci, and the constant difference 2a. I. B]/ Motion (Fig. 67). Fasten one end of a ruler to one focus F' 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 2a, 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 P will describe one branch of an hyperbola. The other branch may be described in the same way by interchanging the fixed ends of the ruler and the string. THE HYPEKBOLA. 169 II. By Points (Fig. 68). Let F, F' be the foci ; then FF' = 2c. Bisect FF' at 0, and from lay off OA = OA' = a. Then AA' = 2a, FA = F'A '. AF'— AF= AF' — A'F' = AA' ^ 2a. A'F— A'F' = A'F— AF = AA' = 2a. Therefore, A and A' are points of the curve. Y X ° 1 1) ~--^1k^. --V'-- A\ /Q\ ^ Fig. 67. Fig. 68. In FF' produced mark any point D ; then describe two arcs, the first with F as centre and AD as radius, the second with F' as centre and A'D as radius ; the intersections P, Q of these arcs are points of the curve. By merely inter- changing the radii, two more points B, 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 BB' J_to FF'-, since the difference of the distances of every point in the line BB' from the foci is 0, therefore the curve cannot cut the line BB'. The locus evidently consists of two entirely distinct parts or branches, symmetrically placed with respect to the line BB', called the right-hand and the left-hand branches. 170 ANALYTIC GEOMETRY. 1 54. The point (J, halfwiiy between the foci, is the Centre. The points A, A', where the line passing through the foci meets the curve, are called the Vertices. The line AA' is the Transverse Axis. The transverse axis is equal to the constant difference 'Ja, and is bisected by the centre (§ 153). Fig. CO. The line BB' passing through perpendicular to AA' does not meet the curve (§ 153) ; but if B, B' are two points whose distances from the two vertices'^, ^' are each equal to c, then BB' is called the Conjugate Axis, and is denoted by 2h. Since AAOB = A A OB', OB=OB'=l>; that is, the con- jugate axis is bisected by the centre. ; In the triangle AOB, OA^=a, OB = h, AB^^i^;^ hence, c" = a' -\- Ir. The chord passing through either focus perpendicular to the transverse axis is the Latus Rectum, or Parameter. Note. Since a and b are equal to the legs of a riglit triangle, a may be greater than or less than b; hence the terms '■'■major'''' and ^^ minor'''' are not appropriate in the hyperbola. THIC HVPKKBOI.A. 17 J 155. By proceeding as in the case of the ellipse (§ 119), using r' — r = ± 2(i instead of /•' + r = 2a, and substituting i^ for c^ — a^, we obtain as the equation of the hyperbola ^'-1?=1. [40] Thus the equations of the ellipse and hyperbola differ only in the sign of h^; that of the ellipse is changed into that of the hyperbola by substituting — V^ for + V^. Hence, Any formula deduced front the equation of the ellipse is changed to the correspondincj formula for the hyperbola by merely changing -\-b- to — Ir, or b to b\l — 1. The lengths r, r' of the focal radii for any point (a*, //) are r ^ ± (ex — a ) , r ' = ± (ex + o ) , in which the upper signs hold for the right-hand branch, and the lower for the left. 156. A discussion of equation [40] leads to the follow- ing conclusions : (i) The curve cuts the axis of x at the two real points {a, 0) and(— «, 0). (ii) The curve does not cut the axis of //. The imaginary intercepts are ± Z* V — 1. (iii) No part of the curve lies between the straight lines x^-\-a and x = — 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 a-. (vii) The curve is symmetrical with respect to the axis of y. (viii) Every chord that passes through the centre is bisected by the centre. This explains wli}- the })oint half- way between the foci is called the centre. 172 ANALYTIC GEOMETRV. 157. An hyperbola whose transverse and conjugate axes are equal is called an Equilateral Hyperbola. Its equation is oc^-y--a-. [41] The equilateral hyperbola bears to the general hyperbola the same relation that the auxiliary circle bears to the ellipse. Fig. 70. 158. The hyperbola that has BB' for transverse axis, and AA' for conjugate axis, obviously holds the same rela- tion to the axis of ?/ that the hyperbola which has AA' for transverse axis and BB' for conjugate axis holds to the axis of X. Therefore its equation is found by simply changing the signs of a^ and b^ in [40], and is 1. (1) The two hyperbolas are said to be Conjugate. THE HYPERBOLA. 173 159. The straight line z/=?na;, passing through the centre of the hyperbola -^— -, = !> meets the curve in two points, the abscissas of which are -\- ah — ah \IU^ — ni^a' V^^ — m^a^ Hence the points will be real, wiarjinavy, or situated at infinity, as h- — m^a? is positive, negative, or zero ; that is, as rn? is less than, greater than, or equal to —• The same line, y = mx, will meet the conjugate hyperbola — r — TZ = — 1 in two points, whose abscissas are a' ¥ -\- ah, — ah Hence these points will be imaginary, real, or situated at h^ infinity, as m^ is less than, greater than, or equal to —^• Whence, If a straight line through the centre meets an hyperhola in iraaginary points, it will meet the conjugate hyjjerhola in real points, and vice versa. 160. An Asymptote is a straight line that passes through finite points, and meets a curve in two points at infinity. We see from § 159 that the hyperbola x"^ // ^ "~ Ij" ~ has two real asymptotes which pass through the centre of the curve, and which have for their equations y = -\--x and b X ; or, 1^ = 0. [42] y = — -x; or ^2 a2 174 ANALYTIC GEOMETRY. Exercise 36. What is the equation of an hyperbola, if : 1. Transverse axis = IG, conjugate axis = 14 ? 2. Conjugate axis = 12, distance between foci = 13 ? 3. Distance between foci = twice the transverse axis? 4. Transverse axis = 8, one point is (10, 25) ? 5. Distance between foci = 2c, eccentricity = V2 ? 6. Prove that the latus rectum of an hyperbola is equal 2lr to Also 2a : 20 ■.•.2b: latus rectum. a 7. The equation of an hyperbola is 9x-^ — 16^/ = 144; iind the axes, distance between the foci, eccentricity, and latus rectum. S. Write the equation of the hyperbola conjugate to the hyperbola 9x-^— !()//-= 144, 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. I'rove that the point (.*•, y) is ivitlumt, on, or ivithin the hyperbola, according as —^ — 'j^—^ is nerjatioe, zero, or positive. 11. Find the eccentricity of an equilateral hyperbola. 12. Find the points that are common to the hyperbola 25x-2— 9/=22r), and tlie straight line 25.^+12^ = 45. 13. The asymptotes of an hyperbola are the diagonals of the rectangle CDEG (Fig. 70, p. 172). 14. Find the foci and the asymptotes of the hyperbola 16a;2 — 9/ = 144. THE HYPERBOLA. 175 15. The asymptotes of an equilateral hyperbola are per- pendicular to each otlier. Hence the equilateral hy})erb()la is also called the rectan- -■ Since ??i and m' are alike involved in [45], it follows that If 0716 diameter bisects all chords par allel to another, the second diameter will bisect all chords parallel to the first. Two such diameters are called Conjugate Diameters. 172. From [45], the slopes of two conjugate diameters must agree in sign ; hence, Ttvo conjxKjate diameters of an hyperbola lie on the same side of the coiijagate axis, and their included angle is acute. Also, if ni in absolute magnitude is less than -, then in' b " must be greater than-- But the slope of the asymptotes is b ^^ equal to ± — Therefore, Two conjugate diameters lie on opposite sides of the asymp- tote in the same quadrant : and of two conjugate diameters, one meets the curve in real points and the other in imaginary jtoints (§ 159). 180 ANALYTIC GEOMETRY. 173. The length of a diameter that 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 tlie conjugate hyperbola iu real points (§ 159); and its length is the length of the chord between these points. But from § 159 we know that if a diameter meet one of the hyperbolas in the imaginary point (AV— 1, k^J—V), it will meet the other in the real point (Ji, k); hence, the length of the semi-diameter, which is V/r + k^, is known from the imaginary coordinates of intersection. 174. The equations of an hyperbola and its conjugate differ only in the signs of a^ and b^. But this interchange of signs does not affect the equation , b' mm = — • Therefore, if tivo diameters are conjugate with respect to one of two conjugate hyperbolas, they ivill be conjugate with respect to the other. Thus, let POP' and QOQ' (Fig. 71) be two conjugate diameters. Then POP' bisects all chords parallel to QOQ' that lie icithin the branches of the original hyperbola and betioeen tlie branches of the conjugate hyperbola; and QOQ' bisects all chords parallel to POP' that lie within the branches of the conjugate liyperbola and between the branches of the original hyperbola. From the above theorem it follows immediately that If a straight line meets each of two conjugate hyperbolas in two real jioints, the two jjortlons of the line contained between the hyperbolas are equal (thus, BD^= B'D', Fig. 71). 175. The tangent di'atvn through the end of a diameter is parallel to the conjugate diameter (§ 143). THE HYPERBOLA. 181 176. Having given the end (^j, ?/i) of a dianiete?-, to find the end, (./•.,, y.,) of the conjugate diameter. Fig. 71. Let (xi, ]/i) be on the given hyperbola, then (arg, 2/2) is on the conjugate. The slope of the tangent at (xi, 1/1) is -7-^; hence, the equation of the diameter conjugate to the diame- ter through (xi, 1/1) is y=^x. (1) a-i/i Now (xo, 1/2) is on the diameter (1) and also on the con- jugate hyperbola ; lience, we have 2/2 = 1. a'l/i a^ 0- Solving equations (2) for Xo and 7/2, we obtain a b (2) X2 '■ l/u y.2 = ±-xy. a 182 ANALYTIC GEOMETRY. The positive signs belong to one end, and the negative signs to the otlier end, of the conjugate diameter. 177. To find flic equation of an liijperlmla referred to any initr of eoirjiigate dimneters as axes of coordinates. From the symmetry of the curve with respect to each, of the new axes, tlie required equation must be of the form Denoting tlie intercepts of the curve on the new axes by a' and b' V— 1 (§ 172), we obtain Whence, ^^--^^l (1) is tlie required equation, in which a' and V are semi-conju- gate diameters. 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 that have been obtained without assuming the axes of coordinates to be at right angles to each other hold good when the axes of coordinates are any two conjugate diameters. Por example, the equation of the asymptotes of the hyperbola represented by equation (1) is 9 and the equation of the tangent is 178. The tangents through tlie ends of two conjugate diameters meet in the asymptotes. The equations of these tangents referred to the conjugate diameters are a; = ± «', y = ±b'. THE HYPKKBOLA. 183 Hence, tlieir intersections are («', b'), (a', — b'), ( — a', b'), and ( — a', — b'). Hut these points evidently lie upon the asymptotes, or the locus of (2) in § 177. 179. //' 6 denotes tlie aiKjle formed by two covj agate semi- diametei's, and a' and b' their lengtlis, then sin ^ = -y-,- Substituting ^V — 1 for b, and i'V — 1 for ^' in equation (1) of § 147 and cancelling the imaginary factorS; we obtain the above result. CoR. 1. Since Irt'i^'sin ^ = 4«^, the parallelogram SES'R' (Fig. 71) equals the rectangle on the axes of the curve. CoR. 2. The length of the perpendicular from upon the tangent SPIi= OF sin OFS ^a' sin e = '-^- b Cor. 3. From §§ 145, 155, 177, we have a'-^ — b'''^a'--b\ 180. If a straight line cats an hyperbola and its asymp- totes, the portions of the line intercepted between the curve and its asymptotes are equal. Let CC (Fig. 72) be the line meeting the asym})totes in C, C'and the curve in. 7*, B', and let the equation of the line be y = vix-\-c. (1) Let M be the middle point of the chord BB' ; then (§ 170) the equation of the diameter through vi is ^ = ^.- (2) By combining equation (1) with the equations of the asymptotes, we obtain the coordinates of the points C and C ; taking the half-sum of these values, we get for the coordi- nates of the point halfway between C and C the values 184 ANALYTIC GEOMETRY. 05 = U^ — m^d^ ' y- V'c U' — m^a^ These values satisfy equation (2) ; therefore, the point halfway between C and C" coincides with M\ therefore, MC= MC. And since MB=MB\ therefore, BC= B'C. Dy Fig. 72. Cor. Let CC be moved parallel to itself till it becomes a tangent at P, meeting the asymptotes in R, S; then the points B, B' coincide at F, and we have PB = PS. Hence, The portion of a tanr/ent intercepted by the asymptotes is bisected by the p)oint of contact. 181. The following method of showing that an hyperbola has asymptotes, and finding their equations, is more general than the method given in §§ 159, 160. THE HYPERBOLA. 185 The abscissas of the points where the straight line y = mx -\- e meets an hyperbola are found by solving the equation x^ (mx -f cy a^ b'' 1, Now, from Algebra we know that as the coefficients of x^ and X in (1) approach zero, both roots of (1) increase without limit. Hence, each root becomes infinity when V^ — nvii? = 0, and Imc = 0, or when m = ± -, and c ^ 0. a Therefore, ij^^±i-x are asijmptotes to the hyperbola. If only b- — m^a^ = 0, then m ^ ± -, the line is parallel to an asymptote, and one root of (1) is infinity, while the , . h' + c" other IS Zmc Hence, a right line parallel to an asymptote meets the hyperbola in only one finite point. 182. To find the equation of an hyperbola referred to the asymptotes as axes of coordinates. Let the lines OB, OC (Fig. 73) be the asymptotes, A the vertex of the curve, and let the angle AOC=^a. Let the coordinates 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 as axes of coordinates. Draw PMl_to OA, PN\\ to CO ; then X = ON cos a -\- NP cos a = {x' + //') cos a, y = NP sin a — ON sin a= (y' — x') sin a. 186 ANALYTIC GEOMKTKY. Fig. 73. Hence, by substituting in [40], we obtain (.rJ -{- //')- cos- a ill' — x'y sin^tt b- 1. But AD cos a: OD Vft2 + i2 OA a OB -sJa'-\-b'^ Substituting these values, and dropping accents, we have 4:XU = H^ + bK [4G] Cor. 1. The equation of the conjugate hyperbola is 4xy=i — (rt- + Z--). 2ab CoR. 2. Sin COT? = sin 2^ = 2 sin « cos a= , , ,, • a' + ¥ It a = b, sin COB = 1 ; therefore, COB = ^ir. Cor. 3. Let {x^, y^) denote P (Fig. 72), referred to the asymptotes ; then OS X OR = 20ffX 2 IIP = 4a-i//i = a^ + b\ That is, tJie product of the intercepts of a tangent on the asymjjtotes is equal to the sum of the squares of the semi-axes. THK HYPERBOLA. 187 CoK. 4. In Fig. 72, the area of tlie triangle ROS equals h OS X OB sin ROS = h (a' + b') ^?^.. = ah. That is, the area of the triangle formed by any tangent and the aaymptotes is equal to the product of the sevii-axes. 183. The polar of the focus («e, 0) is Fig. 74. Hence, if OD is taken so that OF: OA=OA:OD, ^hen DN perpendicular to OF is the polar of F, and is called a Directrix of the hyperbola. In like manner we may construct i>'X', or the directrix corresponding to the focus F'. Cor. As in § 150 we may prove that _PF 188 ANALYTIC GEOMETRY. Whence, the hyperbola may be defined as The locus of a jJoint whose distances from a fixed point and a fixed straight line hear a constant ratio greater than unity. 1 84. To find the polar equation of an hyperbola, the left- hand focus being taken as pole. If X is reckoned from the centre, and we write p=^ex-\- a, (1) p will be positive or negative according as the point is on the right or left-hand branch. Now x^ p cos 6 — c^=p cos 6 — ae. Whence, by substitution and reduction, e cos 6 — 1 "- -■ From (1) we know that a point is on the right or left- hand branch, according as p in [47] is positive or negative ; that is, according as cos ^ > or , as it should, e since in this case the radius vector is || to the asymptote. If 6 = ^TT, p ^ — a (e- — 1)= — semi-latus rectum. If e = ^, p= a — ae = — F'A'. Exercise 39. 1. What is the polar of the point (—9, 7) with respect to the hyperbola 7x- — 12/ = 112 ? 2. Find the equations of the directrices of an hyperbola. 3. Find the angle formed by a focal chord and the line that joins its pole to the focus. THE HYPERBOLA. 189 4. Find the pole of the line Ax-]- B)/-{- C = with re- spect to an hyperbola. 5. Find the polar of the right-hand vertex of an hyper- bola with respect to the conjugate hyperbola. 6. Find tlie distance from the centre of an hyperbola to the point where the directrix cuts the asymptote, 7. If (xi, iji) and (x.2, ijo) are the ends of two conjugate diameters, then X1X2 jh!h _ ^ 8. The equation of a diameter in the hyperbola 2ox^ — 16^/^^400 is 3>/ = x. Find the equation of the conjugate diameter. 9. In the hyperbola 49a'^ — 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 = 3x in the hyperbola 9x- — 4^^ = 36. 11. Two tangents to an hyperbola at right angles intersect on the circle a-^ -[- ?/^ = «^ — b^. 12. Tangents at the extremities of any chord of an hyper- bola intersect on the diameter which bisects that chord. 13. Prove that FQ (Fig. 71) 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. 190 ANALYTIC GKOMKTKY, 18. Find the equation of a tangent at any point (x^, v/i) of the h^^perbola Axt/ = (r-{- Ir. 19. Find the equation of an hyperbola, taking as the axis of ij (i) the tangent through the left-hand vertex ; (ii) the tangent through the right-hand vertex. 20. Find the polar equation of an hyperbola, taking the right-hand focus as pole. 21. Find the polar equation of an hyperbola, taking the centre as pole. 22. To find the centre of a given hyperbola. 23. The distance from a fixed point to a fixed straight line is 10. Find the locus of a point which moves so that its distance from the fixed point is always twice its distance from the fixed line. 24. Show that the locus of x^ — \if — 2x— 16y — 19 = is an hyperbola ; find its centre and axes. 25. If A and B have unlike signs, prove that the locus of Ax^ -\- Bif + I)x + E[i + F^^ is in general an hyperbola whose axes are parallel to the coordinate axes ; and deter- mine its semi-axes. 26. Through the point (— 4, 7) a straight line is drawn to meet the axes of coordinates, and then revolved about this point. Find the locus of the point midway between the axes. 27. A straight line has its ends in two fixed perpen- dicular lines, and forms with them a triangle of constant area d\ Find the locus of the middle point of the line. 28. The base «. of a triangle is fixed in length and posi- tion, 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. 185. The loci represented by equations of the second degree that are not of the first order are called Loci of the Second Order. In the preceding chapters we have seen that the circle, parabola, ellipse, and hyperbola are loci of the second order. We will now inquire whether there are other loci of the second order besides the four curves just named ; in other words, we will determine 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^ -H Bir + Cri/ + Z)a; + % + F= 0, (1) and shall assume that the axes of coordinates are rectan- gular. This assumption will in nowise diminish the gener- ality of our conclusions ; for if the axes were oblique, we could refer the equation to rectangular axes, and this change would not alter the degree of the equation or the nature of the locus which it represents (§ 91). • 186. To find the condition that the rjeneral equation of the second degree may represent two loci of the first order. To do this let us solve (1) with respect to one of the variables. Choosing y for this purpose, we obtain Cx + E 1 , . r y = ^^^^^Lx^ + Mx^N, (2) where L=C^ — iAB, iI/= 2 ( CE — 2BD) , N=^ E- — \BF. 192 ANALYTIC GKOMKTRY, If Lx^ -\- Mx -\- N is a perfect square, then tlie locus of (2), or (1), will be two loci of the first order. Now, from Algebra, we kuow that the condition tliat Lx^ + 3fx + N should be a perfect square is or, substituting the values of L, M, and N, we have (CE-2BDy-{C^ — 4.AB) (E'-4:BF) = 0, or F(C^ — 4:AB)-\-AE^ + BD^—CI)i;=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). This same result was obtained by a more general method in § 57 ; hence, Whenever A = 0, eqiiation (1) represents two loci of the first order. These loci may be readily determined by resolving (1) into two simple equations in x and y. CENTRAL CURVES. 2 NOT ZERO. 187. A centre of a curve is a point that bisects every chord passing through it. Loci are classified as Central and Non-Central, according as they have or have not a definite centre. The circle, ellipse, and hyberbola belong to the first class, the parabola to the second. 188. To find the equation of the central loci represented, by eq^iation (1) referred to their centre. To do this let us change the origin to the point {h, 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-\-k for y, we find that the coefficients A, B, and C remain unaltered, and we may write the transformed equation Ax^ + By"" + Cxy + D'x + E'y = R, (4) LOCI OF THE SECOND ORDER. 193 wh ere D' = 2 A h + Ck + D, E' = 2Bk+Ch-\-E, R=— [A/i" + Bk- + Ckk + Dh + m- + F^. The values of h and k that will make 1)' and E' vanish are evidently found by solving the equations 2Ah+Ck-{-D = 0, 2Bk-\-Ch-\-E=0, CE-2BB j_CD — 2AE and are ^'^-4^^_C2' ''-AAB-C' If 4:AB — C^, denoted by 2, is not zero, these values of h and k are finite and single, and equation (4) may be writttui Ax^ + Bi/^-\-Cxi/ = E. (5) From the form of (5) we see that if (x, y) is a point in its locus, so also is ( — x, — //) ; that is, the new origin (Ji, k) is the centre of the locus. Hence, Whe7i 2 is not zero, equation (1) can he reduced to the form of (5), and represents central curves. When, however, 2 = 0, the values of h and k become infinite or indeterminate, and the locus of (1) lias no defi- nite centre. Hence, When 2 = 0, (1) cannot be reduced to the form of (5), and rejiresents no7i-central curves. The value of R can be reduced to the following useful form, which shows also that R and A vanish together. R = — [Ah' + Bk- + Ch k + l)h + Ek + F^ = -i[(2Ah+Ck + D)h + (2Bk + Ch -\- E)k + Dh + Ek + 2/^] = — i {D'h + Ek + Dh + Ek + 2F) = -^(I)h-\-Ek-\-2F) _ 2BD'—CI)E-\-2AE?—CnE-\-2F(C'-AAB) ~ * C' — 4:AB _ A ~ 2' 194 ANALYTIC GKOMETRY. 189. To reduce (5) to a known form hij rausinf/ the tern„ in xy to disappear. For this purpose we change the direction of the axes through an angle 0, keeping the origin unaltered, and then determine the value of by putting the new term that involves xy equal to zero. The change is made by substituting for x and y, in equation (5), the respective values (§ 86), X cos 6 — y sin 6, X sin 6-]r y cos 6 ; and equation (p) now beco]nes Px''-^ny-+C'.ry = B, where P = A cos- 6 -f- J^ sin^ 0-}- C sin cos 6, (&'^ Q=^ A sin- 6-\- B cos- d — C sin 6 cos 6, (7) C' = 2{B — A) sin 9 cos + C(cos- $ — sin^ 6). (8) Putting C'=^0, we obtain, by Trigonometry, (A — B) sin 26 — C cos 20 = 0, (9) or tan 2^ = —^-. (10) A — x> Since any real number, positive or negative, is the tan- gent of some angle between zero and tt, equation (10) is satisfied by some value of 9 between zero and •^tt. In what follows we shall use the simplest root of (10). I>y this transformation, equation (5) is reduced to the form p^.2^^y._^^ (jl) of which the discussion will be found in the next section. Cor. 1. The values of P and () in terms of A, B, C may be found as follows : From (G) and (7), by addition and subtraction, P^Q = A + B, (12) P — ^ = (^ — J?) cos 2^ + C sin 26. (13) LOCI OF THE .SECOND ORDER. 195 Equation (9) may be written Oz=(A — B) sin 26 — cos 26. (14) Adding the squares of (13) and (14), we have (P-Qf = (A-Br+C', (15) or P-Q^±\/{A-Bf+C'i (IG) Whence, from (12) and (16), P = ^[A + B± V(^ - By + C% (17) Q = i[.l + P rp V(.4 - By + C'l (18) These values of F and Q are evidently always real. Cor. 2. By squaring (12) and subtracting (15), we obtain 4FQ = 4.AB — C = X (19) Hence, I* and Q have like or unlike sir/tis, accordintj as % is positive or negative. Cor. 3. In applying formvdas (17) and (18), the question arises which sign before the radical should be used. If in (13) we substitute for cos 26, its value obtained from (14), we have p ^^ [(^-/?)^+C'^]sin2g C/ Since the numerator of the fraction is always positive, P — Q must have the same sign as 6'; that is, the upper or lower sign in (16) must be taken according as C is posi- tive or negative. Hence, the upper or lower signs in (17) and (18) are to he taken accordinr/ as C is positive or negative. 190. The nature of the locus represented by equation (11) depends upon the signs of /*, Q, and R. There are two groups of cases, according as 2 is positive or negative, and three cases in each group. 196 ANALYTIC GEOMETRY. Group 1. 2 Positive. In this group, P and Q must, by (19), agree in sign. Case 1. If R agrees in sign with P and Q, then, by § 124. the locus is an ellipse whose semi-axes are a/— and \-7^- li P= Q, the locus is a circle. Case 2. If R differs from P and Q in sign, no real values of X and ij will satisfy (11), so that no real locus exists. Case 3. If 2t = 0, the locus is the single point (0, 0). Group 2. % Negative. In this group, P and Q, by (19), must have unlike signs. Case 1. If ^agrees in sign with P, we may, by division (and by changing the signs of all the terms if necessary), put equation (11) into the form of equation [40], page 171. Therefore, the locus is an hyperbola, with its transverse, axis on the axis of x, and havinar for semi-axes «=V] Case 2. If R agrees in sign with Q, we may, by divisioi. (and by change of signs if necessary), put equation (11) into the form of equation (1), page 172. Therefore, the locus is an hyperbola, with its transverse axis on the axis of y. Case 3. If i? = 0, the locus consists of tioo straight lines, intersecting at the origin, and having for their equations NON-CENTRAL CURVES. 2 = 0. 191. To determine the locus of (1) ivhen A and 2 are loth zero. LOCI OF THE SECOND ORDER. 197 If A = 2 =^ 0, then from the first form of A in § 186 we must have ^j,j -2BD = 0. (20) Hence, L = 3I=0, (§186) and (2) becomes 2Bi/ +Cxi-JSzf '\lE^ — 4.BF= 0, (21) which represents two parallel straight lines, two coincident straight lines, or no locus, according as E^ — 4:BF';>, =, or<0. When 4:AB — C^ = 0, if C is not zero, neither A nor B can be zero; if C = 0, either A or B must be zero, but both cannot be, since if A=^B = C = 0, (1) is no longer of the second degree. When C=^B=0, by solving (1) for x, and introducing the above condition, we should obtain, instead of (21) its corresponding equation. 2 Ax -\-Ci/-\-D:f \/l)^—4:AF= 0, (22) whose locus is also two parallel straight lines. Hence, When A and 2 are both zero, equation (1) represents two parallel straight lines, two coincident straight lines, or no locus. Cor. Eliminating B between 2 = and (20), we obtain CD — 2AE=0. (23) In like manner (20) follows from 2 = and (23). From these results, and the values of h and k, we learn that (i) When A=2=0, h and k are both indeterminate, and conversely. (ii) The values of h and k are indeterminate together. 192. To determine the locus of (1) ivhe7i 2 is zero and A is not zero. We simplify (1) by first making the term in ary disappear by proceeding exactly as in § 189; that is, by turning the axes 19/+F=0. (29) And this, by changing the origin to the point 4:PF— U^ U (■ 4.VP IP becomes cc^ = — ^ y. This represents a parabola having the axis of y for its axis, and placed on the j/oslf ire or the neyafire side of the new origin, according as V and P are unlike or like in sign. It should be noted that the value of P or Q, when not zero, is A + P- The values of U and V* can be found from (25) and (26). * We may obtain the values of U and T" iu terms of the origmal coefficients, as follo%YS : From (24) we find, by Trigonometry, -(A- B) ± -J (A - BY- + C2 tan 6 = jf •.• Introducing the condition iAB — C^, we obtain 2A tan 6 = —5 if C is negative ; 2B .. ^ . = -^1 if C IS positive ; whence, if C is negative, 2^ -C sin e V442 -I- (>2 V4^2 + 02 And if C is positive, 200 ANALYTIC GEOMETRY. If C^A = 0, the given equation is of the form of (27), its locus is a parabola and can be found as that of (27). If 6' = ^ = 0, the given equation is of the form of (29), and its locus is a parabola. 193. The main results of the investigation are given in the following Table : Loci Represented by the General Equation of the Second Degree. Ax^ + By^-+ Cxy + Dx + i:y+ F=0. CLASS. CONDITIONS. LOCL'S. I. Loci having a centre. S positive, A not zero. S positive, A = 0. 2 negative, A not zero. S negative, A = 0. Ellipse, or no locus. Point. Hyperbola. Two intersecting straight lines. II. Loci not having a centre. 2 = 0, A not zero. S = 0, A = 0. Parabola. r Two parallel straight lines, ] One straight line, V Or no locus. Thus it appears that there are no loci of the second order besides those whose properties have been studied in the preceding chapters. sm 6 = 2B cos e = V452 + C2 By substitution, we obtain from (25) and (26) if C is negative. V4B2 + C2 if C is positive, U = U = 2AE - CD 2 BE + CD ^JIiF+~c^ v = v = 2AD+ CE CE-2BD ■\J4B^ + C^ (30) (31) • LOCI OF THE SECOND ORDER. 201 194. Examples. 1. Determine t?ie nature of the locus ox^-\-57/-{-2xi/ — 12x — 12ij = 0, (1) transform the equation and construct it. Here A = 5, B=o, C=2, D=-12, E=-12,F=0. Whence, 2 = 96, A ==1152. Hence, the equation represents an ellipse or no real locus. Therefore, the equation of the locus referred to new parallel axes through the centre (1, 1) is (§ 188) 5x'- + 2xi/ + 5f- = 12. (2) To cause the term in x)/ to disappear, we have Whence, 2$ = 90°, or ^ = 45°. (We use the upper signs m the values of P and Q by § 189.) Hence, by § 189 the equation of the ellipse referred to its own axes is 60.2 + 4^^-12, or 1 + 1 = 1. (3) Whence, a = VS, b = V2, and a lies on the axis of i/. To construct the equation, draw the axes OiXi,OiYi (Fig. 75); locate the centre (1, 1). Through this point O2 draw the second set of axes, O^X^, 0^ Y„. Through O2 draw the third set of axes OaA'g, 02^3, making X0O0X3 equal to 45°. Lay off O2A' = O.A = Vs, and 0^0^ = O^B = V2. The ellipse having BO^ and A A' as axes will be the required locus. 202 ANALYTIC GEOMETRY. The equation of the locus referred to the axes OiX^, Oi Yi is (1); its equation referred to O^Xo, O^Y^ is (2); and its equation referred to O^X^, O^Y^ is (3). In constructing the locus it is not necessary to draw the second set of axes C/a JCj, (/a jfa* (1) Fig. 75.' 2. Determine the nature of the locus x^-\-?f- 5x1/ + Sx- - 20^ + 15=0, transform the equation and construct it. Here 2 = -21, A = — 21. Tlierefore, the locus is an hyperbola. i? = A^2 = l, A = -4, k = 0. Hence, the first transformed equation is x^-\-y" — 5xi/ = l. In the second transformation, ^-45°, P = -|, Q = i. Hence, the equation of the curve referred to its own axes is 3x' — 7/ = — 2, from which we see that a = Vf , b = V§, and a, or the trans- verse axis, lies on the axis of y. (2) LOCI OF THE SECOND ORDER. 203 To construct the equation, draw the axes OiXi, OjFi, locate the centre ( — 4, 0), and through it draw the third set Fig. 76. of axes O2X3, O2YS, making A'lOaA's^ 45°. Then lay off 02A'=02A=\/^, and O^B = O^B' = ^ ^ ; and draw the hyperbola having AA' and BB' as its transverse and conju- gate axes respectively. 3. Determine the nature of tlie locus a;^ + 2.ry-y2-f-8.r + 4^-8 = 0, (1) transform the equation and construct it. Here 2 = — 8, A= — 176; hence, the locus is an hyperbola. ^ = 22, h = — 3, /.• = — !; hence, the first transformed equation is x' -\- 2x7/ — if- = 22. _ (2) =22|-°, P=\l2, Q = — \/2; hence, the equation of the curve referred to its own axes is V2a;2— V2// = 22. (3) The hyperbola is equilateral, and its transverse axis lies on the axis of x. (Let the reader construct it.) 4. Determine the nature of the locus x^-\-,f — 2x!i-\-2x — i/ — l=0, (1) transform the equation and construct it. 204 ANALYTIC GEOMETRY. Here 2 = 0, A is not zero. Therefore, the locus is a parabola. = 4:5°, F = 0, Q = 2, U=^-\l2, V= — %\l2- hence, by revolving the axes through an angle of 45°, the equation becomes 2^" + i V2a; — %-^2ij -1 = 0, or t/2-fV2^+(tV2)^=-iV2^ + M. or (2^_3V2)2 = -iV2(x-f|V2). _ (2) Passing to parallel axes whose origin is (f|V2, |V2), (2) becomes y^ = -iV2^, ■ (3) the locus of which is a parabola whose latus rectum, or parameter, is ^\l2. Fig. 77 To construct the equation, draw the original axes O^X^, OxYi, then draw the second set of axes O1X2, O1Y2, making XiC>iXj = 45°. Locate the new origin O3, (f|V2, |V2), and through it draw the third set of axes O^Xz, O3 Pg, to which (3) refers the locus, which is now easily drawn. 5. Determine the nature of the locus x^ — ixy + iy^ — 6x-\- 12y =■ 0. LOCI OF THE SECOND ORDER. 205 Here 2 = and A = 0. Factoririg the first member of the equation, or solving the equation for x or y, we obtain as the equations of these lines x — 2ij = 0, x — 2y — Q = 0. Hence, the locus is two parallel straight lines. 6. Determine the nature of the locus of if' — xy — Q>x- — 3.x + ?/ = 0. Here 2 is negative, and A = 0. Hence, the locus is two intersecting straight lines. Their equations are y + 2a; + 1 = 0, and ?/ — 3x == 0. 195. A Conic is the locus of a point whose distance from a fixed point bears a constant ratio to its distance from a fixed straight line. 196. To find the equation of a conic. Y N Fig. 78. Let F be the fixed point, and YY' the fixed straight line. Through jPdraw XO perpendicular to YY, and use OA'and OFas axes of reference. Let^:> denote the distance OF, and 20G ANALYTIC GEOMETRY. e tlie constant ratio ; then FP -=r NP = e, P being any point (a-, I/) in the curve. Now FP^ = Fir' + MP\ But FP = eX NP = ex, FM=x—p, MP = ij. Therefore, e^x'^ = (x — j'Y + if^ or ( 1 - e^) a:^ + / - 2px + /^^ = Q, (1) which is the equation required. CoR. In equation (1), which is of the second degree, 2 = 4 (1 — tj-), and A = 4/vV. Hence, when the fixed point is without tlie fixed line, A is not zero, and If e 0, and the conic is an ellipse. If e = 1, 2 = 0, and the conic is a parabola. If e >• 1, 2 0, and the conic is the point (0, 0). If e = 1, 2 = 0, and the conic is a right line. If « > 1, 2 < 0, and the conic is two intersecting right lines. If e^O, by § 61, the conic is a cii'cle or a point, accord- ing as p is not or is zero. From §§ 92, 150, 183, it follows that the fixed point is a Focus, the fixed right line a Directrix, and the constant ratio the Eccentricity, of the conic. Exercise 40. Determine the nature of the following loci, transform their equations, and construct them : 1. ^x'-{-2if—2x-\-y-\=0. 2. 3a;2-f-2j7/ + 3/ — ir.// + 23 = 0. LOCI OF TlIK SKCOND ORDER. 207 3. X- — I O.v.v + //" + •'• + // + 1 = 0. 4. x^ -f- xi/ -{- //'^ -\- X -{- 1/ — 5 = 0. 5. if — X- — y = 0. 6. l+2::c + 3^2 = 0. 7. y-~2xi/-\-x^ — Sx-\-16 = 0. 8. ar — 2x1/ -\- 1/'^ — Gx — 6^ + 9 = 0. 9. 7/_2x — 8// + 10 = 0. 10. 4a;2 + V + 8.r + 36// + 4 = 0. 11. 52cc=^ + 72x// + 73/ = 0. 12. 9/ — 4x2 _ 8.^ _|_ 18^ + 41 = 0. 13. y^ — xy — 5x-\- 5i/ = 0. CHAPTER IX. HIGHER PLANE CURVES. 197. An Algebraic Curve is one whose rectilinear equa- tion contains only algebraic functions. A Transcendental Curve is one whose rectilinear equation involves other than algebraic functions. Thus, the loci of 7j=^a^, ?/ = tan x, y = {a — X) tan {jttx -i- a), y = sin ~^x are transcendental curves. Transcendental curves and all algebraic curves above the second order are called Higher Plane Curves. Let the symbol F (x, y) denote any rational integral func- tion of x and y, of the third or higher degree. If F (x, y) breaks up into simple or quadratic factors in x and y, the locus of F (x, y) = consists of lines of the first or second order. If F (x, y) does not break up into rational factors in x and y, the locus of F(x, ?/) ^= is a higher plane curve whose order is of the degree of the equation. Thus, the locus of y^ — x^^{y — x) (y^ + a-y + x^) = 0, consists of the right line y — x = and the ellipse y- -\- xy -\- x^ = ; the locus of y*-\-xy-{-2y^—2x*—5x"—3=(y'-\-2x^+S)(y^—x^-l)=0, consists of the ellipse y- -\- 2x^ + 3^0, and the hyperbola y^ — x^ — 1=0; while the locus of y^ — ax -\- x'^ ^ is a higher plane curve of the third order. In this chapter we shall consider a few of the higher plane curves, some of which possess historic interest from the labor bestowed on them by the ancient mathematicians. 198. The Cissoid of Diodes. Let X^(rig. 79) be a tangent to the circle XSO at the vertex of any diameter HIGHEK PLANE CURVES. 209 OX ; let OR be any right line from to XH cutting the circle at S, and take 0F = ES; then the locus of P, as Oli revolves about 0, is the Cissoid. To find its equation referred to the rectangular axes OX and OY, let OM=x, MP=ij, and OC=CX= CD = r. Now, MP : OM :: NS : ON. (1) Since OP = ES, OM=NX. Hence, 0N= OX— NX= OX— 0M= 2r — x, and NS= -^ONXNX^ \l{2i — x)x. Substituting these values in (1), we obtain £C3 2/^ V(p; iC)X which is the equation sought. or y^ 2r-ac' [48] 210 ANALYTIC GEOMETRY, A discussion of [48] leads to the following conclusions: (i) The curve lies between the lines x = and x = 2r. (ii) It is symmetrical with respect to the axis of x. (iii) It passes through the extremities of the diameter perpendicular to OX. (iv) It has two infinite branches to which x = '2r is an asymptote. CoR. 1. To find the polar equation, let 6 = X0F and P=OF; ^ OX OS then '^'^=0e'''0X' Hence, p= OF = SE= OE- 0S=-^ — 2r cos 6 ^ cos sin'^O cos 9 Therefore, p ^ 2r sin 6 tan 6, which is the equation sought. The cissoid was invented by Diodes, a Greek mathema- tician of the second century b.c, for the solution of the problem of findlvr/ tivo mean proportionals, of which the dupUcation of the cube is a particular case. Cor. 2. To duplicate the cube, in Fig. 79, take CB = 2r, and draw 5X cutting the cissoid in ^; then, smceCX=^CB, EX=^EQ. But from tlie o(] nation of the cissoid, we have W == ^' = r^; therefore, W = 20E'. Let c denote the edge of any given cube ; take ^i so that OE: EQ::c:c,, or OE' : W - «' = «i'- But W = 2 7a£^' ; th eref ore, Cj' = 2c^ ; that is, ci is the edge of a cube double the given cube in volume. In like manner, by taking CB = mr, we can find the edge of a cube m times the given cube in volume. HIGHER PLANE CURVES. 211 199. The Conchoid of Nicomedes. The Conchoid is the locus of a point P such that its distance from a fixed line XX', measured along the line through F and a fixed point A, is constant. A is the Pole, .\'A'' the Directrix, and the constant distance BP, denoted by b, is the Parameter. P>^ Y JM \k' m /r ~^_ Y A Y -■■B Fig. 80. To construct the conchoid by points, through A draw any line AP cutting XX' in P. La}' off PP^^h on both sides of XX'. In like manner locate the points P', and any number of others, and trace the branches through them. To find the equation of the curve referred to XX' and a line through A perpendicular to A'A'' ; let 0M= x, MP = y, AO = a. Now, AB : BP:: RM : MP, and BM= ^'rF - HlF = ^iW^f. Therefore, x \ y-\- a, the lower branch has an oval or loop, as in the figure. If i = a, the lower branch passes through A and is some- what like that in the figure, without the loop below A. If 6 < a, the upper and lower branches are like the dotted lines in the figure. If a = 0, the conchoid becomes a circle. CoR. 1. If A is the pole and A Fthe polar axis, we have p = AB'±B'F' = asec 6 ±h, which is the polar equation of the conchoid. The conchoid, invented by Nicomedes, a Greek mathema- tician of the second century b.c, was, like the cissoid, first formed for solving the problem oi finding ttvo mean propor- tionals or duplicating the cube. It is more readily applicable, however, to the trisection of an angle, a problem not less celebrated among the ancients. Cor. 2. To trisect any angle, as CAP (Fig. 81), on ^C lay off AB any length ; through B draw BK perpendicular to AP, and take KF = 2AB. Construct a conchoid with ^ as a pole, XX' as a directrix, and KP as a parameter. HIGHER PLANE CURVES. 213 At R erect a perpendicuhir to XX' intersecting the cou- choid in D ; then DA will trisect the angle RAF. For bisect DB in S ; then RS = SD = iKF = AR. Therefore, Z RAS= Z ^'S'^ = 2Z RDS= 2Z DAP. Therefore, Z DAP = iZ RAF. / 200. The Lemniscate of Bernoulli. The Lemniscate is the locus of the intersection of a tangent to a rectangular hyperbola with the perpendicular to it from the centre. To find its equation we proceed as follows : The equation of the tangent to the equilateral hyperbola x^ — y^ = ci^ at the point (xi, i/{) is a'la;— yi?/ = «^- (1) The equation of the perpendicular from the origin to (1) is y^-^:r, or - = -^. (2) Xl Xi i/i Solving (1) and (2) for x^ and v/i by multiplying each term of (1) by one of the members of (2), we obtain a"x a'i/ X + V == — = ^1 y\ ^, . ti^x ahi Therefore, cr, = ^r-, — 7/ ?/i ^ ;— r — ^• But, since (a^i, y^) is on the hyperbola, we have Xx — y\=^o^\ hence, by substitution, we obtain (^2 + ^•2)2 = „2(ar;2 _ 2,2)^ ^5QJ which is the equation sought. 214 ANALYTIC (ilOoMlCTHV. From \_iiO^ it follows tliat the curve is syiimietrical with respect to both axes. The form of the curve is given in Fig. 82. Fig. 82. CoR. 1. Substituting p ens 6 for x, anrl p sin 6 for y in [50], and remembering that cos'^ ^— sin^ ^=cos 20, we obtain p^:=a^ cos 20 as the polar equation of the lemniscate. (2) CoR. 2. From (2), p = ±a Vcos 20. Hence, when 0=^0, p = ±a. AsO increases from 0° to 45°, p changes from ± « to ± 0, and the portions in the lirst and third quadrants are traced. As increases from 45° to 135°, cos 20 is negative and p is imaginary. As increases from 135° to 180°, p changes from ± to ± a, and the portions in the second and fourth quadrants are traced. Therefore, (i) The curve consists of two ovals meeting at the pole 0. (ii) The tangents to the curve at are the asymptotes to the equilateral hyperbola. HIGHER PLANP: CURVES. 215 201. The Witch of Agnesi. Let 17/ be a tangent to the circle OKYnt the vertex oi' the diameter OY ; let OB be any line from to Ylf cutting the circle in K; produce the ordinate I>K, and make DP'^^YE; then the locus of P is the Witch. To find its equation, let the tangent OX and the diameter or be the axes; let 0F=2randPbe any point; then OD=ij, DP = x, and OD : OY :: I)K : DP{= YE), (1) 9, '~ or y -.'Zr :: \ly(2r — y)'-x. Therefore, oc-y = 4r-(2/- — y) is the equation of the Witch. [51] CoR. 1. Since a; = ± 2j' V(2/ — y) -r-y, it follows that (i) The curve is symmetrical with respect to the axis of y- (ii) The curve lies between the lines y=0 and y = 2r. (iii) The axis of x is an asymptote to eacli infinite branch. CoR. 2. From (1) it follows that corresponding abscissas of the circle and the Witch are proportional to the ordinate and the diameter of the circle. The Witch was invented by Donna Maria Agnesi, an Italian mathematician of the eiQ:hteenth oontnrv. 216 ANALYTIC GEOMETRY. 202. The Cycloid. A Cycloid is generated by a point P in the circumference of a circle RFC, which rolls along a right line OX. The curve consists of an unlimited number of branches, but a single branch is usually termed a cycloid. The right line 0X\?> called the Base ; the rolling circle RFC the Generatrix ; and P the Generating Point. If 0K=^ KX, the perpendicular KH is the Axis, and JI the Highest Point. OM To find the equation of the curve, first take as the axis of X the base OX, and as the origin the point 0, where the curve meets the base. Let r denote the radius of the generatrix RPC, and $ the angle PCR; then arc PR equals OR over which it has rolled, and 6 ^= a,vG PR -^r. Denote the coordinates of P by x and y ; then X = 0M= OR — MR = arc PR — PN= r6 — r sin d. y = MP = RC — NC=r— r mse. Therefore, X =^r (6 — sin 6) y=^r (1 — cos 6) Equations (1) taken as simultaneous are the equations of the cycloid. To eliminate 6 between these equations, from the second, we obtain (1) HIGHER PLANE CURVES. 21' cos = : therefore, sin ^ = ± — — : /• /• and vers ^=ri —cos ^1 = - ; or^ = vers'^-- L -* r r Substituting these values of 6 and sin 9 in the first of equations (1), we have x = r vers '^ a? y. T ^2ri/ - yS [52] the equation of the cycloid in the more common form. In the value of sin 6, and therefore in equation [52], the upper or lower sign is used according as ^ < or > tt ; that is, according as the point is on the first or second half of a branch. From 1/ = r (1 — cos $) it follows that the locus lies between the lines i/ = and ?/ = 2r. For ?/ = in [52], x = 0, ± 2iri', ± 47rr, ; hence, the locus consists of an unlimited number of branches like OHX, both to the right and to the left of Y. 203. Let the highest point (Fig. 85) be taken as the origin, and OX parallel to the base as the axis of x; then OY, the axis of the curve, will be the axis of y. Let 6 denote the angle HCK. The point K was at Y when P was at 0, and arc KH= YH. Hence, X = 0M= Yir+ BP = re + r sin 6. y = - MP = — XC + BC= — r -\- r cos $. Hence, the equations are a- — ;■ (^ + sin 6) ") ^^^ y=^r (cos ^ — 1) ) ' or a; = r vers ^ ( — •- \ ± V— 2rij — y\ (2) 218 ANALYTIC GEOMETRY, R Y Fig. 85. Tlie invention of the cycloid is usually ascribed to Galileo. After the conies, no curve has exercised the ingenuity of mathematicians more than the cycloid, and their labors have been rewarded by the discovery of a multitude of interesting properties. Thus, the lengtli of the branch ROA is eight times the radius, and the area BOA is three times the area of the s:eneratins[ circle. Exercise 41. 1. Prove that the cissoid is the locus of the intersection of a tangent to the parabola if^= — 8ra; with the perpen- dicular to it from the origin. By § 99, the equation of a tangent to if^ — Srx is (1) y = mx ■ — ) m (2) and the perpendicular to it from (0, 0) is y = X : therefore, ?m = m y ^ Eliminating m between (1) and (2), we obtain a-' •'^"=27^- 2. At the centre of any circle C (Fig. 86), erect CH _L to the diameter OA'; and on A'O produced layoff 0^4 = OC^i'. Let LQii be a rectangular ruler of which the leg QR equals 2r. If the ruler is moved so that the leg LQ HIGHKIl PLANK CURVES. 219 passes through A, while the end li sli.Ies along C'll, ])rove that the locus of P, the middle point of Qli, is a cissoid. Fig. 8fi. Let OA^be the axis of a^, the origin, and AQK any position of the ruler ; then OM^x, MP^=^y, EQ = MN^ CM= X — r, A X= 2 0M= 2x. EP^ = PQ" — JiJif = r' — (.T - ry = 2rx — x'\ (1 ) y = MP=NQ-\-EP. NQ-.AN.-.EQ.EP, or NQ : 2.r : : .r — r : \l2rx — x' From (1), (2), and (,']), we obtain \j2rx — .1 (-0 (3) y + yj2rx-x% or !f- :^r — X This method of deseribing tlie oissoid by continuous motion was invented by Sir Isaac Newton. 220 ANALYTIC GEOMETRY. 3. In Fig. 79 prove that jVS and OJV are two mean pro- portionals between OM and iV7 ; that is, prove OM-.NS:: ON: NI. NS'' = NXX 0N= OMX ON. The right line 01 will pass through K; hence, Z OJN^ Z VOI= ^ arc 0K= ^ arc SX ^ZSOX; therefore, NS: ON:: ON: NI, etc. 4. If in the lemniscate (Fig. 82) 0F'=0F=^a^2, prove that FF X F'F is constant, F being any point on the curve; and hence that the lemniscate may be defined as the locus of a point the product of whose distances from two fixed points is constant. 5. Construct the logarithmic curve y^=a^, or x^log^y. Prove that every logarithmic curve passes through the point (0, 1), and has the axis of x as an asymptote. 6. The Trochoid is the curve traced by any point in the radius of a circle rolling on a right line. If r denotes the radius of the circle, h the distance of the generating point from its centre, and Q denotes the same angle as in § 202, show that the equations of the trochoid are x = r6 — b sin $ > ,.y. y=^r — b cos 6 ) When b r, the Curtate Cycloid. When b = r, the curve is the cycloid. Spirals. 204. A Spiral is the locus of a point whose radius vector continually increases, or continually decreases, while its vectorial angle increases (or decreases) without limit. HIGHER PLANK CURVES. 221 A Spire is the portion of the spiral traced during one revolution of the radius vector. The Measuring Circle is the circle whose centre is the pole and whose radius is the value of p when 6^2Tr. 205. The Spiral of Archimedes. If the radius vector of a moving point has a constant ratio to its vectorial angle, that is, if p = ce, ■ (1) the locus is the Spiral of Archimedes. From equation (1), we have when^ = 0, Itt, |7r, |7r, ir, |7r, |7r, |7r, 27r, |7r, p=0, \itC, lire, fTTC, ttC, ^ttC, &TrC, JttC, 2TrC, ^wC, Hence, to construct the Spiral of Archimedes, draw the radial lines OH', OA, OB, OG, including angles of ^n; on these lay off Oa = J 7rr, Ob = |7rr, Oc = ^irc, , OH^ 27rc, and trace the first spire OabcdefgH through these points. Any number of other spires are easily constructed by noting that the distance between two spires, measured on a radius vector, is equal to 2irc. Thus, by taking a A. bB, cC, dD, eE, fF, gG, HH', each equal to 2irc, we obtain points of the second spire ; and so on. Any number of a(^ditional radial lines may be drawn to locate ]ioints in tlie curve. 222 ANALYTIC GEOMETRY. The spires, being everywhere equally distant along radial lines, aie said to be parallel. The measuring circle is JIMS, whose radius is OH or 2Trc. 206. The Reciprocal or Hyperbolic Spiral. If the radius vector of a point varies inversely as its vectorial angle, that is, if pd=^c, (1) the locus is the Reciprocal Spiral. Since p=^c^6, we have when 6 = 0, lir, lir, ^tt, tt. 4 " ' 2 " ' 4"} ( V V V r, V V V p = CC,i^^ > 4 ^, « — , J — , A — , 4 8 ZtT wTT wTT -iTT ^TT iiTT ZtT ZtT Hence, the radius of the measuring circle is c-f-27r, and its circumference is c. Fig. 88. To construct the curve, draw the radial lines OX, Oa, Ob, Oc, Od, Oe, Of, ();/, including angles of Jtt. Take OH^ e -^ Stt, and lay off Oa = 8 X OJf, Ob ^ 4 X OH, Or =f^X OH, 0(1 = 2 X OH, etc.; and through the points a, b, c, d, etc., trace the curve. In like manner, any number of spires may be drawn. From (1) it is evident that p approaches zero, as approaches infinity ; that is, the curve continually approaches the pole without ever reaching it. Since pd = e, it follows that the arc FK described with the radius vector of any point /'is constant and equal to c. HIGH KM PLANK CUUVKS. 223 Now, as p appi'oaclies infinity, tiiis arc a,|)[n-();i(',lies a perpen- dicular to OX. Hence, the line parallel to OX at the distance of c above it is an as}- niptote to the infinite brancli of the spiral. 207. The Lituus. If the square of the radius vector of a point varies inversely as its vectorial angle, that is, if irO = c, the locus is the Lituus. Fig. 80. Let the student construct tlie curve from its equation and show that (i) The curve continually approaelies the pule without ever reaching it, as 6 increases withoiit limit. (ii) The polar axis is an asymptote to the infinite branch. 208. The Logarithmic Spiral. If tlie radius vector of a point increases in a geometrical ratio, while its vectorial angle increases in an arithmetical ratio ; that is, if p = a^, or 6=:log„p, (1) the locus is the Logarithmic SpiraL Since p = l when 6 = 0, every logarithmic spiral passes through the point (1, 0). To construct a logarithmic s])iral, let r/ = 2; then p = 2". When^ = 0, 1(=57.,r), 2(= 114.0°) '2^. P = l, 2 ,4 _ , , 77.88. 224 ANALYTIC GEOMETRY. In Fig. 90 let XOb = 57.3°, XOc = 114.6°, Oa = l,0b = 2, Oc = 4:, then a, h, c are three points on the spiral. As 6 increases, p increases rapidly, but it becomes infinity only when B does ; and hence only after an infinite number of Fig. 90. revolutions. As 6 decreases from zero, p decreases from unity. Since p approaches zero as 9 approaches negative infinity, the curve approaches the pole without ever reaching it. 209. The Parabolic Spiral. If in the equation if = ^)x, the values of x are laid off from A (Fig. 91) on the circle AH, and those of y on its corresponding radii produced, the locus of the point thus determined is the Parabolic Spiral. To find its equation, denote the radius OA by r, and let P be any point ; then X = AmH ^= rO. y = HF=OF—OH=p — r. HIGHER PLANE CURVES. 225 Siibstitutiug these values of x and // in y^ = 4^jx, we ob- tain the polar equation Fig. 91. The curve consists of two branches beginning at A ; the one determined by the positive values of y is an infinite spiral lying entirely without the circle; the other branch passes through the pole, forms a loop, and passes without the circle when p = — /•, and ^ = ?• -|-^j. Note. Among the ancients no problems were more celebrated than the "Duplication of the Cube" and the "Trisection of an Angle." Hii^pocrates of Chios reduced these two problems to the more general problem of finding two mean proportionals between two given lines. Thus, if c is the edge of the given cube, and x and y are the two mean proportionals between c and 2c, we have c : X = X : y = y : 2c. Therefore, (~^ - - X - X ^^- = i, or x^ - 2c^ \x/ x y 2c Hence, x, the first of the two mean proportionals between c and 2c, is the edge of a cube double the given cube in volume. After years of vain endeavor to solve these problems by the right line and the circle, the ancient geometers began to invent and study other curves, as the conies and some of the higher plane curves. The invention of the conies is credited to Plato, in whose school their properties were an object of special study. PART 11. — SOLID GEOMETRY. CHAPTER I. THE POINT. 210. The position of a point in space may be determined by referring it to three fixed })lanes meeting in a point. The fixed planes are called Coordinate Planes, their lines of intersection the Coordinate Axes, and their common point tlie Origin. In what follows we shall employ coordi- nate planes that intersect each other at right angles. Let XOY, YOZ, ZOX, be three planes of indefinite ex- tent intersecting? each other at risrht anerles in the lines A'A', YY', ZZ'. These coordinate planes are called the planes xy, yz, zx, respectively; the axes XX\ YY\ ZZ^ are called the axes of x, y, z, respectively; and their common point is the origin. TIIK POINT. 227 The coordinate i)laiies divide all space into eight portions, called Octants, wliich are numbered as follows: The First Octant is 0-XYZ, thi- Secomi, 0-YX'Z, the Third, 0-X'Y'Z, the Fourth, 0-Y'A'Z,i\ni Fifth, 0-XYZ', the Sl.rth, 0- YX'Z', the Seventh, 0-X'Y'Z', the Fiffhth, 0-Y'XZ'. The lifth octant is below the first. Let F be any point in space, and througli it pass three planes parallel respectively to the three coordinate planes, thus forming the rectangular parallelopiped F-OJtXM. The position of F will be determined when we know tiie lengths and directions of the lines LP, QP, NP. These three lines are called the Rectangular Coordinates x, y, z, of the point P, which is written {i-, ij, z). A coordinate is jjositive when it has the direction of OX, OY, or OZ; hence, it is ner/atlve when it has the direction of OX', OY', or OZ'. Thus the coordinate x is positive or negative according as it extends to the right or to the left from the plane //.~; 1/ is positive or negative according as it extends to the front or to the rear of the plane zx; and z, according as it ex- tends upward or downward from the ]ilane x//. Hence, the octant in which a point is situated is determined by the signs of its coordinates. Since the first octant has the posi- tive directions of the axes for its edges, tlie coordinates of a point in the first octant are all i)Ositive. If (a, b, e) is a point in the first octant, the corresponding point in the second octant is ( — a, h, r), in the third octant is ( — (t, — b, c), in the fourth octant is (k, — b, r), in the fifth octant is (a, b, — r), in the sixth octant is ( — a, b, — <•), in the seventh octant is ( — a, — b, — c), in the eighth octant is {(i, — b, — c). 228 ANALYTIC GEOMETRY. The point (x, y, 0) is in the plane xy. The point (x, 0, 0) is in the axis of x. The point (0, 0, 0) is the origin. The lines OM, OR, OS, or OM, MN, NP, may be taken as the coordinates of P, for they have the same length and direction as LP, QP, NP, respectively. To construct P, (x, y, z), we take OM^x, draw MN parallel to OY, take MN=y, draw NP parallel to OZ, and take NP = z. 211. The Radius Vector of a point is the line drawn to it from the origin. Thus, OP (=p) is the radius vector of P. From the rectangular parallelopiped in Fig. 92, we have op" = Oil' -f MN"^ + NP". Hence, denoting the coordinates of P by x, y, z, we have That is, the square of the radius vector of a point is equal to the sum of the squares of its rectangular coordinates. 212. By the angle between two non-intersecting straight lines is meant the angle between any two intersecting straight lines that are parallel to them. Thus, any line parallel to OP (Fig. 92) makes the angles XOP, YOP, ZOP with the axes of x, y, z, respectively. The angles which a line makes with the positive direc- tions of the coordinate axes are called its Direction Angles ; and the cosines of these angles are called the Direction Cosines of the line. The direction angles of a line are always positive and cannot exceed ir, or 180°. 213. Let a, fS, y denote the direction angles of OP (Fig. 92), or any line parallel to it, and x, y, z the coordinates of P, and let p = OP; then evidently THE POINT. 229 ac= p cos a, y = p COS p, s = p cos y. [54] Squaring and adding [54], and substituting in [53], we obtain cos2 a + cos2 p + cos2 7=1. [55] That is, the sum of the squares of the direction cosines of a line is equal to unity. Cor. Whatever are the values of x, y, z in [54], if each is divided by p, or Vx^ + 2/^ + «^ the quotients are the direction cosines of the radius vector of the point {x, y, z). Hence, if any three real quantities are each divided by the square root of the sum of their squares, the qiwtients will be the direction cosines of some line. Exercise 42. 1. In what octants may (oc, y, z) be, when x is positive ? when ic is negative ? when y is positive? when ?/ is negative ? when z is positive ? when z is negative ? 2. Inwhatoctantis(— 2,4, 6)? (2,4,-3)? (—2,4,-1)? (-2,-3,-1)? (-2,-3,3)? (2,-3,1)? (2,-1,-3)? Construct each point. 3. In what line is {a, 0, 0) ? (0, 0, c) ? (0, b, 0) ? 4. In what plane is {a, b, 0) ? {a, 0, c) ? (0, b, c) ? 5. Find the length of the radius vector of (3, 4, 5), (2, — 3, —1), (7, — 3, —5). Find the direction cosines of the radius vector of each point. 6. The direction cosines of a line are proportional to 1, 2, 3 ; find their values. What is the direction of the line ? 7. What is the direction of the line whose direction cosines are proportional to A, B, C? Wliat are the values of its direction cosines ? 230 ANALYTIC GEOMETRY. 8. Two direction aiii;los of a lino aro C>0° and 45°, what is tlie third ? If two are (U)° and 30°, wliat is tlie third? If two are 135° and G0°, what is the third ? 214. Projections. The 2)roJectio7i of a point upon a right line is the foot of the perpendicular from the point to the line ; or it is the intersection of the line with the plane throngh the point perpendicular to the line. Thus 31, JR., S (Fig. 92) are the projections of the point P upon the axes of X, y, z, respectively. Here and in the following pages, by projection is meant the orthogonal projection. The 2^^'ojection of a limited right line on another right line is the part intercepted between the projections of its extremities. Thus, 031, OB, OS (Fig. 92) are the projec- tions of OF on the axes of x, y, z, respectively. That is, the coordinates of anypnhU are the i^rojeetlons of its radius vector o7i the three axes. 215. The projections of any line PQ on parallel lines are equal ; for these projections are parallel lines included between parallel planes through P and Q. Now tlie projec- tion of any straight line on another that passes tlirough one of its extremities is evidently equal to the })roduct of its length into the cosine of their included angle. Hence, the projection of any limited strait/ht live on any other straiyht line is equal to its length nmdtiplied hy the cosine of the angle between the lines. 216. Let AT) (Figs. 93, 94) be a straight line, and ABCB any broken line in, space, connecting the points A and D, and let A\ B', C", D' be the projections of A, B, C, D upon OX, whose positive direction is OX. Denote the angle RADhy , the lengths of the lines ^47?, BC, CD by l^, l^, 4, and the angles which they make with the positive direction of OX by ai, tta, Us ; then in Fig. 93 we have TIIK I'OINT. 231 A'jy = .17/' + 7;'c" + c'lf. Therefore, AI>vms^=Ii cosai + /a cos o.^ + ^3 cosa^. [5(j] 111 Fig. 04, CD' is negative ; but ^3 cos a, is also negative, since ug, or SCV, is obtuse 5 hence foriuuhi [oG] holds true in all cases. D ]) B' C D' '" " A' D' D' Fig. 93. Fig. 94. That is, the ulgebralc sum of the ijrnjertiDnx on a r/iren line of the jifU'ts of (1)11/ broken line connecthuj aiii/ two points is equal to the projection on the same line of the straiijht line joining the same two points. 2,\1 . To find the angle between two straigJit lines in terms of their direction cosines. FiR. 95. Let OP and OF' be ])aralh'l respectively to any two given lines in space. Let 6 denote their included angle, and u. /?. y 232 ANALYTIC GEOMETRY. and a, (3', y' their direction angles, respectively. Let OM, MN, NP' be the coordinates of P' ; then the projection of OP' on OP equals the sum of the projections of OM, MN, NP' on OP ; that is, OP' cos 6 = OM cos a + MN COS (i + NP' cos y. But 0M= OP' cos a', MN=OP' cos ^', NP'=^ OP' cos y'. Hence, OP' cos 6 = OP' cos a' cos a + OP' cos /3' cos (3 -\- OP' cos y' COS y, or cos 9 = cos a cos a' + cos p cos P' + cos 7 cos y, [57] which is the required formula. ' 218. To find the distance between two points in terms of their coordinates. Fig. 96. Let Pi be the point (cci, y^, z^), and Pg the point {x^^y^, «2). Through P^ and Pj pass planes parallel to the coordinate planes, thus forming the rectangular parallelepiped whose diagonal is P1P2, and whose edges PiL, LK^ KP^ are parallel to the axes of x, y, z, respectively. Then 1\P' = 1\V ^Tk' + KPI (1) THE POINT. 233 Now P^L is the difference of the distances of P^ and Pg from the plane yz, so that P^L=^x^ — x-^. For like reason, LK= 1/2 — yi, and KP^ = ^2 — ^i- Hence, denoting the dis- tance P1P2 by D, we have, by substituting in (1), D = V(X2 - iCiY' + {»/2 - yif + (^2 - si)S [58] which is the required formula. Cor. 1. Since PiL, LK, KP^ are equal to the projections of the line P1P2 on the coordinate axes, it follows, from [58], that The square of any line is equal to the sum of the squares of its 2)roJections 011 the axes. CoR. 2. If a, ^, y are the direction angles of the line P1P2, we have X2 — Xi^ D cos a, ?/2 — yi=:D cos /?, «2 — «i = -E> cos y. 219. Polar Coordinates. Let XOY be a fixed plane, OX a fixed line in it, and OZ a perpendicular to it at the fixed point 0. To P, any })oint in space, draw OP, and through OP pass a plane perpendicular to A'OY, intersect- ing the latter in OJV; then tlie distance OP and the angles ZOP and WON determine the point P, and are called its 234 ANALYTIC GEOMETRY. Polar Coordinates ; Ol', denoted b^^ p, is the Radius Vector ; and tlie angles ZOP and MON, denoted by 6 and <^, re- spectively, are the Vectorial Angles. The point F is written (p, 6, cf>). <^ determines the plane ZOX, 6 determines the line OF in that plane, and p locates P in OP. CoR. If" XOY is a riglit angle, the rectangular coordi- nates of P are OM, MX, XP. To express these in terms of the polar coordinates of P, we have X = OM = O^'^cos <^ = OP sin 6 cos 4>^P ^i^^ ^ ^''^^ ^■ y = MX= OX&\\ifi>=^ OP sin ^ sin <^ = /J sin ^ sin ^. z = XP^ OP cose=p cos 9. We readily obtain also tan Q = !-^ ^ tan =~- z X 220. The Projection of a p(jint on a plane is the foot of the perpendicular from the point to the plane. The per- pendicular itself is the Projector of the point. Thus, the point iV(Fig. 97) is the i)rojection of P on the plane xy, and PTVis its projector. The jjt'ojection of a limited straight line on a jilane is the straight line joining the projections of its extremities. The Inclination of a line to a plane is the angle it makes with its projection on that plane. The projection of a limited line is evidently equal to its length multiplied by the cosine of its inclination. Thus, 0X= OP cos XOP. The p7-ojectio7i of any curve upon a plane is tlie locus of the projections of all its points. The Projecting Cylinder of a curve is tlie locus of the projectors of all its ])oints. In the case of a right line this locus is the Projecting Plane. THK POINT. 2.35 Ex^cise 43. 1. Find tlie distance between the point.s (1 , 2, 3), (2, 3, 4); (2, 3, 4), (3, 4, 5); (1, 2, 3), (3, 4, 5). 2. Prove that tlie triangle formed by joining tlie three points (1, 2, 3), (2, 3, 1), (3, 1, 2) is equilateral. 3. The lengths of the projections of a line on the three coordinate axes are 3, 4, />, respectively; fin> [59] THE PLANE. 237 which is the equation required. Equation [59] is called the normal equation of a plane. Cor. 1. When the plane is perpendicular to one of the coordinate planes, the plane xyiov example, Oi^lies in the plane xy ; hence, y = •j-ir, cos y =0, and equation [59] becomes X co^ a-\- y CO?, [i^p. (1) CoR. 2. When the plane is parallel to one of the coordi- nate planes, as the plane yz, Oi^lies in the axis of x ; hence, cos a = l, cos ^ = 0, cos y==0, and [59] becomes x=p. (2) CoR. 3. Since OF is perpendicular to the plane ABC, and OX to YOZ, the dihedral angle J-i?C-0 = angle FOX. For like reason, B-CA-0 = FOY, and C-BA-0 = FOZ. 222. The lociis of every equation of the first degree between three variables is a plane. A general form embracing every equation of the first degree between x, y, z is Ax + By+Cz = D, (1) in which D is positive. Dividing both members of (1) by V--1" + B^ + C'\ we obtain A , B , C 'J- = , ^ , (2) in which the coefficients of x, y, z are the direction cosines of some line (§ 213, Cor.). Thus (2) is in the form of [59] § 221 ; hence, the locus of (2), or (1), is a plane. CoR. 1. The length of the perpendicular from the origin upon plane (1) equals the second member of equation (2), and 238 ANALYTIC GEOMETRY. the direction cosines of tliis pfrpcndicular are the coeflficients in (2) of X, I/, z, resi^ectively. These direction cosines are evidently proportional to A, B, C. Hence, to construct equation (1), draw tlie radius vector of the point {A, B, C) ; the plane perpendicular to this line at the distance , from theoritrin is the locus of (1 ). Cou. 2. To reduce any simple equation to tlie ttur/nal form, put it in the form of A.r + Jli/-\- Cz = I), in which U is positive, and divide both niembers by '\/A^-\- J J'^-\- €''■'. CoR. 3. If a simple equation contains only two variables, its locus is perpendicular to the corresponding^ coordinate plane ; if only one variable, its locus is perpendicular to the corresponding coordinate axis. 223. To find tlie e(/'Ucftlo)i of a plane in terms of its inter- cepts on the axes. Let a, b, c denote respectively the intercepts on the axes of the plane whose equation is Ax-\-n!/+Cz = B. (1) Making ?/ = ,^•=:0, and therefore x^^a, (1) becomes Aa = IJ, or A=: D-T- a. Making x = z^^ 0, and therefore // = b, (1) becomes m=D, or B^D-^h. Making a; = //:= 0, and therefore ,t' = (", (1) becomes Cc — I>, or C—D^c. Substituting these values in (1) and dividing by Z>, we have ^+^^+5=1, [60] which is the required equation. Equation [60] is called the sijniinetrical equation of a plane. THK PLANK. 239 224. To find the (nKjJe hetween any two planes. The angle included lietween the two planes Ax-\rB'y^-Cz = D\ Ax-\-By+ Cz = D, is evidently equal to the angle included between the perpen- diculars to them from the origin. l>ut the direction cosines of these perpendiculars are respectively (§ 222, Cor. 1), ABC ylA' + B- + V' V.i- + B"" -[- C ^/A' + B' + C' A' B' C" Substituting these values in [;">"], we have AA'^BIi'^CC cos = [Gl] in which 6 erpials tlie angle included between the planes. Cor. 1. If the planes are iiaralld to each other, 6^0 and cos ^ = 1. Putting cos 6=^1 in [<)1], clearing of frac- tions, squaring, transposing, and uniting, we obtain {AB' - BA 'f + {A a — CA')' -\-(BC- CB'f = 0. Each term being a square, and therefore ])ositivt\ this equation can be satisfied only when each term equals zero, giving us AB'=BA, AC"= CA', BC'= CB\ ^_B^_C ^^ A~ B'~ C' Hence, if two planes are parallel, the coefficients of x, ij, z, in their equations, are proportional, and eonversely. Cor. 2. If the planes are 2)erpendiciilar to each other, cos ^ = ; and hence, AA'-\-BB'+CC' = 0, and converseli/. 240 ANALYTIC GEOMKTKY. 225. To find the perpendlcidar distance of a given 'point from a given pUnie. Let the equation of the given plane be X cos a.-\-y cos /8 + « cos y =-p, (1) and let (cci, y^i ^-^ be the given point. Let the plane X cos a 4" ,'/ cos y8 + 2; cos y =^:»', (2) which is evidently parallel to the given plane, pass through the given point (a^i, ?/i, z^ ; then we have a-i cos a + 2/i cos ^-\-z^ cos y = js'. (3) Hence, x^ cos a + yi cos ^^z-^ cos y — p ^=p^ — p. But ^9' — ^9 equals numerically the distance between the planes (1) and (2), and is therefore the required distance. Hence, to find the distance of any point from the plane x cos a + y cos ^ + « cos y =_p, substitute the coordinates of the point for x, y, z in the expression X cos a-\- y cos /3 + 2: cos y — p. CoR. If the equation of the plane is Ax-\- By-\- Cz=^D, and d denotes the distance of (xi, y^, z-^) from this plane, we have Ja-i + By^ -\-Cz, — D The distance as given by the formulas will evidently be pjositive or negative, according as the point and origin are on opposite sides of the plane, or on the same side. The sign maybe neglected if simply the numerical distance is required. Exercise 44. 1. To which coordinate plane is 3y — 4,? = 2 perpendicular? X — 8-? — 7 = 0? X — 2y = 2? x = mz-\-p? y^nx-\-q? What is the locus of .- = 5? y = — 7? y = 4:? z = — 2? x = 0? y = 0? z = ? THE PLANE. 241 2. Reduce to the normal form 3x — 2y + ;2 = 2; 5iK — 4 // + .- = 4. Wliat is the distance of each of these planes from the origin? What are tlie direction cosines of the perpendicu- lars to each ? Wliich of the eight octants does each truncate? 3. Find the intercepts on the axes of 3x — 2?/ + 4,^ — 12 = 0; of 6x — 4.>/ — 3z + 24: = 0; of 5a; + 7?/ + S.v + 35 = 0. Which of the eight octants does each truncate? Reduce each equation to the symmetrical form. 4. What is the equation of the plane at the distance 7 from the origin, and perpendicular to the line whose direc- tion cosines are proportional to 2, — 3, and V3 ? 5. What is the equation of the plane whose intercojjts on the axes are respectively 4, — 3, — 7? — 1, — 2, — 5? 6. Find the equation of the plane passing through the points (1, 2, 3), (0, 4, —1), and (1, —1, 0). 7. Find the angle between the planes 2x^z-,j = 3, z-\-x-{-2i/ = 5. 8. Find the angle between the planes 3z + 5x — 7>/ = — l, 3z — 2x— y = 0. 9. Find the angle that the plane Ax + Bij -\- Cz = D makes with each of the coordinate planes. 10. Find the distance from (2, —3, 0) to the plane V3;? + 2a; — 3^ = 4. 11. Show that the two points (1, — 1,3) and (3, 3, 3) are on opposite sides of, and equidistant from, the plane 5x + 2>j — lz^Si^0. 242 ANALYTIC GEOMETRY. 12. If, in Fig. 92, OM=a, OE = b, OS=r, find the equation of the phine through the points 31, 1', A*. Find the length of the perpendicular from S ujwu thi.s plane. 13. Prove that the plane A(x - X,) + B(i/ - y,) + Ciz - ,.0 = passes through the point (a-i, t/^, z-^, and is parallel to the plane Ax -f- By -\- Cz = D. 14. Find the equation of the plane passing through the point (3, 4, — 1), and parallel to the plane 2x + 4y — ,^ = 2. 15. What three equations must be satisfied in order that the plane Ax -\- By -\- Cz = D may pass through the two points (xi, yi, z-^), (x^, y^, z^), and be perpendicular to the plane A'x + B'y-{-C'z = I>'? 16. Find the equation of tlie plane passing througli the points (1, 1, 1), (2, 0, — 1), and perpendicular to the plane x-\-y — z = 3. 17. What three equations must be satisfied in order that the plane Ax -f- By + Cz = D may pass through the three points (x^, v/i, z^), (x., y^, z^), (x^, y^, Zs)? 18. Find the equation of the plane that passes through the points (1, 2, 3), (3, 2, 1), (2, 3, 1), and find the distance of this plane from the origin. 19. Find the equation of the jdane through (2, 3, — 1) parallel to the plane 3x — 4// -|- 7z = 0. 20. Find the equation of the plane that passes through the point (1,2,3), and is perpendicular . to each of the planes a; + 2* = 1, y -\- 5z = 1. CHAPTER III. THE STRAIGHT LINE. 226. To find the equations of a straight line. The coordinates of any point on the line of intersection of two planes will satisfy the equation of each of these planes. Hence, any two simultaneous equations of the first degree in a-, y, and ,-.- represent some straight line. Of the Fig. 99. indefinite number of pairs of planes that intersect in, and therefore determine, a straight line, the ecpiations of its projecting planes on the coordinate i)lanes are the simplest, and two of them are taken as the etpiations of the line. Thus, let FPl'21l and W US be the projecting planes of any 244 ANALYTIC GEOMETRY. straight line PP' on the coordinate planes xz and yz, respec- tively; and let the equations of these projecting planes be X = mz -\-p, (1) y = nz-\-q; (2) then are (1) and (2) the equations of the line PP'. Cor. 1. Let RE and SH be the projections of the line PP' on the planes xz and yz, respectively. Since the line RE lies in the plane PP'RE, equation (1) expresses the relation between the coordinates x, z of every point in RE \ hence, (1) is the equation of RE referred to the axes ZZ' and OX. For like reason, (2) is the equation of the pro- jection SH referred to the axes ZZ' and OY. Hence, rti = tan ZA E = slope RE ; ■p = OP = intercept of RE on 0X\ TO = tan ZBff^ slo^e SH; q = 0S=^ intercept of SH on Y. Eem. The locus of (1) in space is the plane PP'ER, while its locus in the plane xz is the line RE. Similarly, the locus of (2) in space is the plane PP'HS, while its plane locus is SH. The locus in space of (1) and (2), con- si9.ered as simultaneous, is the line PP'. CoR. 2. Eliminating z between (1) and (2), we obtain n ( np y=i — x-\-\ q •^ m \^ VI whose locus in space is the projecting plane of PP' on the plane xy, and whose locus in the plane xy is the projection of PP' on that plane. CoR. 3. Making ?,' = in equations (1) and (2), we obtain x=p, y=q\ hence, the line PP' pierces the plane xy in the point (/>, q, 0). This is evident also from the figure. THE STRAIGHT LINK. 245 In like manner, we find tliat the line })ierce.s the planes xz and yz respectively in the points np — mq 0, _ 1 Y ({), ^g ~ '^P^ _ 21 n ^^/ \ ^ ^"' 227. To find the symrtietrical equatioiis of a rujltt line. Let a, (3, y be the direction angles of any right line, (^15 l/u ^i) some fixed point in it, and (x, y, z) any other point of the line. Let r denote the distance between these two points. Then by § 218, Cor. 2, we have X — Xi=^r cos a, y — yi = r cos (3, z — s^ ^ ?■ cos y. (1) TTtn 2C — Xl y—y\ Z — Zl rn.,-, Whence, - = '^ — ^ = > rCol cos a COSp COSY "■ -* which are the symmetrical equations of a right line pass- ing through the point (xi, y^, z^. Cor. If [63] passes through a second point (x.^, y,, '-'2); its coordinates must satisfy [63] ; hence, we have ^2 ^1 1/2 Vl ~2 ^l ,n-. cos a cos /? cos y ^ ^ Dividing each member of [63] by the corresponding member of (2), we obtain xz~x\ 2/2-2/1 zt—zx L -• which ai"e the equations of a right line through the two points (xi, ?/i, Si) and (a-g, y^i z^). 228. If we divide the denominators in any equations of the form x — xi ^ y — yi ^ z — zi L M N ^^ by \l L^-\- ]\P-{- N'^, the denominators will then be the direction cosines of some line (§ 213, Cur.), and the equa- tions will be in the form of [Qo^- 246 ANALYTIC GEOMETKY. Hence, to rcdvice equations in the form of (1) to the sym- metrical form, divide each denominator hij the square root of the sutn of the squares of the denominators. Cor. The locus of equations (1) is the line through (xi,7ji,z{) parallel to the radius vector of the point (L, M, JV). 229. To find the diKjle between the lines L ^ M ^ N ' and — ^ — ^- — -^■ L' M' N' By § 228 the direction cosines of these lines are respectively L M N L' M' N' ^L'-'-^-lI'^+N"" \/L"'-\-M"'-\-N'^ \/Z'2+J/'2+^'2 Substituting these values in [57], we obtain „ LL'+ MM'+ NN< ^^^^ cos e = , --' , [651 Cor. 1. If the lines are parallel, —=-—=—, and con- versely. Cor. 2. If the lines are perpendicular, LL' -f MM< + NN' = 0, and conversely. 230. To find the inclination of the line x — Xi ^ y — ih ^ z — Zy_ L M N ^ > to the plane Ax -f- Bij -j- Cz =^ D. (2) THE STRAIGHT LINE, 247 The equation of the perpeiidiciilar from (x^, iji, z-^) to the plane IS a: — a^i _ y — ?/i _ z — z^ A ~ B ~ C ' ^'^^ Now, the inclination of line (1) to plane (2) is evidently the complement of the angle between the lines (1) and (3). Denote this inclination by v; then sin t> = cos ^, being the angle between the lines (1) and (3). Hence, AL + BM+ C]>f^ ^,.^^ CoR. 1. If the line is parallel to the plane, sin v ^ 0, and, therefore, AL-\- BM-\- C'i\^=0, and conversely. Cob. 2. If the line is perpendicular to the plane, sin v =: 1, and, therefore, ^ = f=f, and conversely. Cor. 3. If line (1) lies in plane (2), then AL-{-BM-\-CN=Q, and Ax^ -\- By^ + Cz^ = D, and conversely. Exercise 45. 1. Determine the position, direction cosines, and direc- tion angles of the intersection of the planes x-{- y — z + 1 = 0, and4a; + y/ — 2,~ + 2 = 0. Eliminating successively y and z between the equations, X 1/ z — 1 we obtain ox — .~+l =0 and 2x — '/ = 0: or 7 = !. = — r; — 1 -J o From the last form we know that the line passes through the point (0, 0, 1), and is parallel to the radius vector of the point (1, 2, 3). The direction cosines are found by divid- ing the denominators 1, 2, 3, by Vl4; and the direction ancrles are found from their cosines. 248 ANALYTIC GEOMETRY. 2. Determine the position and direction cosines of the intersection of x — 2y = 5 and 3x + ?/ — 7z = 0. Here — - — = '- = " ^ , whence the line passes through the point (5, 0, -y), and is parallel to the radius vector of (2, 1, 1). 3. Determine the position of the line 5a; — 4?/ = 1, '6y — bz = 2. 4. Wliat is the position of the line a; = 3, 2/^=4? Of the line y = 4:, z = — 5? Of the line x = — 2, z = 3? 5. Find the equations of the right line passing through the points (1, 2, 3), (3, 4, 1). 6. Find the points in which the line of Example 5 pierces the coordinate planes. 7. Two of the projecting planes of a line are x-\-?/ = 'i and 2x — 5z = — 2 ; find the third. 8. A line passes through (2, 1, —1) and (—3, —1, 1); find the equations of its projections on the coordinate planes. 9. Show that the lines t=9=t and 3-=^^= :|^ are at right angles. 10. Show that the line4a; = 3y = — z is perpendicular to the line 3x= — ?/ = — 4«. 11. Find the angle between the lines X ?/ z X ?/ z 12. Find the angle between the right lines y:=OX-\-3, Z:=3x-\-5, and y^=2x, z = x-\-l. 13. Find the angle between the lines y = 2.r + 2, z = 2x + l, and y^4:X-\-l, z= x-\-5. THE STRAIGHT LINE. 249 14. Show that the lines 3x + 2i/-{-z — 5 = 0,x + y—2z — 'S = 0, and 8x — 4i/ — 4:z = 0,7x-\- 10// — 8z = are at right angles. 15. Find the equations of the line through ( — 2, 3, — 1) parallel to the line y=^ — 2x + 1, z = 3x — 4:. 16. Find the equations of the line through (3, — 7, — 5), its direction cosines being proportional to — 3, o, — 6. 17. Find the equations of the line through (2, — 4, — 6) perpendicular to the plane 3x — 6i/-\-2z = 4:. 18. Find the inclination of the line x — 4 ?/ + 2 z — 5 3 ~ — 2 ~ — 4 to the plane 2x — Ay-\-3z^=l. 19. Reduce the equations, x = mz -\-p, y= nz -\- q, to the symmetrical form, and thus find the direction cosines in terms of m and ?i. 20. Show that the formula for the angle included between the lines x = mz +p, y = nz -\- q, and 2"= vi'x -\-p', y = 7i'z -\- q' mm' -\- nn' + 1 IS cos 6 = — = — ■ • Vm'+?i2_^l Vm'^ + ?i'2 + l 21. Prove that the lines in Example 20 are perpendicular if vim' -\- nn' -\- 1 = 0, and conversely. Prove that they are parallel if vi = m', and n = n', and conversely. 22. Prove that two lines are parallel if their projections are parallel, and conversely. 250 ANALYTIC GEOMETRY. SUri'LEMENTAllY PROrOSlTlONS. 231. The Traces of a plane are its lines of intersection with the coordinate planes. Thus AB, JW, CA (Fig. 98) are the traces of the plane ABC. 232. To find the equations of the traces of the ylane Ax-\-By+Cz = D. (1) For every point in the plane xy, z^O ; hence, putting « = in (1), we obtain Ax^Bu = D, (2) whicli is tlie equation of the trace AB (Fig. 98) on the plane xy. For like reason, By-^Cz=D - (3) and Ax + Cz = D (4) are the equations of the traces BC and CA on the planes yz and xz, respectively. CoR. The perpendicular from the origin to (1) is x // z A~B^C' and its projections on the coordinate planes are Bx = Ay, Cy=Bz, Cx = Az. (5) By comparing coefficients, we see that lines (5) are per- pendicular to (2), (3), and (4), respectively. Hence, If a line in sjxice is 2)(ir2Jendicular to a 2)lane, its projections are jierpendicular to the traces of the plane. 233. To find the conditioii that the right lines X = niz +;y, y — nz + q, and X = m'z + /'', y = n'z -\- q', may intersect, and to find their points of intersection. THE STRAIGHT LINE. 251 Equating the two values of x and y, we have jj' — p q' — q z = -) z=- ■• 7ft — nv n — 71 If the two lines intersect, these two values of z must be p' — /' 7' — Q • • equal ; hence, ^ = -. is the equation of condition m — 111 n — 7i' that the two given lines in si)ace intersect. When this condition is fultilled, the values of x and /j may be found by substituting either value of z in the equations of either line. 234. To pass a plane through the point (.Tj, ?/o, z^) and the right line L ~ 31 ~ N ' ^"^ If the plane Ax + By + Cz = B (1) passes through the point (.Tg, ?/„, .-o), we have Ax^^By.,^Cz. = D; (2) and if line (a) lies in plane (1), we have Ax^^By,-^Cz,=D, (3) and AL^BM\CN=^. (4) The equation of the required plane is found by eliminat- ing A, B, C, D from (1), (2), (3), and (4). To simplify the process of elimination, (1) might be written in the form A'x-\- B'y-\- C'z = 1, but the solution would be less general, as it would not embrace the case when Z> ^ 0. 235. From the forms x = mz -{-j), y = nz-\-q, show that the equations of a line passing through (xi, y^, z^ are \ .r — .r, == m (z — z^), CHAPTER IV. SURFACES OF REVOLUTION. 236. It has been shown that a single equation of the first degree between three variables represents a plane surface, and that two such equations in general represent a right line. It is evident, moreover, that in general three such equations determine a point common to their loci. Thus, if in Fig. 92, OM=a., OE = b, and OS=c, then the equations x = a, y=^h, z^c determine the point P, and are called its equations. We proceed to show that. In general, any single equation of the form f (x. y, s ) = represents a surface of some kind; two such equations represent a curve, and three determine one or more pohits. (i) Let two of the variables be absent ; for example, let the equations be / (x) = 0. Now / (a;) = may be written in the form {x — ai) (a- — a^) {x — a^) (x — «.„) =0, (1) in which a^, Oo, a^, «„ are the n roots of/(a;)=;0. The locus of (1) is evidently the n parallel planes a? = «!, a; = a^, , x=-a^. Similarly, the equations/ (?/) =0,/(s)=:0 repre- sent planes perpendicular to the axes of y and z, respectively, (ii) Let one of the variables be absent ; for example, let the equation be / {x, y) = 0. The locus of /(re, y')=0 in the plane xy is some plane curve. Through P, any point in this curve, conceive a line parallel to the axis of z ; then the coordinates x, y of all points in this line will equal those of P, and hence satisfy the equation / (x, y) = 0. Hence, the SURFACES OF REVOLUTION. 253 locus in space of / (x, y) = is the surface generated by a right line which is always parallel to the axis of z, and which moves along the plane locus oif(x, y) =0. That is, the locus in space of f (x, y) = Q is a cylindrical surface whose elements are parallel to the axis of z, and whose directrix is the pjlane locus off (x, y) = 0. Similarly, the equations / (x, z) = and / (y, z) = rep- resent cylindrical surfaces whose elements are parallel to the axes of y and x, respectively. (iii) Let the equation be / (x, y, z) = 0. If in this equa- tion we putx = a and y=^b, the roots of the resulting equation in z will give the points in the locus that lie on the line through (a, b, 0) parallel to the axis of z. But as the number of these roots is finite, the number of points of the locus on this line is finite. Hence, the locus which embraces all such points for different values of a and b must be a surface and not a solid. (iv) Two equations considered as simultaneous are satis- fied by the coordinates of all the points of intersection of their loci ; that is, they represent the curve of intersection of two surfaces. (v) Three independent simultaneous eqiiations are satis- tied only by the coordinates of the points in which the curve represented by two of them cuts the surface represented by the third ; hence they determine these points. Cor. 1. From (ii) it follows that x"^ -\- if = i^ is the equa- tion of a cylinder whose axis is the axis of z, and whose radius is r. Also, f = ^p:c,~^- = l, and -.-^=1 are equations of cylindrical surfaces whose elements are parallel to the axis of ,~, and whose directrices are respec- tively the parabola, the ellipse, and the hyperbola. 254 ANALYTIC GEOMETRY. CoK. 2. I (■ F (x, i/)=0 is the equation obtained by eliminating z between the two equations / (a-, //, s) = and /i ('^'j l/> ^) = ^j then the locus in space of F (.r, y) = is the projecting cylinder on the plane xi/ of the curve represented by the two equations. The j/lane locus of F(_x, y):=0 is the projection of this curve on tlie plane xi/. If the curve is parallel to the plane of jn-ojection, the curve and its projection are equal. The equation obtained by eliminating x ov y between the two equations has evidently a like interpretation. 237. The Traces of a Surface are its intersections with the coordinate planes. If/(.^;, 11, 0)=0 denotes the equation obtained by mak- ing z = in f(x, y, z) = 0, then the plane locus of / (x, y, 0) = is evidently tlie trace of the surface f (x, y, z) =0 on the plane xy. Surfaces of Revolution. 238. A Surface of Revolution is a surface that may be gen- erated by a curve revolving al)out a fixed straight line as an axis. The revolving curve is called the Generatrix ; and the fixed right line, the Axis of Revolution, or simply the Axis, A section of the surface made by a plane passing through the axis is called a Meridian Section. From these definitions, it follows that (i) Every section made by a plane perpendicular to the axis is a circle, whose centre is in the axis. (ii) Any meridian section is equal to the generatrix. 239. To find the general equation of a surface of revolution. Let the axis of z be the axis of revolution, and let P be any point in the meridian section made by the plane xz. Let PHR be a section through P jterpendicular to the axis of z, and denote the radius CH, or CP, of this circular section by r. SURFACES OF UEVOLUTION. 25/ Now for all points in this circular section, we have »•- + y2=r /-^^ and z = MP. The value of r^, in terms of •:, is ob- tained by substituting r for cc in the equation of the me- ridian section made by the plane zx. Denoting this value of r* \iy f(z), and equating tlie two values of r^, we have Qc'^+y^ = f{z), [G7] Fig. 100. which expresses the relation between the coordinates cr, y, z of all points in the section PHB. But as P is any point in the meridian section NP, [67] is the general equation of a surface of revolution whose axis is the axis of z. 240. Paraboloid of Revolution. A Poraholnid of Perolu- tion is a surface that may be generated by a parabola revolving about its axis. In this case the equation of the meridian section in the plane zx is a;^ = 4/>;s ; hence, ^^ = 4.pz=f(z). 256 ANALYTIC GEOMETRY. Substituting in [07], we obtain X"- + !/- = 4rpz, [68] which is the equation of the j^ctt'cil^oloid of revolution. If in [08] we put x = m, we obtain which is the equation of the projection, on the plane yz, of the section of the paraboloid made by a plane parallel to the plane yz, and at a distance from it equal to m. Now the plane locus of (1), for all values of in, is a parabola ; hence, every plane section of the paraboloid parallel to the plane yz is a parabola. If in [68] we put y^n,we obtain x^ = 4pz — n^. (2) From (2) we learn that all plane sections parallel to the plane xz are also parabolas. From definition, we know that all plane sections parallel to the plane xy are circles. 241. Ellipsoid of Revolution. Xn Ellipsoid of Revolu- tion, or Spheroid, is a surface tliat may be generated by an ellipse revolving about one of its axes. It is called Oblate when the revolution is about the minor axis; and Prolate when about the major axis. (i) "When the revolution is about the minor axis, the equation of the meridian section in tlie plane xz is 7^2 + 7i=l5 lience, r-'^aH l--^ =/(^)- Substituting in [07], and reducing, we obtain S + S + g=l. [69] which is the equation of the ohlate spheroid. Cor. 1. If a = h, [69] becomes x' + lf + ^^a:^, (1) which is the equation of a sphere whose radius is a. SURFACES OF REVOLUTION. 257 Cor. 2. If in [69] we put x^=in, we obtain ^4---!- — . (2-) Since (2) represents an ellipse, a point, or no locus in the plane yz, according as w? a^, the surface lies between the two tangent planes a; = « and a; = — a, and all plane sections parallel to the plane yz are ellipses. . If in [69] we put y = n, we obtain Since (3) represents an ellipse, a point, or no locus in the plane «cc, according as v? <, =, or !> a^, the surface lies between the two tangent planes y^a and y = — a, and all plane sections parallel to the plane zx are ellipses. If in [69] we put z = q, we obtain x-^-\-f = a'^(l-^j\ (4) Equation (4) represents a circle, a point, or no locus in the plane xy, according as ff <, =, or > h"^. Hence, the sur- face lies between the tangent planes z = b and z^ — b, and all plane sections parallel to the plane xy are circles. (ii) When the revolution is about the major axis, the equation of the meridian section in the plane xz is ^2+1=1; hence, ,-=^^(^l-^,^=/<.). Substituting in [67], we obtain f5+|.+ f,= l. [70] which is the equation of the ^jrolate spheroid. If in [69] we interchange a and b, we obtain [70]. Hence, we interchange a and b in the discussion of [69]. 258 ANALYTIC GEOMETRY. 242. Hyperboloid of Revolution. An Htjperholokl of Reiwlutkm is a suriace that ina,y be generated by an hyper- bola revolving about one of its axes. It consists of one or two nappes, or sheets, according as the hyperbola revolves about its conjugate or transverse axis. (i) If in [69] we substitute — V^ for h"^, we obtain ^+'4-f^=l, [71] a^ a- b^ L J which is the equation of the hijperholoid of one nappe. If in [71] we put x^=vi, we have 2 9 9 whose plane locus is an hyperbola for all values of m. Hence, all plane sections parallel to the plane yz are hyper- bolas. The transverse axis of any one of these hyperbolas is evidently parallel to the axis of y or z, according as rn^ ■< or > a^. If m'^ = a- (1) becomes a Hence, the sections of [71] made by the planes a; = ±a are each two intersecting riglit lines. If in [71] we put .y = ?', we have X z ir Hence, all plane sections of [71] parallel to the plane xz are hyperbolas, whose transverse axes are parallel to the axis of X or z, according as «-<[ or > a^- and the sections made by the planes i/=±a are eacli two intersecting right lines. If in [71] we put z = q, we obtain x^ y^_ rf a^ a^ If SURFACES OF REVOLUTION. 259 whose plaue locus is a circle for all values of «/. This cir- cle is smallest when y ^ 0. This smallest circle, which is the trace of the liyperboloid on the plane a"//, is called the Circle of the Gorge. (ii) If in [70] we substitute —V^ for V^, we obtain f?+f?-4=-l, [T2J h- b' a^ '- -■ which is the equation of the hyperholoid of two nappes. The discussion of [72] for parallel plane sections is left as an exercise for the student. 243. The Centre of a surface is a point that bisects all chords passing through it. Central Surfaces are such as have a centre. The ellipsoids and hyperboloids of revolution are central surfaces. For, from their equations, it is evident that, if {x\ y', z') is a point in any one of these surfaces, ( — x', — y', — z') is also a point in the same surface. But the chord joining these two points is bisected by the origin, which is, therefore, the centre of the surface. 244. Cone of Revolution. A Cone of Revolution is a surface that may be generated by a right line revolving about an axis which it intersects. Here the equation of the meridian section in plane xz is z = mx -\- c ; therefore, ?-^ = ( j =f(z). Whence, in-{x- + */-) - {z- <■)- [73] is the equation of the cone of rerolufiun. 260 ANALYTIC GEOMETRY. In this equation c is the distance of the vertex from the origin and m = tan XDB. Fig. 101. If c = 0, [73] becomes im?{oi? -\- ]f^ = z^. (1) From (1) it is evident that the cone is a central surface. If in (1) we put y = n, we obtain z^ 3? ri?if)i} IT? ' whose plane locus is an hyperbola for all values of n. Hence, all plane sections of the cone parallel to the plane zx are hyperbolas whose transverse axes are parallel to the axis of the cone. In lilie manner, we find that all plane sections parallel to the plane yz are hyperbolas. If ^ = 0, s = ± mx, whose locus is two intersecting right lines. Hence, any plane section of a cone parallel to its axis is an hyperbola, and any section containing the axis is two intersecting right lines. A Conic Section is the section of a cone made by a plane. 245. To determine the nature of any conic section that is not parallel to the axis of the cone, we find the equation of any such section referred to axes in its own plane. SURFACES OF REVOLUTION. 261 Let NFJ^ be any section of the cone VB YN passing through the axis of y ; then this section will be perpendic- ular to the plane xz. The cone, and therefore the section, is symmetrical with respect to the plane xz. Refer this section to ON and Y as the axes of x and y respectively. Let (cc, y, z) be the point P referred to the coordinate planes, and {x\ xj) be P referred to ON and OY. Let XON= cf> and 0DV^=6. Draw PJf perpendicular to ON; then it will be perpendicular to the plane xz, and we have y = y', 0B= OM cos , or a; = a:'cos ; BM= Oil/sin <^, or s^cc'sin ^. Substituting these values of x, y, z in [73], we obtain tan^ 6 (cc'2 cos^ <^ + y'^) = (x' sin — cf. Omitting accents and performing indicated operations, we have ?/^tan-^+a'-(cos- tan- ^ for sin- ^, we obtain 2/2 tan2 9 -t-ic^ cos2 <}> (tan2e - tau2 (j>)+ 2cac sin <}> - c- = O, [74] 262 ANALYTIC GEOMETRY. which is the equation of the conic ^ViV/ referred to ON and Y as axes. By giving to c all values between and oo , and to <^ all values between 0° and 90°, equation [74] is made to rep- resent any section of a cone except those parallel to its axis, which have already been considered. Discussion of equation [74]. Here 2=4 cos^ <^ tan^ 9 (tan- — tan- cf>), A= 4c^ [cos- ) -f-tan^ 6 sin- ^]. (i) Fh'.sf su2)j)ose c not equal to zero. Let (f> = tan^ $, 2 = 0, and A is not zero ; hence the section is a parabola. Let 0; then is tan" <^ > tan- ^, 2 is negative, and A is not zero ; hence, tlie section is an hyperbola. Hence, when the cutting ])lane does not pass through tlie vertex of the cone, the section is an, ellipse, a i>arahola, or an hyperholn, according as the angle ibhicli the cutting plane makes with the base of the cone is less than, equal to, or greater than tlutt made hij an element. (ii) If c = 0, A =: ; henee, when the cutting plane passes through the vertex, the elliptical section reduces to a point, the parabolic to a straight line, and the hyperbolic to two intersecting right lines. If f)-l^ = m 7. Find the traces of tlie surface 2x--\-bif — 7.^' = 9; of the surface x^ -\- 3//'^ = 8z. 8. Find the equation of the surface of revolution Avhose axis is the axis of z, and one of whose traces is ,-;= ± 3a' + 5 ; find its trace on the plane xi/. 9. Find the equation of a cone of revolution one of whose traces is 3c^-\-y^ = 9, and whose vertex is (0, 0, 5). 10. Find the equation of the paraboloid of revolution one of whose traces is 2x^^^oz -{- 5. 11. Find the equation of the paraboloid of revolution one of whose traces is i/'-^= 8x. 12. Find the equation of the cone of revolution whose axis is the axis of z, and one of whose traces is 2^/ ^±.~ + 6 ; find its vertex. 264 ANALYTIC GEOMETRY. 13. Find the equation of the surface of revolution whose axis is the axis of z, and one of whose traces is ^3^-\-^z'^ = 36. 14. Find the equation of the surface of revolution whose axis is the axis of z, and one of whose traces is 16y^-\-9z'^ = 144. 15. Find the equation of the surface of revolution whose axis is the axis of z, and one of whose traces is dz"^ — 4?/^ = — 36. 16. Find the equation of the surface of revolution whose axis is the axis of z, and one of whose traces is z^x = l; also when one trace is z^ ^= 2if. 17. Each element of a cone makes an angle of 45° with its axis ; find the semi-axes of the section made by a plane cutting the axis 5 below the vertex and at an angle of 60°. SUPPLEMENTARY PROPOSITIONS. 246. To find the (/eneral equation of the sphere. Let r denote the radius of any sphere, (a, b, c) its centre, and (x, y, z) any point on its surface. Then, since r is the distance between the points {a, b, c) and (x, y, z), we have {X - a)2 +{y- 6)2 + (s - c)^ = »-2, [75] or x'^+f-\-z' — 2ax — 2by — 2cz=^r''—a'' — b^ — c\ (1) which is the general rectangular equation of the sphere. If the origin is at the centre, then a = 5 = c = 0, and [75] becomes ^2 _^ ^2 _^ .2 ^ ^-2^ (2) From (1) it follows that any equation of the form x''-\-y^ + z'^Gx + Hy + Iz = K (3) is the equation of a sphere. Any equation of the form of (3) can readily be reduced to the form of [75], from which the centre and radius of its locus become known. THE SPHERE. 265 Since (3) or [75] contains four arbitrary constants, a sphere may in general be made to pass through any four given points. 247. The intersection of two spheres is a circle. Let the equations of the two spheres be x'+f + z'-^Gx + Hy +Iz=K, (1) and a;2 + y2 _|_ ^2 ^ q^^ _|_ fj,y _|_ jr^^ j^, ^2) Subtracting (2) from (1), we obtain {G —G')x^ (B — H')i/^ (/- T) z = K—K'. (3) Hence, the intersection of the spheres (1) and (2) lies in the plane (3), and is the same as the intersection of (1) and (3). But the plane section of a sphere is a circle. Hence, the intersection of the two spheres is a circle. 248. To find the equation of the tanr/ent plane to a sphere at a given point. Let the given point be (cci, y^, z^; then the equation of the radius to this point, that is, of the line passing through (a, h, c) and (x^, y^, z^)] is x^:^Xi _ y — //i _z — Zi a — Xi h — y^ G — z-^ Now the tangent plane is perpendicular to (1) at the point (ccj, yi, z-^\ but the equation of the plane through (xi, yx, Zi) perpendicular to (1) is (a-x,) (x - X,) + (b- y,) {y - y,) + (c-z,) (z - z,) =0, (2) which is, therefore, t/ie equation of the tangent jdane. If the origin is at the centre, a= ?; = c =0, and (2) becomes 266 ANALYTIC GEOMETRY. Transformation of Coordinates. 249. To cJiange the orirjin of coordinates without changing the direction of the axes. Let (m, n, q) be the new origin referred to the old axes. Let X, y, z be the okl, and .r', y\ z' the new coordinates of any point F; then, evidently, we have x^m-^ x', 1/ = n-\- y', z = q-\-z'. Hence, to find the equation of a locus referred to new parallel axes whose origin is (vi, n, q), substitute m-\-x, n + //, cmd q + z, for x, y, and z, resjjectively. 250. To change the direction of the axes ivithout changing the origin. Let ai, /3i, yi ; ag, /iJg, 72 5 "s, /^s, ys be, respectively, tlie direc- tion angles of the new axes 0X\ Y\ OZ' referred to the Fig. 103. old axes OX, Y, OZ. Let x, y, z be the old, and x', y\ z' the new coordinates of any point P. Draw PN perpendicular to the plane X'OY', and JVJf perpendicular to OX; then OM^^x', MN= y', and NP^z'. Now the projection of QUADRICS. 2G7 OP on OX { = ^) is equal to the sum of the projections of OM, MN, and NP on the same line ; hence, .'C=:x'c0S ai + ^'C0Sa2 + ;s'C0S ttj. (1) In like manner, we obtain ?/ = CC' COS ^1 + i/' cos ySo + s;' COS ySs, (2) and z=^x' cos yi + y' cos yo + z' cos yj. (3) Hence, to change the direction of the axes witliout changing the origin, stibstitiite for x, y, and z their values as given in equations (1), (2), and (3), Since the values of x, y, z are each of the first degree in x', y\ z\ any transformation of coordinates cannot change the degree of an equation. (§ 91.) 251. Quadrics. The locus of an equation of the second degree that contains three variables is called a Quadric. Thus the general equation of a quadric is Ax'^By' + C^ + Dxy + Exz + Fyz -[-Gx+Hy -\-Iz + K=0. (1) Putting z = q in (1), we obtain Ax^ + D^r->, + /,y ^(^Eq + G)x + (Fq +H)y + (tVr + /'/ + A')-0. (2) Since the locus of (2) in the plane xy is a conic, and since the coefficients A, D, B are tlie same for all values of q, all plane sections of the quadric (1), parallel to the plane a-y, are similar conies. Now the axis of coordinates may be so changed that the new plane xy will be one of any system of parallel planes cutting the quadric. But, as this transforma- tion does not change the degree of the equation, it follows that All parallel plane sections of any quadric are similar conies. 268 ANALYTIC GEOMETRY. 252. By transformations of coordinates the general equa- tion (1) of § 251 may be reduced to one of the two following simple forms : * Px'-\-Qu^-\-Rz'=S. (1) Px^+Qy'=Uz. (2) Now whatever be the values or signs of P, Q, R, S, equa- tion (1) evidently represents central quadrics. But the loci of (2) have no centre ; for if tliey had, and the origin were changed to that centre, the first power of z would disappear from the equation. But no expression of the form q -}- z, when substituted for z, can cause z to disappear. Hence, (2) represents non-central quadrics. 253. Central Quadrics. If no one of the coefficients P, Q, R, S is zero in (1) of § 252, we have S^P^ S^Q^ S-^ R which can be written in the form * By changing the direction of the axes the general equation can in all cases be reduced to the form Px2 + Qy^ + Rz^ + G'x -f H'y + I'z- K-O. (1) This transformation is analogous to that in § 189. (i) If no one of the three coefficients P, Q, R is zero, by a change of origin, as in § 188, we obtain Px2 -f Q?/ + Rz"- = S. (2) (ii) If any one of these coefficients is zero, for example R, by a change of origin, we obtain Px2 -I- Qy^ = Uz. (3) If two of these coefficients are zero, (1) can be reduced to a form embraced in (3) by first changing the origin and then the direction of the axes. QUADRICS. 269 5 + f!-|=l. (B) according as S-i-F, S~Q, S-^B are all positive, two positive and one negative, or one positive and two negative. [If all three are negative there is no real locus.] If S is zero, we have Px'-\-Q!/' + Bz' = 0. (D) If P, Q, or B is zero in (1), its locus is a cylindrical surface by (ii) of § 236. 254. A discussion of (A) discovers the following proper- ties of its locus : (i) Its traces on each of the coordinate planes are ellipses, (ii) All plane sections parallel to either coordinate plane are similar ellipses. (iii) The quadric is included between the tangent planes X = dz a, 1/ ^^ ± b, z ^^ ± c. The quadric (A) is called an Ellipsoid. If a =b, the ellip- soid is the oblate or prolate spheroid, according as a ^or — — \. 3. tan <^ = Jg. 4. tan <^ n2 + 2 5. 90°. 6. 135°. 7. 90°. 8. 0°. 9. 30°. 11. j2/ = 5x— 10, 23_ ( ?/ — 3 = m'(x — 2), _ Ix + 5j/=28. ' ( and7n'= — (8±5V3). 12 j2/=5x+ll, [-2/ — 3= ?n'(x— 1),_ ' ( X + 52/ - 3 = 0. 14. ^ ^ , 8 ± 5 ^/3 and m— 22. 2x + .3// — 31 = 0. ^ 11 23. 62x + 312/ -1115 = 0. r x - .3?/ + 26 = 0, 24. 2/-6X-27. 3°- ^5^ + 3^+ 8 = 0, , l2x + 32/- 9 = 0. 25. 2/ = mx ± dVl + ?/i2. 31. X- 6 = 0. 26. £x = ^(2/-6). r 2x- 92/4- 12 = 0, 27. ox — 62/ = a2 - 62. 32. J lOx — 42/ + 63 = 0, 28. (a±6)2/+ (6q:a)(x-a)=0. Il8x - 402/ + 111 = 0. rx— ?/— 6 = 0^ meeting in the point. 33. -J 2x - ?/ - 2 = I (- 4, - 10). ISx — 32/ - 10 = J Distance = V85. — -4±J5tan<^, 3^- '^-^^^= i^±^tan0 (^-^^)- ANALYTIC GEOMETRY. Exercise 15. Page 52. 3. 4. 4. §V5. 5. 0. The learner should construct the given Hnes, and observe how the sign of the required distance gives tlie direc- tion of tlie point from the line. 8. —0, —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. 1. iVio. 2. fVs. 7. -¥,-¥, 16 12 '5"» "5'> -!-, -h 0, +h - hS - hS 9 6 5> 5> -1, 0, + 1- 2 ^/a^ + 62 Exercise 16. Page 54. 1. H. 4. 40. 8. 35. 11. 26. 2. 12. 5. ah. 9. lOf 12. 96. 3. 29. 7. 26. 10. i(xi?/2 - a;22/i)- 13. 41. 14. i(a-c)(&-l). 21. 9a2. 27. ^ 15. |(a-6)(a + 6-2c). g^. f ^g. ,,, 16. i(a2-62). 17. 60°, 00°, 00°; 9V3. 23. 24. ^g _^ 18. 10. 24. 36. 19. j^. 25. 10. 30. 56. 20. H. 26. iab. 31. 10^ 2AB ANSWERS. Exercise 17. Page 56. 3. 2, 00, 90°, 2, 0°. 4. 0, 0, 45°, 0, 135°. 5. iV3-2, 2V3-I, G0°, — 1^^ 150°. 6. 2, f V3, 150°, 1, 00°. 26. 4y = x + 8. 7. 2, -J^/3, 30°, 1, 300°. 27. 4?/ = Ox -24. r- OR i 9x — 20?/ + 0(3 = 0, 8. f Vs. -2, 00°, 1, 330°. 28. / . .3, ^ * ' ' ' ' / 5x — 4y -f 32 = 0. 9. ill'J;+ 2/ = 0, 29. 88x- 121;/ + 371 = 0. I X - 5// + 20 = 0. g^^ 5^ _ y _ 10 = 0, 10. /i V827 ' ^ + '">?/ — -8 = 0. 3x + 4// - 57 = 0, 31. j -•« + 2/ - *^ = <^' 3x + 4(/+ = 0, ' X- 2// -17 = 0. 11. ■{ 12x — 5// — 30 = 0, 32 \ 4x + ?/ - 20 = 0, 12x - 5y + 24 = 0. ' I X — 4?/ - 5 = 0. Area = 63. 12. 43. 13. x=3. jx-y+l=0, U+?/-7 = 0. 15. 5x + G?/ — 39 = 0. 16. 14x - 3?/ - 30 = 0. 17. 4x — 5y + 8 = 0. 18. X + // — 7 = 0. ■ X — X3 X2 — Xi 20. i 2^ = •^' 1-'^ = 5x - 1, 19?/= 5x + 7. 21. 92x +09//+ 102 = 0. 22. x + 4?/ = 34. 23. 3x + 4?/ — 5(1 = 0. 24. 3x + 4?/ = 24. 25. 2/ -2/1= -^(x-Xi). Xi 33. 2x = y, 2?/ = X. 34. 4x + 5//+ 11 ±3 V4r = 0. 35. 2/ = (7 q: 5 -V^) (X + 2). 36. 2x — 5?/ _ I 4x + 3// — 12 V29 5 37. ^ 7X-32/+ 15 = 0, 1 3x + 7// — 03 = 0. 38. \ 8x + 7// -19 = 0, I 16x + 3y +17=0. 39. 135°. 40. 00°. 41. 31 V20 143 42. . hh + al(^ — ah \/a^ + 62 43. c- V/i^ + t-i 44. ±^Vl + m--'. 45. C-. 10 ANALYTIC GEOMETRY. xy represents the two axes. a = 5. x + a = 0, X — b = 0. x + a = 0,y+ b = 0.' The axes and x = y. 2x — y = 0, 7x + y = 0. 62. If h denotes the altitude of the triangle, and the base is taken as the axis of x, the locus is the straight line y = h. 63. The equation of the locus is (x — Xi)2 + {y- yi)- - {X— iCo)^ + {y — y^f. Tills is the equation of the straight line bisecting the line joining (Xi, 2/i) and (X2, y^)-, and _L to it. 64. The two parallel lines represented by 46. fe2 6' 2a2 + 54. 47. • bab + 262 6 57. 58. 48. 49. 17i. 59. 50. 59. 60. 51. (10,5 \)- 61. Ax+By-^C±d ^A^ + i'-' = 0. __ , _, __ Ax + By+C , A'x+ B'y + C , 65. x + y = k. 66. , — I ^- = k. Vyl2 + B^ sJA'-^ + B'2 67. Let h denote the base, k"^ the constant difference of the squares of the other two sides. Taking the base as the axis of x, and the middle point of the base as origin, the equation of the locus is 26x = ± fc^. Exercise 18. Page 64. 1. lx + y- 0. 4 J X - ?/ + 8 = 0, 6. 64x - 23?/ = 59. 2. X + 2?/ - 13 = 0. \ x + ?/_6 = 0. 7. 44x + ?/ = 0. 3. bx + Qy — 37 = 0. 5. 2/ = X + 3. 8. 5x + 2/ - 16 = 0. 9. {A C - A'C)x + (BC - B'C)ii - 0. 10. {BA' - AB')y + CA' - AC = 0. Ax+By+C _ A'x + B 'y + C ■ Axi + Byi+ C~ A'xi +B'yi + C' 12. 472x - 292,' + 174 = 0. 13. 2/ = x Vs + 3 - V3. (4x+32/— 25 = 0, ^_^ — "^^~ ^ ' ( 3x — 42/ + 25 = 0. 'a & ma + 6 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 coordinates in the equation of the third line. ANSWERS. 11 -^ - ^« ,TTi '>^" ~ '>n b" — b 19. m = 1. 20. When — = — • in — m b — b 21. If we choose as axes one side of the triangle and the corre- sponding altitude, we may represent the three vertices by (a, 0), (- c, 0), (0, b). 22. Choosing as axes one side and the perpendicular erected at its middle point, the vertices may be represented by (a, 0), (— a, 0), (6, 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" — p") — 0, (x cos a" + y sin a" — p") — (x cos a + y sin a.— p) = 0. Now, by adding any two of these equations, we obtain the third ; therefore, the three bisectors must pass through one point. • 2g_ ( 2V2, VlO, 2V1O. 29 (x-2/ + 2 = Origin within the A- (x + y — 14 = 0. 26. iJ- VlO, 1 & V.34, If V]3. (X — 1=0 27 (^ + 2/ + 10 = 0, 30. ^_i^o: I 7x - 7?/ + 24 = 0. 28 i ''* ~ 9?/ + 34 = 0, g^ ±(y— ?»x— b) _^ ±(y—jn'x-b') ' \ 9x + ly - 12 = 0. ■ Vl + ?n2 Vl + m'2 Exercise 19. Page 68. 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 C'^ — iAB is positive, negative, or zero. The two lines are _L to each other when A + B = 0. 5. X + y + 1 = 0, and x — 3?/ + 1 = 0. 6. x-2y ±(y - 3) V^ = 0. 7. X — y - 3 = 0, and x — 3y + 3 = 0. 8. 45°. 9. K = 2. 10. K = - 10, or — ?./ . 11. K = 28. 12. A' = V-. 12 ANALYTIC GEOMETRY. Exercise 20. Page 70. 1. Take the point O as origin, and the a.xiss of ij parallel to the given lines. If the ecjiiations 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', jn", the eiiuation of the locus is {b — a) 1/ = vi'b (x — a) — m"a (x — b). 2. Jf a and b are the sides of the right triangle, the equation of the locus is f( 3. Let OA = a, OB = b. Then the equation of the locus is X + y — a + b. 4. Take as axes the base and the altitude of the triangle. Let a and b denote the segments of the base, h the altitude. Then the equa- tion of the locus is 2x 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. Exercise 21. Page 73. 1. X- + y-= — 2rx. 13. (4, 0), 4. 2. x-2 + 2/2 = 2ry. 14. (- 4, 0), 4. 3. x-^ + y'^= - '-iry. 15. (0, 4), 4. 4. (X - 5)- -¥ {y + ;])- = 100. 16. (0, - 4), 4. 5. X--' + (// + 2)-^ = 121. 17. (0, I), 1. 6. (X - r.)-; + y-' = 2.5. 18. (0, 0), 3A;. 7. (X + .5)-' + y- = 25. 19. (0, 0), 2k. 8. (x - 2)2 -\-{y - Zy^ = 25. 20. (0, 0), VoM^. 9. x2 + 2/2 _ .2hx - 2ky = 0. ^l. (-, o\ - VH. 11. (1,2),V5. ^'^ ^ ^. ^ ' 'h k\ VA2 + k^ 12. a, I), ■V(J2. 22. (^,-), 2 ANSWERS. 13 23. When D = D' and E = E' ; in otliiT words, when the two equations differ only in their constant terms. 24. In this case, r = 0. Hence, the equation represents simply the point (a, b). We may also say that it is the equation of an inlinilely small circle, having this point for centre. {{%, I), |V2; r (i) Z>2-4c. 26. <; On OX, 3 and 2 ; 31. \ (ii) E^ = 4C. lOn OY, 6 and 1. y (iii) 4C > D'^ and E^. ( (6, 2), 5 ; 32. x-2 + 2/2 + iqx + 10?/ + 25 = 0. 27. ^ On OX, 6±V21; [On or, imaginary point.s. ^^- ('' "*) =^'"M8, 1). r(2, 4), 2V5; 34. (2, 0)and(^, - ^). 28. -^' On OA', and 4 ; 35. i V5. I On OY, and 8. r(3, -2),3; 36. V a^ + bO 37. •2x - 2/ - 2 = 0. 38. 4.C — 5?/ — 71 = 0. 39. 3x - 5?/ - 34 = 0. 29. -^ On OX, 3 ± V5 ; [On OF, -2. r(-ll, 9), VhB; 30. -^ On OX, - 3 and - 10 ; I On OY, 9±2V6. 40. Let {x, y) be any point in the required locus ; then the distance of {x, y) from (Xi, 2/1) must always be equal to its distance from (jo, 2/2) ; therefore, (x — Xi)2 + {y — 2/1)2 = (x — X2)2 + {y — 7/0)- ; whence 2x (xi — Xo) + 2y (2/1 — 2/2) = (-Ci"^ + yr —x-z^ — 2/2'-). Show that this represents a straight line _L to the line joining (xi, 2/1) and (xo, 2/2) at its middle point. 41. 8x + 62/ + 17 = 0. 42. First Method. Substitute successively the coordinates of the given points in the general equation oi the circle ; this gives three equa- tions of condition, and by solving them we find the values of a, b, r. Second Methoi>. Join (4, 0) to (0, 4) and also to (0, 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 from the centre to either one of the given points will be the radius. A ns. K'- + y' — fix — % + 8 = 0. 43. x2 + y^ — 8x + 62/ = 0. 45. x- + y- + Sax - 6ay = 0. 44. x2 + 2/2 + Ox + 2/ = 0. 46. x2 + 2/2 + 8x + 20 y + 31 = 0. 14 ANALYTIC GEOMETRY. 48. 49. 47. x'- + ?/- — Ox — %+ 14 = 0. 51. x- + y-::f2ax^:2ay + a'^=0. ^ (X - o)--^ + (>j + 8)--2 = 169, 52. x2 + y^ =ax + by. ] (X - 22)-^ + {y- 9)- = 169. 53. (x - 1)2 + {y — 4)^ = 20. < x-2 +2/2 _oo(a; +y) +225 =0, 54- «- + 2/' - 14x - 4?/ - 5 = 0. Ix'+y-— (}(x+y)+ 9=0. 55. x'- + ?/2 ± V2a?/ = 0, 50. x2 + 2/- — 8x — 8y + 16 = 0. X-- + y-± \^ax = 0. 56. r?((x'+?/2)— a6= (ma— 6)x+ (m6 —a)?/. 57. x- + ?/2 = XiX + yiy. 58. (X-Xi)(X-X3)+(2/-yi)(2/-2/2)=0. 3.2_„j,+y2 = r2_ ^" 59. (l + m2)(x2+2/^) -2r(x+m2/) = 0. * ^ 2" Exercise 22. Page 81. 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. 2x + 3y = 26, 3x — 2y = 0; sVlS, 2VT3, -9,-4, -L3 Vl^ Xi2 — r2 r3 „„ rx2 + ?/2 = p2^ ' — ^' • 22. .; , . , Xi2/i ( (p cos a, p sill a). r When C = rV^lM^. 23. J rxTu Aa+Bb-C , ] When — , — = ±r. [ V^2 + J52 24. ax + by = 0. 25. (— a, — b). 26. (2a, b). 27. (0, 6). 28. x2 + y2 = |. 29. m = 0. 30. c = — 36 + 20V6. 14. Ax + By If W^2 + £2 := 0. 31. (x - 5)2 + {y - 3)2 = Jj-y-- 15. 5x-^2/=fWI^T^ = 0. 32 ((X- 2)2+ (.y- 4)2=100, , fi .. - . + .V2 = 0. i (X - 18)^ +{y- li>)' = 100. Xi 5. 9x - 13y = 250. 6. x±Sy= 10. 7. 104 1. 8. x2 + 2/2 = 25*. 9. 14x ±6y = 232. 10. 3x + 2/ = 19. 11. 3x + 42/ = 0. L2. ( 3x + 72/ = 93, ( 3x - 7y = 65. 13. X = r. 16. x-2/±r\/2 = 17. The equation of the two taii- 33. (X- 1)2+ (2/ -6)2 = 25. gents is {h'- - r^)y2 = r2(x - /02. 34. -=-+-■ 18. x + y= ±r\f2. 35 (3.2 -\-y2^a + b + Va2 + 62)2. 19 (a; =10, -2a6(a + 6 + Va2 + &2)(x+y) ( 3x + 42/ = 50. _ -f ^252 = 0. 20. 2/ = 2x + 13 ± 6 V5. 36. x = a + r. 21. -21, -3f. 37. [4r2-2(a-6)2]i. ANSWERS. 15 Exercise 23. Page 84. 1. iV29, (1, - 1). 11- ^' + 2/- - 5x - I2y = 0. 2. |Vrr, (- h f). 12. x2 + 2/2 - 14x - 4?/ - 5 = 0. 3. iVsi, (I, ^). 13. x2 + ?/2 + i4j. + ]4y + 49 = q. / 6 ab \ 14. x2 + ?/■- ;^ 2rx — 2ry + r2 = 0, *■ ^'VVTTa^' VrTa2/ x2 + r' ± 2rx + 2ri/ + r2 = 0. 5. x"- + y^-Sl. 15. x2 + 2/2-2ax-2a2/+a-'--=0. 6. (X- 7)2 + 2/2 =9. ;Lg_ 3.2 + ^2=: 9. 7. (X + 2)2 + (y- 5)2 = 100. o 9.0 o /o /^ \-rt 17. 5(x2+2/2)-10x,+ 30?/+49=0. 8. x2 + 2/2 — 2a(3x + 4?/) =: 0. ^ ^ ' •' 9. x2 + 2/2 + 262 4- c-2 18 H^ ~ ^^^ + ^^ ~ ^)^ ~ ^' = 2 [(6 + c) X + (6 - c)2/] . ■ ( (X + V)' + (y + V)' = f • 10. 3a6(x2 + 2/2) + 2a6(a2 + 62) 19. x2 + y"- — 30x — Uly = 0. = (5a2+262)6x+ {ob-^+2a-)ai/. 20. x2 + 2/2 + SOx + 882/ — 50 = 0. 2^ { x2+ 2/2 — .3Gx- 402/ +324 = 0, I 25x2 + 25^2 _ 80x - 4942/ + 64 = 0. 22. (6,2), 5. 31. x2+2/2 = ±a2/V2,or = ±axV^ 23. - 15. 32. x2 + 2/2 ± 2a(x ± 2/) = 0. 24. — 10. 33. 2(x2 — ax + 2/2 - r-) + a2 = 0. 25. 1V26. 34. X — y = 0. 26. VTO. 35. 4x + 32/=^0. 27. x,x + y,y = x,^ + 2/,2. 36. (18 ± 2V41)x -52/ = 0. 28. (i) Z»2 = A AC, (il) -B2 = 4^(7, ^'^^ a: + "^^ ± ^0 - 0. (iii) Z>2 =:: £2 rr 4^ C. 38. X + ^ - 10 = 0. 29. r2 = 2rmc + c2. 40. +(35 + 24\/30). 43. 135°. 30. k = 40, or - 10. 44. (7, - 5) and (- 6^^, 9||). f (X + 4)2 + (2/ + 10)2 = 85, «. ^/^_514X2 / 670y^_8^ [\ 109/ V^ 169/ 1692 46. The circle (x — Xi)2 + (2/ — Vi)- ~ r". 47. The circle (x - a)- + (2/ — 6)2 = (r +Y)'^, or (X - a)2 + (2/ - 6)2 ^ (r - r')2. 48. The circle (x — rt)2 + (2/ — 6)2 = 7-2 + <2. 16 ANALYTIC GKOMKTKY. 49. Take yl as orijiin, and let the radius of the circle = r; then the locus is the circle x- + //'- = rz. 50. Take A as origin, and let the radius of the circle = r ; then the . , „ , -, 2»irx locus IS the circle x- + y'^ = - — , — m + n 51. Take A as origin, AB as axis of x, and let AB = a ; then the locus is the circle {in- — n-) (x'^ + y'^} — 2am^ + a-m- = 0. 52. Take AB as the axis of x, the middle point oi AB as origin, and let AB = 2a ; then the locus is the circle 2{x^ + y-) — k- — 2(jfi. 53. Using the same notation as m No. 52, the locus is the straight line 4ax = ± Ic^. 54. Taking the fixed lines as axes, the locus isthecircle4(x-+?/-) = fZ-. 55. Take the base as axis of x, its middle point as oi-igin, and let the length of the base = 2a, and the constant angle at the vertex = d. Then the locus is the circle x'^ + y- — 2a cot dy = a'^. 56. Take A as origin, AB an axis of x, and let AB — a, AC = h. Then the locns is the circle (x — },a)'- + y- = —• - ' ■ 4 4r* 57. The circle x- + y- = r-; zzi where I is the length of the chord. 58. The locns is a circle. Exercise 24. Page 97. 1. 7x — 6?/ = 0. 2. X — ?/ = 0. 4. X + ?/ = r, 2x + o// = r, (n + b)x + (a — b) y = r^. 5. l?.x + 2>/ = 40. 6. The tangent at {h, k). 7. (i) 2x+% = 4, (ii) '^x — y = 4, (iii) x — ?/=4. 8. (i) (20, ;]n), (ii) (21, -14), (iii) (.%«, nr>h). 9. (G, 8). 18. 12x + 17;/ - 51 = 0. ^n ( --^ -JJ^1\ 19. x+ 7/ -2 = 0. "• V c ' c ) 11. (4, ± 3), 4x ± ?,y = 25. ^^ ("' ~"^^^^ ~("^ -lfi)y + ac =0. 16. h:2 + /,-2 _ r2. 21. X - 7/ = 0, Vi(a + 6)^ - 4c. 17. 3. 22. (-2, -1). ANSWERS. 17 Exercise 26. Page 109. 1. Writing x + 1 for x, and y — 'l for ?/, and reducing, we have y'^—^x. 2. x- + (/- = ?•-. 5. x'^ + ?/" = '''■'■ 3. x'-^ + y- = 2rx. 6. 2xy = a^. 4. X- + y- = — 2ry. 7. x- — //- + 2 = 0. 8. (i) p = ± ((, (ii) p2 cos 2d - a^. 9. (i) p = ia tan .sec ^, (ii) (a + p co.s ^)- = 4ap sin ^. 10. (i) X-' + //■- = a-, (ii) x- + y- = rtx, (iii) x^ — y- — ofi. 11. x + y = (). 15. x2 — ijxy -\-y- = 0. 12. 2x — 5y + 10 =: 0. 16. xy = .'J. 13. 12x- + Ulzy + 4i/^ +1 = 0. 17. y/- = 2(f(x V2 — a). 14. X- + y- — 25. 18. -Ix// = 25. Exercise 27. Page 111. 1. 6V3. 12. 9x2 + 25(/2 =: 225. 2. 4 sin ^w. 13. 14. 15. p= 8a cos e. p = ± 4a. p2 sin2 e — 5/3 cos = -2:^. 3. Vl3 — 12 cos w. 4. Va- + Ij^ — 2ab cos (^- 0). 5. 2a sin 0. 16. p2 = 49 sec 20. 6. 2(1 cos 0. 17. 18. P' = /i;2 cos 20. xy - a-. 7. aV5-2V3. 9. 2x'-2 + 2x?/ + 2/2 = 1. 19. {x- + 2/2)2 =: 2ixy. 10. 2x2 + 2/2 = 6. 20. ^s— 2/3+ (3x— Sy— 5A;)x2/= =0. 11. 2/ = 0. 21. tan-i |. 22. (i ) tan -(■ -iy (ii) tan-'-- .4 Exercise 28. Page 117. 2. 2/2 = 4/y(x - 7/). 3. 2/- = ^^(x + p). 4. (i) ^2 = lOx, (ii) y- = lOx + 2-'., (iii) 2/2 = lOx — 25. 5. (i) 2/2 =l(3x, (ii) y- = lOx + 04, (iii) y- = lOx — G4. 18 ANALYTIC GEOMETRY. 6. (0, 0), (2, 6). 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 =0 passes through the vertex. The line x = — 2 does not meet the parabola. 11. The line y — 6 meets the parabola in (9, 6). The line y= —8 meets the parabola in (16, — 8). 12. p = 4. 13. (0, 0), (2, 8). 14. (i) y = 0, (ii) x = - 2, (iii) x = 2, (iv) 4x ± 3y-8=0, (v) y= - 2x. 15. (i) 4x - 5?/ + 24 = 0, (ii) x^ + y^ - 20x = 0. 16. 3j3. 17. 8pV3. 24. The latus rectum of each = 4p. The common vertex is at the origin. The axis of x 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 beloiv. We may name them as follows : (i) is a right-handed X-parabola. (iii) is an upward F-parabola. (ii) is a left-handed X-parabola. (iv) is a downward F-parabola. Exercise 29. Page 121. 6. X — 4?/ -I- 20 = 0, 4x + y — 90 = 0. _^ (x-?/ + 3 = 0, , ^=-2- r.« „ ■ n / ^ A^-4C\ A 20. /> numencallv, I — „» r^:: — I' x= — — • . .. / A A-^-^V\ 21. 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{a ± r) X + r- — o-. 22. Let 2a be the given base, ah the given area ; take the base as axis of X, its middle point as origin ; then the locus is the parabola x2 + by = a?. Exercise 32. Page 144. 1. 5, 4, 3, |. 2. V2, 1, 1, iV2. 3. 2, ^/3, 1, \. 1_ ^ J^^ Is- A IB-A ■\Ia ' Vi>'' "^B V^' V^'^^^'"^^^- SWERS. 13. x'-i + 2(/2 = ; 100. 14. 8x2 + y?/2 = = 8tt2. 16. 2 : V3. 17 X = y=± «6 ■ V«' + ^- 18. (1, t)^ (- S — 2\ 31 sr 19. (1 , -0, (1, -2). 21 5. fVe. 6. e = iV3. 7. 4x2 + 9y2 = 144, 8. 25x2 + 1097/2 = 4225. 9. 16x2 + 252/2 = .3600. 10. 16x2 4-252/2= 1600. 11. 25x2 + 1692/2 = 4225. 12. 3x2 + 72/2=115. 20. (:!, l),(3,-l),(-3, l),(-3,- 1). 21. The equation of the locus is x2 + 42/2 = r-. 22. Taking as axes the two fixed lines, and putting AP — a, BP = 6, the acute angle between AB and the axis of x = ^, we find that X — a cos . 12. c2:6-'. 2V3 13. a- — e*Xi2. 4. iV2. 12. c2:6-'. 1 + Vl3 _ 14. V(l- e'^)(a2-e-^Xi2). 6. X + y = ± Va2 + 52. Vl — e'^ b „ 1 , 1. /"TT"; — ;; ,^ 15- tan 4> = = -• 7. &x 4- cy qi oVa^ + c^ = 0. e c 18. The locus is the minor axis produced. X — j + ?/2 = f2; centre is I „, j ; semi-axes ace - and r. 20. The ellipse o'^ ( y — - j + ^^x^ = — — ; centre is (0, - j ; semi- a , b axes are - and -• In 21-23 take the base of the triangle as the axis of x, and the origin at its middle point. 21. The ellipse {.s2 — c2)x2 + s2?/2 = s2(s2 — c2). 22. The ellipse kx- + y~ = kd^. 23. The circle (x + (")- + y- = 4a2. Exercise 35. Page 164. 1. Stt. 2. I Vs. 3. 20x + 63?/,— 36 = 0. b. (^ ^> ^j 8. (1) jni2=-, (u)mi2=-, (m)m^2 = l. ANSWERS. 11. 3x+8y = 4, 2x - 3?/ = 0. 12. Area = — (?n + n), m and n being the two segments (use the 2a polar equation). 13. 2Gx + 33?/ - 92 = 0. 14. x + 2?/ = 8. 15. b'-z + a^y = 0, b^x -a^ = 0, a^y + Wx = 0,hx + ay- 0. 17. ahj^x = b^xiy. 29 ^ + ^ = ^. 23. -T 1 = 0. « ^ _ a(l - e2) 30. /) — 24. 6xVz- — 62 ± a]/\/a;^—l^ = 0. 1 + e cos 25. e=^V6. g^ 62 26. See § 148. ' ^ 1 — e2cos2 6> 32. 1(5x2 + 49?/2 - 128x - 686?/ + 1873 = 0. 33. 2a = 18, 26 = 10. 34. ^' + ^2 = 5^. 144 c2 — 62 a—b 35. cos = \ -^ — 7:7 • 36. tan = p=- 37. Find the ratio of yi to the intercept on the axis of y. 38. h'^hx + aVcy = b^h' + am. ^^ „, ,,. x2 y2 42. The ellipse — + t:; = i. a2 62 - The ellipse 62x2 +a2y2 = 52^2 44. X2 a2 + 2/2 62 = 2. 43. (- - 1 ,1) , a = ^, 6 = 1, 46. The ellipse 25x2 + \C^y^- - 48y = 64. 45. A/^'A/^'inwhich£:=-F+^^%^- Exercise 36. Page 174. s^ _ y^ _ 3. 3x2 — y2 — ;3(i2_ ^^ ^^ 4. G25x2 - 84?/2 = 10,000. 4x2 ^2 _ ^- 25~36~^- 5. 2.c2 - 22/2 = c2. 7. rt = 4, 6 = 3, c = 5, e = |, latus rectum = |. 24 ANALYTIC GEOMETRY. 8. Idy- — Dx- = 144, transverse axis = 0, conjugate axis = 8, dis- tance between foci = 10, latus rectum = A/. 9. a: 6=1: Vs. 11. e = 72. 12. (5,-0^). 14. Foci, (5, 0), (— 5, 0); asymptotes, y = ± |x. 17. b. Exercise 37. Page 176. 1. KJx -Qy = 28, Ox + Uhj = 100; |, %»-. 3. x^ - y^ = 0,(5, 4). 4. The four points represented by x= ■ ' y Va2 - 62 Va2 - 62 9. — =• 10. —^ ^ = 1. 11. When a is less tluin 6. 12. Tlie circle x- + y^ = (j2. Exercise 38. Page 177. 1. 26e, ae2. 12. (0, ± Va-' - 62). 2. 14 and 6 ; (— 8, ± SVS). 13. 62>a2. 3. The sum = 2ex. 14. g4x — Oy — 741 = 0. 8. {a, 6V2), (a, - 6V2). 15. y = ix ± 8\^. 10. They are equal. a'lb- 11. y = xV2 + a. ' a2+~62 Exercise 39. Page 188. 1. 9x + \2y + 10 = 0. 6. a. 8. 75x - Wy = 0. 9. 245x- 122/ -1189 = 0. 10. fVs. 17. See § 140. 18. ^ + ^ = 2. a;i yi 2. x = ±-- e 3. TT 2' 4. / >lffl2 562 X 1. C ' C j 5. X + a = 0. ANSWERS. 25 19. X2 7/2^ 2X_^ _ ^ 5-2 (i') ^-K+'^ = 0. 21. 23. The hyperbola 3x2 — y- + 20x — 100 = 0. The centre is the point ( — L"-, 0). Changing the origin to tlie centre, we obtain 9x2 - 3y2 = 400. 24. Writing the equation in tlie form (x —1)2 — 4(?/ + 2)2 = 4, and x2 ?/2 changing the origin to (1, — 2) we obtain T ~ T — 1- T^^ centre i.s (1, -2), a = 2, 6= 1. 25. Centre is ( -r-^. TTT h semi-axes are -%/ , .*/— , in whith 1)2 ^2 4J. 41? 26. Tlie locus is the curve 2xy — 7x + 4?/ = 0. If we change the origin to the point {h, fc), we can so choose tlie values of h and k as to eliminate the terms containing x and y. Making the change, we obtain 2x1/ + (2A; - 7)x + (2ft + A)y — Ik + 4k + 'Ihk - 0. If we choose h and k so that 2A + 4 = 0, and "Ik — 7 = 0, that is, if we take ft = — 2, fc = .1, the terras containing x and y vanish, and the equation becomes xy = — 7. Hence we see (§ 182, Cor.) that the locus is an equilateral hyperbola, whose branches lie in the second and fourth quadrants, and that the new axes of coordinates are the asymptotes. 27. The equilateral hyperbola 2xy = cfi. 28. Taking the base as axis of x, and the vertex of the smaller angle as origin, the locus consists of the axis of x and the hyperbola 3x2 — ?/2 — 2ax - 0. Exercise 40. Page 206. 1. The ellipse 72x2 + 48^2 =35. 8. The parabola y- = 3x V2. 2. The ellipse 4.x2 + 2//2 = 1. 9. The parabola ?/2 = 2x. 3. The hyperbola 32x2 — 4»y-=d. 10. The ellipse 4x2 + '.)//2 = 3(5. 4. The ellip.'^e dx'^ + ?,y- - 32. 11. The point (0, 0). 5. The hyperbola 4x2 — 4//2 + 1=0. 12. 'I'ho hyperbola 4x2 _ <),/-j == ;j(] 6. The parabola y2 = — jx. 13. TlK'Straightlines.v=x,y=— 5. 7. The parabola y- = 2xV2. 26 ANALYTIC GEOMETRY. Exercise 42. Page 229. 2. 2d ; 5th ; 6th ; 7tli ; ;3d ; 4th ; 8th. 5. 5V2; VTi; V83; O.8V2, 0.4V2, O.5V2 ; iVli, - ^\y/U, - xVVIi ; ^V83,_- A V83, - /^V83. 6. tVVi4, ^Vi4, y\Vl4. The line is parallel to the radius vector of the point (1, 2, 3). 7. Parallel to the radius vector of the point (A, B, C). ABC yjA-^ + ii2 + C2 V^2 4- £2 + c^ V^-^ + B^+ C^ 8. 60° or 120° ; 90° ; 60° or 120°. Exercise 43. Page 235. 1. V3; V3; 2V3. 3. 5V2. 4. - J-^y2, - §V2, \S. 5. Lines parallel to the radius vector of (3, — 2, — 5). ^3^V38, -tVV38, -^\V38. 6. cos-i ||Vl4. 9. (1, V3, 2V3). 7. 90°. 12. (t, - f, I). 8. (4, iTT, iTT). 13. (- -2,9-, - -V«, ¥)• Exercise 44. Page 240. 4. 2x - 3?/ + V3z = 28. 5. ^-f-|=l,T + f + ^=-l,2x-|2/ + 62 = 0. 4 .5 / 1 z 5 6. 6x + 2/ — z = 5. 7. cos-i ^-. 12. - + ^ - - = 1 ; 8. 79° 52'. ■ a 6 c (J — 2abc ^ ^' '^^^"^ Vl^TW^C^' V6%Mr^2c2 + a262 ^ 14. 2x 4- 4?/ — 2 = 23. COS — ^ — — - ' 1 V^2 + ^2 + (72 16. 3x — ?/ -f 2z = 4. i? 18. .c + y + z = ; 2V3. cos-'i , V.42 + i?2 + C2 19. ?,x - iy + 7z + 13 = 0. 10. 2.25. 20. 2x + 5?/ — z = 9. ANSWERS. 27 Exercise 45. Page 247. 3. It passes through (i, 0, — |), and is parallel to the radius vector of (4, 5, 3). x^zA^lL^zl^^^il. 6. (4, 5,0), (0, 1,4), (-1,0, 5). 2 2 — 2 ' 7. 22/ + 5z = 10. 8. 2x — 5?/ + 1 = 0, 2x + 5z + 1 = 0, ?/ + z = 0. 11. cos -1—0.1 x — 'P. _ y+7 _ z + o 12. 14° 57' 45". ' -3 ~ 5 ~ - 6 r- ,„x — 2w + 4z + 6 15. ?/ = - 2x - 1, z = 3a; + 5. 18. sin-i ^%. X — p _ y — Q _ z 19. 1 Vl + »l2 + ?l- Vl + 7)1- + n- Vl + H(2 + ,j2 in which the denominators equal cos a, cos p, cos 7, respectively. Exercise 46. Page 263. 1. The planes x=— 4, x = — 1, x = 2; the planes y = — 2, y =1, y = 3 ; the planes z = 0, z— — m. 2. Answer to the first, the parabolic cylinder whose elements are parallel to the axis of z, and wliose trace on the plane xy is the parab- ola 2/2 = 8x. 3. 5x2 + 10^2 - 5(5^ 5^2 _ j^2 = 8, x2 + 13z2 = 32. 4. X- + ]/-= i, z = i: |V42 ; hence the curves are two circles in the planes z_— ± sV42, whose centres are in the axis of z, and whose radii are Vf- each. 5. a = 3, 6 = 2, e = iV5. _ 6. Two circles whose radii are •\/l3 each. 8. 9(x2 + y"-) = (z - 5)2 ; x2 + 2/2 = 2^5. 9. 25x2 + 252/2 - 9z2 + 90z = 225. 10. x2 + 2/2 — |z — I = 0. 11. 2/2 + 22 - 8x = 0'. 12. x2 + y2 _ J22 - 3z = 9 ; (0, 0, - 6). 13. 9(x2 + 2/2) + 4z2 = 36. 14. lG(x2 + y-) + 9z2 = 144. 15. 4(x2 + 2/2) _ 9^2 = 3(5. 16. x2z* + y-z* = 1, xf + 2/2 - iz3 = 0. 17. a = 5V3, b= jV6. WENTWORTH'S||f^° TRIGONOMETRIES ^pitiqns THE aim has been to furnish in these books just so much of trigonometry as is actually taught in our best schools and colleges. The principles have been unfolded with the utmost brevity consistent with simplicity and clearness, and interesting problems have been selected with a view to awakening a real love for the study. The subjects of Surveying and Navigation are presented in accordance with the best methods in actual use, and in so small a compass that the general student may find time to acquire a competent knowledge of them. List price Plane and Solid Geometry and Plane Trigonometry. i2mo. Half morocco. 655 pages 51.40 Plane Trigonometry. lamn. Cloth. 162 pages 60 Plane Trigonometry, and Tables. Svo. Cloth. 258 pages . . .90 Plane and Spherical Trigonometry. i2mo. Half morocco. 232 pages 85 Plane and Spherical Trigonometry, and Tables. Svo. Half morocco. 32S pages 1.20 Plane Trigonometry, Surveying, and Tables. Svo. Half morocco. 357 pages 1.20 Surveying and Tables. Svo. Cloth. 194 pages So Plane and Spherical Trigonometry, Surveying, and Tables. Svo. Half morocco. 427 pages 1.35 Plane and Spherical Trigonometry, Surveying, and Navigation. i2mo. Half morocco. 412 pages 1.20 Logarithms, Metric Measures, and Special Subjects in Advanced Algebra. i2mo. Paper. 141 pages 25 New Five-Place Logarithmic and Trigonometric Tables. By G. .A. Wentworth and (1. .\. Hill Seven Tables (for Trigonometry and Surveying). Cloth. 7<) pages 50 Complete (for Trigonometry, Surveying, and Navigation). Half morocco. 154 pages i.oo GINN 6 COMPANY Publishers BOSTON NEW YORK CHICAGO LONDON SAN FRANCISCO ATLANTA DALLAS COLUMBUS fVENTfVORTH'S COLLEGE JILGE'BRA REVISED EDITION i2mo. Half morocco. 530 pages. List price, $1.50; mailing price, f 1.65. This book is a thorough revision of the author's "College Algebra." Some chapters of the old edition have been wholly rewritten, and the other chapters have been rewritten in part and greatly improved. The order of topics has been changed to a certain extent ; the plan is to have each chapter as complete in itself as possible, so that the teacher may vary the order of succession at his discretion. As the name implies, the work is intended for colleges and scientific schools. The first part is simply a review of the principles of Algebra preceding Quadratic Equations, with just enough examples to illustrate and enforce these princi- ples. By this brief treatment of the first chapters sufficient space is allowed, without making the book cumbersome, for a full discussion of Quadratic Equations, The Binomial Theorem, Choice, Chance, Series, Determinants, and The General Properties of Equations. Every effort has been made to present in the clearest light each subject discussed, and to give in matter and methods the best training in algebraic analysis at present attainable. ADDITIONAL PUBLICATIONS By G. A. WENTWORTH List Mailing price price Plane and Solid Geometry (Revised Edition) . . . ^1.25 $1.40 Plane Geometry (Revised Edition) .75 .85 Solid Geometry (Revised Edition) .75 .85 Analytic Geometry 1.25 1.35 Trigonometries (Second Revised Editions) .... A list 0/ the various editions of Wentiuorth's Trigonometries will be sent on request. GINN & COMPANY Publishers SCIENCE AND ENGINEERING LIBRARY University of California, San Diego ' DATR nilF. ; 0CT2 8 1S75 ' ^ JDSr 1 4 mo MAY "^ 1QRfi ii\r\{ V