ANALYTIC GEOMETRY BY EDWIN S. CRAWLEY AND HENRY B. EVANS PROFESSORS OF MATHEMATICS IN THE UNIVERSITY OF PENNSYLVANIA E. S. CRAWLEY UNIVERSITY OF PENNSYLVANIA PHILADELPHIA 1918 Copyright, 1918, by EDWIN S. CRAWLEY AND HENRY B. EVANS PRESS OF THE NEW ERA PRINTING COMPANY LANCASTER. PA PREFACE In adding another text-book on analytic geometry to the large list of such books now available, the authors feel that a statement of the reasons for this one is desirable. In compiling the work they have endeavored to keep constantly in view the needs of that large class of students in our colleges and technical schools who, though in a way sympathetic toward mathematical study, are yet lacking in that quick appreciation of the mathematical point of view which is a characteristic of the so-called mathematical mind. The assimilation of new ideas, or even of new combinations and applications of ideas already familiar, in which the study of analytic geometry especially abounds, is not easy even for the trained mind, and for the student approaching the subject for the first time is the source of most of his difficulties. The attempt here is to make this approach by easy gradations so that the student may not lose himself on the way, but may feel sure of his ground as he advances. To try to gain this end by the lack of rigorous demonstration would be to lose sight of an important object of such study, which, rightly viewed, is to assist the student to an acquaintance with correct principles of mathematical reasoning and the accompanying scientific attitude of mind. The aim has been therefore to preserve mathematical rigor throughout. As to details, much of the book necessarily follows well established lines of procedure. The experienced teacher will find, however, here and there, some departure from these lines which it is hoped will be approved. Owing to the increasing use of the imaginary, and its growing importance to the student of both pure and applied mathematics, some elementary disiii iv PREFACE cussion of imaginary elements in geometry has been included, which the authors believe will be of value in accustoming the student to look upon the complex number as a useful member of the number system. The lists of exercises have been prepared with especial care. Throughout the text short lists have been inserted where needed for the immediate illustration of principles or methods. At the conclusion of each chapter longer lists are inserted. These are divided into two parts, " normal exercises," which are direct applications of the text of the chapter, and " general exercises," which are designed to give the student opportunity of testing his grasp of the subject as a whole, and of its underlying principles. There will be found in each set of " normal exercises " one or two examples illustrating each separate point of the theory developed in the corresponding portion of the text, so far as it can be illustrated by exercises. The student may therefore be assured that he has made application of all of the theory when he shall have worked all the normal exercises. While the book is to be regarded as a text-book of plane analytic geometry, the concluding chapter is devoted to solid geometry. Only the barest outline of some of the fundamental principles of this subject is included, enough to enable the student, when he studies calculus, and wishes to apply its principles to problems involving solids and surfaces, to feel that he is not on entirely new ground. UNIVERSITY OF PENNSYLVANIA, July, 1917 TABLE OF CONTENTS PAGE FORMULAS, TABLES, ETC., FOR REFERENCE.................. Vi CHAPTER I. CARTESIAN COORDINATES.............. 1 II. GRAPHICAL REPRESENTATION OF EQUATIO N S................................. 15 " III. THE STRAIGHT LINE...................... 40 IV. THE CIRCLE............................. 65 V. THE CONIC SECTIONS.................... 75 VI. TANGENTS, NORMALS, DIAMETERS, POLES AND POLARS........................... 107 VII. APPLICATIONS OF ANALYTIC GEOMETRY.... 127 " VIII. THE GENERAL EQUATION OF THE SECOND D EGREE.............................. 142 IX. HIGHER PLANE CURVES, PARAMETRIC EQUATIONS........................... 161 " X. POLAR COORDINATES..................... 179 XI. EMPIRICAL EQUATIONS................. 191 " XII. EXTENSION OF COORDINATE GEOMETRY TO SOME SPACE PROBLEMS.............. 205 APPENDIX. CURVES FOR REFERENCE.................. 236 FORMULAS, TABLES, ETC., FOR REFERENCE A few important definitions, formulas, and tables, from elementary algebra and trigonometry, which will be needed in the study of this book, are placed here for convenience of reference. A. Algebraic definitions and formulas I. DEFINITIONS (a) The degree of a term in specified letters is the number of times these letters occur as factors in that term. Thus ab2c3d is of the seventh degree in a, b, c, and d; a2x2y2 is of the fourth degree in x and y. (b) The degree of a polynomial in specified letters is defined as that of the term of highest degree in those letters. Thus ax3 + bx2y + cx2 + dy + e is of the third degree in x and y. (c) A polynomial is said to be homogeneous in specified letters when each of its terms is of the same degree in those letters. Thus ax2 + 5xy - b2y2 is homogeneous and of the second degree in x and y. (d) In an equation containing one or more variables the term which contains no variable factors is called the absolute term. Thus in ax2 + by2 - a2c2 = 0, where x and y are the variables, the absolute term is - a2c2. In x + y = 0, the absolute term is zero. (e) A root of an equation is a value of the variable which, when substituted for the variable, satisfies the equation. II. THE QUADRATIC EQUATION (a) The roots of the quadratic equation ax2 + bx + c = (1) are b + b2 - 4ac- b - b2 - 4ac l = 2a r 2a (2) (b) The roots of the quadratic (1) are real and unequal if b2 > 4ac, real and equal if b2 = 4ac, (3) imaginary if b2 < 4ac. vii viii FORMULAS, TABLES, ETC., (c) The sum of the roots of the quadratic (1) is rl + r2 = -. (4) (d) The product of the roots of the quadratic (1) is rir2 = -. (5) a (e) If the coefficient a in the quadratic (1) varies and approaches zero, the coefficients b and c remaining constant, one root of the equation increases without limit. III. FACTORS OF Ax2 + Hxy + By2 A homogeneous expression of the second degree in x and y, Ax2 + Hxy + By2, (6) can always be factored. The factors are conveniently expressed thus A [2Ax + (H + qH2 - 4AB)y][2Ax + (H - HH2 - 4AB)y]. (7) 4A IV. IMAGINARIES (a) -- 1 is called the imaginary unit, and is usually represented by the letter i.* Thus x + iy means x + y l- 1. (b) An expression containing both real and imaginary terms is called a complex expression. (c) If two complex expressions are equal, the real parts of the two expressions are equal, and the imaginary parts are equal. Thus if x + iy = a + ib, then x = a, and y = b, (8 if x + iy = 0, then x = 0, and y = 0. V. LOGARITHMS (a) If y = ax, x is called the logarithm of y to the base a. This equation may therefore be written in the form x = loga y. (b) In the case of the so-called common logarithms, the logarithms used in computation, the base is 10. * In electrical theory j is used instead of i to represent 1- 1. FOR REFERENCE ix (c) Logarithms used algebraically are usually Napierian, or natural logarithms. The base of this system of logarithms is the incommensurable number e = 2.7182818285**-, called the Napierian base, and defined by the infinite series 1 1 1 1 e = 1 + 1 + + +-+ + + + +. (9) (d) When in algebraic work a logarithm is written without any expressed base, e. g., log (1 + x2), the base is understood to be e. B. Trigonometric formulas If I is the length, and r the radius of the arc subtending the angle of 0 radians, I = rO. (10) 180~ = 7r radians. (11) 1 1 1 csc =, cot = -n sec 0 (12) sin ' tan O' cos 0' sin2 0 + cos2 0 = 1. (13) sec2 0 = 1 + tan2 0. (14) csc2 0 = 1 + cot2 0. (15) sin 0 cos 0 tan 0 = cot 0 (16) cos 0' sin 0 If tan 0, then n sin = m cos. (17) Jm2 + n2 Im2 + -n2 cos 0 = sin (90~ i 0), cot 0 = tan (90 - 0), (18) -tan (900 + 0). sin (180~ - 0) = sin 0, cos (180~ - 0) = - cos 0, tan (180~ - 0) = - tan 0. (19) sin (180~ + 0) = - sin 0, cos (180~ + 0) = -cos 0, tan (180~ + 0) = tan 0. (20) sin (-0) =-sin 0, cos (- ) = cos 0, tan (-0) = —tan 0. (21) x FORMULAS, TABLES, ETC., sin (x z y) = sin x cos y 1 cos x sin y. (22) cos (x h y) = cos x cos y F sin x sin y. (23) t ( ) tan x q= tan y (24) tan (x q- y) =i:- = t(24) 1 =f tan x tan Y' sin 20 = 2 sin 0 cos 0. (25) cos 20 = cos2 0 - sin2 0. (26) tan 20 = 2 tan 0 (27) 1 - tan2 (7 sin 30 =3 sin - 4 sin3, cos 30 =4 cos3 -3 cos. (28) 1 -cos 0 = 2 sin2, 1 +cos =2 cos2. (29) tan 0 = 1 +- os. 0(30) In any plane triangle a b c sin A sin B sin C' a2 = b2 + c2 - 2bc cos A, etc. (31) (32) SIGNS OF THE TRIGONOMETRIC FUNCTIONS First Second Third Fourth Quadrant Quadrant Quadrant Quadrant sin 0 cscin 0 + + - - CseC 0 cos0 sec 0 tan cot 0 FOR REFERENCE xi C. Values of the trigonometric functions of angles I. Certain exact values of these functions Angle Angle in in sin cos tan cot sec csc Degrees Radians 0~ 0 0 1 0 00 1 oo 300 ~ r 1 i3 3 i3 3 2 45~ Ir i1 2 /2 1 1 -12 - 600 ~ i i4 ~. 4 3 2 i43 900 iT 1 0 00 0 0 1 II. Decimal values of the trigonometric functions for each 5~ of the quadrant Angle in Angle in sin os tan cot sec csc Degrees Radians 0~ 0 0.000 1.000 0.000 oo 1.000 o 5~ 0.087 0.087 0.996 0.087 11.430 1.004 11.474 100 0.175 0.174 0.985 0.176 5.671 1.015 5.759 150 0.262 0.259 0.966 0.268 3.732 1.035 3.864 20~ 0.349 0.342 0.940 0.364 2.747 1.064 2.924 250 0.436 0.423 0.906 0.466 2.145 1.103 2.366 300 0.524 0.500 0.866 0.577 1.732 1.155 2.000 350 0.611 0.574 0.819 0.700 1.428 1.221 1.743 400 0.698 0.643 0.766 0.839 1.192 1.305 1.556 450 0.785 0.707 0.707 1.000 1.000 1.414 1.414 500 0.873 0.766 0.643 1.192 0.839 1.556 1.305 550 0.960 0.819 0.574 1.428 0.700 1.743 1.221 60~ 1.047 0.866 0.500 1.732 0.577 2.000 1.155 650 1.134 0.906 0.423 2.145 0.466 2.366 1.103 700 1.222 0.940 0.342 2.747 0.364 2.924 1.064 750 1.309 0.966 0.259 3.732 0.268 3.864 1.035 800 1.396 0.985 0.174 5.671 0.176 5.759 1.015 850 1.484 0.996 0.087 11.430 0.087 11.474 1.004 900 1.571 1.000 0.000 o 0.000 o 1.000 Xii FORMULAS, TABLES, ETC., D. Miscellaneous tables I. Degrees and Radians 1~ = 0.0174533 rad., 1' = 0.0002909 rad., 1" = 0.0000048 rad. 1 rad. = 57~.29578 = 57~ 17' 44".8 = 3437'.75 = 206264".8 0.1 rad. = 5~ 43' 46".5 0.4 rad. =22~ 55' 5".9 0.7 rad. =40~ 6' 25".4 0.2 " =11~27'33".0 0.5 " =28~ 38'52".4 0.8 " =45~ 50'11".8 0.3 " =17~ 11' 19".4 0.6 " =34~ 22'38".9 0.9 " =51~ 33'58".3 II. Mantissas of the common logarithms of numbers N 0 1 2 3 4 5 6 7 8 9 1 000 041 079 114 146 176 204 230 255 279 2 301 322 342 362 380 398 415 431 447 462 3 477 491 505 519 531 544 556 568 580 591 4 602 613 623 633 643 653 663 672 681 690 5 699 708 716 724 732 740 748 756 763 771 6 778 785 792 799 806 813 820 826 833 839 7 845 851 857 863 869 875 881 886 892 898 8 903 908 914 919 924 929 934 940 944 949 9 954 959 964 968 973 978 982 987 991 996 III. Napierian logarithms of numbers N.0.1.2.3.4.5.6.7.8.9 0 -- -2.30 -1.61 -1.20 -0.92 -0.69 -0.51 -0.36 -0.22 -0.11 1 0.00 0.10 0.18 0.26 0.34 0.41 0.47 0.53 0.59 0.64 2 0.69 0.74 0.79 0.83 0.88 0.92 0.96 0.99 1.03 1.06 3 1.10 1.13 1.16 1.19 1.22 1.25 1.28 1.31 1.34 1.36 4 1.39 1.41 1.44 1.46 1.48 1.50 1.53 1.55 1.57 1.59 5 1.61 1.63 1.65 1.67 1.69 1.70 1.72 1.74 1.76 1.77 6 1.79 1.81 1.82 1.84 1.86 1.87 1.89 1.90 1.92 1.93 7 1.95 1.96 1.97 1.99 2.00 2.01 2.03 2.04 2.05 2.07 8 2.08 2.09 2.10 2.12 2.13 2.14 2.15 2.16 2.17 2.19 9 2.20 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29 10 2.30 2.31 2.32 2.33 2.34 2.35 2.36 2.37 2.38 2.39 FOR REFERENCE xiii Xlll IV. Positive powers of e (Napierian anti-logarithms) Exp..0.1.2.3.4.5.6.7.8.9 0 1.00 1.11 1.22 1.35 1.49 1.65 1.82 2.01 2.23 2.46 1 2.72 3.00 3.32 3.67 4.06 4.48 4.95 5.47 6.05 6.69 2 7.39 8.1.7 9.03 9.97 11.0 12.2 13.5 14.9 16.4 18.2 3 20.1 22.2 24.5 27.1 30.0 33.1 36.6 40.4 44.7 49.4 4 54.6 60.3 66.7 73.7 81.5 90.0 99.5 110. 122. 134. 5 148. 164. 181. 200. 221. 245. 270. 99. 0.. 65. 6 403. 446. 493. 545. 602. 665. 735. 12. 98. 92. V. Negative powers of e (Napierian anti-logarithms) Exp. 0 -.1 -.2 -.3 -.4 -.5 -.6 -.7 -.8 -.9 -5 0.01 0.01 0.01 0.00 -4 0.02 0.02 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 -3 0.05 0.05 0.04 0.04 0.03 0.03 0.03 0.02 0.02 0.02 -2 0.14 0.12 0.11 0.10 0.09 0.08 0.07 0.07 0.06 0.06 — 1 0.37 0.33 0.30 0.27 0.25 0.22 0.20 0.18 0.17 0.15 0 1.00 0.90 0.82 0.74 0.67 0.61 0.55 0.50 0.45 0.41 VI. Square roots of numbers N 0 1 2 3 4 5 6 7 8 9 0 000 1.00 1.41 1.73 2.00 2.24 2.45 2.65 2.83 3.00 1 3.16 3.32 3.46 3.61 3.74 3.87 4.00 4.12 4.24 4.36 2 4.47 4.58 4.69 4.80 4.90 5.00 5.10 5.20 5.29 5.39 3 5.48 5.57 5.66 5.74 5.83 5.92 6.00 6.08 6.16 6.24 4 6.32 6.40 6.48 6.56 6.63 6.71 6.78 6.86 6.93 7.00 5 7.07 7.14 7.21 7.28 7.35 7.42 7.48 7.55 7.62 7.68 6 7.75 7.81 7.87 7.94 8.00 8.06 8.12 8.19 8.25 8.31 7 8.37 8.43 8.49 8.54 8.60 8.66 8.72 8.77 8.83 8.89 8 8.94 9.00 9.06 9.11 9.17 9.22 9.27 9.33 9.38 9.43 9 9.49 9.54 9.59 9.64 9.70 9.75 9.80 9.85 9.90 9.95 he Greek alphabet A, a, I, I, iota P, p, rho B, /f, K, K, kappa 2, a, s, sigma r, 7, A, X, lambda T, r, tau A, 8, M,,u, mu T, v, upsilon E, E, N, n, nu, phi Z,, X, i,, xi X, x, chi H., 0, o, omicron pI, A, psi 0, 0,, II, ai, pi Q2, o, omega. e xiv ANALYTIC GEOMETRY CHAPTER I CARTESIAN COORDINATES 1. THEOREM. Let X'OX, Y'OY be two fixed straight lines in a plane, then the position of any point in the plane is determined by its distance from Y'Y measured parallel to X'X, and its distance from X'X measured parallel to Y'Y Let P be a point in the plane to the right of Y'Y, measured parallel to X'X is a, and whose distance above X'X, measured parallel to Y'Y is b. Then the position of P in the plane is determined. Because P must lie in the line XA' MN, parallel to Y'Y at the distance a (as defined above) from it on the right, and also in the line RS, parallel to X'X of X'X, Y'Y, whose distance /Y /N P 0 /K a FIG. 1 at the distance b from it on the upper side. Hence P must be at the intersection of MN and RS. DEFINITIONS. The two lengths or distances which determine the position of a point, as z and b determine P in the preceding demonstration, are called the coordinates of the point. The fixed lines in the plane to which the position of the point is referred are called the axes of coordinates, and their intersection 0 is called the origin of coordinates, or simply the 2 1 2 ANALYTIC GEOMETRY [CHAP. I] origin. The axis X'X is called the axis of abscissas, or the X-axis, and Y'Y is called the axis of ordinates, or the Y-axis. The two coordinates a and N b of the point P, Fig. 1, are called respectively the abR L/ P/ Ls scissa and the ordinate of the point. That is: an~/ X // gThe abscissa of a point is its 0 a /K AX distance from the Y-axis, measured parallel to the X-axis. The ordinate of a point is its i/' M FIG. 1 distance from the X-axis, measured parallel to the Y-axis. The letter x is customarily used to designate the abscissa of a point, and the letter y to designate the ordinate. So also the expression " the x of a point " is often used, meaning its abscissa, or " the y of a point," meaning its ordinate. The coordinate axes are usually taken at right angles to each other, and in this book they will always be considered as so situated unless the contrary is stated. When the axes are not at right angles the coordinate system is said to be oblique. NOTE. The term Cartesian coordinates is applied to the system of coordinates which has just been described because it is essentially the system invented by the French philosopher and mathematician Descartes (1596-1650). Other coordinate systems are often used, one of which, the polar system, will be described in a later chapter of this book. 2. Signs of the coordinates. Notation. In order to be able to locate a point when its coordinates are given we must know not only the lengths of its coordinates but also the direction (right or left, up or down) in which they are to be measured. Thus, Fig. 2, if NO = OM = a, and RO = OS = b, there are four points P1, P2, Ps, P4 at the distance a from Y'Y and the [ART. 2] CARTESIAN COORDINATES 3 distance b from X'X. To avoid this ambiguity we adopt the usual convention of distinguishing between lengths measured in opposite directions by + and - signs. Then, Fig.. 3, P2 S Pi P.2 S Pi X' N M X P3 R P4 YI I FIG. 2 if OM is called + a, ON is - is - b. Hence the coordinates ( +b +b -a +a X' N 0 M X -b -b P3 R P4 Y FIG. 3 - a; and if OS is + b, OR )f P1 are x = + a, y = + b, those of P2 are x = - a, y = + b, those of P3 are x = -a, y = - b, and those of P4 are x = + a, y = - b. In actual practice when an y abscissa or ordinate is + the sign is usually not written. (-3,2)+ In designating the position of a point by its coordinates the X' ii X notation (x, y) is used. That is, the coordinates are written in parenthesis with a comma sepa- +(2,-3) rating them, and the abscissa is Y invariably written first. Thus the point (2, - 3) means the point whose abscissa is 2 and ordinate - 3, the point (- 3, 2) is the point whose abscissa is - 3 and ordinate 2. See Fig. 4. EXERCISES. 1. Draw a pair of coordinate axes and, using any 4 ANALYTIC GEOMETRY [CHAP. I] convenient unit of length, locate the following points (1, 2), (5, -2), (3, 3), (- 3, 1), (-2, 0), (0, 4), (-4, -2), (3, 0), (0, -2), (0, 0). 2. If a point is on the X-axis what is its ordinate? 3. If a point is on the Y-axis what is its abscissa? 4. What are the coordinates of the origin? 5. If the abscissa of a point is 0, on what line will it lie? 6. If the ordinate of a point is 0, on what line will it lie? 3. Directed lengths. By a directed length is meant a length measured in a given direction. As has been explained the coordinates of a point are always directed lengths. The length AB means the distance from A to B in magnitude and direction. I. If A and B are two points, then length AB = -length BA. A B FIG. 5 That is, the magnitude element of the length is the same, whether we think of AB or of BA, but the direction element in the one case is opposite to that in the other. II. THEOREM. If three points A, B, C are arranged in any order on a straight line then AC = AB + BC., I! _ FIG. 6 In the upper line, Fig. 6, the three segments AC, AR, BC are all measured in the same direction, and AC is composed of the other two. Hence AC= AB + BC. Jn the second line of the figure AC = AB - CB, but from I, - CB= + BC, hence AC = AB + BC. In the third line AC = BC - BA = AB + BC, as before. Other possible arrangements can be similarly treated. [ART. 3] CARTESIAN COORDINATES 5 III. PROBLEM. To express the distance from one point to another, when the distances of these two points from a third point on the same line are known. Let the distances AB and AC, Fig. 6, be known in both length and direction, and let the distance from B to C be required. It is then only necessary to express the length of the required segment in terms of the given segments as in II, being careful to write the letters in the proper order in each case. Since it is here required to find the distance from B to C we write BC, and then, according to II, BC = BA + AC, and hence BC = -AB + AC. (1) Had the question been to find the distance from C to B, the same procedure would lead to the equation CB = CA + AB, or CB = -AC + AB. (2) The student should examine this discussion in connection with each arrangement of points in Fig. 6 and satisfy himself that the truth of the result is independent of the order of the points. [ I I I I I I I I I I, O P Q P 0 Q FIG. 7 FIG. 8 ILLUSTRATIONS. In Fig. 7 let OP = 3, and OQ = 7, then PQ = PO + OQ = -3 +7 =4. Again in Fig. 8 suppose OP = - 5, and OQ = 2, then PQ =PO+OQ5) + = - (-5) 2 =7. Similarly with the same data QP = QO + OP = -2 + (- 5) = - 7. 6 ANALYTIC GEOMETRY [CHAP. I] ) Distance between two points. PROBLEM. To find the length of a line joining two points whose coordinates are given, the axes being rectangular. Let it be required to exP2 (X,Y2) press the length of P1P2, in terms of the given coordinates (xi, yl), (x2, y2) of these M, A/ two points. That is we ^xi 0 X have given OM1 = xi, M1P1 /P(x1,/1) N = y, OM2 = X2, M2P2 = y2. Draw P1Q parallel to OX Y ' to meet M2P2 (extended in ~FIG. 9 this case) at Q. Since P1QP2 is a right angle, P1P22 P1Q2 + QP22. (i) But P1Q = P1N + NQ = - OM1 + OM2 = - x + x2, and QP2 = QM2 + M2P2 = - MP1 + M2P2 = - Y1 + Y2. Substituting these values in (i) the result is P1P22 = (x2 - xl)2 + (Y2 - y)2, or since (x2 - x1)2 = (x1 - x2)2, and (Y2 - yl)2 = (y1 - 2)2, PlP2 = (Xl - X22)2 + (y1 - Y2)2. (3) Note carefully that in indicating a line-segment, simply as a length, it is to be written in the positive direction, as PQ,; QP2 in the foregoing demonstration; but in indicating a line-segment as one of the coordinates of a point it must be taken as starting from the axis and going to the point. Thus, above, OM1 (not MjO) is xl, and M1P1 (not P1M1) is yi. 9) Slope of a line. I. DEFINITION. The tangent of the angle which a line makes with the X-axis is called the slope of the line. [ART. 6] CARTESIAN COORDINATES 7 Of the four angles formed by the intersection of a line with the X-axis that one is taken as the " slope " angle which lies between the positive extension of the X-axis and the D upper extension of the line../ t0 Thus the slope of AB is tan 01 and the slope of CD is tan 2 X 2. X II. PROBLEM. To find the slope of a line in terms of the Y coordinates of two points on FIG. 10 the line. The slope of the line determined by the points P1, P2 Fig. 9 is tan XAP2 = tan QP1P2. Now QP2 tan QP1P2 = Q P1Q' and as shown in Art. 4, QP2 = - y1 + Y2, and PQ = - x1 + x2, - y1 q-y2 y1 - y2 tan QP1P2 = mXi 2 (4) - X1 + X2 X1 -- X2 C where m is used to designate the slope. Since in deriving (3) and (4) the principles of directed lengths have been strictly followed the results are true for all positions of the points Pi and P2. 6, Division of a segment of a line in a given ratio. PROBLEM. To find the coordinates of the point which divides a given segment of a line in a given ratio. Some preliminary remarks are necessary. (a) By a given segment of a line is meant a segment the coordinates of whose extremities are known. (b) A segment may be divided internally or externally. Thus in Fig. 11 (a) the point P divides the segment AB in 8 ANALYTIC GEOMETRY [CHAP. I] ternally in the ratio AP: PB; and in Fig. 11 (b) the point P A P (a) B A B P (b) FIG. 11 divides the segment AB externally in the ratio AP: PB. (c) When the division is internal the ratio is positive, because the two segments AP, PB, which are the terms of the ratio, are measured in the same direction, and hence have the same sign; but when the division is external the ratio is negative Y1 A B pf., _. I Q2l because in that case the two segments AP, PB are measured in opposite directions. Our problem then is to find the coordinates of P, having given the coordinates of A and B, and the ratio AP: PB. Let A = (x1, yi), B = (x2, y2), AP: PB = h: k. Draw LA, A L U Y M FIG. 12 MP, NB parallel to OY, and PS, AR parallel to OX. Then AQ AP h PS PB k ' QP AP h. SB PB k AQ LM LO + OM - xi + x PS MN MO + ON - x + x' But (i) (ii) (iii) (iv) QP QM + MP AL + MP -yl+y SB SN + NB PM + NB -y + Y2 Hence, substituting from (iii) in (i) and from (iv) in (ii) -xi+x h - X + x kIc - Y - y _ h - Y + Y2 (v) [ART. 6] CARTESIAN COORDINATES 9 Solving the first of (v) for x, - k + kx = - hx + hX2, or (h + k)x = hx2 + kxl, and hx2 + kXl 1 X h+-k Similarly (5) hy2 + kyl Y= h+k J It should be noted that equations (5) are true whether the axes are rectangular or oblique. Also that when P divides AB externally, the ratio being then negative, it is immaterial, in applying (5), whether the negative sign is attached to h or k. IMPORTANT SPECIAL CASE. The case of bisection is especially important. In this case h: k = 1: 1, and substituting h = k 1 in (5) we have X1 + X2 y1 + y2 2 ' = 2 (6) which are the coordinates of the point bisecting the line joining (x1, yi) and (x2, y.). EXERCISES. 1. Find the length of the sides of the triangle whose vertices are the points (2, 3), (- 2, 0), (1, - 4). Ans. 5, 5, 5^2. 2. Find the length of the sides and diagonals of the quadrilateral whose vertices are the points (5, 2), (2, 6), (- 3, 4), (- 1, - 1). Ans. Sides 5, -29, -129, 3 -5; diagonals 8, 5-8. 3. Find the middle points of the sides of the triangle whose vertices are (5, 2), (- 1, 6), (1, - 4), and find the lengths of the medians of the triangle. Ans. -6, 5, -465. 4. What form does formula (3) take if one of the two given points is the origin? 5. Prove that the quadrilateral whose vertices are (4, 2), (- 2, - 1), (0, - 4), (6, - 1) is a parallelogram. 6. Find the coordinates of both points of trisection of the line joining the two points (1, 6), (4, - 2). Ans. (2, -Y-), (3, 3). 1.0 ANALYTIC GEOMETRY [CHAP. I] 7. A = (- 2, 1), B = (4, 3), and P divides AB externally so that AP: PB = -2 5. Find the coordinates of P. Ans. (- 6, - 1). 8. Trisect each median of the triangle in Ex. 3, so that the longer segment in each case is adjacent to the corresponding vertex. What does the result show? 9. Extend the line joining P1 = (- 1, 3), P2 = (3, 5) each way a distance equal to P1P2, and find the coordinates of the two points thus determined. Ans. (7, 7), (- 5, 1). 10. A parallelogram has two opposite vertices at (1, 2) and (4, 1), and a third vertex at the origin. Find the coordinates of the fourth vertex. Ans. (5, 3). C Area of a triangle. Let the vertices of a triangle be '3 P1 = (xl, yl), P2 = (x2, Y2), P2 P3 = (x3, Y3) P1 Then area P1P2P3 = 0 M1 M3 Mt X M1M3P3P1 + M3M2P2P3 FIG. 13 - M1M2P2P1. (i) Since each of the areas on the right in (i) is a trapezoid we have area M1M3P3P = MM3(MlPi + M3P3) = (X3 - xi)(Y3 + yi), area M3M2P2P3- M3M2(M3P3 + M2P2) = (x2 - X3)(2 + Y3), area MM2P2P1 = M1M2(MlPl + M2P2) = ~(X2 - X1)(y1 + Y2). Hence substituting in (i) area PiP2P3 = ~[(x3 - x1)(y3 + yl) + (x2 - x3)(y2 + y3) - (x2 - xi)(yi + Y22)], from which, expanding and collecting terms, Area P1P2P3 = ~[xl(y2 - 3) + x2(y3 - y) + X3(y1 - y2)]. (7) Formula (7) may also be written thus: [ART. 8] CARTESIAN COORDINATES 11 area P1P2P3 = 2[(X1Y2 - x2YI) + (X2y3 - X3Y2) + (x3Y1 - Xly3)]. (8) When the coordinates of the vertices of a given triangle are substituted in (7) or (8) the resulting number expressing the area of this triangle will be positive or negative according to the order in which the points are taken. The numerical value obtained, irrespective of sign, will always be the required area. EXAMPLE. Find the area of the triangle whose vertices are (- 1, 2), (3, -1), (-4,2). Substituting the given coordinates in (7) the result is Area = [- (-1 -2) + 3(2 - 2) -4(2 + 1)] = 1(3 +0-12) = -4. The required area is therefore 42. The student acquainted with the determinant notation will recognize that formula (8) can be written 1 X1 yl area P1P2P3 = 2 1 x2 y2. (9) 1 X3 y3 8. Area of any polygon. Since any polygon can be divided into triangles by diagonals drawn from one of the vertices, the area of such a figure can be determined by finding the areas of its several parts by means of (7), (8) or (9), and combining the results. EXERCISES. 1. Determine the areas of the following triangles (a) (2, 6), (4, - 1),.(0, 0). (d) (0, 3), (4, - 5), (- 3, - 2). (b) (3, - 2), (- 4, 1), (0, 5). (e) (5, -1), (-1, -5), (2, 4). (c) (1, -3), (4, 5), (- 2, 3). (f) (3, - 2), (0, 4), (5, 1). Ans. (a) 13, (b) 20, etc..2. Determine the area of the quadrilateral whose vertices are (1, 4), (5, - 2), (0, - 3), (- 2,.0). Ans. 252. 3. Determine the area of the quadrilateral whose vertices are (5, 2), (2, - 5), (- 5, 3), (0, 1). Ans. 29. 12 ANALYTIC GEOMETRY [CHAP. I] EXERCISES* ON CHAPTER I Normal Exercises 1. If the abscissa of a point is 4 on what line will it lie? 2. If the abscissa of a point is equal to its ordinate on what line will it lie? 3. What is the distance of the point (2, 4) from the origin? 4. How far is the point (- 3, - 5) from the point (- 1, 2)? 5. What is the slope of the line joining the point (2, 4) to the origin? 6. What is the slope of the line joining the points (2, - 3) and (- 4, - 2)? 7. Find the coordinates of the middle points of the sides of the triangle whose vertices are the points (0; 0), (- 2, - 4), (- 2, 3). 8. If the line from (-2, 3) to (4, - 1) be extended through the latter point until its length is three times its original length, what are the coordinates of the end point? Ans. (16, - 9). 9. Find the area of the triangle of exercise 7. 10. Find the area of the quadrilateral whose vertices are (- 1, 3), (- 2, 3), (- 3, - 3), and (3, - 4). Ans. 22. General Exercises 11. A straight line joining two points is bisected by the origin. One of the points is (2, 3). What is the other? 12. Determine which of the, following sets of points are points of the same straight line (a) (2, 4), (- 1, 0), (5, 8); (c) (2, 4), (3, 5), (5, 7), (0, 2); (b) (- 1, 2), (0, - 1), (2, 4); (d) (- 1, 1), (2, - 1), (0, 0), (-4, 3). 13. Show that (- 1, 2), (2, - 2), (4, 7), and (9, 5) are the vertices of a trapezoid. * The exercises at the end of each chapter are divided into two groups, " Normal Exercises," and " General Exercises." Those given under the first heading require for their solution only the direct application of the methods or formulas developed in the chapter. Under the second heading are given other exercises of like character together with more varied problems requiring indirect, combined, or extended applications of the chapter's results. CARTESIAN COORDINATES 13 14. Show that (- 1, 2), (3, - 1), (4, 7), and (8, 4) are the vertices of a parallelogram. 15. Show that (2, 4), (7, 3), (-2, - 3), (- 1, 2) are the vertices of an isosceles trapezoid. 16. Determine which of the following sets of points are the vertices of a parallelogram (a) (2, 5), (0, 0), (- 1, - 2), (2, - 4); (b) (1, 2), (7, 5), (3, 3), (5, 4). 17. Show that the middle points of the sides of the quadrilateral whose vertices are the points (1, - 2), (2, 4), (- 1, - 4), and (3, - 5) are the vertices of a parallelogram. 18. Find the coordinates of the fourth vertex of each of the parallelograms three of whose vertices are the points (-1, 2), (- 3, 4), (2, 1). Ans. (0, 3), (-6, 5), (4, - 1). 19. Find the lengths of the sides of the triangle whose vertices are the points (2, 1), (5, 5), and (- 5, 0). Ans. 5, 52, 515. 20. Show that the points (- 1, 3), (- 2, - 4), (6, - 4), (2, - 6), and (7, - 1) are all on one circle whose center is the point (2, - 1). 21. Show that the diagonals of the quadrilateral whose vertices are the points (0, 3), (3, 5), (5, 2), and (2, 0) are equal. Show that this quadrilateral is a rectangle. 22. Find the point on the X-axis equally distant from the points (2, 4) and (-2, 6). Ans. (-, 0). 23. Find the point on the Y-axis equally distant from the points (2, 4) and (6, 5). Ans. (0, 20~). 24. Find the point equally distant from the points (-2, - 4), (6, - 4), and (- 1, 3). Ans. (2, - 1). 25. Show that the points (- 1, 2), (4, - 3), (5, 3) are the vertices of an isosceles triangle. 26. Find the third vertex of one of the equilateral triangles two of whose vertices are the points (1, 2), (4, 6). Ans. ( 2 + 2, 4 - 2 -3). 27. The coordinates of the points A and B are (- 1, 2) and (4, - 2) respectively. P is a point on AB produced, and is twice as far from A as it is from B. Find its coordinates. Ans. (9, - 6). 28. The point (4, 7) is on the line joining (- 1, 2) to the point B, and is one third as far from (- 1, 2) as from B. Find the coordinates of B. Ans. (19, 22) or (- 11, - 8). 14 ANALYTIC GEOMETRY 29. Find the area of the triangles the coordinates of whose vertices are: (a) (0, 0), (0, 4), (4, 0); (b) (- 1, -4), (-5, -5), (-2, -1). 30. Determine two values of x so that the area of the triangle whose vertices are the points (x, 2), (2x, 1), and (4, 5) is one half the area of the triangle whose vertices are the points (x, - 1), (2, 3), and (5, 2). Ans. - -, ~. 31. Show that the area of the triangle whose vertices are (- 3, 5), (3, 7), (5, - 1) is four times the area of the triangle whose vertices are the mid-points of its sides. 32. Find the area of the quadrilateral whose vertices are (4, 1), (- 1, 2), (- 2, - 3), and (1, 3). Ans. 16~. 33. Find the area of the polygon whose vertices are (2, 4), (5, 1), (4, - 3), (1, - 5), (- 2, - 3), (- 2, 4); first by combining the areas of the triangles into which it is divided by diagonals from one vertex, and secondly, by combining the areas of the triangles formed by joining the vertices to the origin. 34. Find the coordinates of the middle points of each of the diagonals of the parallelogram whose vertices are (0, 0), (a, 0), (b, c), and (a + b, c), and thus prove that the diagonals of a parallelogram bisect each other. NOTE. Exercises 35 to 40 inclusive are propositions in elementary plane geometry which are to be proved by the use of coordinates in a manner similar in general to that indicated in exercise 34. 35. Prove that the diagonals of a rectangle are equal to each other. 36. Prove that the mid-point of the hypotenuse of a right triangle is equally distant from each of the vertices. 37. The line joining the mid-points of two sides of a triangle is parallel to the third side and is half the length of the third side. 38. The sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. 39. The lines joining the mid-points of opposite edges of any quadrilateral bisect each other. 40. The medians of a triangle intersect in one point which trisects each of them. CHAPTER II GRAPHICAL REPRESENTATION OF EQUATIONS 9. Variables and constants. I. DEFINITION. A variable is a number whose value may change, arbitrarily, or in accordance with some law. The speed of a train changes as it gathers headway on leaving a station, hence the number which expresses the speed is a variable. A stone thrown into the air changes its distance from the ground from moment to moment, hence the number which expresses this distance is a variable. II. DEFINITION. A number which does not change in value in the course of any discussion is called a constant. In analytic geometry the equations employed contain variables and constants. The constants are either definite numbers, as 3, 8, r, log 2, -2, etc., or they may be represented by letters which stand for quantities whose values are assumed to be known in the problem under discussion. (. Graphical representation of equations. Examples. 1. Let x + y= 2 be an equation in two variables x and y. The equation states in algebraic language that two variable numbers, x and y, are so related to each other that their sum is always 2. Evidently any value may be assigned to either x or y, and then the value of the other is determined. Thus if x = 3, y = -1; if x = 1, y = 1; if x = -4, y = 6; if x =, y = -; etc. If this be done systematically and the results arranged in a table we have, taking at first only integral values: 15 16 ANALYTIC GEOMETRY [CHAP. II] x=-4 -3-2-1 0 1 2 3 y= 6 5 4 32 1 0 -1 -2 Draw a pair of coordinate axes and locate the points which have these corresponding pairs of values of x and y as abscissas and ordinates respectively, thus obtaining the points Pi, P2, * *, Pg, Fig. 14 (a). Through the points thus obtained draw a line, as in Fig. 14 (b). This is found to be a straight line. As lY ~.Pi Pz * P3 QS P3, P4 * * X. PP5 6....PP6 P X 0 *P8 X X' O Q\P X P9 PP (a) (b) FIG. 14 many additional pairs of values of x and y as we please can be found from the equation, and the corresponding points located. Thus x = a, y = I satisfy the equation and give the point Qi; x = 1 + y2, y =1 - -/2 satisfy the equation and give the point Q2; x = - ]7, y = 2 + #7 satisfy the equation and give the point Q3. If it were possible to construct all of the infinitely many points whose coordinates satisfy the equation, we should find that they all lie on the line which has been drawn. This straight line is therefore a graphical representation (called briefly the graph) of the equation x + y = 2. 2. Construct the graph of x2 - 6x - 4y - 5 = 0. [ART. 10] GRAPHS OF EQUATIONS 17 First solve the equation for y, thus y = ( - 6x - 5). Assigning values to x, and computing the corresponding values of y, the results are as follows: x=-2 -1 0 1 2 3 4 5 6 7 8 Y= -- -- - -F 3 - 3 - - 5 1 -point P1 P2 P3 P4 P6 P6 P7 P P9 P10 P11 Plotting the points thus determined it is evident that they do not lie on a straight line as was the case in example 1. If through the points thus located a curve be sketched smoothly the result is as shown in Fig. 15. FIG. 15 3. Construct the graph of y = x3 - 62 + 1lx - 3. x = - 1 0 1 2 3 4 5 y = - 21 - 3 3 3 3 9 27 point * P1 P P3 P4 P5 * * The points (- 1, - 21) and (5, 27) are omitted because the numerical values of y are too large to go on the diagram conveniently. 3 18 ANALYTIC GEOMETRY [CHAP. II] The five points P1, P2, "**, P5 are shown by heavy dots in Fig. 16 (a). From these alone the true shape of the graph is uncertain, and some intermediate points must be found. When y Y op5 PS P2 P3 P4P4 O X 0o X YeI (a) (b) FIG. 16 the additional points Qi(0.7, 2.1), Q 2(2X ), Q3(, ), Q4(, ) are plotted in Fig. 16 (b), and the curve drawn through them all, the result is as shown in the diagram. If still greater accuracy is required, especially in that portion of the curve between P2 and P4, more values of x must be taken between 1 and 3, and a larger scale must be used. Thus: x = 1.2 1.4 1.75 2.25 2.6 2.8 y = 3.29 3.38 3.23 2.77 2.62 2.71 point R1 R2 R3 R4 R5 R6 These with the points P2, P3, P4, Q2, Q3 already found are [ART. 10] GRAPHS OF EQUATIONS 19 plotted on a larger scale in Fig. 17 and the corresponding portion Y T, Ji I 0 t. 0 X FIG. 17 of the curve sketched through them. EXERCISES. 1. Construct the graph of each of the following equations (a) y = 2x + 3, (b) 3y = 4x + 5, (c) x - 2y - 3 = 0, (d) x - 2y + 3 = 0, (e) y = x, (f)Y = — x. 2. Construct the graph of each of the following equations (a) y = 42 + 7, (b) 2 + 4y - 2x - 3 = 0, (c) y2 =x - 2, (d) y = x3, (e) y = - 6x2 + 11x - 6, (f)y =4- x2 + 2. 3. Construct on one diagram the graphs of 3x + 4y - 5 = 0 and 3x + 4y + 2 = O. What relation do these graphs bear to each other? 4. Construct on one diagram the graphs of 4y = x2 and 4y = x2 + 8. In what respect do these two graphs differ? 5. Construct on one diagram the graphs of 4y = x2 and 8y = x2. In what respect do these two graphs differ? Graphical representation of equations, continued. e preceding examples show that the construction of the graph 20 ANALYTIC GEOMETRY [CHAP. II] of an equation in two variables x and y consists fundamentally in finding pairs of corresponding values of the variables which satisfy the equation. Points having these coordinates are then marked on the diagram, and the curve or graph is drawn through them. Care must be taken to use enough values to give the curve completely and correctly. The actual labor involved may often be considerably reduced by a careful examination of the form of the equation before any numerical values are substituted in it. The following additional examples will illustrate some of the ways in which this can be done. 1. Construct the graph of 4x2 + 9y2 = 36. It is often well to begin by examining the equation to see whether the curve passes through the origin, and by finding where it cuts the coordinate axes. In this case the curve does not pass through the origin, because the coordinates (0, 0) do not satisfy the equation. To find where the curve cuts the X-axis, make y = 0 in the equation and solve for x, because at each point where the curve cuts the X-axis the ordinate is zero. Doing this in the equation given above we have x2 = 9, or x = i 3. Hence the curve cuts the X-axis at two points A, A', three units on either side I I Y l I I I I j_ fA_ B l_ ___ X' A' 0 A X -I^Z^Z-~~~ II _ Y'l FIG. 18 [ART. 11] GRAPHS OF EQUATIONS 21 of the origin. By the same reasoning, making x = 0, y = +- 2, and hence the curve cuts the Y-axis at B, B', two units above and below the origin. If the equation be solved for y, y = + ~9 - x (i) from which the following conclusions are drawn: (a) For every value of x there are two equal values of y with opposite signs. Hence the curve is symmetrical with respect to the X-axis. (b) Since x occurs only in the second degree, equal positive and negative values of x lead to the same values of y. Hence the curve is also symmetrical with respect to the Y-axis. (c) If x2 > 9, y is imaginary. Hence the only values of x which make y real lie between - 3 and + 3 and the curve does not extend beyond these limits. If the equation be solved for x, the result is x = if 2 ]4 - y2, (ii! from which, in addition to (a), (b), (c) above, it appears that if y2 > 4, x is imaginary, and hence the curve is included between y = - 2 and y = + 2. To construct the curve some pairs of corresponding values of x and y must be computed. For this purpose either (i) or (ii) may be used. From the former x = - 3 - 2.7 - 2 - 1 0+1 + 2 + 2.7 +3 y = 0 0.87 - 1.49 4 1.89 -2 + 1.89 i1.49 - 0.87 0 and the resulting curve is drawn in Fig. 18. It may be asked what will happen if a value of x greater than 3 or less than - 3 be substituted in the equation. It has been shown that the curve does not extend beyond these limits, yet 22 ANALYTIC GEOMETRY [CHAP. II] it is evident that any value we please may be substituted for either x or y. Let x = 4 in (i), then the resulting values of y are - + 2- 7, a pair of imaginary numbers. Hence the corresponding points cannot be constructed. Nevertheless, since the two pairs of coordinates (4, + 1- 7), (4, - - J- 7) both satisfy the equation they are said to be the coordinates of two imaginary points on the curve. 2. Construct the graph of x2 - y2 = a2. GENERAL DEDUCTIONS. The curve does not pass through the origin; it cuts the X-axis at x = + a, and x = - a; it does not cut the Y-axis, because when x = 0, y is imaginary. Since x and y are involved only in the second power the curve is symmetrical with respect to both axes. Since y = - i2 - a2, all values of x between - a and + a make y imaginary, hence the curve does not exist in the part of the plane between the two lines parallel to Y'Y at a distance a on either side of it. Since x = i A-a2 + y2, all values of y lead to real values of x. The curve therefore consists of two distinct portions, one lying on the positive side of x = a, and the other on the negative side of x = -a. This equation differs in one important respect from those treated up to this point, in that it contains a general constant a2. Moreover the equation x2 - y2 =a2 is homogeneous (See A, I, (c), p. vii) in the three letters x, y, a, and therefore since x and y represent lengths, a is a length also. When an equation contains a general constant, and is homogeneous in x, y and the constant, the best practice in plotting the graph is to assume some arbitrary length for the constant, and then take the values of x or y as multiples of the constant. Thus if x = 2a, y = ~ aV3, and if y = 2a, x = -Ia a5. It is easily seen that with different assumed lengths for the constant the resulting figures will be alike in every respect except size, for the effect [ART. 11] GRAPHS OF EQUATIONS 23 of a change in the length assumed for the constant will be merely to change all linear distances in the figure proportionally. Taking the equation in the form x = d a-a2 + y2, the following table of values is computed. y= 0 ~a a -a 2a 3a etc. x = 4 a -.1.12a 1.41a 4 1.80a ~4 2.24a ~4 3.16a For negative values of y the resulting values of x are the same as for the corresponding positive values of y. The portion of I \ 1 I I Y I l I 1 I ' / A' o A I -/: J= = - 1 ' 1 I I Y'l I I I1 FIG. 19 the curve between y = - 3a, and y = + 3a is drawn in Fig. 19. 3. Plot the graph of 3cx cx2 + 2 This equation like that of example 2 contains a general constant c, but the equation is not homogeneous in the constant and variables. This will perhaps be more clearly seen if the equation be cleared of fractions, cx2y + 2y - 3cx = 0. Hence c is not a length and some numerical value must be assigned to it in order to plot the graph. Let c = -, then the equation becomes 5x2y + 4y - 15x = 0. 24 ANALYTIC GEOMETRY [CHAP. II] Putting x = 0 the only resulting value of y is y = 0, hence the graph passes through the origin and does not have any other intersection with the Y-axis. Putting y = 0, the equation gives x = 0, showing that the graph cuts the X-axis only at the origin. It should be noted, however, that as the numerical value of x increases indefinitely, the value of y approaches zero as a limit. To show this solve the equation for y, thus 15x y '^Tw(i) 5x2 + 4() Then dividing numerator and denominator by x 15 4 = 4 i) 5x + - x From (ii) it is clear that as the numerical value of x increases indefinitely y approaches zero as a limit. Changing the sign of x, but not its value, changes only the sign of y. Hence the graph is symmetrical with respect to the origin. Using (i) the following table of values for x and y is computed. x=-5 -4 -3 -2 - -1 -_ 0 Y =- -j- -T -- -- 9 - - - -0 etc. = - 0.58 - 0.71 - 0.92 - 1.25 - 1.48 - 1.67 - 1.43 0 The resulting graph is drawn in Fig. 20. FIG. 20 [ART. 11] GRAPHS OF EQUATIONS 25 It will be a valuable exercise to assume different values for c in this equation and construct the corresponding curves, noting the changes thus produced. These will be especially conspicuous if both positive and negative values of c are taken. The student will do well, however, not to consider negative values of c until after he has studied example 4, immediately following. 4. Plot the graph of x(x - 2) (x + 2)(x -) Making x = 0, the equation gives y = 0, showing that the graph passes through the origin. Making y = 0, the equation gives x(x - 2) = 0, or x = 0 and x = 2, showing that the graph cuts the X-axis at (2, 0) as well as at the origin. Thus two points 0 and A are located on the curve. For values of x which make one or three of the four factors x, x - 2, x + 2, - 1, negative, y,_, I 1Y_ A _ N _ will be negative, otherwise y will be positive. For all values of x < - 2 these four factors are all negative, hence y is positive. To the left of M'M there- - fore the curve lies above X'X. If -2 < x < 0 (read "x is X' I _ _ZOA -7 1 between -2 and 0") x +2 is positive, but the other three factors are negative, hence y is negative, and the curve lies below X'X between M'M and 11 IT Y'Y. If < x < 1, yis positive, II - iy' because two factors x and x + 2 FIG. 21 26 ANALYTIC GEOMETRY [CHAP. II] are positive and the other two negative, hence between Y'Y and N'N the curve lies above OX. Similarly if 1 < x < 2, y is negative, so that between N'N and A the curve is below OX, and if x > 2, y is positive, and hence to the right of A the curve is above OX. Special attention must be given to values of x in the neighborhood of - 2 and + 1. As x approaches these values, from either side, the corresponding numerical values of y increase without limit, and when x = - 2 or + 1 the denominator of the fraction vanishes, and hence the value of y is undefined. This is customarily expressed under such circumstances by writing y = oo. Without any further information the curve can now be roughly sketched in, but in order to construct it with accuracy the coordinates of a few points must be computed. X =-4- 2 -3 -2 - - + + +2k +3 +4 y V= M M _- _ _f i _ A ^ ^ =2.4 3.75 6.43 - 4.2 - 1.5 — 0.56 0.6 - 0.43 0. 0.3 0.44 _- - L- -Y- - - 1IY FIG. 21 FIn. 22 I-I [ART. 13] GRAPHS OF EQUATIONS 27 The part of the curve between these limits is drawn in Fig. 21. 5. Let the student show that the curve in Fig. 22 is the graph of x(x - 2) (x +2)(x - 1) 1. The fundamental principle of analytic geometry. From the examples discussed in Arts. 10, 11, it is clear that the graph of an equation has the nature of a locus. It is in fact the locus of points whose coordinates satisfy the equation. Hence the expressions graph of an equation, and locus of an equation will be used interchangeably in this book. DEFINITION. The graph or locus of an equation in two variables is that straight line or curve upon which lies every point whose coordinates are pairs of values of the variables which satisfy the equation; and the coordinates of every point on which satisfy the equation when substituted for the variables. This definition is the foundation upon which the whole structure of analytic geometry is built. The measure of the student's apprehension of the principle here set forth will be the measure of his success in mastering the subject. 13. Practical Suggestions.-In following the discussion of the examples in Arts. 10, 11, the student will have seen that the work does not consist exclusively in determining definite corresponding pairs of values of x and y, but that it is desirable by careful examination of the equation to learn as much as possible about the general nature of the graph before undertaking the systematic computation of pairs of values of x and y. Attention was directed in some of the examples to definite methods of procedure by which this can be done. These consist in determining whether the locus passes through the origin, where it cuts the axes, whether it has symmetry with respect to either axis, or with respect to the origin, the range of 28 ANALYTIC GEOMETRY [CHAP. II] values of one variable (if any) which make the other variable imaginary, and the values of one variable (if any) which make the other infinite. Before undertaking to substitute numerical values for x and y in the equation it is usually best to solve the equation for one of the variables, choosing the one for which this can be done more easily. It will sometimes happen that the equation cannot be solved for either variable. For example, x3 + y3 - 3xy = 0. Values of x or y may then be substituted in the equation as it stands, and the resulting equation solved for the other variable. From what has been said it may be concluded that no general rules of procedure can be formulated which apply to all, or even to a majority of cases. The student must depend to a great extent upon his own ingenuity, and precisely for this reason he will find the plotting of loci, or curve tracing as it is called, a valuable exercise, and one which will abundantly repay considerable effort to secure some mastery of it. EXERCISES. 1. Construct the graph of each of the following equations (a) xy = 4, (b) 4x2 + y2 = 16, (c) 4x - y2 = -36, (d) X2 + y2 = 25a2, (e) ay2 = x3, (f) xy = c + ax + y, (g) y3 + x3 = 8, (h) y4 + X4 = 16, (i) X3 + y3 - 4x + 4 = 0, a3 a4 X2 - 4x W = + ' (Ick) y -= + (1) y2 = Cj) Y Z2t~a2x -]-4a' x - i 2. Construct on one diagram the graphs of the following equations, and note the differences in the graphs due to the differences in sign in the equations x2 + y2 = 25, x2 - y2 = 25, - x2 + y2 = 25. 3. Construct on one diagram the graphs of the following equations, and note the differences in the graphs due to differences in the values of the numerical coefficients x2 + y2 = 36, x2 + 4y2 = 36, 4x2 + y2 = 36, x2 + 4y2 = 25. 14. Intersection of loci.-Since the graph of an equation is the locus of points whose coordinates satisfy the equation, it [ART. 15] GRAPHS OF EQUATIONS 29 follows that, if two loci are plotted on the same figure, the coordinates of the points where they intersect must satisfy both equations. Hence RULE. To determine the coordinates of the points of intersection of two loci solve their equations simultaneously for x and y. EXAMPLES. 1. Find the intersection of the loci of 3x - 7y + 2 = 0 and x + 2y - 6 = 0. Solving the equations for x and y the results are x = A, y = 2T, the coordinates of the required point. Construct the loci and verify the results from the figure. 2. Determine the intersections of the loci of x2 + y2 = 25 and 3x + 4y = 25. From the second equation y = ~(25- 3x), and substituting this value in the first the result is X2 + a (25 - 3x)2 = 25, which reduces to x2 - 6x + 9 = 0, or (x - 3)2 = 0, a quadratic with two equal roots, x = 3. Substituting x = 3 in x2 + y2 = 25 gives y = -t 4, and making the same substitution in 3x + 4y = 25 gives y = 4 only. The latter is therefore the only value of y which satisfies both equations when x = 3. Each of the two coincident values x = 3 leads therefore to the same value of y, viz., y = 4, and the loci are said to meet in two coincident points (3, 4). Draw the figure and show that the line 3x + 4y = 25 is tangent to the 'orcle x2 + y2 = 25 at (3, 4). ~5 Solution of equations by intersection of loci.-The approximate solution of a pair of simultaneous equations in two variables can be obtained graphically from the loci of the equations. This method is useful when the algebraic solution cannot be conveniently performed. For example, let it be required to solve for x and y the equations x2 + y2 - 7x + 2y - 3 = 0 and xy + 2x - 4y - 2 = 0. 30 ANALYTIC GEOMETRY [CHAP. II] The loci of these two equations are drawn in Fig. 23. The points of intersection are P1 and P2, whose coordinates are found by measurement to be ap_Y_ ____ _ proximately (2.8, 2.9) and (- 0.4, P___ _ ~ ----- - 0.6). From the principles of al_r7 _I' __gebra it is evident that there are X _ ~i two other pairs of values of x and y,.=_ _ _ __ __ _ but since the loci intersect visibly _ ___ _ __i -- at P1 and P2 only, the other solu_ _t- \-_ _ -I _ tionare imaginary. _ /_ _g e eCtry1 -Imaginaries in analytic [__ __geometry.- In several of the FIo. 23 preceding discussions of this Chapter reference has been made to imaginary values of x or y. See Ex. 1, p. 21, Ex. 2, p. 22, and the end -of Art. 15. These serve to show that imaginary points can exist on real loci, which is only a way of stating in geometric language that in general an equation in two variables can be satisfied by imaginary values of the variables. There are also equations whose loci are wholly imaginary, or which contain at most only a limited number of real points. DEFINITION. An equation which is satisfied by only imaginary values of the variables, or by at most a finite number of real values, is called the equation of an imaginary locus. Three simple types of equations fall under this classification. (a) Equations one or more of whose coefficients are imaginary numbers. For example y2 - (2 + i)x + (3 - 2i)y + 2 = 0. [i = - 1 Putting any real or imaginary value for one variable in this equation and solving for the other variable, the result will in general be imaginary. If, however, the equation be written in the form y2 - 2x + 3y + 2 - (x + 2y)i = 0 [ART. 16] GRAPHS OF EQUATIONS 31 it will evidently be satisfied for values of x and y which make y2-2x + 3y + 2 = 0, and x + 2y = O simultaneously.* Solving these two equations therefore, for x and y, we find the two pairs of real values x = 7 + 41, y = -7- 41, and x = 7- 41, y= --- + 41, both of which satisfy the original equation, and therefore determine two real points on the imaginary locus. Moreover, these are the only pairs of real values of x and y which can satisfy the given equation. (b) Equations all of whose coefficients are real, but which are not satisfied by any real values of the variables. Equations of this type consist of, or can be reduced to an even power of a variable term, or the sum of two or more such even powers, equated to a negative constant. For example (x - 2)2 + y2 = - 3. As the square of a real number is always positive it is clear that an equation of this type cannot be satisfied by any real values of x and y. (c) Equations all of whose coefficients are real, but which can be satisfied by only a limited number of pairs of real values of x and y. Equations of this type can be reduced to a sum of even powers of variable terms equated to zero. For example x2 + y2 = 0 is satisfied by the single pair of real values x = 0, y = 0. Similarly (x2 - y2)2 + (2x - y + 4)2 = 0 is satisfied by those values of x and y which make x2 - y2 = 0 and 2x - y + 4 = 0 simultaneously. Solving these two equations for x and y, we find x = -4, y = -4, and x = -, y = 4, which determine the only two real points on the locus. * See IV (c) p. viii. 32 ANALYTIC. GEOMETRY [CHAP. II] EXERCISES. 1. Find from the equations the coordinates of the points of intersection of the graphs of each of the following pairs of equations. Check the results graphically. (a) x - 2y + 5 = 0, x2 + y2 =25; Ans. (3, 4), (-5, 0). (b) 2y2 = x, 4x2 + 9y2 = 25; Ans. (2; i- 1). (c) 16y - 15x + 99 = 0, x2 -4y2 = 16; 2. Show by finding the coordinates of their points of intersection that the graphs of the following pairs of equations are tangent to each other. Check graphically. (a) x2 + y2 = 25, 3x + 4y -25 = 0; (b) 4x2 + 9y2 = 25, 8x + 9y + 25 = 0; (c) xy = 4, x + y - 4 = 0; 3. Find approximate values graphically for the coordinates of the points of intersection of the graphs of the following pairs of equations (a) 2 - 4y2 =4, y 4 (c2 -) y = ) 8y = X3. 4 + X2; x2_ 4, (b) x2 + y2 = 25, y2 -3 = 0; 4. Find the coordinates of the real points in each of the graphs whose equations are (a) y2 + (3 + i)x + (7i - 4)y - 3 - 23i = 0; Ans. (2, 3), (- 131, 22). (b) (2 + i)x + (3i- 4)y + 10 - 5i = 0; Ans. (- 1, 2). (C) (1 + i)X2 + (1 -- i)y2 - 13 + 5i = 0. Ans. (4 2, ~- 3). 5. Show that there is but one real point on each of the graphs whose equations are (a) (3 + q- 4)x + (5 — 1 - 4)y - 6 + 19 1 —i = 0; (b) x2 + i(y2 + x2) + 2ixy - 4x + 4 = 0; (c) x2 + y2 - 4x + 6y + 13 = 0. 6. Show that there is no real point on the graph of any of the foilowing equations (a) 2x2 + y2 + ix - 2(2 + i)y + 40 - 4i = 0; (b) X2 + y2 + 4 = 0; (c) x2 + 4y2-2x + 5 = 0. [ART. 18] GRAPHS OF EQUATIONS 33 7. Show graphically that the roots of the following pairs of equations are imaginary (a) x2 + y2 = 4, x + y = 4; (b) y2 - 4x +4 = 0, y2 + 4x +4 = 0; /j Functions.-The notation f(x, y) is often used to denote in general an expression containing the variables x and y. It is read: " function of x and y," or " f of x and y." Hence f(x, y) = 0 is a short way of representing in algebraic language any equation in the two variables x and y all the terms of which have been transposed to the first member. In a specific problem or discussion f(x, y) may stand for an expression of definite form, and when so used it represents the same form of expression throughout the discussion. Thus in one discussion we may have f(x, y) = X2 + y2 _ a2, in another f(x, y) -3x + 2y - 6, etc. If f(x, y) - x2 - 3y2 + 6x - 4, then in the same discussion f(a, b) means the expression obtained by substituting a for x, and b for y, hence f(a, b) = a2 - 3b2 + 6a - 4. Similarly f(2, 3) = 22 - 3(3)2 + 6(2) - 4 = 4- 27 + 12 - 4 = - 15, f(1, I) - 12 - 3(0)2 + 6(1) -4 = 3, f(O, 0) = - 4, etc. 8) Composite loci. THEOREM. If the expression f(x, y) is factorable, the locus of the equation f(x, y) = 0 consists of as many distinct lines and curves as there are variable factors of f(x, y). Let f(x, y) consist of three factors, f1(x, y), f2(x, y), f3(x, y). Then the equation f(x, y) = 0 can be written in the form fl(x, y)'f2(x, y)' f3(, y) = 0. (i) Any values of x and y which satisfy the equation fl(x, y) = o will also satisfy equation (i), and similarly with reference to f2(x, y) = 0 and f3(x, y) = 0. Hence all points on the three 4 34 ANALYTIC GEOMETRY [CHAP. II] separate loci of fl(x, y) = 0, f2(x, y) = 0, f3(x, y) = 0 will also be points on the locus of f(x, y) = 0. The same argument holds no matter what number of factors f(x, y) may have. The theorem is therefore proved. If f(x, y) has equal factors, the corresponding component of the complete locus is counted as many times as the factor occurs. DEFINITION. A locus which is thus made up of separate parts is called a composite locus. r19) Equations containing only one variable.-Equations wch contain only one variable explicitly, for example 2x - 3 = 0, may always be interpreted as containing the other variable also with coefficient zero. For all values of y in the equation 2x - 3 = 0, or 2x + Oy - 3 = 0, x has the same value, x =. Hence the locus consists of a straight line parallel to the Y-axis and 1~ units on the positive side, because this line contains all points having the abscissa 3, and any ordinate whatsoever. Similarly y2 - 5y - 6 = 0, or (y - 6)(y + 1) = 0 is the equation of the two lines parallel to the X-axis, one six units above and the other one unit below this axis. EXERCISES. 1. Construct the complete locus of each of the following equations (a) x2 = y2, (d) X4 - y4 = 0, (b) x2 - 4xy + 4y2 - 4 = 0, (e) xy - 3x + 5y - 15 = 0, (c) X + xy2 - 4x = O, (f ) x3 - 6x2 +ll - 6 = 0. 2. Show that the locus of the equation Ax2 + Bx + C = 0 is a pair of parallel lines, a pair of coincident lines, or a pair of imaginary lines, according as B2 - 4AC is positive, zero, or negative. 4. What is the form of the equation of a line parallel to the X-axis? the Y-axis? 5. What is the equation of the X-axis? the Y-axis? 6. Find for each of the following sets of equations a single equation [ART. 20] GRAPHS OF EQUATIONS 35 whose locus will be the combination of the loci of the separate equations. (a) x = y, x = -y. Ans. x2 - y2 = 0. (b) x = 0, x = 4, x = 6. Ans. X3-10X2 + 24x = 0. (c) x = y, x2 + 2 = 4, (d) x2 + y2 = 8, xy + 4 = 0. 20. Determination of the equation of a given locus.So far in this chapter we have discussed the construction of the loci of given equations. The inverse problem of finding the equation from the locus, or from its definition, is a vital part of the subject. EXAMPLES. 1. A point moves in a plane so as to be always equally distant from the points (2, 6), (4, - 2). Find the equation of the locus of the moving point. \ B Let P1 = (2, 6), and P2= \ (4, - 2), then we know from elementary geometry that the required locus is AB, the per- o \, X pendicular bisector of P1P2. To determine the equation of AB take any representative point P(x, y) on it. Then by (3), p. 6, the distance PPi = (x - 2)2 + (y - 6)2, and PP2 = A(x - 4)2 + (y + 2)2. By the conditions stated these are equal, hence -/(x - 2)2 (y6)2 = (x - 4)2 + (y + 2)2 (i) which is the required equation. Equation (i) may be simplified. Thus squaring and expanding, it becomes X2 - 4x + 4 + y2 - 12y + 36 = x2 - 8x + 16 + y2 + 4y + 4, which reduces to x - 4y + 5 = 0. 36 ANALYTIC GEOMETRY [CHAP. II] 3. A point moves in a plane so that the sum of the squares of its distances from two fixed points in the plane is constant. What locus will it describe? The problem is stated without reference to any particular lines as axes of coordinates. Any two lines in the plane may be used for this purpose. The form of the resulting equation will depend upon the choice of axes, but the geometric properties expressed by the equation will be the same in all cases. P(x,y^) It is best to choose the axes -,oT ~ so that the resulting equation /,, \\ \ will be as simple as possible. -By" D A \ ~In this problem take the X (-c, ) Q (c, ) X line joining the two given points as X-axis, with the origin half way between them. Hence let A(c, 0) and B(- c, ~Yf -0) be the two given points, FIG. 25 and P(x, y) any point on the required locus. Then since it is given that AP2 + BP2 = k2, where k2 is a constant, we have by (3) p. 6 (x - C)2 + y2 + (x + c)2 + y2 = k2, which reduces to x2 + y2 = k2 - c2. If this be plotted for definite values of k2 and c it will be found that the locus is a circle with center half way between the given points. EXERCISES. In each of the following exercises find the equation of the locus of a point which moves in a plane according to the conditions stated: 1. So as to be always equally distant from the origin and the point (2, 4). Ans. x + 2y = 5. [ART. 20] GRAPHS OF EQUATIONS 37 2. So that the sum of its distances from (- 2, 0) and (2, 0) is always equal to 6. Ans. 5x2 + 9y2 = 45. 3. So that its distance from the origin is always equal to the rlope of the line joining it to the origin. Ans. y2(1 - x2) = x4. 4. So that its distance from a fixed point is twice its distance from a fixed line. 5. So that its distance from the point (2, 4) is twice its distance from the point (- 1, 4). Ans. x2 + y2 + 4x - 8y + 16 = 0. EXERCISES ON CHAPTER II Normal Exercises 1. Construct the graph of each of the following equations (a) x2 - y2 = 4a2, (d) y = ax+ 4, (b) y2 = x(x2 - 9), (e) x2y2 -4 = 0, x(x - 4) (f) x2 - 6x + 5 = 0, (c) y= - (=) y=4. e2 - 4 ) (g) y = 4. 2. Find from the equations the coordinates of the points of intersection of the graphs of each of the following pairs of equations (a) x2 + y2 = 10, x2 + y2 - 1Ox = 0; Ans. (1, i- 3). (b) 2 - y2 = 0, y + 3 - 6 = 0. Ans. (3, -3), (, ). 3. Find from their graphs approximate values of the roots of the following pairs of simultaneous equations (a) y(x2 + 5) = 8, y = X4 - 42 + 2; (b) X2y + xy2 = 4, 8x2 - y3 = 0. 4. Determine for each of the following pairs of equations the coordinates of the points of intersection of their graphs and hence show that the graphs are tangent to each other (a) y2 = 4x, 2x + 4y + 8 = 0; (b) 2 -y2 = 4, 2 + y2 + 2x - 8 = 0. 5. Find the coordinates of the real points on the graphs of each of the following equations (a) (2 + 3i)x + (4 - i)y - 6i = 0, Ans. (12, ). (b) (1 + i)x + (2 - 3i)y - 6 - i = 0, Ans. (4, 1). (c) x2 + 2y2 - 4x - 12y + 22 = 0, Ans. (2, 3). (d) 4x2 + y2 + 4y + 4 = 0. Ans. (0, - 2). 38 ANALYTIC GEOMETRY [CHAP. II] 6. Show that there are no real points on the graph of any of the following equations (a) x2 + y2 + 4 = 0, (b) 2x2 + 3y2 + 4x - 12y + 16 = 0, (c) (4 + i)x2 + 3y2 + (3 + 4i)x + 4 + 5i = 0, (d) 2 + 2y2 + ix- 3iy + 6 - 7i = 0. 7. A point moves in a plane so that the ratio of its distances from the points (2, 4) and (- 3, 7) is always as 1 to 3. Find the equation of its locus. Ans. 4x2 + 4y2 - 21x - 29y + 61 = 0. 8. Find the equation of the locus of a point which moves in a plane so that the slope of the line joining it to the point (2, - 4) is always equal to the slope of the line joining it to the point (3, - 5). Ans. x + y + 2 = 0. General Exercises 9. Construct the graphs of the following equations (a) x = 2, (f) x2 - 4y2 = 36, (k) x3 - y = 0, (b) x = 2y, (g) x2 -4y = 36, (1) yx = a2, (c) x = 2y + 2, (h) x2 - 4y2 = 0, (m) yx2 = a3, (d) x2 + y2 = 36, (i) 2 - 4 = O, (n) y2(x2 -4) = x4. (e) x2 + 4y2 = 36, (j) y2(x2- 3x + 2) = x(x + 4), 10. Construct on one diagram the graphs of the following equations x + y = 2, x2 + y2 = 4, x3 + y3 = 8, 4 + y4 = 16. 11. Construct on one diagram the graphs of the following equations y =4, y + 2x = 4, y + x = 4, y + 2x = 4. 12. Construct on one diagram the graphs of the equation y = x + b, for b equal to 0, 2, 4, and - 2 respectively. In what respect do these graphs differ from each other? 13. Construct on one diagram the graphs of the equation xyn = 1, for n equal to - 2, - 1, - ~, 0, ~, 1, and 2 respectively. What is common to all these graphs? What properties are common to the graphs with negative values of n? What for positive values of n? 14. Construct the graph of the equation y2 = (x - a)(x - l)(x - 4) (i) for a = - 1, (ii) for a = 0, (iii) for a = + 1. 15. Find the value of A so that there will be only one real point on the graph of the equation x2 - 4x + 4y2 - 16y + A = 0. Ans. A = 20. GRAPHS OF EQUATIONS 39 16. For what range of values of A will the graph of the equation of exercise 15 be entirely imaginary? Ans. A > 20. 17. Find the coordinates of the real points of the graph of each of the following equations (a) x2 + y2-4x + 4 = 0, Ans. (2, 0). (b) (1 + i)x + (1 - i)y = 10, Ans. (5, 5). (c) (1 + 2i)x2 + (8i - 4)xy + (4 + 2i)y2 = 0, Ans. (0, 0). 18. Determine which of the graphs of the following equations are tangent to each other 5x -4y = 9, x2 - y2 = 9, 2 + y2 = 9, 2 + y2 - 12x + 10y + 20 = 0. 19. A point moves so that the difference of its distances from the points (2, - 4) and (2, 4) is always equal to 6. Determine the equation of its locus, and construct the locus from the equation. Ans. 9x2 - 7y2 - 36x + 99 = 0. 20. A point moves so that the slope of the line joining it to the point (0, 2) is twice the slope of the line joining it to the point (0, - 2). Determine the equation of its locus, and construct the locus from the equation. Ans. y = - 6. 21. A = (2, 4) and B = (- 1, 3) are fixed points. The variable point P moves so that the area of the triangle ABP is always equal to 6. Find the equation of its locus, and construct the ltcus from the equation. Ans. x - 3y - 2 = 0, and x - 3y- 10 = 0. 22. A point moves so that the slope of the line joining it to the point (2, 4) is always equal to the square of its distance from the Y-axis. Find the equation of its locus and construct the locus from the equation. Ans. y = x- 2x2 + 4. 23. Let A = (0, - 2), and let AP, any line through A, cut the X-axis at M. If P moves so that MP always equals 3, find the equation of the locus of P, and construct the locus. Ans. 2y2 + (y2 - 9)(y + 2)2 = 0. 24. A point moves so that its distance from the origin is always equal to four times the cosine of the angle which the line joining it to the origin makes with the positive extension of the X-axis. Find the equation of its locus, and construct the locus. Ans. x2 + 2 - 4x = 0. CHAPTER III THE STRAIGHT LINE 21. Conditions which determine a straight line.-In plane geometry a straight line is said to be determined when its position in the plane is definitely fixed. Elementary geometry teaches that a straight line is thus determined by two conditions, properly chosen. For example, a straight line is determined when two points on it are given. In this chapter several standard forms of the equation of the straight line, determined by prescribed sets of conditions, will be derived. 22. Problem.- To derive the equation of a straight line A\ in terms of the coordinates of p(o t) two given points on the line. \N. ~ Let AB be the straight P1(:Y,/1) line determined by the two P2(,,y,) points P1(xi, y1), P2(x2, y2),...._ and let P(x, y) be a repreX X sentative point on AB. Since P, P1, P2 are all on the same FIG. 26 line AB, the slopes of PP1 and of P1P2 are equal. Hence by (4), p. 7, the required equation is - 1 1 - y (1) X -X X1 -X2 which may also be written in the form Y - Y - (X - X1). (2) 4 X 40 [ART. 23] THE STRAIGHT LINE 41 Either (1) or (2) is known as the two-point form of the equation of the straight line. If one of the given points, say (xl, y1), is the origin (0, 0), equation (2) takes the form y = 2 x(3) X2 which is the equation of the straight line through the origin and another given point. The student familiar with the determinant notation can verify the fact that equation (1) can be written in the form Xi Y1 1 X2 y2 1 = 0. (4) x y 1 23. Problem.-To derive the equation of a straight line in terms of the slope and the coordinates of a given point on the line. In equation (2), Art. 22, the factor yl- 2 on the right is XI -X2 the slope m, see (4), p. 7. Hence we may write y - y1 = m(x - xi), (5) which is the required equation. This is known as the slope and one-point form of the equation of the straight line. If the given point is the origin, xl = 0, y1 = 0, and hence (5) becomes y = mx, (6) which is the equation of the line through the origin having the slope m. Compare this result with (3) and show by a figure that they are consistent. DEFINITION. The distances from the origin to the points where a line cuts the coordinate axes are called the intercepts. The x-intercept, OA, Fig. 27, is designated by a, and the y-intercept, OB, by b. The signs of the intercepts are positive 42 ANALYTIC GEOMETRY [CHAP. III] or negative according to the location of the intersections A and B with respect to the origin. In the figure the intercept a = OA is negative, and the interY cept b = OB is positive. 24. Problem.- To derive the / B equation of a straight line in terms of the slope and y-intercept. ___/ _Having given the y-intercept, X' /A O X' OB, Fig. 27, is equivalent to having given the point (0, b) on the FIG. 27 line, hence the result can be obtained from (5) by substituting x1 = 0, yj = b in that equation. This gives y - b = mx, or y mx + b. (7) This is known as the slope and y-intercept form of the equation of the straight line. 25. Problem.-To derive the equation of a straight line in terms of its intercepts. The coordinates of A and B, Fig. 27, where the line cuts the X- and Y-axes are respectively (a, 0) and (0, b). Substituting these for (x1, yi) and (X2, Y2) respectively in equation (1) we have y -b x -a a which reduces to bx + ay = ab, or dividing by ab + = 1. (8) a b This is known as the intercept form of the equation of the straight line. 26. Problem.-To derive the equation of the straight line in [ART. 26] THE STRAIGHT LINE 43 terms of the length of the perpendicular from the origin to the line and the angle which this perpendicular makes with OX. The length OD of the perpendicular from the origin to any line AB is designated by p, and the angle XOD by a. The length /B p is taken as positive for all posi- D tions of AB, and a is always measured around from OX to OD, so that, according to the position XA 0 X of AB, a may have any value from o0 to 360~. FG To derive the equation of AB in terms of p and a we shall use equation (5), Art. 23, because both the slope m, and the coordinates of D, a point on the line, can easily be expressed in terms of p and c. Thus m = tan OAD = cot DOA = -cot a, (i) and if D = (x1, yi) then for all positions of AB xi = p cos a, y1 = p sin a. (ii) Substituting the values from (i) and (ii) in (5), we have y - p sin a = - cot a(x - p cos a). Multiplying by sin a this becomes y sin a - p sin2 a = - cos a(x - p cos a), = - x cos a + p cos2 a. x cos a + y sin - p(sin2 a + cos2 a) = 0 or xCO a + ysina - p = 0, (9) which is the required equation. This is known as the normal form of the equation of the straight line. 44 ANALYTIC GEOMETRY [CHAP. III] 27. Recapitulation.-For convenience of reference the principal forms of the equation of the straight line are here repeated. Two-point form y - yl = (x - Xi). (2) X1 - X2 Slope and one-point form y - y1 = m(x - xi). Slope and y-intercept form y = mx + b. (5) (7) Intercept form x +y a = 1. a b (8) (9) Normal form x cos a + y sin a - p = 0. EXERCISES. In each part of exercises 1 to 4 first construct the line determined by the given quantities. Then write its equation, using the appropriate formula from those given above. Finally reconstruct the line from the equation, and see whether it coincides with the figure first drawn. 1. (a) Two points (2, 3), (-5, - 1). Ans. 4x - 7y + 13 = 0. (b) a = -3, b = 2. (c) m = -2, b = 1. (d) m = 2, one point (2, - 5). (e) p =3, a = 60~. 2. (a) m = - 4, one point (- 2, - 1). (b) p = 3, a = 210~. (c) m = —, b = -2. Ans. x + 2y + 4 = 0. (d) Two points (- 6, 2), (3, - 4). (e) a = 2, b = 3. 3. (a) a = 6, b = 5. (b) m = - 1, one point (3, 0). (c) p 3, a = 135~. Ans. x-y + 3f2 = 0. (d) Two points (0, 0), (- 4, 7). (e) m =, b = - 3. 4. (a) m = 1, b = 3. (b) Two points (5, 3), (2, - 3). (c) p = 3, a = 315~. (d) m = 5, one point (0, 0). (e) a = -3, b = -5. Ans. (b) 5x + 3y + 15 = 0. 5. Construct the line for which p = 2, and a = 0~. What are the values of m, a and b for this line? Ans. m= oo, a = 2, b oo. 6. Construct the line for which p = 2, and a = 180~. What are the values of m, a and b for this line? Ans. m = oo, a = - 2, b = oo. 7. Construct the line for which p = 42, and a = 225~. What are the values of m, a and b for this line? Ans. m = - 1, a = b = -2. 8. Derive the slope and y-intercept form of the equation of a straight line directly from a figure. [ART. 29] THE STRAIGHT LINE 45 9. Derive the intercept form of the equation of a straight line directly from a figure. 28. The parameters of a straight line. DEFINITION. The constants m, a, b, p, a used in equations (5), (7), (8) and (9) are called parameters or determining constants of the straight line. For every straight line these parameters have definite values. The following equations express some of the relations connecting them. I. m = - cot a. For proof see p. 43, equation (i). b II. m = -. This is proved by Fig. 27, because a OB b AO a since OB = b, and AO = - OA = - a. a III. tan a =. This follows from I and II. These relations between the parameters serve to emphasize the fact, otherwise evident, that the five parameters of a straight line are not independent quantities. In fact arbitrary values can never be assigned to more than two of them. 29. The straight line referred to oblique axes. Let AB, Fig. 29, be a straight line determined by the points Pl(x1, yl), P2(x2, y2) referred to oblique aies, and let P be a representative point on AB. Draw MP, M1Pi, M2P2 parallel to OY, and P1K, P2K1 parallel to OX. Then the triangles P1KP and P2K1P1 are similar, and KP KIP, -- = --- (i) PlK P2K1i But KP = y - y1, P1K = x - x, (ii) KiP1 = y - Y2, P2K1 = xi - x2. (iii) 46 ANALYTIC GEOMETRY [CHAP. III] Hence, substituting from (ii) and (iii) in (i), y yl Y - Y2 X - Xi Xl - X2 from which it is seen that the two-point form of the equation of the straight line is the same whether the axes are oblique or P2 M2 M1 M X FIG. 29 rectangular. The same therefore is true of the intercept form, which is only a special case of the other. In other words equations (1), (2), (3), (4), Art. 22, and (8), Art. 25, can be used with both rectangular and oblique axes. The same is not true with regard to those forms of the equation of the straight line which involve the slope. Let o be the angle between the axes and 0 the angle which AB makes with OX. Then from the triangle P1KP KP sin KP1P P1K sin P1PK' where KP1P = 6, and P1PK = c -0, (v) hence, substituting from (ii) and (v) in (iv), Y - y_ sin 0 x - x sin (co -0)' [ART. 30] THE STRAIGHT LINE 47 or sin 0 ( Y - 0(x - xX1) (10) sin (w - 0J ) which is the form of the equation corresponding to (5) when the axes are oblique. Similarly y = nx + b, where n (11) sin (w - o)' will be the form corresponding to (7). 30. Two fundamental theorems. I. To every straight line corresponds an equation of the first degree in two variables. This has been proved in the foregoing discussion. Each form of the equation of the straight line has been found to be of the first degree in two variables. II. Conversely, every equation of the first degree in two variables is the equation of a straight line. Let Ax + By + C = 0 be any equation of the first degree in two variables, where A, B, C are constants which can have any finite value, including zero, except that A and B cannot both be zero. If B + 0 the equation can be transformed into A C y = -x B by transposing and dividing by B. It is now in the form y = mx + b, hence it is the equation of the straight line whose slope m is - (A/B), and whose y-intercept b is - (C/B). If B = 0, the equation takes the form Ax + C = 0, or x = - (C/A), which is the equation of a straight line parallel to the y-axis. See Art. 19. 31. Reduction of the general equation Ax + By - C. = 0 to the forms (7), (8), (9). 48 ANALYTIC GEOMETRY [CHAP. III] It has been shown immediately above how to reduce Ax + By + C = 0 to the form (7), y = mx + b. Similarly to reduce the same equation to the form (8), x y + = 1, a b transpose the absolute term to the right, and divide by - C. Ax By This gives - + =1. -C -C Then dividing numerator and denominator of the two fractions by A and B respectively, x Y +, = C C A B which is in the required form. EXAMPLE. The equation 3x - 2y + 8 = 0 when solved for y becomes y = -x + 4. It is now in the form y = mx + b, so that m =,, b = 4. To reduce the same equation to the form (8) we have successively 3x - 2y = -8, - x + -y = 1, and + =1, so that a = -, b = 4. THEOREM. To reduce the equation Ax + By + C = 0 to the form (9), x cos a + y sin a - p = 0, divide by + 4-A2 + B2 if C is negative, or by - A2 + B2 if C is positive. Trigonometry teaches that two numbers are the sine and cosine of an angle when, and only when, the sum of their squares is unity. Hence to reduce Ax + By + C = 0 to the form x cos a + y sin a - p = 0 it must be divided by a number such that the sum of the squares of the resulting coefficients of x and [ART. 31] THE STRAIGHT LINE 49 y shall be unity. Let k be this number. Then A B C kx + Y+ = (i) where k is to be chosen so that (A)2 (B)2 1. (ii) C ) - 1. Solving (ii) for k it is found that k = i~ -A2 + B2. (iii) Hence, substituting from (iii) in (i), A B C x- + y + - 0, (iv): V/A2 + B2 A2 + B2: tA2 + B2 which is in the form x cos a + y sin a - p = 0, provided the sign of the radical AA2 -B2 is chosen so as to make ~ /IA2+B2 a negative number. That is, if C is positive, JA2 + B2 must be taken with the negative sign, and vice versa. Thus the theorem is proved. The student should make careful note of this theorem, as the normal form of the equation of the straight line is the form most conveniently used in solving an important class of problems. EXERCISES. 1. Construct the locus of each of the following equations. Reduce each equation in succession to the forms of equations (7), (8), (9). Determine the values of m, a, b, p, a, and verify the results so obtained by reference to the figure. (a) 2x - y - 5 = 0, (f) 3y - 4x - 5 = 0, (b) 3x +4y- 15 = 0, (g) y +2x +7 = 0, (c) x - 2y + 6 = 0, (h) 5y - 5x + 12 = 0, (d) 5x + 12y + 13 = 0, (i) 5y- 12x- 13 = 0, (e) 6x - 5y + 2 = 0, (j) y + 3x + 2 = 0. x _ 2x - y - 5 Ans. (a) y = 2x - 5, + = 1, =0, m =2, a =, b = -5, p= -5, sina = -}5, cos = -5. 5 50 ANALYTIC GEOMETRY [CHAP. III] 2. Show that y = m(x - a) is the equation of a straight line in terms of the parameters m and a. 3. The vertices of a triangle are (2, 7), (5, - 3), (- 3, 2), find the equations of the three sides. Ans. 10x + 3y - 41 = 0, x - y + 0 5 0 x + 8y - 1 = 0. 4. The equations of the three sides of a triangle are 7x - 4y - 1 = 0, 8x + 3y -39 = 0, x + 7y + 15 = 0, find the coordinates of the vertices, and the equations of the medians. Ans. (3, 5), (- 1, -2), (6, -3), 15x - y - 40 = 0, 6x- 1ly - 16 = 0, 9x + 10y - 24 = 0. 5. What series of straight lines is represented by equation (8) if a remains fixed in value, and b takes all possible values? 6. If in equation (9) p remains constant and a takes all possible values what series of lines will be produced? 7. What relation will hold between the coefficients A, B, C of the equation Ax + By + C = 0, if (a) the x- and y-intercepts are equal? Ans. A = B. (b) the line makes an angle of 45~ with OX? Ans. A = - B. (c) the line is tangent to the circle of radius r with its center at the origin? Ans. C2 = r2(A2 + B2). (d) the line passes through the origin? (e) the line passes through the point (- 1, 1)? (f) the line makes an angle of 135~ with OX? (g) the x-intercept = 2p? Y B /D 32. The angle between two straight lines. DEFINITION. The angle which a straight line makes with a second line is the angle through which the sec/ '\2 / \V ond line must be turned, in O6 / Q - - ---- X the positive direction of rotaC/ ttion, to bring it into coinciA dence with the first line. FIG. 30 According to the definition [ART. 33] THE STRAIGHT LINE 51 the angle which AB makes with CD is the angle DPB (or the angle CPA), but the angle which CD makes with AB is the angle BPC (or APD). THEOREM. If ) be the angle which line AB, with slope mi, makes with line CD, with slope m2, then tan P = lq- m2 (12) 1 + mlm2 Let y = m1x + bl be the equation of AB, and y = m2x + b2 that of CD, and let the angles which these lines make with X'X be 01 and 02 respectively. Then ml = tan 01, and m2 = tan 02. From the figure f = CPA = 01 - 02, hence tan, - tan 2 mi -m2 tan ) = tan (01 - 02) = - = tan0 1 + tan 01 tan 2 1 + mimr2 which was to be proved. 33. Theorems relating to parallel lines. I. If two straight lines are parallel their slopes are equal. II. The equation of the straight line parallel to a given line y = mx + b, and passing through a given point (xi, yl) is y - yi = m(x - xi). (13) Compare with (5), Art. 23. III. The lines Ax + By + C = 0, and A'x + B'y + C' = 0 are parallel if A/B = A'/B'. The slopes of these two lines are respectively - (A/B) and - (A'/B'), see Art. 30. The lines being parallel these are equal. It follows that two equations which differ only in their absolute terms represent parallel lines, for example Ax + By + C = 0, and Ax + By + D = O. IV. The equation of the straight line parallel to Ax + By + C = 0, and passing through (xi, yl) may also be written in the form A. - Xi)- ~ - _= \ 1. A X ~ 'A(x - xi) +:(y - yl) = U. t14) 52 ANALYTIC GEOMETRY [CHAP. III] Equation (14) represents a line parallel to Ax + By + C = 0 because the coefficients of x and y in the two equations are the same, and it passes through the point (xl, yi) because the equation is satisfied when these values are substituted for x and y. 34. Theorems relating to perpendicular lines. I. If two straight lines are perpendicular to each other, the slope of one is the negative reciprocal of the slope of the other. If two lines are perpendicular the angle 0 in equation (12), p. 51, is 90~, hence tan o = oo, and therefore the denominator of the fraction m- m2 is zero. That is 1 + mlm2 = 0, and 1 + mlm2 hence 1 1 m2 = -, and ml = - (15) mif 1Um2 From this it follows that: II. The equation of the straight line perpendicular to y = mx + b, and passing through (xl, yl) is y - y = ( - ( - x1). (16) III. The two lines Ax + By + C = 0 and A'x + B'y + C' = 0 are perpendicular if AA' + BB' = 0. The slopes of these two lines are - (A/B) and -(A'IB'), respectively, see Art. 30, hence if they are perpendicular we have by (15) - (A/B) = (B'/A'), or AA' + BB' = 0. (17) It follows from (17) that two equations Ax + By + C = 0, Bx - Ay + D = 0, in which the coefficients of x and y are interchanged and the sign of one of them is changed, always represent perpendicular lines. IV. The equation of the line perpendicular to Ax + By + C = 0 and passing through (xi, yl) is B(x - xi) - A(y - yi) = 0. (18) The formal proof is left to the student. [ART. 35] THE STRAIGHT LINE 53 EXERCISES. 1. Determine the tangent of the angle between each of the following pairs of lines. (a) 3x - y + 16 = 0, x + 2y - 7 = 0. Ans. ~ 7. (b) y-x = 0, 7x + 12y + 2 = 0. Ans. - 1-9. (c) x + 3y + 6 = 0, 5x + 2y + 1 =0. 2. (a) Find the equation of the line which passes through the point (1, 2) and is parallel to 3x - y - 7 = 0. Ans. 3x - y - 1 = 0. (b) Also through (0, 0) parallel to 7x + 3y + 8 = 0. Ans. 7x + 3y = 0. (c) Also through (7, - 1) parallel to 3x - y - 6 = 0. 3. With the data of Ex. 2, find the equation of the straight line which passes through each given point perpendicular to the given line. Ans. (a) x + 3y - 7 = 0, (b) 3x - 7y = 0, (c) x + 3y - 4 = 0. 35. Perpendicular distance to a given point from a given line. THEOREM I. The perpendicular distance to the point (x', y') from the line x cos a + y sin a - p = 0 is numerically equal to the expression x' cos a + y' sin a - p. Let P(x', y') be the given point, and AB the given line, x cos a + y sin a- p = 0. Then OL = p, and XOL = a. (i) Draw MN through 0 parallel to AB. There are two cases. I. When the point P and the line AB are both on the same side of MN, as in Fig. 31. Through P draw CD parallel to AB. (x', y') FIG. 31 54 ANALYTIC GEOMETRY [CHAP. III] The parameters of the equation of CD in the normal form are a = XOK, the same as for AB, and OK = p'. Hence the equation of CD in the normal form is x cos a + y sina - p' = 0. (i) Since CD passes through the point P the coordinates of P will satisfy equation (ii), hence x' cos a + y' sin a - p' = 0, or p' = x' cos a + y' sin a, (iii) hence p' is known, since x', y', and a are known. The length of the required perpendicular is QP, and from Fig. 31 QP = LK = LO + OK = OK - OL, hence from (i) and (iii) QP = p' - p = x' cos a + y' sin a - p. (iv) II. When the point P and the line AB are on opposite sides of MN, as in Fig. 32. As before draw CD through P parallel to AB, and LOK through 0 perpendicular to CD. Then QP = LO + OK. (v) Y L M FIG. 32 [ART. 35] THE STRAIGHT LINE 55 Denote the length of OK by p'. The figure shows that the length of QP is numerically equal to the sum of the lengths LO and OK, and therefore since from (i) LO = - p, OK must also be taken with the negative sign, hence from (v) QP = - p - '. (vi) Since angle XOK = 180~ + a, where a = XOL, the parameters of CD are 180~ + a and p', and the equation of CD is x cos (180~ + a) + y sin (180~ + a) - p' = 0, or - x cos a - y sin a - p' = 0. (vii) The point P is on CD, therefore its coordinates (x', y') will satisfy (vii), and hence - ' cos a - y' sin a - p' = 0, or p = - (x' cos a + y' sin a). (viii) Finally from (vi) and (viii) QP = x' cos a + y' sin a -p, In every case therefore the numerical value of the length d of the perpendicular to a point (x', y') from the line x cos a + y sin a - p = 0 is d = x' COs a + y' sina - p. (19) SIGN OF THE PERPENDICULAR. From (iv) under Case I it is seen that QP is + or - according as p' > p, or p' < p. Therefore QP is positive when P is on the opposite side of AB from the origin, negative when P is in the part of the plane between AB and MN, in other words when the origin and P are on the same side of AB. When MN is between P and AB, QP is always negative, as appears from (vi). In this case the origin and P are always on the same side of AB. In other words: 56 ANALYTIC GEOMETRY [CHAP. III] THEOREM II. The perpendicular d from the line AB to the point P is positive when P and the origin are on opposite sides of AB, and negative when P and the origin are on the same side of AB. THEOREM III. The perpendicular distance to the point (x', y') from the line Ax + By + C = 0 is Ax' + By' + C - IA2 + B2 where the sign of the radical is determined as in Art. 31. The equation of a line in the form Ax + By + C = 0 is reduced to the normal form by dividing by i IA2 + B2, where the radical is to be given the sign opposite to that of C, as shown in Art. 31. When this division has been performed, we have from (19) d = Ax' + By' + C (20) d=A B. (20) - /A2 + B2 EXAMPLE. What is the perpendicular distance from the line 3x -4y + 10 = 0 to the point (2, - 1)? By formula (20) the result is 3(2) -4(- 1) + 10 6 +4 + 10 _ -5 - 54. -5 -5 Note that we divide by - /A2 +B2 = - 9 + 16 = - 25, because C = + 10; also the result being - 4, the distance from the given line to the given point is 4 units, and the point is on the same side of the line as the origin. THEOREM IV. Ax + By + C = 0 being any straight line, (x1, yi), (x2, y2), any two points not on the line, will lie on the same side or on opposite sides of the line according as the two numbers Ax1 + Byl + C and Ax2 + By2 + C have like or unlike signs. The proof is left to the student. EXAMPLES. 1. Are the points (6, 2), (- 5, 1) on the same or opposite sides of the line 3x - 2y + 7 = 0? [ART. 36] THE STRAIGHT LINE 57 Substituting the coordinates of the given points in the left-hand side of the given equation the results are 3(6) - 2(2) + 7 = 21, 3(- 5) - 2(1) + 7 = - 10, hence the points are on opposite sides of the line. (Theorem IV.) In the exercises which follow draw the figure in every case. 2. Find the length of the perpendicular from (a) 3x - 4y - 7 = to (2, 3), (c) x - 5y - 6 = to (-3, 0), (b) 12x + 5y + 8 = 0 to (1, - 4), (d) 2x - 3y + 7 - 0 to (0, 4). Ans. (a) - —, (b) 0, (d) + q5-13. 3. In each of- the following cases determine whether the given points are on the same or opposite sides of the given line. (a) (2, 3), (8, - 6), 3x - 4y - 7 = 0. (b) (1, - 2), (-2, 1), 7x- 5y + 6 = 0. (c) (5, 1), (-2,3), x + 2y -9 = 0. 4. Find the lengths of the three altitudes of the triangle whose vertices are (6, 2), (3, - 5), (- 1, 7). Ans. 2 158, 417 P, ~i-0. 36. Lines through the intersection of two given lines. THEOREM. The equation Ax + By + C + k(A'x + B'y + C') = 0, (21) where k is any constant, is the equation of a straight line passing through the intersection of the two lines Ax + By + C = 0 and A'x + B'y + C' = 0. Equation (21) is the equation of a straight line because it is of the first degree in the variables x and y. Let (x', y') be the point of intersection of the two given lines, then their equations will both be satisfied for x = x' and y = y'. Hence Ax' + By' + C - 0 and A'x' + B'y' + C' - O. Therefore equation (21) is satisfied for x = x' and y = y', and hence is the equation of a line which passes through (x', y'). The number k in equation (21) is called the undetermined parameter of the line. By assigning the proper value to k, the line can be made to satisfy one other condition besides passing through the intersection of Ax + By + C = 0 and A'x + B'y + C' = 0. The exercises worked below illustrate this. 58 ANALYTIC GEOMETRY [CHAP. III] EXERCISES. 1. Determine the equation of the line which joins the origin to the point of intersection of the two lines x - 5y - 6 = 0 and 7x + 2y + 3 = 0. From (21) the equation of any line through the intersection of the two given lines will be x-5y - 6 + k(7x + 2y + 3) = 0. (i) Since this line is to pass through the origin (0, 0) the values x = 0, y = 0 must satisfy its equation. Substituting in (i) - 6 + 3k = 0, or k = 2. This value of k in (i) gives 15x - y = 0 for the required equation. 2. Find the equation of the line passing through the intersection of 2x - 5y - 8 = 0 and 3x + 2y - 9 = 0, and perpendicular to 5x - 3y + 12 = 0. By (21) the required equation has the form 2x - 5y - 8 + k(3x + 2y - 9) = 0, or (3k + 2)x + (2k - 5)y - 9k - 8 = O. (ii) The slope of the line whose equation is (ii) is - [(3k + 2)/(2k - 5)], and the slope of 5x - 3y + 12 = 0 is 3. Hence by (15), p. 52, if the two lines are to be perpendicular 3k + 2 3 -2k -5 1 - from which k = - ~Substituting this value of k in (ii) the result is 57x + 95y - 153 = 0. 3. Determine the equations of the lines passing through the intersection of 7x + 2y - 9 = 0, x - 3y + 2 = 0, and also (a) passing through the point (2, 3), Ans. 2x - y - 1 = 0. (b) having the slope 2, Ans. 2x - 3y + 1 = 0. (c) parallel to 2x + 5y + 6 = 0, (d) perpendicular to 2x + 5y + 6 = 0. 37. Imaginary points and lines. I. DEFINITION. Two imaginary points whose coordinates differ only in the signs of the imaginary parts are called conjugate imaginary points. [ART. 37] THE STRAIGHT LINE 59 For example, (2 + 5i, 3 + Si) and (2 - 5i, 3 - 8i), are a pair of conjugate imaginary points. II. THEOREM. If an imaginary point lies on a real line Ax + By + C = 0, then the conjugate imaginary point lies on the same line. If (a + bi, c + di) is a point on Ax + By + C = 0, then A(a + bi) + B(c + di) + C = 0, or Aa + Be + C + i(Ab + Bd) = O, (i) hence Aa + Bc + C = 0, and Ab + Bd = O. (ii) See IV (c), p. viii. Then from (ii) it must also follow that Aa + Bc + C - i(Ab + Bd) = 0, or A(a - bi) + B(c - di) + C = 0. Hence the point (a - bi, c - di), conjugate to (a + bi, c + di), also satisfies the equation Ax + By + C = 0. III. THEOREM. On every imaginary line there is one and only one real point. Let (pi + qli)x + (p2 + q2i)y + p3 + q3i = 0 be the equation of an imaginary line. This equation may be written in the form pix + p2y + p3 + i(qlx + q2y + q3) = 0. (iii) In general there will be one and only one pair of values of x and y which satisfy simultaneously the two equations plx + p2y + p3 = O, qlx + q2y + q3 = O, (iv) and these values are always real. They will also satisfy (iii), hence they are the coordinates of the only real point on the 60 ANALYTIC GEOMETRY [CHAP. III] imaginary line (iii). This point is the intersection of the two real lines whose equations are given in (iv). SPECIAL CASES. If in (iii) p = p2= k and P- * k ql q2 q3 the two corresponding equations (iv) will not be satisfied by finite values of x and y. In that case the real point on the imaginary line (iii) is said to be at infinity, and the real lines (iv) are parallel. IfP P2P k, If p = p2= p3= k, q1 q2 q3 then pi = kq1, p2 = kq2, p3 = kq3. These values substituted in (iii) give k(qlx + q2y + q3) + i(qix + q2y + q3) = 0, from which, dividing by k + i, we have qlx + q2y + q3 = 0, so that in this case the line (iii) is not an imaginary line. IV. DEFINITION. Imaginary lines whose equations differ only in the signs of their imaginary terms are called conjugate imaginary lines. For example (pI - qli)x + (p2 - q2i)y +.p3 - q3i = 0 (v) is the conjugate of the line whose equation is (iii). The coordinates of the real point which satisfy (iii) will also obviously satisfy (v), hence: V. THEOREM. A pair of conjugate imaginary lines meet in a real point. By III above, this is the only real point on either of them. 38. Higher equations which represent straight lines. [ART. 38] THE STRAIGHT LINE 61 It was shown in Art. 18 that if the equation f(x, y) = 0 can be expressed in the form fl(x, y)'f2(x, y)'f3(x, y). =, (i) the locus corresponding to f(x, y) = 0 consists of all the separate loci corresponding to fl(x, y) = 0, f2(x, y) = 0, f3(x, y) = 0, etc. (ii) Hence if f(x, y) = 0 is of the nth degree in x and y, and the equations (ii) are all of the first degree, the equation f(x, y) = 0 will represent n straight lines. These n lines will not necessarily be all distinct nor all real for some of the factors in (ii) may be equal to each other, and some may be imaginary. For example the equation (x - 3y + 2)2(X2 - y2)(x2 + y2) = 0 corresponds to two coincident real lines x - 3y + 2 = 0, two other real lines x + y = 0, x - y = 0, and the conjugate imaginary lines x + iy = 0, x - iy = 0. EXERCISES ON CHAPTER III Normal Exercises 1. Find the equation of the straight line determined by each of the following sets of quantities (a) Two points (-2, 4), (3, - 7), (d) a = -3, b = 5, (b) m = -, b = -3, (e) p = 2, a =30. (c) m = - 2, one point (- 2, 5), 2. Find the equation of the straight line determined by each of the following sets of quantities, the angle between the axes being 60~ (a) Two points (- 2, 4), (3, - 7), (c) 0 = 30, b = 4, (b) 0 = 30~, one point (- 2, 5), (d) a = - 3, b = 5. Ans. (b) x - y + 7 = 0. 3. Find the parameters a, b, p, m, and a of the following lines (a) 2x + 4y +7 = 0, (c) + y = 0, (b) 5x - 6y = 10, (d) y = 2. 62 ANALYTIC GEOMETRY [CHAP. III] 4. Find the angle which line (a) of Exercise 3 makes with each of lines (b), (c), and (d). 5. Find the equation of the two lines which pass through the point of intersection of the lines 3x - 4y - 18 = 0 and 2x + y - 1 = 0 and which are respectively parallel and perpendicular to the line x + 3y - 9 = 0, (a) by using the coordinates of the point of intersection, (b) without finding the point of intersection. 6. Find the distance of each of the points (- 2, 0), (0, 4), (2, 1), (- 2, - 4), (2, 4), from the line 3x - 4y- 8 = 0. Which of these points are above the given line? Which below? Which are on the same side of the line as the origin? 7. Find the coordinates of a pair of conjugate imaginary points lying on the line 4x - y + 2 = 0. 8. Find the coordinates of the real point which lies on the imaginary line (4 + i)x - (3 - 2i)y - (20 - 6i) = 0. What is the equation of the conjugate imaginary line? What are the coordinates of the point of intersection of these two imaginary lines? 9. Construct the loci of the following equations, and find the equation of each separate part of each locus (a) x2 - y2 - 4x + 4 = 0, (b) x2 - 4y2 = 0, (c) x4- y4 = 0. General Exercises 10. Given the vertices (4, 3), (2, - 2), (- 3, 4) of a triangle, find the equations of the sides and of the medians, and the coordinates of the point of meeting of the latter. This point is called the centroid of the triangle. Ans. Centroid = (1, ~). 11. Determine b so that the line y = mx + b shall pass through the point (xl, yl), and hence derive equation (5), p. 41. 12. Given the vertices (1, 9), (11, - 3), (- 3, - 1) of a triangle, find the equations of the perpendiculars from each vertex to the opposite side. Find also the lengths of these perpendiculars, and the coordinates of their point of intersection. This point is called the orthocenter of the triangle. Ans. Orthocenter = (- 37, 3). 13. Given kx + 3y + 8 = 0, where k is an undetermined parameter. Determine k so that the line shall (a) pass through (- 2, 3), (b) be parallel to OX, (c) have slope 2. Ans. (a) k = -. 14. Given 7x + 2y + k = 0, where k is an undetermined parameter. THE STRAIGHT LINE 63 Determine k so that the line shall (a) pass through point (3, - 1), (b) have a = b + 3, (c) have p = 2. Ans. (b) k = 42-. 15. Given the vertices (5, 3), (- 3, 1), (2, - 6) of a triangle, find the equations of the perpendicular bisectors of the three sides; and the coordinates of their point of intersection. This point is called the circumcenter of the triangle. Ans. Circumcenter = (19, - h). 16. Given the two lines 3x - 5y + 6 = 0, 2x + y - 9 = 0. Find the equation of the line which passes through the point of intersection of these lines (without finding the coordinates of this point), and which (a) passes through the point (- 6, 1). (c) is parallel to OY. (b) is perpendicular to 5x + 2y - 7 = 0. (f) has p = 3. (c) is parallel to 3x + y - 17 = 0. (g) has b = 4. (d) has its x-intercept twice its y-intercept. (h) has a = 90~. 17. The three sides of a triangle are 5x - 3y + 22 = 0, 6x + 5y - 8 = 0, x + 8y + 13 = 0, find the equations of the perpendiculars from the vertices to the opposite sides, and the coordinates of the orthocenter (see Ex. 12), without finding the coordinates of the vertices of the triangle. Ans. Orthocenter = (- 1, 2j). 18. Show that the equations of the bisectors of the two pairs of vertical angles formed by the lines Ax + By + C = 0, A'x + B'y + C' = 0 are Ax +By +C A'x +B'y +C' = 0, A2 +B2 A,2 + B'2 and prove that these two bisectors are perpendicular to each other. 19. The three sides of a triangle are 3x + 4y - 10 = 0, 8x - 6y + 3 = 0, 12x + 5y + 15 = 0. Find the equations of the bisectors of the three angles of the triangle, and the center of the inscribed circle. Ans. (- s, sl). 20. Prove that the medians of any triangle meet in one point. [Use oblique axes, taking two sides of the triangle as axes of coordinates.] 21. Prove that the perpendicular bisectors of the sides of a triangle meet in one point. [Take the base and the perpendicular to it at one extremity as axes of coordinates.] 22. Prove that the perpendiculars let fall from the vertices of a triangle to the opposite sides meet in one point. [Take the axes as in Ex. 21.] 64 ANALYTIC GEOMETRY 23. Find the locus of the point whose distance from 12x - 5y - 16 = 0 is twice its distance fr6m 3x + 4y - 5 = 0. Ans. 138x + 79y - 210 = 0, 18x + 129y - 50 = 0. 24. Find the locus of the point whose distance from Ax + By + C = 0 is k times its distance from A'x + B'y + C' = 0. 25. Determine how many real points lie on each of the following lines and where possible find their coordinates: (a) (4 - 2i)x + (3i - 3)y - 2 + 4i = 0, (b) (4 - 8i)x + (6i - 3)y - 2 = 0, (c) x + iy = 1, (d) (3 - 9i)x - (6i - 2)y + 5 - 15i = 0. 26. Prove that 5x2 + 13xy - 6y2 = 0 is the equation of two straight lines intersecting at the origin. See III, p. viii. 27. Prove that Ax2 + Hxy + By2 = 0 in which A, B, and H are real constants represents a pair of straight lines passing through the origin, and that these two lines are real and distinct, real and coincident, or imaginary, according as H2 - 4AB is positive, zero, or negative. See III, p. viii. CHAPTER IV THE CIRCLE 39. Equation of the circle in rectangular coordinates. DEFINITION. A circle is the locus of a point which moves in a plane so as always to be at a constant distance from a fixed point of the plane. The fixed point is called the center of the circle, and the constant distance is called the.P, y) radius. PROBLEM. To find the equation C L of the circle with radius r, and center (d e) at the point (d, e). Let C(d, e) be the center, and, P(x, y) any point on the circle. 0 N M X Then from the definition CP= r. By (3), p. 6, the distance CP is FIG. 33 (x - d)2 + (y - e)2. Hence (x - d)2 + (y - e)2 = r or (x - d)2 + (y - e)2 = r, (1) which is the required equation. By inserting the appropriate values of d, e and r, equation (1) will give the equation of any circle. Conversely any equation in x and y which is reducible to the form (1) is the equation of a circle in rectangular coordinates. (a) The equation of the circle, the origin being at the center. See Fig. 34. In this case d = e = 0, and hence equation (1) becomes x2 + y2 =. (2) 6 65 66 ANALYTIC GEOMETRY [CHAP. IV] (b) The equation of the circle through the origin whose center is on the positive portion of the X-axis. See Fig. 35. Here we have d = r, and e = 0, hence equation (1) becomes (x - r)2 + y2 = r2 or, y2 = 2rx - x2. (3) y 0' M 0 C M A X FIG. 34 FIG. 35 40. The Equation Ax2 + Ay2 + Gx + Fy + C = 0. I. THEOREM. The equation of a circle can always be put in the form Ax2 + Ay2 + Gx + Fy + C = 0, where A + 0. Expanding equation (1), Art. 39, it takes the form x2 + y2 - 2dx - 2ey + d2 + e2 - r2 = 0, or 2 + y2 + Lx + My + = 0, (4) where L = - 2d, M -2e, N = d2 + e2 - r2. If equation (4) be multiplied by any constant A, and if in the result we put AL = G, AM = F, and AN = C, the equation takes the form Ax2 + Ay2 + Gx + Fy + C = 0. (5) Note that equation (4) is just as general as equation (5), because, since A + 0, the latter can always be reduced to the former by dividing by A. [ART. 40] THE CIRCLE 67 II. THEOREM. Equation (4) always represents a real circle when N < 1(L2 + M2). Equation (4) can be put in the form (x2 + Lx + 1L2) + (y2 + My + 1M2) = 4(L2 + M2) - N, or (x + ~L)2 + (yQ + M)2 = (L2 + M2) - N. (i) This equation is in the form (1), Art. 39, if -(L2 + M2) -N > 0, or N < -(L2 + M2), and it then represents a circle with center at the point whose coordinates are d = - 2L, e = — Ml and radius (6) r = a/-(L2 + M2) -. J III. POINT CIRCLE, IMAGINARY CIRCLE. (a) If N =!(L2 + M2), r = 0, and equation (i) takes the form (x + L)2 + (y + M)2 = 0, or (x-d)2 + (y- e)2 = 0. (ii) The most obvious interpretation of this equation is that of a circle of zero radius, the point circle (- L, - M) or (d, e), since these are the only real coordinates which satisfy the equation.* (b) If N > 4(L2 + M2), r2 is negative, and equation (i) is that of an imaginary locus with no real points. See Art. 16 (b). The locus is called an imaginary circle. The equation x2 + y2 + Lx + My + N = 0 therefore represents a real circle, a point circle, or an imaginary circle according as N R(L2 + M2). Hence: IV. THEOREM. The general equation of the second degree in x and y, Ax2 + Hxy + By2 + Gx + Fy + C = 0, always repre* Since the equation (x - d)2 + (y - e)2 = 0 may be written in the form [(x - d) + i(y - e)][(x - d) - i(y - e)] = 0 another interpretation is that the equation represents a pair of imaginary lines meeting at the real point (d, e). See Arts. 16 (c), 37 V, and 38. 68 ANALYTIC GEOMETRY [CHAP. IV] sents a real circle, a point circle, or an imaginary circle, when H = 0, and A = B: 0. When in practice it becomes necessary to examine an equation of the form (4) or (5) to determine the nature of the locus represented, it is best actually to reduce it to the form (1). The following example illustrates the process. EXAMPLE. Determine whether the equation 3x2 + 3y2 - 18x + 12y - 69 = 0 represents a real circle, and if so determine its center and radius. Dividing the given equation by 3 x2 + y2- 6x + 4y -23 = 0, which is in the form (4). Adding 9 and 4 to both sides in order to complete the squares of the terms in x and y, x2- 6x + 9 + y2 + 4y + 4 = 36, or (x - 3)2 + (y + 2)2 = 36, which is in the form (1). By comparison d = 3, e = - 2, r = 6. In other words the given equation is that of a real circle whose center is at the point (3, - 2), and whose radius is 6 units. EXERCISES. 1. Write the equation of each of the following circles, and construct the curve. (a) Center at (3, 4), radius 6. (b) Center at (- 2, - 5), radius 4. (c) Center at (4, - 2), radius 7. 2. The radius of a circle is 5. Write its equation, and construct the circle for each of the following positions of the center, (3, 4), (- 3, 4), (0, 0), (0, 5), (0, - 5), (5, 5). 3. Determine the special forms assumed by equations (1), p. 65, and (4), p. 66, for circles whose positions are defined as follows. (a) Passing through (0, 0), with center on the negative portion of the X-axis. Ans. y2 = - 2rx - x2. (b) Passing through (0, 0), with center on the Y-axis. (c) Tangent to the X-axis, and center above the X-axis. Ans. x2 + y2 - 2dx - 2ry + d2 = 0. (d) Tangent to both axes, with center in the second quadrant. [ART. 41] THE CIRCLE 69 4. Determine which of the following equations represent real circles. Find the center and radius of each of these and construct the curve. (a) 2 + y2-4x + 6y - 3 = 0. (c) x2 + y2+ 3x + 7y + 15 = 0. (b) X2 + y2- 6x - 2y + 10 = 0. (d) 92 + 9y2 + 12x.- 6y - 31 = 0. Ans. (a) d = 2, e =- 3, r = 4. 5. Determine the equations of the following circles, and construct the curves. (a) Passing through the point (3, 4), center at the origin. (b) Passing through the origin, center at (3, 4). (c) Passing through (2, 4), center at (- 1, 0). Ans. (b) 2 + y2 - 6x - 8y = 0, (c) 2 + y2 + 2x - 24 = 0, 41. The circle through three given points.-Three points not in the same straight line determine one and only one circle. The process of finding the equation of the circle will be illustrated by an example. Find the equation of the circle passing through the three points (5, 7), (2, - 1), (- 3, 3). Either equation (1) or equation (4) may be used. The latter is usually the more convenient and will be used here. The process consists in finding the values of L, M, N for which equation (4) will be satisfied when x and y have each of the three given pairs of values. Hence substituting in turn these values for x and y in (4), three simultaneous equations in L, M, and N are obtained: 74 + 5L + 7M + N = 0, 5+2L- M + N = 0, 18 - 3L + 3M + N = 0. Solving these for L, M, and N, the results are L = -4, M = -, N = - and substituting these in (4) x2 + y2 - x_ 96 75 = 0 x2 + y2 TY - T= 70 ANALYTIC GEOMETRY [CHAP. IV] or 13x2 + 13y2 - 43x - 96y - 75 = 0, which is the equation of the required circle. Let the student verify the result by substituting the -coordinates of the given points in this equation. The general solution of this problem is most conveniently performed with the aid of determinants. Let (xi, yl), (x2, y2), (x3, y3) be three given points not in the same straight line. Substituting these in turn for x and y in the equation x2 + y2 + Lx + My + N = 0, we have X12 + Y12 + Lx + Myl + N = 0, X22 + y22 + Lx2 + My2 + N = 0, X32 + y32 + Lx3 + My3 + N = 0. If the last three of these equations be solved for L, M, N, and the results so obtained be substituted in the first equation, the result will be the equation of the required circle. This is the process which was used in the example above. By determinants the same result is obtained by expressing the fact that all four equations are satisfied simultaneously by the same values of the three quantities L, M, N, which gives 2 + y2 x y 1 xi2 + y12. y 0. (7) =0. (7) X2 + y22 X2 Y2 1 32 + y32 x3 Y3 1 42. Distance from a point to a circle. THEOREM I. If P(x1, y1) is any point in the plane of the circle (x - d)2 + (y - e)2 = r2, outside of the circle, the distance from P to the circle, measured on one of the two tangents through P, is ((xi - d)2 + (yl - e)2 - r2. (8) [ART. 42] THE CIRCLE 71 In Fig. 36 let C = (d, e), and P = (xi, yl), then CP2 = (xi - d)2 + (yi - e)2. The triangle PCN is right- p angled at N, and CN =r,-/ hence I ---- ' PN -= CP2 - CN2 = (xl - d)2 + (y - e)2- r2. C Similarly if the equation of the circle is o /x x2 + y2 + Lx + My + N =O the distance from P to the FIG. 36 Fro. 36 circle is 1i2 + yi2 + Lxi + Myi + N. (9) Let the student prove this and also the following theorem. THEOREM II. The point (xi, yl) is outside, on, or inside the circle x2 + y2 + Lx + My + N = 0 according as the expression X12 + y12 + Lxl + My1 + N is positive, Zero, or negative. EXERCISES. 1. Find the equations of the circles determined by the following sets of points. Construct each circle from its equation, and check the result by noting whether the circle passes through the given points. (a) (5, 7), (2, 8), (7, 3). (c) (1, 5), (9, - 7), (- 4, 6). (b) (0; 0), (3, 1), (- 2, - 4). (d) (0, 0), (0, 6), (3, - 3). Ans. (a) x2 + y2 - 4x - 6y - 12 = 0. 2. Given the circle 2 + y2 + 4x - 6y + 4 = 0, and the points (0, 0), (- 2, 1), (1, - 3), (- 1, 3), (1, 3), (- 4, 2). Determine which of the points are inside and which are outside the circle. Check the results graphically. 3. For each point of Exercise 2 which is outside the given circle, find the length of the tangents from the point to the circle. 72 ANALYTIC GEOMETRY [CHAP. IV] 43. Equation of a circle through the points of intersection of two circles.-Let X2 + y2 + Lx + -Mly + N1 =, (i) x2 + y2+ L2 + M2y + N2 = (ii) be the equations of two circles. Form a third equation by adding one of these to k times the other, where k is any constant. The result is (x2 + y2 + L1x + M1y + N1) + k(x2 + y2 + L2x + M2y + N2) = 0, or (1 + k)x2 + (1 + k)y2 + (Li + kL2)x + (Ml + kM2)y + (N1 + k2) = 0. For all values of k, except k = - 1, this equation is in the form (5), p. 66, and hence Y. represents a circle. If k = - 1, (10) is an equaXi/ x i\f ^tion of the first degree, and c;. / | \ therefore represents a straight line. ( C2 THEOREM. For all values -- (xz2,Y2) / of k, equation (10) is the equation of a locus (circle 0 o x or straight line) which passes through the intersecFIG. 37 tions of the circles given by equations (i) and (ii). Let Ci, C2 be the centers of the circles represented by equations (i) and (ii), and P,(xi, yl), P2(x2, Y2) their points of intersection. Considering the point P1, we have from (i) and (ii) xl2 + yL2 + L1x1 + M1y1 + N1 0, xl2 + y12 + L2x1 + M2y1 + N2 0, [ART. 43] THE CIRCLE 73 and hence the coordinates of P1 will satisfy equation (10). In other words the locus represented by this equation passes through P1. Similarly it passes through P2. If the circles represented by equations (i) and (ii) meet in imaginary points (that is if they do not visibly intersect), equation (10) will still represent a locus passing through their imaginary points of intersection, for the argument above holds good whether P1 and P2 have real or imaginary coordinates. EXERCISES ON CHAPTER IV Normal Exercises 1. Write the equations of the circles with radius 4, whose centers are respectively (- 3, - 2), (0, - 5), (4, 0), (0, 0), and construct each circle from its equation. 2. Determine which of the following equations represent real circles, and find the center and radius of each of these. (a) 2 + y2 -2x + 4y - 4 = 0, (c) x2 + y2 + 2x + 4y + 10 = 0, (b) 4x2 + 4y2 = 25, (d) 36x2 + 36y2 - 48x + 108y + 97 = 0. 3. Find the equations of the circles passing through the following sets of points. (a) (2, 2), (3, 1), (-5, -5), (b) (2, 1), (3, -2), (-4, 0). Ans. (a) x2 + y2 + 2x + 4y - 20 = 0. 4. Find the length of the tangents from each of the points (0, 1), (2, 3), (4, 2), (- 3, - 1) to the circle x2 + y2 - 4x + 8y - 5 = 0. Ans. 2, 26, etc. 5. Determine which of the points (0, 0), (1, - 1), (- 2, 3), (- 1, 3), (0, 5) are inside the circle 4x2 + 4y2 + 12x - 20y + 25 = 0. 6. Write the equation which represents all circles passing through the points of intersection of the circles x2 + y2 - 2x - 2y - 23 = 0, and x2 + y2 - 16x - 9y + 68 = 0. 7. Determine the equation of the circle which passes through the intersections of the two circles of Exercise 6, and also through the origin. Ans. 13x2 + 13y2 - 72x - 49y = 0. General Exercises 8. Find the equations of the circles determined by the following sets of conditions. 74 ANALYTIC GEOMETRY [CHAP. IV] (a) Center at the intersection of x + 3y + 7 = 0 and 2y - 3x + 12 = 0 and passing through (- 1, 1). Ans. x2 + y2 - 4x + 6y - 12 = 0. (b) Center at (1, 3), and tangent to 2x + y + 5 = 0. (c) Center on x + y - 5 = 0, and tangent to both coordinate axes. Ans. 4x2 + 4y2 - 20x - 20y + 25 = 0. (d) Circumscribed about the triangle whose sides have the equations x+y-4 = 0, 4x - y -6 =0, 2x-y-8 = 0. (e) Inscribed in the triangle whose vertices are (0, 0), (0, 6), (8, 0). Ans. x2 + y2 - 4x - 4y + 4 = 0. (f) Having the line from (2, - 3) to (- 4, 7) as a diameter. 9. What system of circles is represented by each of the following equations. (a) X2 + y2 + Lx + 2y = 0, (c) 4x2 + 4y2 + 4Lx + L2 - 16 = 0, (b) X2 + y2 + Lx + Ly = 0, (d) X2 + y2- 2rx - 2ry + r2 = 0. Ans. (a) Circles passing through the origin, and having their centers on y + 1 = 0. 10. Of all circles which pass through the intersections of x2 + y2 + 6x + 8y- 144 = 0 and 2 +y2- 10x - 8y + 16 = 0, find the equation of the one which (a) passes through the origin, (b) has its center on the X-axis, (c) has its center on the line x - 2y - 2 = 0. Ans. (a) 5x2 + 5y2 - 42x - 32y = 0, (b) x2 + y2- 2x - 64 = 0. 11. Prove by the methods of analytic geometry that an angle inscribed in a semi-circle is a right angle. 12. Show by means of their equations that a straight line can not cut a circle in more than two points. 13. Find the equation of the locus of a point whose distance from the origin is always twice its distance from (2, 3). What curve is it? Ans. 3x2 + 3y2 - 16x - 24y + 52 = 0. 14. Find the locus of a point whose distance from 3x + 4y - 1 = 0 is always equal to the square of its distance from (2, 3). Ans. 5x2 + 5y2 - 23x - 34y + 66 = 0. 15. Find the locus of a point the square of whose distance from x - y - 2 = 0 is equal to the area of the rectangle formed by drawing perpendiculars from the point to the axes. Ans. x2 + y2- 4x - 4y + 4 = 0. CHAPTER V THE CONIC SECTIONS 44. Historical note.-The Greek geometers, after studying. the geometry of the right line, plane, circle, sphere, cylinder and cone, turned their attention to three curves called the ellipse, parabola, and hyperbola, which they derived by cutting right circular cones by planes. For this reason these curves are called conic sections, or briefly, conics. In applying to them the methods of analytic geometry, however, we prefer to disregard their connection with the cone and define them as plane curves, using for this purpose certain characteristic properties from which their equations can be easily derived. 45. Definitions. I. CONIC, FOCUS, DIRECTRIX. A conic is the locus of a point which moves so that the ratio of its distance from a fixed point, called the focus, to its distance from a fixed line, called the directrix, remains constant, the point being always in the'plane of the focus and directrix. II. ECCENTRICITY. The constant ratio in I is called the eccentricity of the conic. It is represented by e. III. ELLIPSE. If e < 1 the conic is called an ellipse. IV. PARABOLA. If e = 1 the conic is called a parabola. V. HYPERBOLA. If e > 1 the conic is called a hyperbola. 46. Problem.-To construct a conic having given focus, directrix, and eccentricity. I. THE ELLIPSE. Let F, Fig. 38, be the given focus and DD' the given directrix of a conic with eccentricity e < 1. Through F draw EG perpendicular to DD'. Construct the angle GEH, so that tan GEH = e, and draw GEH' = GEH. Since e < 1, the 75 76 ANALYTIC GEOMETRY [CHAP. V] angle GEH < 45~. Draw a series of parallels perpendicular to EG. On any one of them, as KK', mark with dividers the FIG. 38 points P, P', so that FP = FP' = LK. Then P and P' are points on the ellipse, because, EL being the distance of P from the directrix, FP LK = = tan GEH = e. EL EL The same is true of P'. As many points of the ellipse as may be desired can be constructed in this way. As this construction is continued with successive parallels farther and farther from F a point will be reached beyond which the construction ceases to be possible. For example on M'M there is no point whose distance from F equals NM, and [ART. 46] THE CONIC SECTIONS 77 hence M'M does not intersect the ellipse. The same is true of a line such as I'I sufficiently far from F on the other side. In other words the ellipse is limited in extent in both directions from F. It is obvious also that the ellipse lies wholly between the lines EH, EH', and that it touches these lines at Q and Q', where the perpendicular to EG through F cuts them. To find the points where the ellipse cuts EG draw RR' through F, making angle EFR = 45~, cutting EH, EH' in R, R' respectively. From R and R' let fall perpendiculars to EG, meeting it at A and A'. These will be the required points, because AF AR EA = -tan GEH =e, EA EA and FA' R'A' EA' EA = tan GEH' = e. EA' EA' From this discussion it is clear that the ellipse is a closed curve symmetrical with respect to EG, and lying in that part of the plane between A and A'. D EXERCISES. 1. Construct the ellipse for which e = ~, and EF = 2 inches. 2. Construct the ellipse for which _ e = 2, and EF = 1 inches. E F\ 3. Construct the ellipse for which e =, and EF =1 inch. \ II. THE PARABOLA. The D' method of construction is the B' same as for the ellipse. Let F be the focus and DD' the directrix. Since e = 1 the angles GEH XH' and GEH' are each 45~, and the FIG. 39 78 ANALYTIC GEOMETRY [CHAP. V] line RF is parallel to EH'. The point A is on the curve, but there is no corresponding point A'. In making the construction the following facts with reference to the parabola become apparent. (a) The point A bisects the distance between the focus and directrix. (b) The distances FQ, FQ' are equal in length to EF, and hence to 2AF. (c) The parabola is an open curve symmetrical with respect to EG, and extending indefinitely from A, on through B and B'. EXERCISE. Construct the parabola for which EF = 1 inch. FIG. 40 [ART. 48] THE CONIC SECTIONS 79 III. THE HYPERBOLA. Let F be the focus and DD' the directrix. The construction is the same as for the ellipse and parabola. Since e > 1, GEH = GEH' > 45~, and the line RF cuts EH and EH' on opposite sides of the directrix. Hence A and A' are also on opposite sides of the directrix. The curve consists of two parts extending indefinitely to right and left from A and A', and no part of it lies between the two lines drawn through A and A' parallel to the directrix. Like the ellipse and parabola it is symmetrical with respect to EG. EXERCISES. 1. Construct the hyperbola for which e = 4, and EF = 4 inch. 2. Construct the hyperbola for which e = 2, and EF = ~ inch. 47. Definitions. I. The perpendicular to the directrix through the focus of a conic, the line EG in Figs. 38, 39, 40, is called the principal axis. II. The points A, A' in the ellipse and hyperbola, and the point A in the parabola, where these curves respectively cut the principal axis, are called the vertices. III. In the ellipse and hyperbola the point C which bisects AA' is called the center of the curve. IV. The chord through the focus perpendicular to the principal axis, Q'Q in Figs. 38, 39, 40, is called the latus rectum. Note especially Art. 46, II (b). 48. The equation of any conic.-By suitably choosing the coordinate axes the equation of any conic can be derived by translating the defi- - F N X nition Art. 45, I, into the language of algebra. The principal axis is taken as the X-axis, and the directrix as Y' the Y-axis. The constants FIG. 41 80 ANALYTIC GEOMETRY [CHAP. V] to be used are the given distance OF = p, and the eccentricity e. Let P be any representative point on the conic, then from the definition Art. 45, I, FP FP D O = e, (i) DP OM but FP = 4(x - p)2 + y2, from (3), p. 6, since P = (x, y), and F = (p, 0); and OM = x. Hence substituting in (i) (x - p)2 + y2 x or ( - p)2 + 2 = e2. (1) 49. Transformation of coordinates.-It is sometimes necessary or desirable to change the axes of coordinates to which a locus is referred. This change of axes has no effect upon the locus itself, but it changes the equation. PROBLEM. To derive the formulas for transforming the equation of a locus from one pair of axes to Y Y P another pair, when the new X-and Y-axes are parallel respectively to the original X- and Y-axes. o' I X Let OX, OY be the original axes of coordinates, and O'X', L — M — X - O'Y', the new axes. The coordinates of 0', the new origin, FIG. 42 referred to the original axes are FIG. 42 OL = xo, LO' = yo. Let P be any point in the plane. The coordinates of P in the two systems are OM = x, MP = y; and O'M' = x', M'P = y'. The formulas required are those which express the original coordinates of P in terms of the new coordinates and the con [ART. 50] THE CONIC SECTIONS 81 stants Xo, ye. From the figure OM = OL + LM = OL + O'M', MP = MM' + M'P = LO' + M'P, hence, x = x' + xo, y = y' + yo. (2) EXAMPLE. Transform the circle -x2 + y2- 6x + 4y - 3 = 0 to new axes parallel respectively to the old, the new origin being at (3, - 2). By (2) the formulas of transformation are x = x' + 3, y = y'-2. Substituting in the given equation we have (x' + 3)2 + (y' - 2)2 - 6(x' + 3) + 4(y' - 2) - 3 = 0, which reduces to x'2 + y'2 = 16. Let the student construct the figure. EXERCISES. 1. Determine the new equation of each of the following loci when transformed to new axes parallel respectively to the old, the new origin being taken at the point specified in each case. (a) x + y = 1, (-2, 1); Ans. x' + y' - 2 = 0. (b) x2 + y2 + 4x = 0, (- 2, 0); Ans. x'2 + y'2 = 4. (c) 42 + 92 + 8x + 18y - 23 = 0, (-1, -1); (d) 42 + 9y2 + 8x + 18y - 23 = 0, (1, 1); 2. Find the coordinates of the point so situated that when the circle x2 + y2 -4x + 8y - 5 = 0 is transformed to new axes through this point, parallel respectively to the original axes, the new equation shall be x2 + y2 = 25. Ans. (2, - 4). 3. Derive the general equation of a conic with the principal axis as X-axis, and the origin at the focus. 4. Derive the equation of the conic whose eccentricity is 2, focus (4, 0), and directrix the Y-axis. Ans. 3x2 + 4y2 - 32x + 64 = 0. 50. The parabola.-The eccentricity, e, is 1. Therefore equation (1), p. 80, becomes after reduction y2 = 2px - p2, (3) 7 82 ANALYTIC GEOMETRY [CHAP. V] which is the equation of the parabola referred to the axes OX, OY, Fig. 43, the latter being the directrix. To obtain a simpler form of equation take the origin at the vertex V, with the tangent at V for Y-axis. The coordinates of the new origin are (2p, 0), since V bisects O V F.M X o F M X FIG. 43 FIG. 44 OF, hence by (2) the formulas of transformation are x = x' +p, y = y'. Substituting these values in (3) and reducing, y'2 = 2px'. Hence, dropping accents, the equation of the parabola referred to the principal axis as X-axis, and the tangent at the vertex as Y-axis, Fig. 44, is y2 = 2px. (4) 51. Properties of the parabola, I. The perpendicular from any point of a parabola to the axis of the curve is the mean proportional between the latus rectum and the distance from the vertex to the foot of the perpendicular. This theorem is the translation of equation (4) into the language of geometry, because y = MP, 2p = L'L, and x = OM. [ART. 51] THE CONIC SECTIONS 83 II. The parabola y2 = 2px lies wholly on the positive side of the Y-axis. Since p is intrinsically positive x cannot be negative in this equation without making y imaginary. III. Similarly y2 = - 2px represents a parabola lying wholly on the negative side of the Y-axis. IV. The equations x2 = - 2py represent parabolas symmetrical with respect to the Y-axis, with their vertices at the origin, and lying respectively above and below the X-axis. Since 2p is simply a positive distance the equations y2 = 2px, and x2 = - 2py can be put in the form y2 = cx and x2 = cy, where c is any constant, positive or negative, and the following figures show the different positions of the parabola according to the form of the equation and the sign of c. Y Y Y 0 o~F X -FX A /Y 0o X y2 = x, c > 0 y2 = X, c < 0 x2 = cy, c > 0 x2 = cy, c < 0 FIG. 45(a) FIG. 45(b) FIG. 45(c) FIG. 45(d) EXERCISES. 1. Make careful drawings of the following parabolas, locate the focus and directrix of each one, and draw its latus rectum. (a) y2 = 6x; (b) x2 = 10y; (c) y2 = -8x; (d) x2 = - 3y. 2. Make careful drawings of the following pairs of parabolas, and determine the coordinates of the point of intersection of each pair. (a) 4y2 = 3x, 2X2 = 9y; (b) y2 + x = 0, 2 + y = 0. 3. Derive the equation of each of the parabolas determined by the following sets of conditions. The vertex in each case is at the origin. 84 ANALYTIC GEOMETRY [CHAP. V (a) Passing through (4, 1), axis coincident with OX. (b) Passing through (- 2, 3), axis coincident with OX. (c) Passing through (1, - 2), axis coincident with OY. (d) Passing through (h, k), axis coincident with OX. Ans. (a) 4y2 = x, (b) 2y2 = - 9x, (c) 2x2 = - y. 52. The ellipse.-Let OY be the directrix, and OX the principal axis of an ellipse ^y —^_^ ~with focus F, vertices A, A', and center C. I. The distances OA, OA' 0 AF c A X from the directrix to the vertices. Since A, A' are on the Xaxis and 0 is the origin, the FIG. 46 required distances are obtained by making y = 0 in equation (1), p. 80, and solving for x. This gives (x - p)2 = e2x2, or x - p = -- ex, from which x = 1 - c Separating the two values of x, we have therefore = OA = -' X = OA'= xx =0.4- 1 + e' xI = OA- 1-e' (5) Since e is positive and less than unity it follows from (5) that xl and x2 are both positive, x1 < x2, and 2p < X1 < p. The last statement shows that A lies nearer to F than to O. II. The distance AA' between the vertices. From the figure AA' = OA' - OA.'. from (5) p p 2ep AA' -e 1 -2 1 - e 1 + e 1-e~" [ART. 53] THE CONIC SECTIONS 85 For convenience let AA' = 2a, then a = AC = CA', and ep a = -- 1 - e2. III. The distance OC from the directrix to the center. From Fig. 46, and (6), p. 9, OC = (OA + OA') = ( 1 e+ 1 -e2' (6).'. from (6) above a OC =-. e (7) IV. The distance FC from the focus to the center. From the figure p e2p FC = OC - OF - - 2P e2 - -~~~~ - e2P 1 e2 ).'. from (6) FC = ae. (8) 53. The ellipse (continued PROBLEM. To find the equa tangular axes, with the principal axis for X-axis, and the origin at the center of the ellipse. This equation is conveniently derived by transforming equation (1) to the new axes here defined. The coordinates of the new origin C, referred to XOY, are (a/e, 0), of the ellipse referred to rec I. I I I. AF C FIG. 47 A' X see (7). Hence the formulas of transformation are a x = x' + e y = y'l Substituting in (1), p. 80, (x' + ae P -_ 2 = e2 x' -( a ) 86 ANALYTIC GEOMETRY CHAP. V] but (ale) - p = OC - OF = FC = ae, by (7) and (8), hence the equation becomes (x' + ae)2 + y2 =e2 (X' + a) e ' which reduces to (1 - e2)x'2 + y2 = a2(1 e2), and finally, dividing by a(1 - e2), to x,2 /2 x, y -+ -=1. (i) a2 a2(1 - e2) Dropping accents, and writing b2 = 2(1 — e2) (9) as a2(1 - e2) is positive, equation (i) becomes a.2 2 xa y a2 + 2 =1 (10) the form in which the equation of the ellipse is most commonly used. See Fig. 48. If we substitute x = 0 _B in equation (10) and solve for y, the result is y = th b, D - - A D, D showing that OB = b and A F F0 F' A' X OB' = -b. DEFINITIONS. I. The ^ B/^"^- 'length AA' = 2a is called ~FIG. 48 the transverse axis, or FIG. 48 major axis of the ellipse. II. The length B'B = 2b is called the conjugate axis, or minor axis, of the ellipse. An ellipse is fully determined when its semi-major and semiminor axes, a and b, are given. Equation (9) solved for e gives [ART. 53] THE CONIC SECTIONS 87 the value of the eccentricity ]a2 — ~b2 e = a- (11) in terms of a and b. Then FO = ae, and DO = a/e, are also known. Equation (10) shows that the ellipse is symmetrical with respect to both axes, and therefore with respect to the center. Hence there is another focus F', and another directrix, through D', Fig. 48, where OF' = FO, and OD' = DO. If in the foregoing discussion the line through the foci be taken as the Y-axis, and the perpendicular to it through the center as the X-axis, the resulting equation of the ellipse will evidently be b2 + a = 1. (12) Thus (x2/16) + (y2/9) = 1, and (x2/9) + (y2/16) = 1 both represent the same ellipse with the origin at the center, but in the first the major axis and the foci lie in the X-axis, and in the second they lie in the Y-axis. It is customary therefore, when discussing in a general way the equation (x2/a2) + (y2/b2) = 1, to disregard the condition implied in definitions I and II above, that a > b, and to say that this equation represents an ellipse for all values of a and b which are positive and not zero. The larger of the two numbers a and b is then the length of the semi-major axis, and the other, the length of the semi-minor axis. Hence: THEOREM. The equation Ax2 + By2 = C, where A, B, C all have the same sign and none of them is zero, always represents an ellipse with center at the origin, and axes coincident in direction with the coordinate axes. 88 ANALYTIC GEOMETRY [CHAP. V] If this equation be divided by C it takes the form AX2 By2 x2 y2 - +-C = 1, or C + 1, C C C C A B which is in the form of (10) or (12), because by hypothesis C/A and C/B are both positive. If C/A > C/B the major axis of the ellipse coincides with the X-axis, but if C/A < C/B the major axis coincides with the Y-axis. 54. A property of the ellipse. DEFINITIONS. I. A chord of B a conic through a focus is called a focal chord. "fSA II. The distance of a point on F', F x a conic from a focus is called the focal radius of the point with respect to that focus. FIG. 49 III. THEOREM. The sum of the two focal radii of any point on the ellipse is constant and equal to 2a. Let P be any point on the ellipse, Fig. 49. Since P = (x, y), and F = (ae, 0), the distance FP = (x - ae)2 + y2. (i) From equation (10) y2 (b2/a2)(a2 - x2), and from (9) b2/a2 = (1 - e2), hence y2 = (1 - e2)(a2 - x2). Substituting this value of y2 in (i), we have FP = i(x - ae)2 + (1 - e2)(a2 - x2) = a2 - 2aex + a - e = a x. (ii) Only the positive square root is taken because the length of FP as a positive distance is desired, and a > ex. Similarly F'P = a + ex. (iii) [ART. 55] THE CONIC SECTIONS 89 Adding (ii) and (iii) we have FP + F'P = 2a. (13) It follows from this that FB = F'B = a, which gives a convenient way of locating the foci of an ellipse when the axes are given. EXERCISES. In exercises 1 and 2 the X- and Y-axes are to be taken coincident with the major and minor axes of the curve respectively. 1. Find the equations of the ellipses determined in each case as stated below. Make careful drawings of the curves, and locate their foci and directrices (a) a = 5, b = 3; (b) a = 4, e = -3; (c) b = 2, e = a. Ans. (a) 9x2 + 25y2 = 225, e = 4, ae = 4, ale = —. 2. Find the equations of the ellipse and of its directrices for which a = 6 and the foci bisect the semi-major axes. Ans. 3x2 + 4y2 = 108, x = i 12. 3. Show that in any ellipse the length of the semi-latus rectum is b2/a. 4. For each of the following ellipses find the length of the semi-axes, the eccentricity, the coordinates of the foci, and the equations of the directrices. (a) 4x2 + 9y2 = 36. (c) 49x2 + 25y2 = 1225. (b) 12x2 + 16y2 = 192. (d) 9X2 + 5y2 = 180. Ans. (a) a = 3, b = 2, e = 3 - (5, (4 -5, 0), x = - f5. (d) a = 6, b = 2T5, e = -, (0, - 4), y = - 9. 5. For each of the two ellipses 16x2 + 25y2 = 400, and 9x2 + 5y2 = 180, find the distance between (a) center and focus, (c) focus and end of minor axis, (b) center and directrix, (d) focus and nearer vertex. Draw the curves carefully and check the distances found by measurements of the graphs. 55. The hyperbola.-Let OY be the directrix, and OX the principal axis of a hyperbola, with focus F, vertices A', A, and center C. The distances OA, OA' are found from equation (1) 90 ANALYTIC GEOMETRY [CHAP. V] in the same manner as for the ellipse, Xi = OA _= 1-, Xe 2 = OA' = 1- e (14) \ Y Since e is positive and greater than unity, xi is positive and x2 is negative, as shown in Fig. 50. Also I x2!* > x1, and xl < ~p. - A' COQA F - -X That is, A is nearer to the directrix than it is to the focus. Let A'A = 2a as in the ellipse. From Fig. 50 A'A = A'O + OA. FIG. 50.'. using (14) A'A =2a-p _ p 2ep AA +e 1 -e -e2- 1 ep or a= e2- (15) The student will now be able to show that for the hyperbola, as for the ellipse, center to directrix = CO = - (1.6) e and center to focus = CF = ae. (17) PROBLEM. To find the equation of the hyperbola referred to rectangular axes, with the principal axis as X-axis, and the origin at the center of the hyperbola. This equation will be derived by transformation of coordinates from equation (1), as was done in Art. 53 for the ellipse. The coordinates of C the new origin, Fig. 50, referred to the old system are [- (a/e), O], hence the formulas of transformation * The notation I X2 1 is used to indicate the absolute value of the expression between the bars, thus I - 3 = 3. [ART. 55] THE CONIC SECTIONS 91 are a X = --- y=y. e Substituting these in equation (1) we have - _ P)2 + y2 = e2 (x' a) e — e But - -p - CO- OF= - (CO + OF) - CF= - ae, e hence (x' - ae)2 + y'2 = e2 ( - a) e which, after reduction, becomes (e2 _ 1)'2 _ y2 = a2(e2 1), or, dividing by a2(e2 - 1), x12 y12 a2 a2(e2 - 1) ( Dropping accents, and writing b2 = a2(e2 - 1), (18) since e2 - 1 is positive, equation (i) becomes a2 b2 (19) Equation (19) shows that the Y hyperbola is symmetrical with respect to both X'X and Y'Y, and hence with respect to the center O. There are therefore two foci, F, F', symmetricallysituated with respect x' F' ' o A FX to 0, and two directrices, one corresponding to each focus. The directrices cut the principal axis at points D, D', so that D'O = \D.FI. 1 D'O = OD. FIG. 51 92 ANALYTIC GEOMETRY [CHAP. V] A hyperbola is fully determined when the values of a and b are given. The value of the eccentricity in terms of a and b, obtained from (17) is 5a2 + b2 a (20) Equation (18) shows that b may be larger or smaller than a. If e2 > 2, b > a; but if e2 < 2, b < a. 56. Conjugate hyperbolas.-If in equation (19) the letters x and y, and a and b be interchanged respectively the result is 2Y" X 1 2 - = 1, x2 Y2 or = - 1. a b2 (21) Y N B H P A A' B' H' I 52 FIG. 52 This change transforms the hyperbola into one whose principal axis lies in the Yaxis, and whose vertices are at the points (0, Ofb). Hence (21) is the equation of a hyperbola in this position. The two hyperbolas are drawn in Fig. 52, where A'A = 2a, and B'B = 2b. DEFINITIONS. I. The two hyperbolas x2 y2 x2 y2 (A) -2 1, (B) a2b2 -1, a2 2a2 b2 in which a and b have the same value in both equations, are called conjugate hyperbolas. II. The transverse axis of a hyperbola is the line joining its vertices. Thus, Fig. 52, A'A = 2a is the transverse axis of hyperbola (A), and B'B = 2b is the transverse axis of hyperbola (B). [ART. 57] THE CONIC SECTIONS 93 III. The conjugate axis of a hyperbola is the transverse axis of its conjugate hyperbola. Thus B'B = 2b is the conjugate axis of hyperbola (A), and A'A = 2a is the conjugate axis of hyperbola (B). 57. The asymptotes of a hyperbola. THEOREM. As x increases the hyperbola (x2/a2) - (y2/b2) = 1 continually approaches the lines y = t (b/a)x, but never meets them. The rectangle whose sides are parallel to the coordinate axes, and pass respectively through the points A, A', B, B', Fig. 52, has for its diagonals the lines y = i (b/a)x, since AH = b, AH' = - b, and OA = a. Considering only one of these lines, the line OL, its equation may be written in the form bx - ay = 0. (i) Let P(x, y) be in the first quadrant, on that part of the hyperbola whose vertex is at A, and draw PQ perpendicular to OL. Then by formula (20), p. 56, PQ = bx - ay (ii) la2 + b2 From the equation of the hyperbola, ay = + b x2 - a2. Using the positive sign for the point P, since it is in the first quadrant, we have, substituting in (ii), bx - b X2 - a2 b(x - x2 - a2) (x+ -x2-a2) Pa2 + b2 /a2+ b2 (x +.lx2 - a2) a2b or PQ= a2b _. (ii) qa2 + b2(x + x2 - a2) Thus in (iii) the length of PQ is expressed as a fraction whose numerator is constant, and whose denominator increases in 94 ANALYTIC GEOMETRY [CHAP. V] definitely as x increases. That is PQ approaches the limit zero as x increases indefinitely. Similarly it can be shown that in each quadrant the hyperbola approaches one of the two lines OL or OL'. The lines to which the hyperbola is thus related are called the asymptotes of the curve. In the same manner it can be shown that the conjugate hyperbola (x2/a2) - (y/b2) = - 1 approaches one of the lines OL or OL' in each quadrant. In other words: The two conjugate hyperbolas x2/a2- y2/b2 = 1 and x2/a2 -y2/b2 =- 1 have for asymptotes the same pair of lines y -= - x. (22) 58. A property of the hyperbola. THEOREM. The difference of y\.~ / the focal radii of any point on \the hyperbola is constant and equal to 2a. Let P = (x, y) be any point on -F'~ \ F X ~~the curve, then we are to prove F'P - FP = 2a. The details of the proof which are the same as in Art. 54, III, FiG. 53 are left to the student. FIG. 53 The results are F'P = ex + a, FP =ex - a, and hence F'P - FP = 2a. (23) 59. The equilateral or rectangular hyperbola. DEFINITION. If the semi-axes a and b of a hyperbola are equal the hyperbola is called an equilateral or rectangular hyperbola. [ART. 59] THE CONIC SECTIONS 95 The equation of an equilateral hyperbola, referred to its axes as axes of coordinates, is found by putting b = a in equation (19), p. 91, and is therefore xz - y2 = a2. (24) THEOREM. The asymptotes of an equilateral hyperbola are perpendicular to each other. By (22) the equations of the asymptotes to the curve of (24) are y = x and y = - x, which are perpendicular lines. EXERCISES. In exercises 1 and 2 the X- and Y-axes are to be taken coincident with the transverse and conjugate axes of the curve, respectively. 1. Find the equations of the hyperbolas determined in each case as stated below. Make careful drawings of the curves, and locate their foci and directrices. (a) a = 4, b = 3, (b) a = 6, e = 2, (c) b = 3, e =. Ans. (a) 9x2 - 16y2 = 144, e = -, ae = 5, a/e = -J-. 2. Find the equations of the hyperbola, and of its asymptotes, and the coordinates of its foci, for which a = 6, and the directrices bisect the semi-transverse axes. Ans. 3x2 - y2 = 108, y = -4- 3x, (- 12, 0). 3. Show that in any hyperbola the length of the semi-latus rectum is b2/a. 4. For each of the following hyperbolas find the lengths of the semiaxes, the eccentricity, the coordinates of the foci, and the equations of the directrices and asymptotes. (a) 16x2 - 9y2 = 144, (c) x2 - y2 = 4, (b) 9x2 - 16y2 = 144, (d) 9X2 - 16y2 = - 144, Ans. (a) a = 3, b = 4, e = -, x = + 9, y = 1 Ix. (d) a = 3, b = 4, e =, y = 1, y = + 3x. 5. Show that the four foci of two conjugate hyperbolas, and the four points of intersection of the tangents at their vertices, all lie on one circle whose center is at the common center of the two hyperbolas. 6. For each of the two conjugate hyperbolas 144x2 - 25y2 = - 4 find the distance between 96 ANALYTIC GEOMETRY [CHAP. V] (d) center and focus, (c) focus and asymptotes, (b) center and directrix, (d) focus and nearer vertex. Check graphically, using a large scale. 60. Transformation of coordinates. PROBLEM. To find the formulas for transforming the equation of a locus from one pair of axes to another pair having the same origin, the original system being rectangular. Let XOY, Fig. 54, be the original or old system, and X'OY' the new system. Let OX' p 0' x' make the angle 0 with OX, and Y' the angle O' with N M' R OX. Let P be any point in yp of v / the plane, then the old coordinates of P are x = OM, ___ \ Y_ ~y = MP, and the new coorM L M1 dinates of P are x' = OM', y' = M'P. We must express FIG. 54 x and y in terms of x', y' and the two angles 0 and 0'. Draw M'L perpendicular to OX, and NR through M' parallel to OX. Then, giving attention to the directions of the different lengths, and x = OM = OL + M'N = x' cos 0 + y' cos ', (25) and (25) y = MP = LM' +NP =x'sin0 + y'sin '.J If the new system is rectangular, as well as the old, then o' = 90~ + 0, and formulas (25) take the form x = x' cos 0 - y' sin 0, ( y = x' sin 0 + y' cos. J When 0' = 90~ + 0 the effect produced is the same as that of rotating the system of coordinates through the angle 0, because [ART. 61] THE CONIC SECTIONS 97 the new X- and Y-axes make the same angle 0 with the old X- and Y-axes respectively. This transformation is generally described as rotation of the axes through the angle 0. 61. The hyperbola referred to its asymptotes. THEOREM. The equation of a hyperbola referred to its asymptotes as axes of coordinates is xy = c, where c is a constant. To transform equation (19) p. 91 to the system of axes X'OY' use (25), where 0 = XOX' = -tan-l-, a so that by (17) p. ix 4a2 + b2 b sin O' = b ~a2 + b2 Y xYI 9 FIG. 55 FIG. 55 x and 0' = XOY' = tan-1b, a a cos 0 = la2 + b2' (i cos 0' = a /a2 + b2 and hence, substituting in (25) ax' + ay' X - Aa2 2+ b2 - bx' + by' y ^= a2 + b2 (ii) Finally, substituting these values of x and y in (19), we have (x'+ y')2 (-x' + y')2 x'y =+. 1 a2+ b2 a2 + b2 x'y' = }(a2 + b2). or, reducing, (iii) If OX", OY", Fig. 55, be taken as the new X- and Y-axes 8 98 ANALYTIC GEOMETRY [CHAP. V] respectively, instead of OX' and OY', the values of sin 0, cos 0, sin 0', cos 0' will be the same as before, except that sin 0 will be positive and cos O' negative. Hence instead of (ii) we have ax' - ay' bx' + by' X,- y xa2 + b2 y a2 +b2 and substituting in (19) and reducing the result is x'y - -(a2 + b2). (iv) The form of (iii) and (iv) shows that the product of the coordinates of any point on the hyperbola, referred to the asymptotes, is a constant. Dropping the accents, and replacing 4i (a2 + b2) by c, a constant, the equation takes the form xy = c. (27) If c is positive the signs of x and y must be alike, so that the curve lies in the first and third quadrants, but if c is negative the signs of x and y are unlike, and the curve lies in the second and fourth quadrants. If the coordinate axes (the asymptotes) are at right angles the hyperbola is equilateral. The equilateral hyperbola is the form of this curve which most commonly occurs in problems of physics and engineering, and the equation referred to the asymptotes as axes of coordinates is the one most often employed. EXERCISE. Make careful drawings of the two hyperbolas xy = 4 and xy = - 4, referred to rectangular axes. 62. Equations of the conics in more general form. I. Equation of the parabola with vertex at (xo, yo), and axts parallel to one of the coordinate axes. Let V be the vertex of a parabola, whose axis VX' is parallel to OX. Referred to X'VY' as coordinate axes, with ', y' as [ART. 62] THE CONIC SECTIONS 99 variables, the equation of the parabola will be y,2 = 2px'. (i) Let the coordinates of V, re- y ferred to XOY, be OM = xo, MV = yo, then the coordinates of 0, referred to X'VY', will be MO = - Xo, VM = - yo, and - hence by (2), p. 81, the formulas for transforming from the system X'VY' to the system XOY o -M will be x' = x -x, y' = y -yo. (ii) FIG. 56 Substituting these values in (i) we have (y - yo)2 = 2p(X - Xo), (28) which is the required equation. In more general form (28) may be written (y - yo)2 = c(X - Xo), (29) where c is any constant positive or negative, and the focus of the parabola will lie on the positive or negative side of V according as c is positive or negative. Similarly (x - xo)2 = c(y - yo) (30) is the equation of a parabola with vertex at (xo, yo) and axis parallel to the Y-axis. Compare Art. 51, IV. II. Equations of the ellipse and hyperbola with center at (Xo, yo). If the coordinates of C, Fig. 57, referred to the system XOY, are (xo, yo), then as in I, by transformation of equation (10), p. 86, from the axes X'CY' to the system XOY, the equation of the ellipse with center at (xo, yo), and axes parallel to the coordinate 100 ANALYTIC GEOMETRY [CHAP. V] axes is found to be (x - Xo)2 (y - yo)2 a2 b2 (31) Y Y C In connection with equation (31) refer to the latter portion of Art. 53. Similarly by transformax tion of equations (19), p. 91, and (21), p. 92, the equations of the hyperbolas x with center at (x0, yo), and axes parallel to the coordinate axes are found to be 0 FIG. 57 FIG. 57 (X - Xo)2 (y - yo)2 2 - = 4-. a2 b2 (32) In like manner the equation of the hyperbola with center at (xo, yo), and axes parallel to the asymptotes is derived from (27). (x - xo)(y - yo) = C. (33) 63. The parabolic arch.-Let OVA, Fig. 58, represent the outline of an arch in the form of a parabola, and Y I let the distance OA, called.. -. ---the span, be represented 0 A by s, and the height MV, M " - x called the rise, by r. ' I, Taking the axes as indi- FIG. 58 cated in the figure, the vertex of the parabola is at the point (~s, r) and its axis is parallel to OY. Hence from (30) the equation will be (x - ~s)2 = c(y - r), (i) [ART. 64] THE CONIC SECTIONS 101 in which c is to be determined so that the curve shall pass through the origin. Substituting x = 0, y = 0 in (i), we have s2 =s2 - cr, or c = -4 Substituting this value of c in (i) and reducing 4rx2 - 4rsx + S2y = 0. (34) 64. The equation Ax2 + By2 + Gx + Fy + C = 0. This equation, except in certain cases easily recognized, always represents a conic. It will now be shown, by means of examples, how to treat an equation of this form in order to determine the nature and position of the locus to which it corresponds. 1. 2x2 + 5y2- 8x + 7y - 12 = 0. By rearranging the terms the equation becomes 2(x2 - 4x) + 5(y2 + 7y) = 12, and completing the squares in the two parentheses 2(x2 - 4x + 4) + 5(y2 + jy + -4) = 12 + 8 +, or 2(x- 2)2+ 5(y + -)2 = 449 which reduces to (X - 2)2 (y + TTU) 449 + 449 -Tly - TT Comparing this with (31) it is seen that the figure is an ellipse with center at (2, — A), a = -49, and b = -4 The major axis is parallel to OX. 2. y2 + 6x - 7y + 15 = 0. The term in x2 is missing, and the operation is therefore modified accordingly. We write first y2- 7y = - 6x - 15, and then y2 - 7y + 4 = - 6x - 15 + -, 102 ANALYTIC GEOMETRY [CHAP. V] or (y- _ )2 = -6(x + 2 T). By comparison with (29) this is seen to be a parabola with vertex at (- Ix, j), and axis parallel to OX. The focus is 1 units on the negative side of the vertex, and the length of the latus rectum is 6. The student should construct the curves in the two preceding examples. The examples which follow illustrate exceptional cases which arise. 3. 2x2 + y2 - 8x + 4y + 12 = 0. This equation reduces to 2(x - 2)2 + (y + 2)2= 0. Hence the only real coordinates which will satisfy it are x = 2, y = - 2. It therefore represents the point to which an ellipse shrinks as its semi-axes a and b simultaneously approach zero. Since the equation can be written in the form [ 4(x - 2) + i(y + 2)][ 2(x - 2) - i(y + 2)] = 0 it may also be interpreted as representing a pair of imaginary lines meeting in the real point (2, - 2). See footnote, p. 67. 4. 2x2 - y2 -8x + 4y + 4 = 0. This equation reduces to 2(x - 2)2 - (y - 2)2 = 0. It therefore represents the two straight lines 42(x - 2) + (y - 2) = 0, 42(x - 2) - (y - 2) = 0. This is a limiting form of the hyperbola as its semi-axes simultaneously approach zero. 5. 2x2 +y2 - 8x + 4y + 16 = 0. This equation after transformation takes the form x - 2)2 (y + 2)2 2- 4 - which is not satisfied by any real values of x and y. It is therefore the equation of an imaginary locus. THE CONIC SECTIONS 103 EXERCISES ON CHAPTER V Normal Exercises 1. Construct accurately the conics determined as follows (a) e = 1, distance from focus to directrix = - inch. (b) e = 3, distance from focus to directrix = 1 inches. (c) e = a, distance from focus to directrix = 1 inch. 2. Derive the equations of the conics in Ex. 1, using the directrix as Y-axis, and the principal axis as X-axis. 3. Find the equations of the conics determined by the following sets of conditions, and draw each curve. (a) The parabola, with vertex at the origin, p = 2, and focus on the positive extension of the X-axis. (b) The parabola, with vertex at the origin, and focus at (0, - 4). (c) The ellipse, with center at the origin, a = 4, b = 2, and major axis in the X-axis. (d) The ellipse, with center at the origin, one focus at (0, 3), and e = 4 -(e) The hyperbola, with center at the origin, a = 2, b = 4, and transverse axis in the X-axis. (f) The hyperbola, with center at the origin, a = 2, e = 4, and transverse axis in the Y-axis. (g) The hyperbola, with a = b = 4, referred to its asymptotes as axes of coordinates. 4. Find the equations of the conics determined by the following sets of conditions, and draw each curve. (a) The parabola with vertex at (- 2, 4), p = 4, the axis parallel to the X-axis, and the curve lying on the positive side of the vertex. Ans. y2 - 8x - 8y = 0. (b) The parabola passing through (0, 4), with vertex at (2, - 3), and axis parallel to the X-axis. Ans. 2y2 + 49x + 12y - 80 = 0. (c) The ellipse, with center at (2, 1), a = 4, b = ~, and major axis parallel to the X-axis. (d) The ellipse passing through the origin, with center at (- 2, - 3), one vertex at (-2, 7), and major axis parallel to the Y-axis. Ans. 91x2 + 4y2 + 364x + 24y = 0. 104 ANALYTIC GEOMETRY [CHAP. V] (e) The hyperbola, with center at (- 1, - 1), a = 3, b = 7, and transverse axis parallel to the Y-axis. (f) The hyperbola passing through (8, 0), with center at (4, 2), b = 2, and transverse axis parallel to the X-axis. Ans. x2 - 5y2- 8x + 20y = 0. 5. Determine the nature and position of the conics having the equations given below. If the curve is an ellipse or hyperbola find its eccentricity and semi-axes, the coordinates of its center and foci, and the equations of its axes and directrices. If the curve is a parabola find the coordinates of its vertex and focus, and the equations of its axis and directrix. (a) y2 = - 1x, (f) 4x2 - y2 = 9, (b) 92 =4y, (g) 4x2 y2- 12x - 6y + 23 0, (c) X2 + 9y2 = 9, (h) 9X2 + 42 + 18x-27 = 0, (d) 4x2 + y2 = 9, (i) 4y2 - 7x- 8y + 1 = 0, (e) 2 - 9y2 = 9, (k) 3x2 + 222 + 12x- 12y + 34 = 0. 6. Find the lengths of the focal radii of those points on the ellipse 7x2 + 16y2 = 28 whose abscissa is 1. Ans. a, -l-. 7. Find the lengths of the focal radii of those points on the hyperbola 5x2 - 4y2 = 5 whose abscissa is - 3. Ans. 7, -1-. 8. Write the equations of the asymptotes of the following hyperbolas. Construct each curve and its asymptotes. (a) 4x2- 4y2 = 9, (c) 25x2 - 4y2 = 49, (b) 36x2 - 9y2 = - 16, (d) 9x2 - 36y2 = -16. 9. Find the equations of the hyperbolas in Ex. 8 referred to their asymptotes as axes of coordinates. 10. (a) Transform x2 + 4y = 8 to new axes parallel respectively to the old axes, with the new origin at (2, 4). Ans. x'2 + 4x' + 4y' + 12 = 0. (b) Transform 5x2 + 4y2 - 20x + 8y + 23 = 0 in a similar manner, with the new origin at (2, - 1). Ans. 5x'2 + 4y'2 = 1. (c) Transform x2 - y2 = 4 to new axes through the same origin, the new axes making angles of 30~ and - 30~ respectively with the original X-axis. Ans. x'2 + 4x'y' + y/2 = 8. (d) Transform x2 + 4y2 + 4xy - 36 = 0 in a similar manner to new axes having the slopes 2 and - ~ respectively. Ans. 5x'2 36 = 0. THE CONIC SECTIONS 105 General Exercises NOTE. In exercises 11-19 the equations of the curves are to be found from the data, and the results checked graphically. 11. The parabola with vertex at (- 4, 0), and directrix x = 2. Ans. y2 - 24x - 96 = 0. 12. The parabola with axis parallel to the Y-axis, which passes through the points (0, 0), (2, 4), (- 2, 4). Ans. x2 - y = 0. 13. The parabola with axis parallel to the X-axis, which passes through the points (2, 4), (4, 8), (1, - 1). Ans. y2 + 3y - 30x + 32 = 0. 14. The ellipse which has the line from (2, 4) to (8, 4) for major axis, and one focus at (4, 4). Ans. 8x2 + 9y2 - 80x - 72y + 272 = 0. 15. The ellipse with center at (10, - 5), one focus at (4, - 5), and x = 2 for the corresponding directrix. Ans. x2 + 4y2 - 20x + 40y + 152 = 0. 16. The ellipse with one focus at (4, 2), one vertex at (4, - 10), and the minor axis in the line y = - 2. Ans. 4x2 + 3y2 - 32x + 12y - 116 = 0. 17. The hyperbola with center at (0, 0), one focus at (0, 6), and one vertex at (0, 4). Ans. 5y2 - 4x2 = 80. 18. The hyperbola whose transverse and conjugate axes are in the X- and Y-axes respectively, whose eccentricity is 2, and which passes through the point (2, 3). 19. The hyperbola with transverse axis in the Y-axis, which has 3y - 2x = 0 for one asymptote, and one focus in the line y = - 4. Ans. 117y2 - 52x2 = 576. 20. Regarding the circle as an ellipse in which b = a, locate its foci and directrices, and determine its eccentricity. 21. Show that the equation of an ellipse referred to its major axis as X-axis, and the tangent at the left-hand vertex as Y-axis is y2= (2ax - x2), or y2 = (1 - e2)(2ax - x2). 22. The asymptotes of the hyperbola xy = 4 are inclined to each other at an angle of' 60~. Find a and b. Ans. 243, 2. 23. Using the definition of Art. 45 find the equation of (a) the parabola with focus at (2, 4) and x + y = 1 for directrix, Ans. x2 - 2xy + y2 - 6x - 14y + 39 = 0. 106 ANALYTIC GEOMETRY [CHAP. V] (b) the ellipse with focus at (- 2, 2), eccentricity = ~, and 3x - 2y + 4 = 0 for directrix, Ans. 43x2 + 12xy + 48y2 + 184x - 192y + 400 = 0. (c) the hyperbola with focus at (-1, - 2), eccentricity = -, and 4x - y = 0 for directrix. Ans. 76x2 - 72xy - 59y2 - 136x - 272y - 340 = 0. 24. Derive an equation which will represent all ellipses whose foci are the points (4, 0) and (- 4, 0). Ans. 2 + - 2 a = 1, a > 4. 25. Derive an equation which will represent all hyperbolas whose foci are the points (4,-0) and (- 4, 0). Ans. X + 2 16 = 1, a < 4. 26. Find for what values of c the straight line 3x + 4y = c cuts the circle x2 + y2 = 25 in (a) two real and distinct points, (b) two coincident points, (c) two imaginary points. Ans. (a) - 25 < c < 25, (b) c = 25, (c) c2 > 625. 27. Find the coordinates of the real intersections of the following pairs of curves. Draw the figure for each case. (a) x2 + y2 = 16, and y2 = 6x. Ans. (2, 4- 2). (b) x2 - y2 = 10, and xy = 12. Ans. (3 12, 212), (- 3-2, - 2-2). (c) x2 + 6y2 = 1, and x2 + 5y = 0. (d) 9x2 + 25y2 = 225, and x2 - 16y2 = 16. 28. Find the coordinates of the intersections of the parabola y2 = 3x with the curve 4y2 = x(2x - 3)(2x - 7). Check graphically. Ans. (0, 0), (2, t i46), (, 4- i-6). 29. Find the coordinates of the real points where the ellipse 16x2 + y2 = 64, and the curve y2 = x(x - 2)(x - 4) intersect. Check the results graphically. Ans. (2, 0). CHAPTER VI TANGENTS, NORMALS, DIAMETERS, POLES AND POLARS 65. Definition of a tangent. I. If a secant intersects a curve in two points, and then moves in such a way as ultimately to bring these two points into coincidence, this limiting position of the secant is called a tangent to the curve. Thus if the secant AB1, Fig. 59 (a), moves so that the point A ^Qp \ 3/A3 B.5 B FIG. 59(a) FIG. 59(b) remains fixed while B1 travels on the curve through the positions B2, B3, etc., until it comes into coincidence with A, then PQ the limiting position of the secant is the tangent to the curve at A. Or, if the secant AiB1, Fig. 59 (b), moves so as to remain parallel to its original position, until its two points of intersection come into coincidence at A, then PQ the limiting position of the secant is the tangent to the curve at the point A. Adopting this conception of a tangent to a curve the definition may be stated more concisely as follows: II. A straight line is said to be tangent to a curve when it meets the curve in two coincident points. 107 108 ANALYTIC GEOMETRY [CHAP. VI] 66. Tangents to the conics* in terms of the slope. The question to be investigated may be stated thus: Given the equation of a conic to derive the equation of a straight line, with given slope, which shall be tangent to the curve. I. The tangent to the ellipse (x2/a2) + (y2/b2) = 1 in terms of the slope. Let AB be a straight line having the given slope m. Its equation will then be y = mx + k. (i) If y be eliminated between this equation and that of the ellipse the result is x2 (mx + k)2 + 1, a2 b2 or, after reduction, (a2m2 + b2)x2 + 2a2mkx + a2k2 - a2b2 =. (ii) The roots of this quadratic in x are the abscissas MP, NQ of the points where AB cuts the ellipse. If AB be moved so as /B2 Y /B /B N P FIG. 60 to remain parallel to its original position, m will remain constant in (i) and (ii), but k will vary, and when AB takes either of the positions ARB1, A2B2, tangent to the ellipse, the roots of equa* The word " conic " as used here includes the circle, as well as the ellipse, hyperbola, and parabola. [ART. 66] TANGENTS, NORMALS, DIAMETERS 109 tion (ii) will become equal. When these roots are equal (see II (b) p. vii) 4a4m2k2 - 4(a2m2 + b2)(a2k2 - a2b2) = 0, or b2k2- a2m2b2 - b4 = 0 from which k = IaW2m2+ b2. These are therefore the values of k for which the straight line y = mx + k will be tangent to the ellipse, and hence these tangents are y = mx ca2m2 + b2. (1) It may be observed that the first form of this result corresponds to the tangent A2B2, Fig. 60, whose y-intercept is positive, and the second form corresponds to A1B1, whose y-intercept is negative. II. The equation of the tangent to the hyperbola x2 y2 _ _ - = 1 a2 b2 1 in terms of the slope m, is y = mx a2m2 - b2. (2) This result is obtained simply by changing the sign of b2 throughout the foregoing demonstration. III. The equation of the tangent to the circle x2 + y2 = r2 in terms of the slope m is y = mx r lm2 +1. (3) This result is obtained from (1) by substituting a = b = r. The student should for practice derive (2) and (3) independently, as (1) is derived. IV. The equation of the tangent to the parabola y2 = 2px in terms of the slope. Let y = mx + k be a straight line having the given slope. 110 ANALYTIC GEOMETRY [CHAP. VI] As in I eliminate y between the equation of the line and that of the parabola, then m2x2 + 2(mk - p)x + k2 = 0. This equation will have equal roots if 4(mk - p)2 - 4m2k2 = 0, or - 2mpk + p2 = 0, from which k= P 2m' Hence the required equation is y =mnx + 2 (4) Since only one value of k is found, there is in the case of the parabola only one tangent having any given slope. EXERCISES. The complete solution of the following exercises includes in every case a well-drawn figure. 1. Write the equation of the tangents (a) to x2 + y2 = 25, having the slope + ^3; (b) to (x2/9) - (y2/25) = 1, having the slope - 2; (c) to y2 = 12x, having the slope 3; (d) to (x2/16) + (y2/9) = 1, having the slope - 1; (e) to (x2/5) + (y2/2) = 1, having the slope 2; (f) to (x2/36) - (y2/25) = 1, having the slope 1. Ans. (a) y = 3'x:i 10, (b) y = - 2x 4t ii1, (c) y = 3x + 1. 2. Show that the equations of the tangents to (x2/b2) + (y2/a2) = 1, with slope m, are y = mx i ]a 2 + b2m2; and write the equations of the tangents to 7x2 + 3y2 = 21 having the slope - 3. Ans. y = - 3x -- 4. 3. Show that the equation of the tangent to x2 = 2py, having the slope m, is y = mx - ~pm2; and write the equation of the tangent to x2 = 10y having the slope - 2. 4. Show that the equations of the tangents to xy = c in terms of the slope m are y = mx -- 2 - mc; and write the equation of the tangent to xy = 3, having the slope -. [ART. 67] TANGENTS, NORMALS, DIAMETERS 111 5. Show by transformation of coordinates that the equations of the tangents to the circle (x - h)2 + (y - k)2 = r2 with the slope m are y - k = m(x - h) -- ril + m2. 6. Deduce from the result of Ex. 5 the equations of the tangents to 2 + y2 - 4x + 2y - 11 = 0 having the slope 3. Also solve the problem independently. Ans. 2x - 3y - 7 t- 4-i3 =0. 7. Derive the equation of the tangent to x2 - 10x - 8y + 41 =0 having the slope 2. Ans. y = 2x - 16. 8. Derive the equation of the line with slope -- tangent to 4y2- 16x + 4y + 17 = 0. Ans. x + 2y + 4 = 0. 9. Deduce from equation (2), and from the results of Ex. 4 the values of m for which the tangents will be imaginary in each case, and draw figures illustrating the results. 67. Tangents to conics in terms of the coordinates of the points of tangency. I. PROBLEM. To derive the equation of the tangent to the conic Ax2 + By2 + Gx + Fy + C = 0 (i) at the point (x1, yi). Let AB be a portion of the conic whose equation is (i), Pi(xi, yi) the point of tangency, and P2(xl + h, yi + k) any other point on the conic, then P1N = h, and NP2 = k. The slope of the chord P1P2 is tan NP1P2 = k/h. As P2 moves on the curve Q so as to approach P1, the / chord P1P2 approaches the B tangent P1Q, and at the same time k and h approach the N limit zero. Hence the slope of the tangent P1Q is the limit of the ratio k/h as k 2X and h approach the limit zero. The value of this limit A may be determined as follows. FIG. 61 112 ANALYTIC GEOMETRY [CHAP. VI] Since Pi and P2 are both on the conic their coordinates will satisfy equation (i), hence Axl2 + By,2 + Gxl + Fyi + C =, (ii) and A(xi + h)2 + B(yi + k)2 + G(x1 + h) + F(yi + k) + C = O. (iii) Expanding (iii) and subtracting (ii) from it the result is 2Axlh + Ah2 + 2Bylk + Bk2 + Gh + Fk = 0, or h(2Axi + Ah + G) + k(2Byi + Bk + F) = 0, from which k _ 2Axi + Ah + G h 2Byi + Bk + F' Hence from (iv), as h and k approach the limit zero, m k 2Axi+ G h 2Byl + F' which is the required slope of the tangent at (xi, yl). By equation (5), p. 41, the equation of the tangent is therefore 2Ax1 + G - x W y-yl= -2Byi + (x-x) (v) Clearing this equation of fractions and rearranging the terms it becomes 2Axx + 2Byly + Gx - Gxl + Fy - Fyl - 2Ax,2 - 2Byi2 = 0. (vi) Adding to (vi) twice equation (ii), the result is 2Axlx + 2Byly + Gx + Gx1 + Fy + Fyl + 2C = 0, and dividing by 2, we have finally Axix + Byly + ~G(x + xi) + F(y + y) + C 0, (5) [ART. 67] TANGENTS, NORMALS, DIAMETERS 113 which is in more convenient form than (v) or (vi), and is the required result. II. By comparing (5) with the equation of the conic (i) it will be seen that the two equations have well-defined resemblances. It is easy therefore, when the equation of a conic is given, to write the equation of the tangent at a given point. The following example shows how this can be done. Let the conic be 2x2 - 3y2 + 7x - 4y - 19 = 0. Rewrite the equation thus 2xx - 3yy + 7(x- +) - 2(y + y) -19 = 0, and affix subscripts to one of the variables in each term where they are multiplied, and in each pair of terms in parenthesis, thus obtaining 2x1x x +x +- xi) - 2(y + y1) - 19 = 0, which is the equation of the tangent at the point (x1, yl). If the point (3, 2) be taken as the point of tangency, the equation of the tangent to the conic at this point will be 6x - 6y + 2(x + 3) - 2(y + 2) - 19 = 0, or 19x - 16y- 25 = 0. III. From equation (5) it is easy to write the equations of the tangents to the conies in the following special cases by substituting the appropriate values of A, B, G, F, C. Thus the equation of the tangent at the point (xl, yl) to the circle x2 + y2 = r2 is xix + yly = r. (6) parabola y2 = 2px is yly = p(x + xi). (7) X2 y2 X ellipse 2++ =1 is + = 1 (8) a2 b2 a2 b2 x2 y2 xix lyy hyperbolas - 1 is - = 1. (9) 9 a2 2 9 114 ANALYTIC GEOMETRY [CHAP. VI] The student should derive each of these equations independently by the process used in deriving (5). IV. To derive the equation of the tangent at the point (x1, yl) to the hyperbola xy = c, with either rectangular or oblique axes, FIG. 62 take two points on the curve, Pl(Xl, yl), P2(X1 + h, yi + k), and draw the secant AB, through them. Its equation by (2), p. 40, and Art. 29, will be k y - yl = (x - Xl). y Y (i) As P2 approaches Pi on the curve the secant P1P2 approaches the tangent at P1, and h and k approach the limit zero. To find the limit of k/h substitute the coordinates of P1 and P2 in the equation xy = c, obtaining x1yl = c, and (xl + h)(yl + k) = c. Expanding the second of these and subtracting the first from it the result is kxi + hyl + hk = 0, [ART. 68] TANGENTS, NORMALS, DIAMETERS 115 from which k y1 + k h x1 Hence as h and k approach the limit zero,.k y1 Lim- =h x1 and therefore, substituting in (i), we have y - yi = (x - X1). x1 Reducing, and remembering that x1y1 = c, the result is ylx -+ xly = 2c. (10) 68. The normal at any point on a conic. DEFINITION. The perpendicular to the tangent to any curve at the point of tangency is called the normal to the curve at that point. From the results given in (6), (7), (8), (9), p. 113, and from equation (16), p. 52, the student can easily verify that the equation of the normal at (xl, yl) to Y1 hi1 (11y x2 + y2 = r2 is y - yl = -(x - x) or y = Y-. (11) XI XI y2 = 2px is y y ( - xi (x - ). (12) 2 2yl a2+ 1 is y- =;(-)yl. (13) x2 y2 a2y1 -x a2 is -2 = -- x1). (14) Note that (11) shows the normal at any point of the circle to be a line through the center, in agreement with a well-known property of the circle. 116 ANALYTIC GEOMETRY [CHAP. VI] EXERCISES. The complete solution of the following exercises includes in every case a well-drawn figure. 1. Write the equations of the tangent and normal (a) to x2 + y2 = 25 at the point (- 3, 4), (b) to y2 = 8x at the point (2, - 4), (c) to 4x2 + 9y2 = 25 at the point (2, 1), (d) to x2 - y2 = 16 at the point (- 5, 3), (e) to xy = 8 at the point (4, 2), Ans. (a) Tan. 3x-4y + 25 = 0, Nor. 4x + 3y = 0. (b) Tan. x + y + 2 = 0, Nor. x - y - 6 = 0. (c) Tan. 8x + 9y - 25 = 0, Nor. 9x-8y- 10 = 0. 2. The point (- 1, 2) is on each of the following curves. Find the equations of the tangent and normal to each curve at that point. (a) 42 + y2 - 24x + 4y - 40 = 0, Ans. Tan. 4x - y + 6 = 0, Nor. x + 4y - 7 = 0. (b) 42 - y2 - 24x + 4y - 32 = 0, Ans. Tan. x + 1 = 0, Nor. y - 2 = 0. (c) 4x2 - 24x + 4y - 36 = 0, 3. Derive the equations of the tangents to the conics in Ex. 1, at the points indicated for each, by using the process of Art. 67, I. 4. Derive the equation of the tangent to the curve y = x3, at the point (2, 8), by the method of Art. 67, I. Ans. y = 12x - 16. 5. Derive the equations of the tangents to the curves X2 + y2 = 25 and 3y2 = 16x at one of their points of intersection, and determine the angle between the tangents. Ans. tan-l 1_-. 6. Using equation (6) determine the equations of the two tangents to the circle x2 + y2 = 25 which pass through the point (1, 7). Ans. 4x + 3y = 25, -3x + 4y = 25. 7. Using equation (6), show by transformation of coordinates that the equation of the tangent to (x - h)2 + (y - k)2 = r2, at the point (x1, Yl) is (x - h)(xi - h) + (y - k)(y - k) = r2. 69. Diameters of conics. I. DEFINITION. A diameter of a conic is the locus of the middle points of a system of parallel chords. [ART. 69] TANGENTS, NORMALS, DIAMETERS 117 The definition of the diameter of a circle in elementary geometry, a line passing through the center, is included in the definition just given, because a diameter of a circle bisects all N chords perpendicular to it. II. THEOREM. A diameter of Y a parabola is a straight line parallel to the axis of the curve, K and conversely, every line par- allel to the axis of a parabola is 0 X a diameter. Let MN be a line with slope m, meeting the parabola y2 = 2px in the points P1, P2. The equa- M tion of MN may be written FIG. 63 y = mx + k. Let (x1, yl), (X2, Y2) stand for the coordinates of P1, P2, respectively, and let P(xo, yo) be the middle point of the chord. Then by (6), p. 9, X = ~(xi + X2), Yo I '( + Y2). (i) To express the values of xl, x2, y1, Y2 in terms of the given constants, p, m, and k the equations y2 = 2px and y = mx + k must be solved simultaneously for x and y. Eliminating x we have the quadratic my2 -2py + 2pk = 0, (ii) whose roots are y1 and y2, the ordinates of P1 and P2 respectively. Hence from (i), (ii) and II (c), p. viii, 2 Z\M~ 7 ~2)(iii) yo = 2(y2 + 2b2) = p (iii) Since the value of yo in (iii) does not contain k, it follows that the middle points of all chords having the slope m are at the same distance p/m from the axis of the parabola. Hence there. 118 ANALYTIC GEOMETRY [CHAP. VIi is no necessity for finding the values of x, and the line y = P (15) m parallel to the axis of the curve is the diameter corresponding to this system of parallel chords. To prove that every line y = c is a diameter of the parabola y2 = 2px, it is only necessary to write c = p/m, from which m = p/c, which will be the slope of the system of chords bisected by the line y = c, as appears from (15). Hence y = c is a diameter of the parabola. III. THEOREM. Every diameter of an ellipse is a straight line passing through the center. Let y = mx + k be the equation of the line MN, with given slope m, meeting the ellipse x2/a2 + y2/b2= 1 at Pi, P2, N whose coordinates will be represented by (xl, yl), (x2, y2). Let P(xo, yo) be the middle 0 X point of P1P2. Then as in the demonstration of II -7^Pt Zxo = ~ (xi + X2), FIG. 64 Yo = I (Y1 + Y2). To find the values of xl, x2, y?, Y2 in terms of a, b, m, and k solve the equations of the ellipse and line simultaneously. Eliminating y (a2m2 + b2)x2 + 2a2mkx + a2(k2 - b2) = 0, hence as in II a2mk o = 1 + 2) = -a2m2 + b2 (i) Substituting this value of xo for x in y = mx + k b2k yo = a2m2 + b2 (ii) a2m2 + b2' [ART. 70] TANGENTS, NORMALS, DIAMETERS 119 Dividing (ii) by (i) k is eliminated, and therefore the result Yo b2 XO a2m is true for the middle point of every chord which has the slope m. Hence the equation of the locus of these middle points, that is the equation of the diameter bisecting the system of chords whose slope is m, is b2 Y= -a x (16) a 2m This equation shows (1) that the diameter of an ellipse is a straight line, and (2) that it passes through the origin. As the origin is at the center of the ellipse the theorem is proved. IV. THEOREM. Every diameter of a hyperbola is a straight line passing through the center. This theorem is proved in the same manner as III, and the result is obtained simply by changing the sign of b2 throughout. Hence b2 Y= a2- (17) is the equation of the diameter of the hyperbola (x2/a2) - (y2/b2) = 1 which bisects the system of parallel chords which have the slope m. V. The points where the diameter of an ellipse or hyperbola meets the curve are called the extremities of the diameter. 70. Conjugate diameters of ellipse and hyperbola. I. DEFINITION. A pair of diameters of an ellipse or hyperbola, each of which bisects the system of chords parallel to the other, are called conjugate diameters. II. THEOREM. If m and m' are the slopes of two diameters of an ellipse, they will be conjugate diameters if b2 mm = -2 (18) 120 ANALYTIC GEOMETRY [CHAP. VI] Let m' be the slope of the diameter of the ellipse which bisects the system of parallel chords having the slope m, then from (16) b2 I b2 m a= - 7 or mm' =- a2m a III. THEOREM. If m and m' are the slopes of two diameters of the hyperbola (x2/a2) - (y2/b2) = 1, they will be conjugate diameters if b2 mm' = -. (19) a2 The proof is left to the student. EXERCISES. 1. Find for each of the conics below the equation of the diameter which bisects the system of parallel chords whose slope is the number given, and determine the point or points where the diameter meets the conic. (a) y2 = 6x, m = 2; (c) 42 + 9y2 =36, m= 1; (b) y2 = 4x, m = 1; (d) 42 - 9y2 = 36, m = 2. Ans. (a) (6, 6), (c) (3 i1-3, - 4 13), (- -93 3, 4 41-3). 2. Construct each of the following conies and mark on it the point whose coordinates are indicated. Find (i) the equation of the diameter to the point indicated, and (ii) the slope of the chords which the diameter bisects. Construct three chords of the system. (a) y2 = 4x, (4, 4). Ans. (i) y = 4, (ii) 2. (b) 4x2 + 9y2 = 40, (1, - 2). Ans. (i) y = -2x, (ii) ]. (c) x2 -4y2 = 9, (5, 2). (d) x2 - y2 = 16, (5, -3). 3. In Exs. 1 (c) and 2 (b) find (i) the equation of the diameter conjugate to the one already found, (ii) the coordinates of the points where this conjugate diameter meets the curve. 4. For each hyperbola of Exs. 1 and 2 find (i) the equation of the diameter conjugate to the one already found, (ii) the coordinates of the points where this conjugate diameter intersects the conjugate hyperbola. Ans. to Ex. 2 (c), y = Ix, (4, -), (- 4, - 5). 5. What is the diameter conjugate (a) to the major axis of an ellipse, (b) to the transverse axis of a hyperbola? [ART. 71] TANGENTS, NORMALS, DIAMETERS 121 71. Pole and polar with respect to a conic. In the following discussion the results will be worked out in detail only for the circle. From these it will be easy to infer the corresponding results for the conics in general, or to work them out independently. I. DEFINITION. The line joining the points of contact of two langents to a circle (or conic) is called the chord of contact of the two tangents. II. THEOREM. If two tangents to the circle x2 + y2 = r2 meet at the point (xo, yo) the equation of the chord of contact is XoX + YoY = r2. The coordinate axes are omit- i ted from the figure to save useless complication. Let the equation of the circle be x2 + y2 = r2, and let the points /PO(oYo) \( 2) of tangency of the tangents FI 65 which meet at (xo, yo) be (xl, yi) and (x2, Y2) Then the equations of these tangents will be Xix + yiy = r2, x2x + y2y = r2. (i) Since both of these lines pass through (xo, yo) we have identically X1Xo + yiyo = r2, x2Xo + y2yo = r2. (ii) Then it is clear that xox + yoy = y (20) is the equation of PiP2, because, being of the first degree it is the equation of a straight line, and if the coordinates of P1 and P2 be substituted for x and y it is evident from (ii) that it is satisfied. Hence the line (20) passes through P1 and P2. III. THEOREM. If through a fixed point Po(xo, ye) in the plane of a circle (or conic) a straight line be drawn meeting the 122 ANALYTIC GEOMETRY [CHAP. VI] circle (or conic) in the points A and B, and if tangents be drawn to the circle (or conic) at the points A and B, the locus of the intersection of these tangents will be a straight line. Let the equation of the circle be x2 + y2 = r2, as in II. The demonstration applies to both figures. Let the two tangents Po (X0 VO) FIG. 66(a) FIG. 66(b) meet at Pi(xl, yl) then AB is the chord of contact of Pi, and hence by (20) the equation of AB is x1X + yly = r2. (i) Since the line (i) passes through (xo, yo) X1Xo + ylyo = r2. (ii) Now let AB move, but always in such a way that it passes through P0, then it is clear from (ii) that Xox + yoy = r (21) is the equation of a straight line which is satisfied by the coordinates of P1 in all of its positions, hence this line (21) is the locus of P1. Since the locus of P1 is a straight line, it may be constructed for any given position of Po, by drawing two secants through [ART. 72] TANGENTS, NORMALS, DIAMETERS 123 Po, constructing the two pairs of tangents at the points where the secants intersect the curve, and connecting the two points where the two pairs of tangents meet. In Figs. 66 (a) and (b) the construction is completed with dotted lines, and PiP1' is the required locus. IV. DEFINITION. The line whose equation has been derived in (21) is called the polar of the point Po with respect to the curve, and the point Po is called the pole of the line. 72. Discussion and extension of the results of Art. 71. Equations (20) and (21) have the same form as that of the tangent to the circle x2 + y2 = r2 at the point (xo, yo). Hence the tangent to a circle and the chord of contact of a point are special cases of the more general relation of pole and polar. In all these cases the form of the equation is the same, the only difference being in the location with respect to the circle of the fixed point (xo, yo) whose coordinates appear in the equation. If in Art. 71 the equation of a conic had been used instead of that of a circle the nature of the results would have been the same. EXERCISES. 1. Write the equations of the polars of the points (6, 4), (4, 4), (2, 4) with respect to each of the conics given below, and construct separate figures for each curve with its poles and polars. (a) x2 + y2 = 32, (b) 3x2 + y2 = 64, (c) y2 = 4x, (d) 4x2 - y2 = 48, (e) x2 = 4y, (f) xy = 16. Ans. (a) 3x + 2y = 16, x + y = 8, x + 2y = 16. (c) 2y = x + 6, 2y = x + 4, 2y = x + 2. 2. Show that for any conic the polar of a focus is the corresponding directrix. EXERCISES ON CHAPTER VI Normal Exercises 1. For each conic below find the equation of the tangent or tangents having the slope indicated. Check the results graphically. (a) 81x2 + 175y2 = 63, m = -; (b) y2 = 6x, m = 2; 124 ANALYTIC GEOMETRY [CHAP. VI] (c) 16x2- 25y2 = 16, m = 1; (d) 9x2 + 9y2 =25, m = 4. Ans. (a) 3x +- 5y i 4 = 0, (b) 8x - 4y + 3 = 0. 2. Determine the equations of the tangent and normal to each conic below at the point indicated. Check the results graphically. (a) 2 + 4y2 + 2x + 8y - 20 = 0, (2, -3); (b) X2 - y2-4x + 6y + 7 = 0, (4, - 1); (c) 162 + 16y2 - 25 = 0, (1, - ); (d) y2 +x =0, (-2, -4); (e)2 + 9y2 = 40, (-2,2); (f) x2- 2 + 16 = 0, (-3,5); (g) 4x2 - 9y2 = 7, (2, -1); (h) xy = 6, (2, 3). Ans. (a) Tan. 3x - 8y = 30, Nor. 8x + 3y = 7. (b) Tan. x+ 2y - 2 =0, Nor. 2x - y - 9 =0. 3. For each conic below find the equation of the diameter which bisects the system of parallel chords having the given slope. Construct the figures. (a) y2 = 6x, m = - 3; (b) 4x2 + y2= 36, m=3; (c) 4-2 -y2 = 36, m = 2. 4. For each conic below find the equation of the diameter to the point indicated, and the slope of the chords which it bisects. Construct the figures. (a) 2x2 + 3y2 = 50, (- 1, - 4); (b) y2 = 6x, (2, 243); (c) 4x2 -y2 = 32, (-3, 2). Ans. (a) y = 4x, -. 5. For each ellipse and hyperbola in exercises 3 and 4 find the equation of the diameter conjugate to the one indicated. 6. For each conic below find the equation of the chord of contact of the point indicated. Construct the figures. (a) x2 + y2 = 36, (2, 6); (b) 4x2 + 9y2 = 100, (3, 4); (c) y2 + 4x =0, (2, -3); (d) x2 - 4y2 = 16, (1,3); (e) xy = 5, (2, 2); (f) 4X2 -y2 +16 =0, (1,3). 7. For each of the following conics determine the equation of the polar of the point indicated. Construct the figures. (a) x2 + y2 = 49, (- 1, 1); (b) 4x2 + 9y2 = 36, (2, 1); (c) 4X2 - 9y2 = 36, (4,1); (d) x2 -y2 + 16 = 0, (8,1) (e) y2 =8x, (2, -3); (f) xy =4, (-4, -2). TANGENTS, NORMALS, DIAMETERS 125 General Exercises 8. Find the equations of the tangents to 4x2 + 16y2 = 25 which are parallel to 4x - 5y + 7 = 0. 9. Find the equations of the tangents to x - 4y2 = 9 which are perpendicular to 2x - 4y - 3 = 0. 10. Using equation (7) determine the equations of the two tangents to y2 = 4x which pass through the point (- 4, 3). Ans. x + y + 1 = 0, x - 4y + 16 = 0. 11. Derive the equations of the lines with slope 22, which are tangent to 15x2 - 8y2 + 30x + 32y - 137 = 0. Ans. y = 12x + 3 + -2, y = fZ2x +1 + t2. 12. Find the equation of the normal to xy = 4 which has the slope 4. Ans. x - 4y a 15 = 0. 13. Find the equation of the normal to y2 = 4x which is perpendicular to 3x - 4y = 0. Ans. 36x + 27y = 136. 14. Find the equation of the normal to 8x2 + 9y2 = 72 whose slope is 1. Ans. -17x - -17y + 1 = 0. 15. Prove that the line joining the center of a circle to any point is perpendicular to the polar of the point with respect to the circle. 16. Two perpendicular lines are tangent to y2 = 4x, one of them touching the curve at (1, - 2). Find the equations of these tangents, and show that they meet on the directrix. 17. Two perpendicular lines are tangent to x2 + 4y2 = 20, one of which touches the curve at (4, 1). Find the equations of these tangents and show that they intersect on the circle x2 + y2 = r2, where r2 is equal to the sum of the squares of the semi-axes of the given ellipse. 18. Find the equations of the normals to the ellipse (x2/a2) + (y2/b2) = 1 which pass through the center. Use equation (13), p. 115. 19. Find the coordinates of those points of the ellipse (x2/a2) + (y2/b2) = 1 the normals at which pass through a focus. 20. If in an ellipse the normal at one extremity of the latus rectum passes through the opposite end of the minor axis, what is the eccentricity of the ellipse? Ans. ~1 -. 21. Given the coordinates (x1, yi) of one extremity of a diameter of the ellipse (x2/a2) + (y2/b2) = 1, determine the coordinates of the extremities of the conjugate diameter. Ans. (-(a/b)yl, (b/a)xi) and ((a/b)yl, - (b/a)xl). 126 ANALYTIC GEOMETRY [CHAP. VI] 22. Using the results of exercise 21 show that if a' and b' are the lengths of any two conjugate semi-diameters of an ellipse, then a'2 + b = a2 b2. 23. Given the coordinates (xi, yi) of one extremity of a diameter of the hyperbola (x2/a2) - (y2/b2) = 1, determine the coordinates of the extremities of the conjugate diameter. Ans. ((a/b)yi, (b/a)xi) and (- (a/b)yl, - (b/a)xi). 24. Using the results of exercise 23 show that if a' and b' are the lengths of any two conjugate semi-diameters of a hyperbola, then a2 - b'2 = a2-b2. 25. Prove that the tangent at any point of (i) a parabola, (ii) an ellipse, (iii) a hyperbola is parallel to the chords bisected by the diameter drawn to that point. 26. For each of the conics below determine the equation of the chord whose middle point is the point indicated (a) y2 = 4x, (2, 1). (b) 2 + 4y2 = 16, (1, 1). (c) y2 = 6x, (3, -3). (d) X2 -4y2 = 16, (1, -1). Ans. (a) 2x - y - 3 = 0, (b) x + 4y - 5 = 0. 27. Find the equations of the two conjugate diameters of the ellipse (x2/a2) + (y2/b2) = 1 whose lengths are equal. Ans. y = =- (b/a)x. 28. Find for the following conics the equations of the two tangents to the curve from the point indicated in each case. (a) y2 = 4x, (2, - 3). (b) 8X2 + 21y2 = 221, (- 1, - ) (c) xy = 2, (2, - 3). (d) 9x2 - 25y2 = 225, (5, 2). Ans. (a) x + y + 1 = 0, x + 2y + 4 = 0. (b) 16x - 63y - 221 = 0, 40x - 21y + 221 = 0. 29. Using the method of Art. 67, derive the equations of the tangent and normal to each of the following curves at the point indicated: (a) y = X3 + 2x, (1,3). (b) y =2x3 + 3, (-2, -13). (c) y = 4- 1, (2, 15). (d) y = x3, (1, 1). Ans. (a) Tan. 5x - y - 2 = 0, Nor. x + 5y - 16 = 0. (b) Tan. 24x- y + 35 = 0, Nor. x + 24y + 314 = 0. CHAPTER VII APPLICATIONS OF ANALYTIC GEOMETRY 73. Remarks and definitions.-Analytic geometry is, in the hands of the mathematician, a powerful instrument for proving theorems relating to, and discussing the properties of, curves and loci whose equations are known. This chapter will be devoted to an explanation of some of the methods used. They will be illustrated by the discussion of properties of the conics. DEFINITIONS. I. The distance on the X-axis from the point where a tangent to a curve cuts this axis, to the foot of Y the ordinate at the point of tangency, is called the subtangent of the point. Thus TM is the subtan- gent of the point P. T M N X II. The distance on the X-axis from the foot of the A ordinate at any point on a FIG. 67 curve, to the intersecticn with the X-axis of the normal to the curve at that point, is called the sub-normal of the point. Thus MN is the subnormal of the point P. In other words the subtangent and subnormal of a point on a curve are the projections on the X-axis of the portions of the tangent and normal respectively which lie between the point of tangency and the intersection with the X-axis. 127 128 ANALYTIC GEOMETRY [CHAP. VII] 74. Subtangent and subnormal of a point on the parabola. I. The length of the subtangent of P(xl, yi) on the parabola y2 = 2px is 2x1. The equation of the tangent TP to the parabola y2 = 2px, at the point P(xi, yl) is (7), p. 113, yyi = p(x + xi). /T o F M N X FIG. 68 Making y = 0, we have x = - xl, which is therefore the intercept OT of TP on X'X. But if OT = - x, then TO = x, and hence the length of the subtangent TM is TM = TO + OM = xi + xi = 2x. (1) II. The length of the subnormal of any point on the parabola y2 = 2px is constant and equal to p, the semi-latus rectum. The equation of the normal PN to y2 = 2px at P(xi, yi) is (12), p. 115, y -yl = - (x - x). P Make y = 0, then x = xl + p, which is the length of ON, the x-intercept of PN. Hence the length of the subnormal MN is MNr = MO + ON = - x + (xi + p) = p. (2) [ART. 76] APPLICATIONS 129 Let the student deduce from I and II two methods for constructing the tangent to a parabola at a given point. 75. Deductions from Art. 74, I. I. The tangent at any point of a parabola makes equal angles with the axis of the curve and the focal radius to the point of tangency. The notation is the same as in Art. 74, and also the figure, with the addition of the focal radius FP, and the directrix, which intersects the axis of the parabola at D. See Fig. 69. From the definition of the parabola FP = DM = DO + OM = ~p + x1, and from Art. 74, I TF = TO + OF = xl + 2p, hence FP = TF. The triangle TFP, therefore, is isosceles, and angle FTP = angle FPT. (3) Hence, also: II. The normal PN at any point on the parabola bisects the angle between the focal radius FP, and the diam- Y eter PL through the point. 76. Additional proper- \ ties of the parabola. \ L I. T h e.perpendicular from the focus of a parabola to any tangent meets the lat- /T D 0 FM N X ter on the tangent at the vertex. Let y2 = 2px be the par- FIG. 69 abola, and draw FQ perpendicular to the tangent TP, Fig. 69. To prove the theorem either form of the equation of the tangent, (4), p. 110, or (7), 10 130 ANALYTIC GEOMETRY [CHAP. VII] p. 113, may be used. Let us choose the equation in terms of the slope y = mx + (i) 2m' Hence FQ, the line through F(~p, 0), perpendicular to PT, has the equation 1 y = - — (x -p) (ii) Eliminating y from (i) and (ii), and solving for x, we have after reduction, x = 0. Therefore, the intersection of PT and FQ lies on the Y-axis which is the tangent at the vertex. II. The perpendicular from the focus of a parabola to any tangent meets the diameter through the point of tangency on the directrix. If (xl, yl) be the point of tangency the equation of the tangent is yyl = p(x + x1) and that *^ ^R> ^ J of the diameter is y = yl. ' / aThe details of the proof are / "^^ / >-.T left to the student. 77. Theorems on the ellipse. I. The normal at any point of the ellipse bisects the angle ~FIG. 70 ~ between the focal radii drawn to that point. Let (x2/a2) + (y2/b2) = 1 be the ellipse, then the equation of the normal at the point P(x1, yi) is a2yl (x x) Y - Y1 - (- X1). (i) b2X1 [ART. 77] APPLICATIONS 131 It is proved in elementary geometry that if a line PN drawn from the vertex P of a triangle FPF' divides the opposite side F'F in segments F'N, NF which are proportional to the adjacent sides F'P, FP, then this line bisects the angle F'PF. Hence, PN being the normal to the ellipse at P, if the proportion F'N F'P =_- (ii) NF FP is true, PN bisects the angle F'PF. To show this let y = 0 in (i), then a2 - b2 x = ON = - x = e2xl, by (11), p. 87, a2 and, since F'O = OF = ae, F'N = F'O + ON = ae + e2xl = e(a + exi), NF = OF - ON = ae - e2zx = e(a - exi), F'N a + exl ~~~NF ~~a-ex1 e(iii) NF a - ex, Also by (ii) and (iii), Art. 54, F'P a + exl FP a - ex l Hence from (iii) and (iv) it follows that (ii) is true, and therefore the theorem is proved. Since PN bisects F'PF it follows also that the tangent PT bisects the angle FPR. This property connecting the tangent and normal at a point of the ellipse with the focal radii drawn to that point affords a convenient method for constructing the tangent and normal to an ellipse at a given point. II. DEFINITION. The circles concentric with a given ellipse, and having for diameters the major and minor axes of the ellipse, 132 ANALYTIC GEOMETRY [CHAP. VII] are called respectively the major auxiliary circle and the minor auxiliary circle of the ellipse. Y P A' B' O M B A T X D FIG. 71 Thus the circle ADA'D' is the major auxiliary circle of the ellipse ACA'C', and the circle BCB'C' is the minor auxiliary circle. III. DEFINITION. A pair of points P, P', Fig. 71, on an ellipse and on its major auxiliary circle, determined by a perpendicular to the major axis, are called corresponding points. IV. THEOREM. The ordinates to a pair of corresponding points on an ellip-e and its major auxiliary circle are to each other respectively as b is to a. Taking the coordinate axes as indicated in Fig. 71 the equation of the ellipse ACA'C' is x2 y2 a2 b2 I (i) and that of the major auxiliary circle ADA'D' is x2 + y2 = a2 (ii) The corresponding points P, P' have the same abscissa OM = x. [ART. 78] APPLICATIONS 133 Designate MP by y, and MP' by y'. Then from (i) b MP = y = a2 -x2, (iii) a and from (ii) MP' = ' = a2 - 2. (iv) Dividing (iii) by (iv) y b!^ ~ — ~ a~ ~ (4) 78. The eccentric angle of a point on the ellipse. I. DEFINITION. The eccentric angle of a point on the ellipse Y is the angle between the major axis and the line connecting @ ' the center with the corresponding point on the major auxiliary circle. Thus, the angle AOP' =, 0 M A X is the eccentric angle of the point P on the ellipse. II. THEOREM. The coordinates of a point on the ellipse in terms of the eccentric angle FIG. 72 are x = a cos 4, y = b sin 4. (5) From the right triangle MOP' OM = OP' cos 4, or x = a cos 4, since OP' = OA = a. Also, MP' = OP' sin 0, or, letting MP' = y', y' = a sin 4,. Hence from (4), MP = y = b sin 4. When the coordinates of a point on a locus are thus expressed 134 ANALYTIC GEOMETRY [CHAP. VII] in terms of a third variable, this variable is called a parameter, and the equations (5) are called the parametric equations of the locus. 79. Additional methods of constructing an ellipse. I. Construction of an ellipse by two pins and a string. The ~~B ~ major and minor axes, AA', ' P BB', are supposed to be given. x,,___ ^ \ _Locate the foci F, F' by making BF = BF' = CA. A'F C F /A Fix two pins firmly at F and F', and with a piece of string B' make a loop over the pins FIG. 73 of such length that, when stretched taut with a pencil in the loop, the point of the pencil will be at B. Move the pencil, keeping the string taut. Then the pencil will describe the ellipse BA'B'A, because if P be any position of the pencil the sum of the two distances F'P + FP = F'B + FB = A'A. See Art. 54. II. Another method of construction. Upon the straight edge AB of a piece of paper or cardboard mark off PR equal to the semi-major axis of the re- y quired ellipse, and PQ equal B to the semi-minor axis. Draw D X'X and Y'Y at right angles, and place the edge of/ the marked paper so that R X /Q C X will fall on Y'Y, and Q on X'X. Then P will be a R point on the ellipse which has the given semi-axes and F. 4 center at 0. To prove that this construction will give points on the ellipse [ART. 80] APPLICATIONS 135 with semi-axes RP = a, QP = b, take X'X, Y'Y as axes of coordinates, and let angle XQB = 0. Then the coordinates of P are x = DP = RP cos 0 = a cos 0, y = CP = QP sin = b sin 0. Since these coordinates have the form (5), p. 133, the locus of the point P(x, y) is an ellipse. 80. Theorems on the hyperbola. I. The tangent to a hyperbola at any point bisects the angle between the focal radii drawn to that point. The proof, which is similar to that of Art. 77, I, is left to the student. II. The portion of any tangent to the hyperbola, which is included between the asymptotes, is bisected at the point of tangency. R,Q x FIG. 75 Taking the asymptotes as axes of coordinates, the equation of the hyperbola will be xy = C, (i) 136 ANALYTIC GEOMETRY [CHAP. VII] and that of the tangent T1T2 at P(x', y'), see (10) p. 115 y'x + x'y = 2c. (ii) From (ii) making y = 0, the x-intercept OT1 of the tangent is 2c 2x'y' x- - - 2x'. yI y That is OT1 = 2x' = 20M. Hence MP which is parallel to OT2 bisects T1T2, since it bisects OT1, which proves the theorem. 81. Loci by elimination of parameters. I. Perpendicular tangents to the parabola meet on the directrix. Let y2 = 2px be any parabola, then,P 1 mp y =mx +, y= —x- (i) 2m' m 2 will be the equations of any two perpendicular tangents. Equations (i) may be written in the form my - m2x = p, my + x -m2p, from which, subtracting the first from the second, (m2 + l)x = - p(m2 + 1), or x = - p. (ii) This result gives the value of the abscissa of the point of intersection of the two perpendicular tangents (i). Since it does not contain m, this same value will result for all pairs of perpendicular tangents. Hence x - p is the equation of the locus of the point of intersection of all pairs of perpendicular tangents. It is the equation of the directrix, and therefore the theorem is proved. [ART. 81] APPLICATIONS 137 II. VARIABLE PARAMETERS. When in an equation a quantity, other than the coordinates x and y, is assumed to be variable, this quantity is called a variable parameter. Thus in equations (i) m is a variable parameter if we take these equations to represent not a particular pair of perpendicular tangents, but all possible pairs of perpendicular tangents as m takes in succession all possible values. When the equations of two loci are given containing the same variable parameter, and this parameter is eliminated between them, the resulting equation in x and y is that of the locus of the intersection of the two given loci, because the new equation thus derived expresses a relation between x and y which holds true simultaneously for the coordinates (x, y) of all the points of intersection. Thus in I above by eliminating the variable parameter m between the two equations (i) there results an equation (ii) which is satisfied by the coordinates of the points of intersection of all the pairs of tangents represented by equations (i), and is therefore the equation of the locus of their intersection. This process is further illustrated in III and IV below. NOTE. The fact that y is eliminated simultaneously with m in I is purely incidental to this particular problem. In cases where more than one variable parameter occur the same general principles apply. One or more additional equations between the parameters are then necessary, however, in order to effect the elimination of the parameters. III. The major auxiliary circle of an ellipse is the locus of the intersection of any tangent to the ellipse with the perpendicular upon it from the focus. The equation of the tangent to the ellipse (x2/a2) + (y2/b2) = 1 in terms of the slope is, by (1), p. 109, y = mx i~ a2m2 + b2, (i) 138 ANALYTIC GEOMETRY [CHAP. VII] The perpendicular to (i) through the focus (ae, 0) is y = - -(x - ae). (ii) To obtain the locus of the intersection of these two lines the variable parameter m must be eliminated from (i) and (ii). From (i) y - mx = ~ a2m2 + b2 (iii) and from (ii) my + x = ae = aa2- b2. (iv) Squaring (iii) and (iv) y2 - 2mxy + m2x2 = a2m2 + b2, (v) m2y2 + 2mxy + x2 = a2 - b2. (vi) Adding (v) and (vi) (1 + m2)y2 + (1 + m2)x2 = a2(1 + m2) or x2 + y2 = a2, the equation of the required locus. IV. Perpendicular tangents to the ellipse meet on the circle x2 + y2 = a2 + b2. The equations of a pair of perpendicular tangents to the ellipse (x2/a2) + (y2/b2) = 1 are y = mx a2m2 + b2, (i) and y =- - 1 + b2, (ii) from which m must be eliminated. From (i) y - mx =- a a2m2 b2, (iii) and from (ii) my + x = -t aa2 + m2b2. (iv) [ART. 81] APPLICATIONS 139 Squaring and adding (iii) and (iv) (1 + m2)y2 + (1 + m2)x2 = (1 + m2)a2 + (1 + m2)b2, or x2 + y2 = a2 + b2 V. DEFINITION. The circle x2 + y2 = a2 + b2 is called the director circle of the ellipse (x2/a2) + (y2/b2) = 1. EXERCISES ON CHAPTER VII 1. Draw FP from the focus F to any point P on a parabola, and FL perpendicular to FP at F. Show that FL and the tangent to the parabola at P meet on the directrix. 2. From any point P of a parabola let fall PD, perpendicular to the axis of the curve at D. Let K be another point on the parabola at a distance from the axis = 'DP. Draw DK, and produce it to intersect the tangent at the vertex V, at R. Prove VR = 3DP. 3. The tangent at a point P on a parabola meets the directrix at R, and the perpendicular to the axis through the focus F, at Q. Prove FR = FQ. 4. Prove that in any parabola the tangent at one extremity of the latus rectum is parallel to the normal at the other extremity. 5. Let the tangents at any two points P and Q on a parabola meet at R, and let S be the middle point of the chord PQ. Draw RS. Prove (i) that T, the middle point of RS, is on the parabola, (ii) that RS is parallel to the axis of the parabola, (iii) that PQ is parallel to the tangent at T. 6. Prove that the line from the focus of a parabola to the intersection of any two tangents bisects the angle between the lines from the focus to the points of tangency. 7. Prove (i) that the tangents at the extremity of any focal chord of a parabola meet on the directrix, (ii) that these tangents are perpendicular to each other, (iii) that the chord is perpendicular to the line joining the focus and the intersection of the tangents. 8. Prove that the circle constructed on any focal chord of a parabola as diameter is tangent to the directrix. 9. Prove that if a circle be constructed with any focal radius FP of a 140 ANALYTIC GEOMETRY [CHAP. VII] parabola as diameter, it will be tangent to the tangent at the vertex of the parabola. 10. Given a parabola with its axis and focus, describe six different methods of constructing geometrically a tangent to the parabola at any point on the curve. 11. Given a parabola, describe a method for constructing geometrically its axis, focus and directrix. 12. Prove that the perpendicular from the focus of an ellipse or hyperbola to any tangent, meets the line passing through the center and the point of tangency, on the corresponding directrix. 13. Prove that tangents to an ellipse and to its major auxiliary circle intersect on the major axis produced. 14. Prove that any two tangents to an ellipse or hyperbola intersect on the diameter which bisects the chord joining the points of tangency. 15. Prove that if a straight line intersects a hyperbola, the segments on this line included between the curve and its asymptotes, are equal. 16. Tangents are drawn to an ellipse or hyperbola from any point on one of the directrices. Prove that the line drawn through the points of tangency passes through the corresponding focus. 17. A perpendicular is erected to any focal chord of an ellipse or hyperbola at the focus. Prove that this perpendicular meets the tangent at either extremity of the chord, on the corresponding directrix. 18. Prove that in any ellipse or hyperbola the product of the perpendiculars from the foci upon any tangent is equal to the square of the semi-conjugate axis. 19. Perpendiculars are erected to the transverse axis of any ellipse or hyperbola at the focus F, and any other point M, meeting the curve in R and P respectively. Draw the tangent at R, and let it meet MP (produced) in K. Show that MK = FP. 20. If the vertex of a hyperbola is twice as far from the center as from the focus, find the slopes of the asymptotes. 21. Let the tangent at one vertex of a hyperbola cut the conjugate hyperbola at P and P'. Prove that the tangents at P and P' pass through the other vertex. 22. Prove that an ellipse and hyperbola having the same foci cut each other at right angles. 23. Let P be any point on a hyperbola, V the vertex, and C the APPLICATIONS 141 center. The tangent at P meets CV at M, and the tangent at V meets CP at S. Prove that MS and VP are parallel. 24. Prove that in an equilateral hyperbola any two perpendicular diameters are equal. 25. Prove that in any hyperbola all triangles formed by a tangent and the asymptotes are equal in area. 26. Prove that in any hyperbola, (i) the distance from a focus to an asymptote equals the semi-conjugate axis, (ii) the distance from the center to the foot of the perpendicular from a focus to an asymptote equals the semi-transverse axis. 27. Prove that in any ellipse or hyperbola the polar of any point of a diameter is parallel to the conjugate diameter. 28. Given an ellipse, with its axes, foci and directrices, describe three methods for constructing a tangent to the ellipse at any given point. 29. Given a hyperbola, with its axes, foci and directrices, describe three methods for constructing a tangent to the hyperbola at any given point. 30. Given an ellipse (the curve only) show how to determine geometrically, its center, axes, foci and directrices. CHAPTER VIII THE GENERAL EQUATION OF THE SECOND DEGREE 82. Introductory note.-From equations (28)-(32), pp. 99-100, it can be seen that the equation of every conic whose principal axis is parallel to one of the coordinate axes can be expressed in the form Ax2 + By2 + Gx + Fy + C = 0, (1) that is, with no term in xy; and, conversely, from the discussion of Art. 64 it follows that every equation of the form (1) represents either a conic, a pair of straight lines, or an imaginary locus. To avoid unnecessary repetition the locus corresponding to equation (1) will always be referred to as a conic. This term will therefore include pairs of straight lines, which will be called degenerate conics,* and imaginary loci, which will be referred to as imaginary conics, since their equations are of the same type as those of real conics. Conies which are not degenerate or imaginary are called proper conics. 83. Theorem.-The general or complete equation of the second degree in two variables Ax2 + Hxy + By2 + Gx + Fy + C = 0 (2) always represents a conic. It should be observed that the coefficients A, H, B in equation (2) cannot all be zero, as the equation would then cease to be one of the second degree; and also that if H = 0 no proof is necessary, as the equation is then in the form (1). It is assumed therefore that H + 0. * Sometimes also called composite conics. 142 [ART. 83] EQUATION OF THE SECOND DEGREE 143 Apply to (2) the formulas (26), p. 96, for turning the axes through the angle 0, without changing the origin, x = x' cos -y' sin 0, y = x' sin 8 + y' cos 6. Substituting in (2) we have, after reduction x'2(A cos2 0 + H sin 0 cos 0 + B sin2 0) + x'y'[H(cos2 6 - sin2 0) + 2(B - A) sin 0 cos 0] 3 + y'2(A sin2 0- H sin 0 cos 0 + B cos2 0) + x'(G cos 0 + Fsin 0) + y'(F cos 9 - G sin 6) +C = 0. The term in x'y' will disappear from this equation if 0 has a value which will make the coefficient of x'y' equal to zero. Equating this coefficient to zero, and solving for 0, we have H(cos2 0 - sin2 0) + 2(B - A) sin 0 cos 8 = 0, or H cos 20 + (B - A) sin 20 = 0, from which tan 20 = B (4) This always gives a real value of 20, and hence, by turning the axes through the angle 0 thus determined, equation (2) can always be reduced to the form (1), which proves the theorem. EXERCISES. Determine in each of the following cases the value of the angle 0 through which the axes must be turned to reduce the equation to a form without the term in xy. 1. 2x2-4xy + 5y2 + 6x -3y- 1 = 0. 0 = tan-. 2. 3x2 +2xy- y2 +5x + 3y +5 = 0. 8 = tan-l. 3. x2 + 3xy + y2-6x + 2y + 7 = 0. = tan- oo = 45~. 4. 42 - 4xy + y2 + 2x - 6y - 10 = 0. 0 = tan-1 (- ). 5. 6x2-5xy - y2 + 4x + 10y - 7 = 0. 0 = 2 tan-1 (- ). 6. x2 + 6xy +9y2-2x-6y -10 = 0. 0 = ~ tan-1 (- ' ). NOTE. No attempt is made at this point to go further and determine the complete form of the new equation, as it would be tedious and difficult to work out the values of the new coefficients by means of 144 ANALYTIC GEOMETRY [CHAP. VIII] equation (3). A better method for reaching this result will be given below. See Art. 86, II. 84. Central and non-central conics. I. A conic is said to have a center when there is a single point in its plane which bisects all chords of the conic which pass through this point. Such a point is a center of symmetry of the curve, and is called its center. II. A conic which has a center is called a central conic. Such are the ellipse and hyperbola. III. A conic which has no center is called a non-central conic. Such is the parabola. IV. THEOREM. An equation of the second degree which contains no terms of the first degree in x and y, is the equation of a conic with center at the origin. If equation (2) contains no terms of the first degree in x and y it becomes Ax2 + Hxy + By2 + C = 0. (i) If this equation is satisfied by the coordinates (x1, yi), it is also satisfied by (- xl, - yl). That is if P(xi, yi) is on the conic, P (xiIy,) so also is Q(- xi, - y), and every chord PQ of the conic is bisected at 0. Hence 0 is the f ^0 /5 x center of the conic.,,'I / V. THEOREM. The equation Y o' / yAx2 + Hxy + By2 + Gx + Fy (-X1X~- ^ 1+-{- C = 0 represents a central conic if 4AB - H2 1 0, and a noncentral conic if 4AB - H2 = 0. Transform equation (2) to new axes OX', OY' parallel to the original axes, letting (xo, yo) represent the new origin referred to the original axes. The formulas of transformation are x = x' + Xo, y = ' + yo, [ART. 85] EQUATION OF THE SECOND DEGREE 145 and substituting in (2) the result, after reduction, is Ax'2 + Hx'y' + By'2 + (2Axo + Hyo + G)x' + (Hxo + 2Byo + F)y' (5) + (Axo2 + Hxoyo + Byo2 + Gxo + Fyo + C) = O. This equation is to be reduced, if possible, to one without the terms in x' and y'. This requires that 2Axo + Hyo + G = O, Hxo + 2Byo + F = 0. (6) Solving for Xo, yo, -HF - 2BG GHI - 2AF Xo 4A B- H2' y = 4AB - 2 (7) These values of xo and yo are determinate if 4AB - H2 * 0, but if 4AB - H2 = 0 they are either indeterminate or infinite, according as the values of the numerators in (7) are zero or not. Hence if 4AB - H2 * 0 equations (7) determine a point, to which if the origin be transferred, equation (2) will reduce to a form having no terms of the first degree in x' and y'. It will therefore be the equation of a central conic with the new origin for center (see IV,.above). But if 4AB - H2 = 0 no such determinate point exists, and hence the conic is non-central. It should be carefully noted, as appears from (5), that when this transformation is made the coefficients of x'2, x'y', and y'2 in the new equation are the same as those of the corresponding terms in the old equation, and that the new absolute term is obtained by substituting the coordinates of the new origin for x and y in the left-hand member of the original equation. 85. Transformed equation of a central conic.-By Art. 84, V, the equation of a central conic referred to coordinate axes passing through the center is in the form Ax'2 + Hxy' + By'2 + C' = 0, (8) 11 146 ANALYTIC GEOMETRY [CHAP. VIII] where C' = Axo2 + Hxoyo + Byo2 + Gxo + Fyo + C, (9) Xo and Yo having the values given by (7). I. To express the value of C' exclusively in terms of the coefficients A, H, B, G, F, C of equation (2). From equation (9) 2C' = 2Axo2 + 2Hxoyo + 2Byo2 + 2Gxo + 2Fyo + 2C, = Xo(2Axo + Hyo +-G) + yo(Hxo + 2Byo + F) + Gxo + Fyo + 2C. But by (6), the expressions in parentheses vanish,. C' = Gxo + Fyo + 2C. (10) Substituting in (10) the values of Xo, yo from (7) the result is HF - 2BG OH -2AF 2C' = G +F-F + 2C, 2 = 4AB - H2 4AB - H2 or 4ABC + FGH - AF2 - BG2 - CH2 = AB -- (11) II. The numerator of the fraction in (11) is called the discriminant of Ax2 + Hxy + By2 + Gx + Fy + C. Hence representing the discriminant by D, we have D 4ABC + FGH - AF2 - BG2 - CH2. (12) EXERCISE. Show that of the six equations on p. 143, nos. 1, 2, 3, 5 represent central conics. Find the coordinates of the center of each of these four, and show that the equations reduce to the forms given below, when transformed to new axes through the center, parallel to the original axes. 1. Center at (- 2, - ); 2x'2 - 4x'y' + 5y2 - 2 = 0. 2. ( l (- 1, 2); 3x'2 + 2x'y - y12 + -1- = 0. 3. " " (- - 2-, -5-); x'2 + 3x y' + y,2 + 11 = 0. 5. " (, -, 20); 6x'2 _ 5x'y -- y,2 + 9 = 0. [ART. 86] EQUATION OF THE SECOND DEGREE 147 86. Equations of central conics. Final reduction.When the general equation (2) represents a central conic it has been shown in Art. 84 how to reduce it to the form (8), Ax'2 + 2Hx'y' + By'2 + C' = 0 (i) by transforming to new axes passing through the center. We can next remove from (i) the term in xy by turning the axes through the angle determined by equation (4), p. 143, as shown in Art. 83. The equation is thus reduced to the form A'x"2 + B'y"/2 + C' = 0 (13) where, as appears from (3), p. 143 A' = Acos20 + H sin 0 cos 0 + B sin2, (14) B' = A sin2 0 - H sin 0 cos 0 + B cos2 0, and C' remains unchanged. I. Selection of the value of 0.-Since equation (4), p. 143, determines two values of 20 less than 360~, which differby 180~, there will always be one available value of 20 between - 90~ and + 90~. It will be found convenient always to use this value of 20. Hence when tan 20 is positive, 0 will be a positive angle between 0~ and 45~; and when tan 20 is negative, 0 will be a negative angle between 0~ and 45~. II. Determination of the values of A' and B'.-In (14) make the trigonometric substitutions cos2 0 = (1 + cos 20), sin2 0 = ~(1 - cos 20), sin 0 cos 0 = 4 sin 20. Rearranging the terms, the results are A' = ~(A + B) + (A - B) cos 20 + ~H sin 20,1 B' = ~(A + B) - ~(A -B) cos 20- ~H sin 20. 2\ TU1- \ -U/Cj~2L 211bl 148 ANALYTIC GEOMETRY [CHAP. VIII] But tan 20 = H/(A - B), and hence from trigonometry sin 20 = + 4H2 + (A - B)2 (iii) cos 20 A1 {H2 + (A - B)2 Substituting these values in (ii) the results, after reduction, are either A' = [A + B + 2 + (A- B), (15a) B' = '[A + B- 2 + (A - B)2, or A' = '[A + B - WH2 + (A - B), (15b) B' = '[A + B + + (A - B)2, according as the sign of I/H2 + (A - B)2 is taken as positive or negative. To eliminate the ambiguity in the choice of values for A' and B', due to the double sign of H2 + (A - B)2, it will be seen from I that, since 20 is always to be taken in the first or fourth quadrants, cos 20 is always positive. Hence the sign of the radical is to be taken to agree with that of (A - B). The following rule should therefore be used. RULE. If A - B is positive use (15a) for determining A' and B', but if A - B is negative use (15b). 87. Discussion of equation (13), p. 147, A'x"2 + B'y12 + C' =. It has been shown in Art. 86, II, how the values of A' and B' can be determined. Two other relations involving A' and B' are important The first is obtained by adding the two equations (14), which gives A' + B' = A + B. (16) [ART. 87] EQUATION OF THE SECOND DEGREE 149 Also, multiplying the equations (15a) or (15b) the result is 4A'B' = 4AB - H2. (17) Since by hypothesis 4AB - H2 4 0, equation (17) shows that neither A' nor B' can be zero. Hence in discussing equation (13) there are only two cases to consider, according as C' is zero or not. The latter will be taken first. I. C' 4 O. Equation (13) can then be put in the form A'x"2 + B'y"2 = - C', or + C= 1. (18) A' B' If - (C'/A') and- (C'/B') are both positive, (18) represents an, ellipse. This will be the case when A' and B' have like signs, and C' the opposite sign. If - (C'/A'), and - (C'/B') have opposite signs, (18) represents a hyperbola. This will be the case when A' and B' have unlike signs. If A', B', C' all have the same sign, equation (18) represents an imaginary locus, called an imaginary ellipse, since the equation is in the ellipse form. From equation (11), p. 146, C' = D/(4AB - H2), where D is the discriminant (12). Hence when C' * 0, it follows that D 0. This discussion shows that equation (13), when C' * 0, represents an ellipse, real or imaginary, or a hyperbola, according as the product A'B' is positive or negative. Hence from (17), it follows that the general equation (2) represents an ellipse, when 4AB -.-2 > 0, and D:= 0, (1 a hyperbola, when 4AB - H2 < 0, and D 0. (19) 150 ANALYTIC GEOMETRY [CHAP. VIII] II. C' = 0. When C' = 0 equation (13) takes the form A'x"2 + B'y"2 = 0. (20) If A' and B' have unlike signs the left-hand member of this equation can be factored in real factors of the first degree in x" and y". Hence it represents a pair of real intersecting lines meeting at the origin. Also from (17) 4AB - H2 < 0. If A' and B' have like signs the only real values of x" and y" which will satisfy the equation are x" = 0, y" = 0. Hence the equation may be regarded as that of a point ellipse, the limit to which the locus of equation (13) approaches when A' and B' have like signs and C' approaches the limit zero. Or, since in this case the left-hand member of equation (20) can be factored in imaginary factors of the first degree, the equation may be said to represent a pair of imaginary lines intersecting at a real point, the origin. In this case from (17) 4AB - H2 > 0. When C' = 0, D = 0 also, as appears from (11). Hence equation (2) represents real intersecting lines, when 4AB — H2 < 0, and D = 0, imaginary intersecting lines, or a point ellipse, when 4AB - H2 > 0, and D = 0. EXAMPLES. 1. Discuss fully the equation 3x2 + 2xy - y2 + 5x + 3y + 5 = 0. The results already found for this equation in Ex. 2, p. 143 and p. 146, will be used in this discussion. First step. 4AB - H2 = - 16, and by (12) p. 146, D = -52. Hence by (19) the locus is a hyperbola. Second step. Determine the center by (7), p. 145, and transform the equation to new axes through this point, parallel-to the axes XOY. This is done by using (8) and (10), pp. 145, 146. On p. 146 the center [ART. 87] EQUATION OF THE SECOND DEGREE 151 has been found to be (-1, 2), \ i f and the transformed equation is 3'2 + 2x'y' - y'2 + -13 = 0. The coordinate axes are now X'O'Y'. ' Third step. By (4) 0= / tan-1 '. Turning the axes _ through this angle they take the X position X"O'Y". The equation now takes the form (13) where, i — 'since A -B is positive, the values of A' and B' are determined from (15a). Thus A' = ~[2 +- 4 + 16] = 1 + 5, B' = [2 - j4 + 16] = 1 - 5. FIG. 77 Hence the transformed equation is (1 + 4)x"2 + (1- -5)y,2 + 1- = 0, or (%5 + 1)x"2 - (- - 1)y"2 = - This reduces to X,2 y,2 13 13 4(1 + ^5) 4(45 - 1) It is now in the form corresponding to (21), p. 92, and therefore represents a hyperbola with transverse axis in the Y-axis. The semi-transverse axis = - _ - (5 +1 = 1.62, 13(4 - 1 ) and the semi-conjugate axis = ( -1) = 1.00. 4(5 + 1) 1 16 The curve can now be drawn as shown in the figure. 2. Discuss the equation 322 - 8xy + 4y2 - 2x - 4y - 8 = 0. First step. 4AB - H2 = - 16, D = 0. Hence by (21) the locus is a pair of intersecting real lines. 152 ANALYTIC GEOMETRY [CHAP. VIII] Second step. The coordinates of the center (the intersection of the lines) are found by (7) to be x = - 3, yo = - 2; and hence the equation, referred to the new axes X'O'Y', with the point (- 3, - -) as the new origin, is by (8) and (11) 3x2 - 8x'y' + 4y'2 = 0. Y'l Y /B 0 X Third step. The left-hand side of this equation is a homogeneous expression of the second degree in x' and y'. Such an expression can always be separated into linear factors. See III, p. viii. The factors in this case are 3x' - 2y', and x' - 2y'. Hence the two real lines which the given equation represents are 3x' - 2y' = 0, and x' - 2y = 0, 0' FIG. 78 the lines AB, CD, respectively in the figure. NOTE. Since every equation in the form (8), p. 145, in which C' = 0, can be treated as in the " Third step " above, it is unnecessary with such an equation to revolve the axes to remove the term in xy. 3. Discuss fully the equation of exercise 1, p. 143, making use of the results given on pp. 143 and 146, and draw the curve. 4. In the same manner discuss the equation of exercise 3, p. 143. 5. Discuss also the equation of exercise 5, p. 143. 88. Equations of non-central conics.-The general equation Ax2 + Hxy + By2 + Gx + Fy+ C = 0 (i) represents a non-central conic when 4AB - H2 = 0, as has been shown Art. 84, V. If H = 0 then either A or B must also equal zero, and no further investigation is necessary, as the equation can be treated as in Art. 64, Ex. 2. In the general case, when H is not zero, A and B must have the same sign, and hence may [ART. 88] EQUATION OF THE SECOND DEGREE 153 always be taken as positive. The condition 4AB - H2 = 0 also shows that the first three terms in (i) form a perfect square. To simplify the subsequent discussion equation (i) is multiplied by 4B, and 4AB the coefficient of x2 is replaced by its equal H2. The equation may then be written (Hx + 2By)2 + 4B(Gx + Fy) + 4BC = 0. (ii) Equation (ii) is now transformed to new axes, taking Hx + 2By = 0, and 2Bx - H? = 0, (iii) for the new X- and Y-axes respectively. These are seen to be perpendicular to each other by (17), p. 52. The slope of the first. is - (H/2B), hence this change of axes is equivalent to turning the original axes through the angle 0= tan-' (- ). Since B is positive; 0 will be a negative angle, between 0~ and - 90~ in value, if H is positive, and a positive angle, between 0~ and 90~ in value, if H is negative. Y Y Y.XP ),,.U>P <? /,/ / X 00/ X X / ____ x x (a) (b) H positive. H negative. FIG. 79 154 ANALYTIC GEOMETRY [CHAP. VIII] The new abscissa x' of any point P is the perpendicular distance from OY' to P, and the new ordinate y' is the perpendicular distance from OX' to P. Therefore by (20), p. 56 2Bx - Hy Hx + 2By, y =, (iv) /4B2 + H2 = 4B2 H2 where the sign of the radical is to be taken positive. NOTE. For every point on OX', Fig. 79 (a), the original abscissa x is positive, and the original ordinate y is negative. Therefore since both B and H are positive the new abscissa of every point on OX' is positive, as appears from the value of x' given in (iv). Hence OX' is the positive portion of the new X-axis. Similarly it can be shown that OY' is the positive portion of the new Y-axis in the same figure, and that in Fig. 79 (b), where H is negative, points in the quadrant X'OY' have both coordinates positive. Solving equations (iv) for x and y, we have 2Bx' + Hy' - Hx' + 2By' x = -, y = 4B. (v) 44B2 + H2 ^4B2 + H2 Substituting from (v) in (ii) and reducing (4B2 + H2)y'2 + 4B [(2BG - HF)x' 4B2~ + H2 (vi) + (GH + 2BF)y'] + 4BC = 0. Restoring to H2 its value 4AB, and dividing the equation by 4B, the result is (2BG - HF), (2BF + GH) y' = (A + B)y'2 +(2BG-F) X+ (2BF + G) y + C = O. (vii) 2 4B2 + AB 2 4B2 +AB This equation is now in the form B'y'2 + G'x' + F'y' + C = 0, (22) [ART. 89] EQUATION OF THE SECOND DEGREE 155 where B' = A + B, G -2BG - HF Ft 2BF + GH (23) 2 JB2 AB 2 +B2 +2 AB By the process used in Art. 64, Ex. 2, equation (22) is easily reduced to the form F,.^'y 7 2 G - 4B'C y( + ) = _ X - - ' (24) provided G' is not zero. The equation is now in the form (29), p. 99, and therefore represents a parabola, with principal axis parallel to the X-axis, and vertex at the point ((F' - 4B'C)/G', - (F'/2B')). The length of the latus rectum is 2p = I G'/B' 1. 89. Degenerate non-central conics. I. If G' = 0, equation (22) takes the form B'y'2 + F'/ + C = 0. (25) Multiplying by 4B', and adding and subtracting F'2 equation, (25) takes the form (2B'y' + F')2 - (F'2 - 4B'C) = 0, or factoring, (2B'y' + F' + IF'2 - 4B'C)(2B'y' + F' - VF'2- 4B'C) = 0. Equation (25) therefore represents the two straight lines, 2B'y' + F' + JF'2 - 4B'C = 0 2 2B'y' + F' - F'2 - 4B'C = 0 (26) parallel to the X-axis. These lines will be real and distinct, coincident, or imaginary according as F'2 - 4B'C >, =, or <0. II. When 4AB - H2 = 0, the condition G' = 0 in (22) is equivalent, in the general equation (2), to D = 0. See (12), p. 146. In the last term of the expression for D, substitute for H2 its 156 ANALYTIC GEOMETRY [CHAP. VIII] value 4AB, then D = FGH - AF2 - BG2. (i) Also, from (23), if G' = 0, 2BG 2BG - HF = 0, or F =. (ii) H Substituting this value of F in (i) the result is AB2G2 D = 2BG2 -4 — 2 BG2 or, since by hypothesis H2 = 4AB, D = 2BG2 - BG2 - BG2 =. (iii) III. Conclusion. The general equation (2) represents a parabola, when 4AB - 12 = 0 and D 0, parallel or coincident lines, when 4AB - 2 = 0 (27) and D = O. In the latter case the lines are real and distinct, coincident, or imaginary according as F'2 - 4B'C>, =, < 0. (28) EXAMPLES. 1. Determine the nature and position of the locus of the equation 4x2 - 4xy + y2 + 2x - 6y - 10 = 0. First step. 4AB - H2 = 0, D = - 100. Hence by (27) the locus is a parabola. Second step. By (iii), p. 153, the lines -4x + 2y = 0, 2x + 4y = 0 are taken as new axes of coordinates, and hence by (v), p. 154, the formulas of transformation are x' - 2y' 2x' + y' i) x= /, =y -. (i) Substituting in the given equation and reducing, the result is 2 y2 _ (x' + y') - 2 = 0, which is the equation of the parabola referred to the axes X'OY'. [ART. 89] EQUATION OF THE SECOND DEGREE 157 Third step. The last equation is easily transformed as in Art. 64, Ex. 2 into y, 1 2 x, 11, (yt_5>) 2 =w ( )15 / -//x showing that the vertex is at the point/ (- -) = (-2.46, 0.45), (ii) The focus is at a distance F 41F. 20 FIG. 80 on the positive side of the vertex, that is at the point ( —12 + 2'-' ) = (- 2.24, 0.45). (iii) To determine the position of vertex and focus referred to the original axes substitute from (ii) and (iii) in (i). Thus it is found that the vertex is at (- 3, - 2) and the Y Y' f "~/Y "focus at (-, - 9). The slope of the axis is 2. A^^^^^ / 2. Investigate the equation x2 + 6xy + 9y2 - 2x - 6y - 10 = 0. -Ca ---- Xf First step. 4AB - H2 = 0, 70^^^ B and D = 0. Hence by (27) the / ' ' locus is a pair of parallel or / - coincident lines. By (23) F'= - (20 //- ) =FIG. 81 22-T0, B'= 10, and F'2- 4B'C =440. By (28) the lines are therefore real and distinct. Second step. By (iii), p. 153, the new X- and Y-axes are respectively x + 3y = 0, and 3x - y = 0. The transformed equation can be written at once from (25), p. 155. 10y'2 - 2 00y' - 10 = 0. 158 ANALYTIC GEOMETRY [CHAP. VIII] Solving this equation for y', =, or y' = 1.36 and -0.73. The two lines are shown in Fig. 81 in the positions AB and CD. 90. Summary.-The results obtained in this Chapter are summarized in the following table: D=0O D=0 Real ellipse if C' differs in sign from A' and B'. Arts. 85, 86, 87, I. A pair of imaginary lines 4AB - H-2> 0 Imaginary ellipse if A', meeting at the real point B', C' have the same given by (7), p. 145. See sign. Arts. 85, 86, Art. 87, II. and 87, I. A pair of real lines meeting 4AB- H2 < 0 Hyperbola. Arts. 85, at the point given by (7), 86, and 87, I. p. 145. See Art. 87, II. Real parallel lines, when F'2 - 4B'C > 0. Coincident lines, when 4AB- H2 = 0 Parabola. Art. 88. F'2 - 4B'C = 0. Imaginary lines, when F'2 - 4B'C < 0. See (23), p. 155, and Art. 89. The procedure in treating an individual equation has been illustrated in the exercises worked. The steps may be summarized as follows: A. Compute 4AB- H2, and D (12), p. 146. The results will determine the species of the conic, and whether it is proper or degenerate. See (19), p. 149, (21), p. 150, and (27), p. 156. B. (i) If 4AB - H2 * 0, find the coordinates of the center by (7), p. 145, and transform the equation to new axes through this point parallel to the original axes. See Art. 85. [ART. 90] EQUATION OF THE SECOND DEGREE 159 (ii) If D t 0, turn the axes through the angle H 0 = tan-1A, 0 = 2 tanA - B' and the equation will then be in the form (13), p. 147. See Art. 86. (iii) If D = 0 the first transformation will reduce the equation to a factorable form, and further transformation is unnecessary. C. (i) If 4AB - H2 = 0, choose new axes as stated in Art. 88, and transform the equation to these axes. (ii) If D * 0 the new equation will be in the form (22), p. 154, and the parabola can be constructed from this equation; or by a second transformation the origin can be moved to the vertex, the axes remaining parallel to the second set. (iii) If D = 0, the equation after the first transformation will contain only one variable. It can therefore be readily factored, so that further transformation is unnecessary. EXERCISES ON CHAPTER VIII Normal Exercises In exercises 1-12 determine the species of the conic whose equation is given, and its location with reference to the axes. In the case of each central conic determine the position of the center, the slope of the principal axis, and, if the conic is real and not degenerate, the lengths of the semi-axes. If the conic is non-central and not degenerate determine the position of the vertex, the slope of the axis, and the value of p. Draw each locus. 1. 5x2 - 5xy + 3y2 - 5x + 6y - 3 = 0. Ans. A real ellipse, center at (0, - 1), slope of major axis 1.48, semi-axes 2.14, 0.95. 2. 5x + 2xy - y2 + 6x - 4y - 12 = 0. Ans. A hyperbola, center at (-, - —, slope of transverse axis 0.16, semi-axes 1.26, 2.65. 160 ANALYTIC GEOMETRY [CHAP. VIII] 3. 9x2 + 12xy + 4y2 + 10x - 54y - 68 = 0. Ans. A parabola, vertex (2, - 2), p = T 3-. 4. x2 + 6xy + 9y2 + 2x + 6y - 25 = 0. Ans. Parallel lines with slope -. 5. 2x2 + 6xy + 5y2 - 10x - 6y + 53 = 0. Ans. Imaginary lines, meeting at (16, - 9). 6. 9x2 - 6xy + y2 + 12x - 4y + 4 = 0. 7. 3x2 - 6xy + 5y2 + 5x - 10y + 12 = 0. 8. x2 - 4xy + 3y2 + 3x - 7y + 2 = 0. 9. 3x2 + 6xy + 4y2 + 4x - 8y - 12 = 0. 10. 3x2 - 2xy - 5y2 + 3x + 7y - 8 = 0. 11. 4x2 - 4xy + y2 - 4x + 2y + 4 = 0. 12. 3x2 - 12xy + 2x - 3y - 20 = 0. General Exercises 13. Prove that x + zy = -a is a parabola which touches the coordinate axes at the extremities of the latus rectum. 14. Prove that -b + ~ y = 4ab is a parabola which touches the X-axis at x = a, and the Y-axis at y = b. 15. For what value of c will x2 + 2xy - y2 + 6x - 2y + c = 0 represent a pair of real lines? Ans. c = 1. 16. For what value of c will 2xy - 4x + 7y + c = 0 represent a pair of real lines? Ans. c = - 14. 17. For what value of c will x2 + 4xy + 4y2 + 4x + 8y + c = 0 represent a pair of coincident lines? Ans. c = 4. 18. For what value of c will a2x2 + 2abxy + b2y2 + c = 0 represent a pair of coincident lines? 19. Prove that hxy + gx + fy + c = 0 is in general an equilateral hyperbola with hx + f = 0, and hy + g = 0 for asymptotes, but that if fg = ch it represents a pair of real lines. 20. If in the general equation (2), p. 142, 4AB - H2 = 0, and if the term in xy be removed by turning the axes through the angle 0 [Eq. (4), p. 143], show that the resulting equation will always be one in which the coefficient of either x2 or y2 is zero. CHAPTER IX HIGHER PLANE CURVES, PARAMETRIC EQUATIONS 91. Introductory remarks. DEFINITION. The locus corresponding to any equation not in the form Ax + By + C = 0, or Ax2 + Hxy + By2 + Gx + Fy + C = 0, and not reducible to one of these two forms, is called a higher plane curve. For example, y3 = cx2, x3 + y3 + 3axy =0, y = a sin bx, y = log x, y = tan-l x, y = ea are all equations of higher plane curves. To determine whether an equation containing radicals or fractional exponents, is the equation of a higher plane curve, it must first be reduced to a form without radicals or fractional exponents. A few curves of this kind will be discussed. 92. Loci of equations of the form xny = c.-The curves corresponding to equations of the form xny = c, (1) where c is a constant, and n has any numerical value, constitute an important family. Equations of this form arise in the study of the expansion of gases and vapors, and the corresponding curves are used to represent graphically the changes of pressure and volume which take place under different conditions, as for example, in the cylinder of a gas or steam engine. I. For certain values of n equation (1) represents a straight line or a conic. Thus, for n = 0 it is the equation of the straight line y = c, n = 1 it is the equation of the hyperbola xy = c, n = - ~ it is the equation of the parabola y2 = c2X, n = - 1 it is the equation of the straight line y = cx. n = - 2 it is the equation of the parabola y = cx2. For other values of n the locus is a higher plane curve. 12 161 162 ANALYTIC GEOMETRY [CHAP. IX] Two important general conclusions with reference to equation (1) may be noted. II. For any given value of n, c may be determined so that the locus shall pass through any one given point in the plane. For example, let n = 4, then the locus of xiy = c will pass through the point (4, 3) if 4. 3 = c, that is if c = 24. III. For all values of n the locus will pass through the point (1, c). For equation (1) is always satisfied if x = 1, y = c. xny=l FIG. 82 In Fig. 82 are drawn several curves of this family exhibiting the shape and position of the curves for different values of n. The value c = 1 is used in all the curves. Hence they all pass through the point (1, 1). IV. In connection with the curves discussed in this article a device frequently employed is to use what may be called logarithmic plotting. If the logarithms of both sides of equation [ART. 93] HIGHER PLANE CURVES 163 (1) be taken the result is n log x + log y = log c. (i) This equation is of the first degree in log x and log y. Hence if log x and log y be plotted as coordinates instead of x and y the resulting figure is a straight line, with slope - n and y-intercept log c. Therefore to each curve of the family xny = c there corresponds a straight line of the family (i) with the same value of n. In applying equation (1) to the study of problems involving pressures and volumes of gases and vapors this method is conveniently employed. All of the properties derivable from the curves themselves can be obtained equally well from the corresponding straight lines in the logarithmic diagram. Paper ruled logarithmically is made for this purpose. The lines on the paper are spaced proportionately to the logarithms of the natural numbers instead of the numbers themselves. By its use the transformation from the actual values of x and y to the logarithmic values is made mechanically without any reference to tables of logarithms. It should be noted that this transformation of these curves into straight lines applies only to those portions which lie in.the first quadrant, where x and y are both positive. NOTE. The equations xny = c, and y = cxn are essentially the same; as by changing the sign of the exponent the one form passes into the other. Irrational values of the exponent n are sometimes met with. In such cases an approximate rational value of the exponent is used. For example if n = 2, the approximate value n = 1.41 may be used. 93. The sine and cosine curves. I. The sine curve or sinusoid. The equation of this curve is y = sin x. (2) In plotting this curve the values of x are expressed in radians, 164 ANALYTIC GEOMETRY [CHAP. IX] so that graphically x will be the length of the arc, radius = 1, which subtends the angle whose sine is y. In this way both x and y are expressed in terms of the same unit of length, as shown in trigonometry. The same remark applies to all equations in which either of the variables stands for an angle. The following table gives a few pairs of simultaneous values of x and y. x = 0 ir = 0.52 1Tr = 0.79 ~2r = 1.57 r = 4.71 etc. y = 0 ~ = 0.50 22 = 0.71 1 - 1 etc. For an accurate construction of the curve a table of natural sines should be used. In the figure the solid curve is the sine y -tXL X 2R ^>c\ 0% 3r - -4',.. w '-,/ 2w'., _.q.^7, ~~. 3..FIG. 83 curve. It extends indefinitely to right and left along the X-axis, and cuts this axis at an infinite number of points, whose abscissas are x = nwr, where n is any whole number, positive or negative. The value of y is never greater than + 1, nor less than - 1. II. THE COSINE CURVE. The equation of this curve is y = cos x. (3) This curve is constructed in the same manner as the sine curve. It is the same curve shifted to the right or left along the X-axis a distance ~2r. It is drawn dotted in Fig. 83. 94. Modified forms of the sine curve.-The curve whose equation is (2) is the fundamental form of the sine curve. Some modifications of it are important from their application in the theory of wave motion. I. y = a sin x. (4) [ART. 94] HIGHER PLANE CURVES 165 The curves which correspond to equations of this form for different numerical values of the constant a, differ from the locus of equation (2) only in the fact that the highest and lowest points on the curve are at distances + a and - a respectively from the X-axis, and the ordinate of any other point on any of the curves is to the corresponding ordinate on the curve of (2) as a: 1. They all cut the X-axis where sin x = 0, that is where x = nr, n being any whole number, positive or negative. II. y = a sin bx. (5) In this equation a and b have any constant numerical value. The effect of the constant a has been pointed out in the discussion of equation (4) in I. The value of the constant b affects the location of the points where the curve cuts the X-axis, for these points occur where sin bx = 0, which gives bx = nir, or x = (n/b)7r, where n is any whole number, positive or negative. Hence for b > 1 these points are nearer each other than in the curve of (2) or (4), and for b < 1, they are farther apart. Taking a =, b = 2, the equation is y = | sin 2x, and the corresponding curve is Fig. 84a. If a = 3, b = 1, the equation is y = 3 sin 3x, and the corresponding curve is Fig. 84b. Y Y 1~ ja X 0 2 27 3w7P*(a) =_3sin 2 (b) y= sin3X FIG. 84 III. y = a sin (bx + c). (6) The curve corresponding to this equation, in which c, as well as a and b, has any numerical value, differs from that of equation (5) only in the location of the points in which it cuts 166 ANALYTIC GEOMETRY [CHAP. IX] the X-axis. Let a = 3, b = 2, c = 0.52, then the equation is y = 2-sin (2x + 0.52). (i) To determine where the curve cuts the X-axis let y = 0, then 2x + 0.52 = nr, or x = I(nr - 0.52). (ii) Hence every integral value of n in (ii) determihes the abscissa of a point where the curve cuts the X-axis. Thus: n = -1 gives x = - r -0.26 = - 1.83, n = 0 gives x = - 0.26, n = 1 gives x = Tr - 0.26 = 1.31, n = 2 gives x = r - 0.26 = 2.88, etc. etc. etc. The curve is therefore the same as that in Fig. 84a moved a distance 0.26 to the left, as shown in Fig. 85, where OA = - 0.26. In order to determine the value of y corresponding to any given value of x, proceed as follows. Let x = 0.40, then 2x + 0.52 = 1.32, then y = 2- sin (1.32 radians) = 3 sin 75~ 38' = 4(0.969) = 1.45. Hence (0.40, 1.45) is a point on the curve. From this it may be conY ^"\<~ y^s eluded in general that the curve of y = a sin (bx + c) 0 - /-\ ---.-/ X- - cuts the X-axis at points determined by bx + c = nir, or x = (nr - c) /b, where n is any FIG. 85 integer, and hence that the curve differs from that of y = a sin bx only in being shifted a distance - (c/b) along the X-axis, [ART. 95] HIGHER PLANE CURVES 167 95. Logarithmic curves. I. The curve y = loga x. (7) The general shape of this curve is the same for all values of the base a, provided a is positive and greater than unity, the only cases which are usually met with. For all such values of a we have log a = 1, log 1 = 0, and log O = - oo. Hence the corresponding curves all cut the X-axis at x = 1, they all approach the negative end of the Y-axis without cutting it, and as x increases from x = 1, y also increases, but more slowly Y Y A A 0 X B FIG. 86 FIG. 87 than x. In Fig. 86 the curve drawn full is the locus of y = loge x and the dotted curve is the locus of y = logio x. II. The curve y- =ex. (8) This curve, Fig. 87, is the logarithmic curve in a different position, as y = eX, and x = loge y are the same equation in different forms. When x = 0, y = 1. As x increases, y increases also, and as x approaches - oo, y approaches zero. 96. y = eax sin bx. (9) This equation and the corresponding curve are practically 168 ANALYTIC GEOMETRY [CHAP. IX] important on account of their connection with the theory of alternating currents. The letters a and b stand for numerical constants, and e is the Napierian base, 2.718-.. (V (c), p. ix.) For all finite values of a and b, y = 0 when x = 0, hence all curves of this family pass through the origin. Since eal cannot vanish for any finite value of x, the points where the curve cuts the X-axis are determined by making sin bx = 0, from which, as in Art. 94, bx = nr, or x = (n/b)7r, where n has in turn all integral values. These points are therefore the same as for the curves of equation (5). Also, since sin bx cannot be numerically greater than unity, the curve lies between the two curves y = i ea, touching y = eax when sin bx = 1, and y = - eax when sin bx = - 1. Let a = -, b = 6, then the equation is y = eOx sin 6x. (i) This curve will cut the X-axis at the points determined by x = Wnr, where n is any integer, that is at x = 0, -ry, -w, -r, etc. Let x = 0.2 in (i), then i/ y = e0-3 sin (1.2 radians) = e03 sin 68~ 45'. - / /- I/ \ /| ~Using logarithms the result is 0 a log y = 0.09971, or y= 126. 3- W I Ir The ordinates of other points e;, s j | are similarly computed. \ \ | A portion of the curve on ~\ \ | the positive side of the origin is drawn in Fig. 88. The horizontal unit used in the drawing is ten times the verFIG. 88 tical unit. [ART. 97] HIGHER PLANE CURVES 169 The points A, B, C, D, etc., midway between the points where the curve cuts OX, are the feet of the ordinates to the points where the curve touches the two curves y = a: ea. EXERCISES. 1. Construct on a diagram similar to Fig. 82 the loci of the family xny = c for which c = 1, and n = 3, 3, - ~, - 3 respectively. 2. Construct on one diagram the curves of the family xny = c, for which c = 2, and.n = - 2 - 1, - 2, 0, 2, 1, 2 respectively. Point out the general differences between these curves and those of Fig. 82. 3. Name some characteristics common to all curves of the family xny = 1, (i) in which the value of n is positive, (ii) in which the value of n is between 0 and - 1, (iii) in which the value of n is less than - 1. 4. Construct on one diagram the loci whose equations are y = tan x, y = 4 tan x, y = tan 4x, y = tan (x + 4). 5. State in what definite way each of the loci y = a tan x, y = tan bx, y = tan (x + c), y = a tan (bx + c) differs from the curve y = tan x. 6. What is the effect on the graph of y = a tan (bx + c) of a change in a? in b? in c? 7. Construct the following loci (a) y = sin 2x, (b) y ='sin 7rx, (c) y = sin2 x, (d) y = x + sin x, (e) y = 1 -cos x, (f) y = tan2x, (g) y = sin-' x, (h) y = tan-' x, (i) y = 4 sin-1 (x2), (j) y = x sin-l x, (k) y =x + sin-' x, (1) y = ex sin 2x, (m) y = 4ex, (n) y = ex4, (o) y = 4 log x, (p) y = log (x + 4), (q) y = e log x, (r) y = x log x, 97. Parametric equations.-When the coordinates of points on a curve are expressed separately in terms of a third variable, or parameter, the two equations are called the parametric equations of the curve. This was done in the case of the ellipse, Art. 78, where the coordinates of any point on the ellipse are expressed in equations (5), p. 133, in terms of the eccentric angle 4, x = a cos X, y = b sin (. (i) 170 ANALYTIC GEOMETRY [CHAP. IX] When this can be done it is merely an alternative to that of expressing the relation between x and y by means of a single equation. The latter can be derived from the parametric equations by combining them so as to eliminate the parameter, as shown in Art. 81. Thus eliminating q between the two equations (i) by writing them in the form x y - = cos, -= sin, a b and then squaring, and adding, the result is x2 2 a2 b2 the ordinary equation of the ellipse. EXAMPLES. 1. Show that x = a sec t, y = b tan t are parametric equations of the hyperbola. The two given equations can be expressed in the form - = sec t, = tan t. a b Hence, squaring and subtracting the second from the first X2 y2 a2 b2 1, which is the equation of a hyperbola. 2. Construct the locus whose parametric equations are x = It2 - 1, y = t+2. To determine the coordinates of points on the locus assign values to the parameter t, and compute the corresponding pairs of values of x and y. Thus t=-4 -3 -2 -1 0 1 2 3 4 x = 3 0 - - 1 - O 3 etc, y =-2 -1 - 1 2 3 4 5 6 The curve is drawn in Fig. 89. It looks like a parabola. To determine [ART. 98] HIGHER PLANE CURVES 171 whether it is a parabola, eliminate the y parameter t between the two given equations, thus obtaining the equation =3 t-2 x=- (- 2)2 - 1, or t= y -4y - 4x = 0. t=o which is the equation of a parabola. 3. Construct the locus whose para- =t metric equations are x = aO, y = b t=-2 (1- cos ). X0 Assigning values to 0, and com- t3 -puting the corresponding values of x =-4 and y the results are tabulated as follows: FIG. 89 0=0 1ir T Ir 1trr 7T it x=0 I-ra 1~ra ~ 3ra 2ra 7ra 7ra {7ra etc. y=0 b(l1- 3) lb b b b(l+ ~ ) 2b b(l+ 3) The curve is drawn in Fig. 90. Y / b \ I / Ib b -'-ra 0 — Tra Xra x27ra 27ra i2 2 FIG. 90. 98. The cycloid. DEFINITION. If a circle rolls on a straight line the curve described by any fixed point on the circle is called a cycloid. Let r be the radius of the rolling circle, and OA, Fig. 91, the straight line on which it rolls. Let the point P of the circle describe the curve OHA, where 0 is the position of the describing 172 ANALYTIC GEOMETRY [CHAP. IX] point where it touches the given line, and A the position it reaches after one complete revolution. The curve OHA is then a cycloid, sometimes called one arch of the cycloid, since as the circle continues to roll the point P will describe a succession of curves identical with OHA in form. The line OA is called the base of the cycloid. As the circle rolls every point of it comes successively into contact with OA, hence the length of OA is the same as the Y H L OM N A X FIG. 91 circumference of the circle, 2-r. For the same reason when the circle has turned through the angle NCP = 0, as shown in the figure, distance ON = arc PN = rO. (i) Take OA for X-axis, and 0, the initial position of P for origin, then for the coordinates of P, x = OM, y = MP. (ii) But OM = ON + NM = ON + LP = rO - r sin 0, (iii) because by (i) ON = rO, and from the figure LP = -r sin 0. The negative sign is used for LP because it is measured in the negative direction for abscissas. Also, MP = NL = NC + CL = r - r cos 0 (iv) [ART. 99] HIGHER PLANE CURVES 173 because CL = - r cos 0. The negative sign is used for CL because it is measured in the negative direction for ordinates. Hence from (ii), (iii) and (iv) x = r(O - sin 0), y = r(l - cos ), (10) which are the parametric equations of the cycloid. 99. The epicycloid and hypocycloid. I. THE EPICYCLOID. DEFINITION. If a circle roll on the outside of a fixed circle the curve described by any point of the rolling circle is called an epicycloid. Let P be the generating point on the rolling circle, and A one of its positions where it touches the fixed circle. Take the center 0 of the fixed circle for origin, and OA for the Xaxis. Let AQ be an arc of the epicycloid. The line OC, join- X ing the centers of the two circles passes through K, their point of C L contact. Let AOK = 0, KCP P = 4, OK = R, CK = CP = r. Draw NC, MP perpendicular to OX, and BCL parallel to OX, AJ N M X meeting MP produced at L. I Angle BCO = 0. Then for the FIG. 92 coordinates of P, x = OM = ON + NM = ON + CL, (i) where ON = OC cos NOC = (R + r) cos, (ii) and CL = - CP cos BCP = - r cos (4 + 0). (iii) 174 ANALYTIC GEOMETRY [CHAP. IX] The negative sign is written in (iii) because CL is positive, while cos (4 + 0) is negative... x = (R + r) cos (- + 0). (iv) Also, y = MP = ML + LP =NC + LP, (v) where NC = OC sin NOC = (R + r) sin 0, (vi) and Y LP = -CP sin BCP = -r sin (0 + 0). (vii) The negative sign is written in (vii) because LP is negative, C \( $WL while sin (/ + 0) is positive. K p.b. y = (R + r) sin 0 - r sin (0 + 0). (viii) iO/ -\ Afn rj N - M To eliminate 0 from equations (iv) and (viii) we have FIG. 92 arc KP = arc AK, hence r0 = R0, or 4 = 0. r Substituting this value of q in (iv) and (viii) the results are x = (R + r) cos0 - rCos ( + ~') ( y = (R + r) sin 0 - r sin ( + r) 1) which are the parametric equations of the epicycloid. II. THE HYPOCYCLOID. DEFINITION. If a circle roll on the inside of a fixed circle the curve described by any point on the rolling circle is called a hypocycloid. [ART. 100] HIGHER PLANE CURVES 175 Using the same notation as for the epicycloid the student will be able to show that the parametric equations of the hypocycloid are FIG. 93 x = ( - r) cos + rCosr - ),1 y = ( - r) sin 0 - r sin ( -r.O) (12) These equations may also be obtained by changing the sign )f r in the equations of the epicycloid (11). 100. Special forms of the hypocycloid. I. When the diameter of the rolling circle is half the diameter of the fixed circle, the hypocycloid is a straight line. Making R = 2r in equations (12) the results are x = r cos 0 + r cos 0 = 2r cos 0, y = r sin 0 - r sin 0 = 0, from the second of which it appears that the generating point travels on the X-axis, which proves the theorem. II. To find the equations of the hypocycloid when the diameter of the rolling circle is one fourth the diameter of the fixed circle. 176 ANALYTIC GEOMETRY [CHAP. IX] Making R = 4r in (12) we have x = 3r cos 0 + r cos 30, y = 3r sin 0 - r sin 30, Y\B or, substituting for cos 30 and sin 30 /" A"',\ their values in terms of 0 [see (28) / y\ \ x], 7^ \N C\ -o / > x = 4r cos30, y = 4r sin3 0, \ \/ / which are the required equations. Replacing 4r by its value OA = R, D these equations become FIG. 94 xC = R cos3, y = R sin3 0. (13) The four points A, B, C, D in the figure are called cusps, and the curve is known as the four-cusped hypocycloid. From (13) x1/3 = R1/3 cos 0, y113 = R1/3 sin 0, hence squaring and adding X2/3 + y2/3 = R2/3. (14) which is the equation in non-parametric form. 101. The involute of the circle. DEFINITION. Imagine a cord to be wound tightly around a circle, then as it is unwound, keeping it taut, the end of the cord describes the involute of the circle. Let AQ be an arc of the involute generated by the point P, whose initial position was at A. LP is tangent to the circle at L. Take the center of the circle 0 for origin, and OA for the X-axis. Draw MP and NL perpendicular to OX, and KP parallel to OX. Let AOL = 0 and OL = r. Since the triangles NOL, KLP are similar, the angle KLP = 0. Then for the coordinates of the point P x = OM = ON + NM = ON + KP, (i) y = MP = NK = NL + LK, () [ART. 101] HIGHER PLANE CURVES 177 FIG. 95 where ON = OL cos 0 = r cos 0, NL = OL sin 0 = r sin 0, (iii) and KP = LP sin 0, LK = - LP cos 0 (because LK is negative). But LP = arc LA = r9, hence KP = rO sin 0, LK = - rO cos 0. (iv) Hence finally substituting from (iii) and (iv) in (i) and (ii), x = r cos 0 + rO sin, } y = r sin 0 - rO cos 0, which are the required parametric equations. EXERCISES ON CHAPTER IX 1. Construct the loci determined by the following pairs of parametric equations, as in Exs. 2 and 3, Art. 97. (a) y = 4t, x = 3t + 5; (b) x = 8t, y = 4t2; (c) x = 3 sin t, y = 4 cos t; (d) x = 4t, y = -6t + 4; (e) x = 4t2, y = 3t3- 1; (f)y = 4 sec t, x = 3 tan t. 2. Determine the equation in x and y for each of the loci in Exercise 1, by eliminating the parameter between the given equations. Ans. (a) 4x - 3y - 20 = 0, (b) 16y = x2, (c) 16x2 + 9y2 = 144. 3. Express the equation of the circle x2 + y2 = r2 in parametric form. 13 178 ANALYTIC GEOMETRY [CHAP. IX] 4. Obtain parametric equations for the ellipse (X - )2 (y )2 a2 + b2 5. When a circle rolls on a straight line the curve traced by a fixed point on a radius of the circle (or on the radius produced) is called a trochoid. If a is the radius of the circle, and b the distance of the tracing point from the center, show that the parametric equations of the trochoid are, with axes as in Art. 98, x = a - b sin, y = a - b cos0. 6. Construct from their equations the trochoids (Ex. 5) for which a = 2, and b = 1, 2, 3, respectively. 7. Construct the loci determined as follows: (a) x = 4(0 - cos 0), y = 4(1 + sin 0). (b) y = 2 sin 0-sin 20, x = 2 cos 0- cos 20. (c) y = 3 sin0 - sin 30, x = 3 cos 0 + cos 30. (d) x = 2 cos 0 + 20 sin 0, y = 2 sin 0-20 cos 0. 8. Find the equation in x and y for the locus in Ex. 7 (a) by eliminating 0. Ans. x = 4 sin-1 [- (y - 4)] - 8y - y2 9. Obtain equations in parametric form for the parabola y2 = 2px. 10. Prove that x = a(0 + sin 0), y = - a(l - cos 0) are the parametric equations of the cycloid when the origin is at the highest point on the curve, and the X-axis is the tangent at this point. 11. What does the trochoid (see Ex. 5) become when b = a? When b = O? How do the trochoids in which b < a differ from those in which b > a? 12. A straight line revolves about the point (0, - b) intersecting the X-axis in the variable point M. On this line the point P is taken at the constant distance a from M. Prove that the parametric equations of the locus of P (called the conchoid) are x = a cos + b cot 0, y = a sin 0. CHAPTER X POLAR COORDINATES 102. Polar coordinates of a point in a plane. THEOREM. The position of a point in a plane is determined by its distance and direction from a fixed point in the plane. Let A be a fixed point in the plane, and AB a fixed line drawn from A determining in the plane a fixed direction. Let P be any point in the plane. Then the position of P is determined if the \ A B length AP and the angle BAP are FIG. 96 given. The distance AP and the angle BAP are coordinates of the point P, since they locate the position of the point in the plane. DEFINITIONS. Coordinates of the kind here described are called polar coordinates. The fixed point A is called the pole of the system, and the fixed line AB is called the polar axis, or the initial line. The coordinates AP and BAP of any point P are designated respectively by r and 0, and the point P is the point (r, 0). The distance r is called the radius vector of the point, and the angle 0 is called the vectorial angle. If in Fig. 96 AP = r = 5 and BAP = 0 = 120~, the point P is the point (5, 120~) or (5, -3r). 103. Signs of the coordinates.-The sign of the vectorial angle 0 is determined as in trigonometry. It is therefore positive when measured counter-clockwise from AB, and negative when measured in the opposite direction of rotation. The sign of the radius vector r requires some explanation. 179 180 ANALYTIC GEOMETRY [CHAP. X] Every value of 0 determines a line through A, as NM. One part of this line AM forms the M terminal side of 0. The other portion, AN is the terminal side 0\~ ^ of 0 produced through A. On every such line NM there is a A\ B positive and a negative direction. The positive direction is always from A on the terminal side of 0, and the opposite direcFIG. 97 tion is negative. Thus in Fig. 97 for 0 = BAM, AM is the positive direction on the line NM, and AN the negative direction. A point P on AM, whose vectorial angle is 0 = 120~, Fig. 98, therefore has a positive radius vector. Let AP = 3 units, then P is the point (3, 120~), or (3, ]2r). The same value 0 = 120~ may also be used as the vectorial angle of a point p Q on AN, but AQ the radius vector of Q being measured on the terminal side of 0 produced through A, is negative. s=3oo0 / B If the length of AQ is 3 units =-60~ the coordinates of Q will be (- 3, 120~), or (- 3, ~Tr). Instead of taking 0 = 120~, Q the value 0 = -60~ may be N used, since this also determines FIG. 98 the line MN in the same position as before. In this case AN is the terminal side of 0. Hence AN is now the positive, and AM the negative direction, and the coordinates of P and Q [ART. 104] POLAR COORDINATES 181 are (- 3, - 60~), and (3, - 60~) respectively. Also since the angles - 60~ and + 300~ are the same, we may write P(- 3, 300~), Q = (3, 300~). Any of the infinitely many ways whereby the angle which MN makes with AB can be expressed, may be used as the value of 0, and the radius vector of any point on MN will then be positive or negative according as the point lies on the terminal side of 0, or on the terminal side produced through A. EXERCISES. 1. Locate each of the following sets of points, each set on one diagram (a) (2, 30~), (4, 120~), (2, ~ir), (3, fTr), (4, ar), (3, 281~), (5, 7), (2, T7r), (3.7, 47~ 18'), (1.47, 15~ 14'), (0, 0), (0, 47r). (b) (4, 75~), (4, - 75~), (-4, 750), (-4, - 75~). (c) (3, -w-,), (3, - 1-7r), (- 3, l), (- 3, - o7r-). (d) (4, 1r), (4, - ), (- 4, -617r) (-4, - ), (4, 1-). 2. What is the distance between the points (4, ^T) and (4, i7r)? between (4, ir) and (- 4, Jr)? between (4, -r) and (- 4, lr)? 3. What angle with the polar axis is made by the line connecting (4, ir) and (4, - ~r)? 4. What is the direction of the line connecting (6, ~7r) and (6, 27r)? 5. On what curve are situated all points whose radius vector is 4? whose radius vector is a? What is the equation in polar coordinates of the circle whose radius is a, and whose center is at the pole? 6. What is the locus of all points whose vectorial angle is ~r? whose vectorial angle is a? What is the equation in polar coordinates of the straight line which passes through the pole, making an angle a with the polar axis? Ans. 0 = a. 104. Curve plotting in polar coordinates. 1. Plot the curve r = 2a cos 0. By assigning values to 0 in the equation, and computing the corresponding values of r, the coordinates of points on the curve are obtained and plotted on the figure. The curve is then traced through these points. This should be done in an 182 ANALYTIC GEOMETRY [CHAP. X] orderly manner beginning usually with 0 = 0~, and proceeding with values of 0 at regular intervals. The curve proceeds from point to point in the same order as that in which the points are obtained as 0 increases. The smaller the interval between successive values of 0 the greater the number of points obtained, and hence the more accurately the curve can be traced. It is sometimes desirable to make this interval only 5~ or 10~. Usually, however, careful examination of the equation will reveal how r varies as 0 increases, and a smaller number of points will serve the purpose. In this example values of 0 differing by 30~ will be used. The following table shows how the work may be conveniently arranged.,__ p0 r Point 0~ 2a P1,1// / \ 30~ 1.73a P2 600 a P3 _____.-'__, /900 0 A A 2a P1 B 1200 -a P5!i~~\\ "~ ) 150~ - 1.73a P6 ~ \\ \^. /~ ~180~ - 2a Pi Ij~ iPS-^^, ^ ~2100 - 1.73a P2 ~I \ --- —--- r6~ ^etc. etc. etc. FIG. 99 It should be noted that the point (- 2a, 180~) is the same as (2a, 0~), and that as 0 increases beyond 180~ the series of points obtained is the same as that determined by the values of 0 from 0~ to 180~. This recurrence of points often happens in plotting equations containing trigonometric functions, and is a result of the periodicity of the values of these functions. The place in the series of values of 0 where the recurrence begins [ART. 104] POLAR COORDINATES 183 is not always 180~, but depends upon the form of the equation and upon the trigonometric functions involved in it. The curve in Fig. 99 is a circle. (See Art. 105). 2. Plot the curve r = a sin 30. Taking values of 0 at intervals of 10~ the table of corresponding values of 0 and r is as follows: 0 r 10 r0 00 0 1000 - 0.87a 10~ 0.5a 110~ - 0.5a 20~ 0.87a 120~ 0 300 a 130~ 0.5a 400 0.87a 140~ 0.87a 500 0.5a 150~ a 60~ 0 160~ 0.87a 700 - 0.5a 170~ 0.5a 80~ -0.87a 180~ 0 900 - a 190~ - 0.5a Pa FIG. 100 The curve is completely traced as 0 changes from 0~ to 180~. The points Pi, P2, P3 are given respectively by the coordinates (a, 30~), (-a, 90~), (a, 150~). 3. Plot the curve r = 2 - sin 20. Since the value of the sine of an angle is never greater than + 1 nor less than - 1, the values of r in this case are always positive and cannot be less than 1, nor greater than 3. Therefore the curve does not pass through the pole. The minimum value of r is +1, corresponding to values of 0 for which sin 0 =1. Similarly the maximum value of r is + 3, corresponding to values of 0 for which sin 30 = - 1. 184 ANALYTIC GEOMETRY [CHAP. X] The table of corresponding values of 0 and r is now computed. 0 Pt. 0 r tPt 0 r Pt. 0 r Pt. 0~ 2.0 1 210~ 2.7 8 420~ 3.0 15 630~ 2.7 22 30~ 1.3 2 240~ 2.0 9 450~ 2.7 16 660~ 3.0 23 600 1.0 3 2700 1.3 10 4800 2.0 17 6900 2.7 24 90~ 1.3 4 300~ 1.0 11 510~ 1.3 18 7200 2.0 1 1200 2.0 5 3300 1.3 2 5400 1.0 19 7500 1.3 2 150~ 2.7 6 360~ 2.0 13 570~ 1.3 20 etc. etc. etc. 1800 3.0 7 390~ 2.7 14 600~ 2.0 21 6 14 19 B 2 O 2i 22 FIG. 101 In plotting this curve it must be noted that although when 0 = 360~, r has the same value as when 0 = 0~, the curve is not complete until 0 = 720~, after which the points recur in regular order. 4. Trace the curve r = aG. This is one of a large class of curves called spirals. This is known as the spiral of Archimedes. The values of 0 are taken in radians. The values of r evidently increase indefinitely as 0 increases. [ART. 1041 POLAR COORDINATES 185 The portion of the curve drawn solid is that given by positive 6 0 1 7r 3 57r T7 r 0 0.79a 1.57a 2.36a 3.14a 3.93a etc. FIG. 102 values of 0, and the dotted portion is given of 0. by negative values 5. Plot the curve r2 = a2 cos 20. In this equation if cos 20 is negative r is imaginary. Hence the curve does not exist for values of 20 between 90~ and 270~, that is for values of 0 between 45~ and 135~. For every value of 0 which makes cos 20 positive r has two values numerically equal with opposite signs. Thus: 00 10~ 20~ 300 400 450 r =- a i 0.97a ~ 0.88a ~- 0.71a ~ 0.42a 0 FIG. 103 Values of 0 between 0~ and - 45~, and between 135~ and 225~, give the same series of values. of r as those between 0~ and 45~. This curve is called the lemniscate of Bernouilli. EXERCISES. Plot the following curves in polar coordinates. 1. r = 2a sin 0. 2. r = a sin 20. 3. r = a csc 20. 4. rO = a. 5. r =a sin 10. 6. r = 2a tan 0. 7. r = acos20. 8. r = acos30. 9. r2 =a2cos0. 10. r 100. 11. r = 2a sec 0. 12. r a tan 20. 186 ANALYTIC GEOMETRY [CHAP. X] 13. r = a(l - cos 0). 15. r2= 4 4 2 - COs 6 ' 17. r = 1 i - cos 0 19. r2 = a2(1 + cos 0). 21. r = a(1 + 2 sin 0). 14. r = a(1 + sin ~0). 4 16. r = 1 -4 cos 8 4 18. r2 = 1 - cos 0 1 - cos 20 ' 22. r = a(1 + 2 sin 0). 105. Transformation of coordinates.-If the origin of a rectangular system of coordinates, and the positive extension of the Xaxis be taken as the pole and polar 7 axis, respectively, of a polar system, the following relations exist between the coordinates (x, y) and the two systems. FIG. 104 (r, 0) of any point P in x = r COS 0, r = Vx~ + /2, y = r sin 0, 0 = tan-1', X (1) (2) Only the positive value of /x2 + y2 is taken in (2) and in the following equations, because a positive value for r may always be taken. See Art. 103. From (1) and the first equation in (2) we have also x a X cos 0 = -= r x2 + /y2 sin 0 - y, (3) r /x+ + y2z' which are often useful. To illustrate the application of these formulas take the equation y2 = 2ax - X2 representing the circle with radius a, and center at (a, 0). To find the polar equation of this circle write it in the form x2 + y2 = 2ax, [ART. 106] POLAR COORDINATES 187 and substitute x2 + y2 = r2 from (2), and x = r cos 0 from (1). The result is r2 = 2ar cos 0, or r = 2a cos 0, which is therefore the polar equation of the circle of radius a which passes through the pole, and has its center on the polar axis. See Fig. 99, p. 182. As another example take the equation r = a(cos 0 - sin 0), and transform it to rectangular coordinates. Using formulas (2) and (3) the result is x2X2 + Y2 2 + y2 or x2 + 2 - a(x - y) = 0, which is the equation of a circle passing through the origin, with its center at (~a, - a). See Art. 40. 106. Polar equation of the straight line.-Let HK be any straight line referred to the polar system in which A is the pole and H P AB the polar axis. Draw AM from M the pole perpendicular to HK, and let AM = p, and the angle BAM // = a. These constants, p and a B determine the position of the line, A as in Art. 26. Let P be any FIG. 05 point on the line with coordinates AP = r, BAP = 0. Then in the right triangle AMP, the angle MAP = 6 - a, and hence the required equation is r cos (6 - a) = p. (4) 188 ANALYTIC GEOMETRY [CHAP. X] 107. Polar equation of the conics with focus at the pole. Let P be any point on a conic whose focus is at the pole A, and whose principal axis coincides with the polar axis AB. Let DN be the directrix. Then DA = p. From the definition N P of a conic, Art. 45, AP = e NP, (i) / Dbut AP =r, NP = DM= DA D B + AM = p + r cos 0. Hence \ substituting in (i) r = e(p + r cos 0), FIG. 106 from which ep (5) r 1 - e os0 (5) If the conic is a parabola e = 1, and equation (5) becomes r= os (6) EXERCISES ON CHAPTER X Normal Exercises 1. Locate on one diagram each of the following points: (4, 17r), (5, 245~), (- 5, 245~), (4, T7r), (4, -r), (- 4, 7r), (0, 7ir), (4, 7r). 2. Write four other pairs of polar coordinates each of which represents the same point as (2, 60~). 3. Find the rectangular coordinates of the points of Ex. 1, the origin and X-axis coinciding with the pole and polar axis respectively. 4. Find the polar coordinates of the points whose rectangular coordinates are respectively (- 2, - 2), (4, 3), (- 3, 4), (5, 2), (0, 6), (0, - 6), (6, 0), (- 6, 0), the pole and polar axis being as stated in Ex. 3. 5. Find the rectangular equation of the cardioid r = a(1 + cos 0), the pole and polar axis being taken as stated in Ex. 3. Plot the curve from each equation on separate diagrams. Ans. a2(x2 + y2) - (2 + y2 - ax)2 = 0. POLAR COORDINATES 189 6. Find the polar equation of the semi-cubical parabola ay2 = x3, the pole and polar axis being taken as stated in Ex. 3. Plot the curve from each equation on separate diagrams. Ans. r = a tan2 0 sec 0. 7. Plot the loci whose polar equations are (a) r =a sin |0, (b) r = 1 + cos 80, 4 4 (c) r 4 - cos2 ' (d) r =2 cos 432 6 (e)r2 4 -cos2 (f)r = 1 -2 cos 0 ' 144 (g) r cos ( - 60~) = 2, (h) r2 =__ 14 - 9 sin2 08 - 16 cos2 08' 8. Find the polar equation of each of the following loci, and plot the graph. (a) The lines whose inclination to the polar axis is 60~, and whose distance from the pole is 2. (b) The parabola whose focus is one unit from the directrix. (c) The ellipse in which p = 2 units, and e = 1. (d) The hyperbola in which p = 2 units, and e = 3. General Exercises 9. Each of the equations which follow contains a general constant or parameter. Plot the locus of each equation on one diagram for several different values of the parameter which it contains, and note the differences in the resulting graphs due to the change in the value of the parameter. (a) r = a cos 0, (b) r = cos m0, (c) r cos 0 = a, (d) r = 1 + cos m0, (e) r = a + cos 0, (f)r cos ( - a) = 4, (g) r(l - cos 0) = p, 10. Plot each of the following straight lines. (a) r cos (0 - 7r) = 4, (b) rcos (0 + Xi) = 2, (c) r cos (0 - 327~) = 3, (d) r cos 0 = 2. 11. Transform each of the equations of Ex. 10 to rectangular coordinates with the origin at the pole and the X-axis coincident with the polar axis, and plot each line from its resulting equation. Ans. (a) x + y3 -8 = 0, (b) y + 2 = 0. 190 ANALYTIC GEOMETRY [CHAP. X1 12. Find the equation in polar coordinates of the line which is (a) perpendicular to the polar axis, and passes through (4, 0). Ans. r cos 0 = 4. (b) perpendicular to the line 0 = 31r, and passes through (4, 0). Ans. r cos (0 - 60~) = 2. (c) parallel to the polar axis, and passes through (6, ~ir). 13. Derive equation (4), p. 187, by transforming equation (9), p. 43, to polar coordinates. 14. Find the equations of the six conics determined by the following data, the focus in each case being at the pole, and the principal axis of the conic coincident with the polar axis. (a) p =2, e = 1; (b) p = 4, e = ~; (c) p =3, e =2; (d) p = 1, e = ~; (e) p = 4, e = 1; (f) p = 3, e = 3. 15. Find the polar equation of the circle whose center is at (4, 60~), and radius 2. Ans. r2 - 8r cos (0 - 60~) + 12 = 0. 16. Show that 02 = a2 represents two straight lines intersecting at the pole, and symmetrical to each other with respect to the polar axis. (See Ex. 6, p. 181.) 17. Develop a method for finding the distance between two points whose polar coordinates are given, and find the distances between the following pairs of points: (a) (4, ~Tr), (5, Tr); (b) (2, - 4T), (2, w); (c) (ri, 01), (ri, 02); (d) (4, 1w), (4, ~w). Ans. (a) 4i1, (b) 2 43, (c) 2r1 sin ~(01 - 02). 18. Find the lengths of the semi-axes, the eccentricity, and the coordinates of center and foci of each of the following conics 4 4 (a) r = _ cos (b) r= 2 co 1 -2 cos 0' 2 cos O' 10 4 (C) 10 (d) 4 (c) r = 2-5 cos ' (d) r = 1-3cos Ans. (a) a = 4, b =, e = 2, C = (, r), foci = (0, 0), (16, 7). (b) a = 8, b = 4]3, e = C = (4, 0), foci = (0, 0), (,, 0). 19. Determine the polar equation (a) of the ellipse, (b) of the hyperbola, the pole being at the center, and the polar axis coincident with the transverse axis. CHAPTER XI EMPIRICAL EQUATIONS 108. General remarks.-In the latter portion of Chapter II the problem of determining the equation of the locus of a point moving under prescribed conditions was briefly discussed, and some illustrations of the processes of solution were given. Elsewhere at numerous places throughout the preceding pages the same principles were employed, in deriving the equations of the straight line, circle, and conies, and of different loci connected with them. A problem of somewhat different character will now be considered. The coordinates of a series of points are given. These will usually have been obtained from some practical investigation, as a result of some form of measurement, or from some statistical tabulation, as opposed to theoretical determination, and hence must be assumed to be subject to some degree of error, small in comparison with the given values themselves. It is required to find the equation of a curve which, as nearly as possible, passes through the given points. Sometimes in a problem of this kind it is known from theoretical considerations what is the general form of the equation of the curve on which the given points are supposed to lie, and sometimes nothing at all is known except the given coordinates. Two distinct varieties of the problem are therefore to be considered. In either case an equation of a curve derived from data of this kind is called an empirical equation. 109. Empirical equations, the law of the curve given. The method of procedure will be illustrated by examples. 1. The results of a test made upon a piece of steel at the 191 192 ANALYTIC GEOMETRY [CHAP. XI] U. S. Arsenal, Watertown, Mass., to determine the elongation under different loads were as follows. oads, lbs. Elongation, Load, lbs. linchies Load Elongation Load Elongation 5,000 0.0013 35,000 0.0118 44,000 0.0148 10,000 0.0030 40,000 0.0135 45,000 0.0151 20,000 0.0060 41,000 0.0139 46,000 0.0156 30,000 0.0100 42,000 0.0142 47,000 0.0160 43,000 0.0145 It is required, using the corresponding pairs of numbers as coordinates of points to derive the equation of the locus on which the points lie. It is known from practical and theetical cn theoretical considerations that, within the elastic limit of the material, the elongation under different loads is proportional to the load. Hence the locus of these points is theoretically a straight line. Therefore in seeking its equation the proper form to use is y = mx + b, (i) where m and b are to be determined so that the resulting line shall pass as near as possible to all of the given points. Since the difference in the order of magnitude of these two sets of numbers is so great, it is obvious that if corresponding pairs were used as coordinates of points the resulting diagram would be of no practical value. Hence the scales used in the figure should be so adjusted that the numbers expressing abscissas and ordinates shall be of the same order of magnitude. This may be done by making the unit for the first column 1,000 lbs., and that for the second column 0.001 inch. Thus, Load, unit = Elong., unit = 1,000 lbs. 0.001 in. Load Elong. Load Elong. Load Elong. 5 1.3 35 11.8 42 14.2 45 15.1 10 3.0 40 13.5 43 14.5 46 15.6 20 6.6 41 13.9 44 14.8 47 16.0 30 10.0 [ART. 109] EMPIRICAL EQUATIONS 193 The next step is to plot the points whose abscissas and ordinates are the corresponding pairs of numbers in the first and second columns of this table, respectively. See Fig. 107. I I I I I I I 20- - - - 13 15 5 10 4 - 5 /- - - ol ~,~'"" / —l 15 110 15 20 2 5 30 35 45 50 X FIG. 107 Any two of the thirteen given points will determine a line which will approximate more or less closely to the required line. It is desirable to use points that do not lie close together, because if the points are close together a small error in the location of either of them will cause greater displacement of the line than would be produced by a similar error when the points are farther apart. Hence we choose the first and last points of the series, the points (5, 1.3), (47, 16). Substituting these values in (i) the results are 5m + b = 1.3, 47m + b = 16.0, from which m = 0.35, and b = - 0.45. Hence the equation of the line through these two points is y = 0.35x - 0.45. (ii) To test this equation compute the value of y for each given value of x, and compare the result with the corresponding given value of y. The differences are placed in the column A. 14 194 ANALYTIC GEOMETRY [CHAP. XI] Y Y y y x Given Computed A x Given Computed A 5 1.3 1.3 0 42 14.2 14.25 - 0.05 10 3.0 3.05 - 0.05 43 14.5 14.60 - 0.10 20 6.6 6.55 + 0.05 44 14.8 14.95 - 0.15 30 10.0 10.05 - 0.05 45 15.1 15.30 - 0.20 35 11.8 11.80 0 46 15.6 15.65 - 0.05 40 13.5 13.55 - 0.05 47 16.0 16.0 0 41 13.9 13.90 0 The table shows that all of the points except five lie below the line determined by the first and last. This result is not so satisfactory as it should be, and another line should be computed. For this purpose use the second point and the one next to the last, the points (10, 3) and (46, 15.6). Computing m and b as before the results are found to be m = 0.35, b = - 0.50. The equation of this line is therefore y = 0.35x - 0.50. (iii) The student should compute a table of values for this equation similar to the one above. It will be found that the line (iii), determined by the second and twelfth points, passes through the fourth, sixth and eighth also, while of the rest of the points five lie above the line and three below it. It may be noted also that the sum of the positive A's thus obtained is numerically equal to the sum of the negative A's. Equation (iii) may therefore be accepted as a satisfactory result. A systematic method for handling data of this kind for the purpose of arriving at the most probable result is based upon what is known as the method of least squares. It is, however, beyond the scope of this book. 2. When a gas is compressed, or allowed to expand, under such circumstances that heat is neither gained nor lost during the process, the gas is said to be compressed or to expand adiabatically. The equation in this case connecting the pressure [ART. 109] EMPIRICAL EQUATIONS 195 and volume of the gas is pvn = k, (i where p = pressure, v = volume, and n and k are constants. A certain body of air was compressed adiabatically, and corresponding pressures and volumes were observed to be as follows: p = 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 v = 8.78 7.17 6.12 5.38 4.82 4.38 4.03 3.74 Determine the values of n and k. which will give the curve passing as nearly as possible through the eight ~ points which have these pairs of values - of p and v as coordinates. First plot the points using the value of p as abscissas, and the cor - --. responding values of v as ordinates. __ The first attempt to reach a satisfac- tory solution may be made by finding - the curve whose equation has the form (i) which passes through the first and last points of the series, the points - - (1.2, 8.78) and (4.0, 3.74). Substitut- ing these pairs of values in (i) the twoFIG 108 resulting equations are 1.2(8.78)" = k, 4.0(3.74)" = k, (ii) which are to be solved for n and k. Dividing the first by the second the result is 8.78 n 4.0 3.74 - 1.2' 196 ANALYTIC GEOMETRY [CHAP. XI] and taking the logarithms of both sides n(log 8.78 - log 3.74) = log 4.0 - log 1.2 or 0.3706n = 0.5229, from which n = 1.411. Substituting this value of n in the first equation of (ii) the value of k is k = 1.2(8.78)1'4, or log k = log (1.2) + 1.411 log (8.78). log k = 1.4103, from which k = 25.73. Hence the curve of the required type which passes through the first and last points of the series has the equation PV411 = 25.73. (iii) To test this result substitute in it the other six given values of p and compare theteiuiing values of v computed from the equation with the corresponding given values. The results appear in the following table p = 1.6 2.0 2.4 2.8 3.2 3.6 v (given) = 7.17 6.12 5.38 4.82 4.38 4.03 v (computed) = 7.16 6.11 5.37 4.82 4.38 4.03 The computed values of v agree so closely with the given values that equation (iii) may be accepted as a satisfactory result. Had it not been so our next attempt would have been to find the curve which passes through the first and seventh points, then through the second and eighth. If the results obtained from these two curves should still be unsatisfactory, the means of the values of n and k from the last two determinations could be used, or the means taken from all three determinations. 110. Empirical equations, the type of curve not given. This problem arises when a series of corresponding pairs of [ART. 110] EMPIRICAL EQUATIONS 197 numbers is given, which can be used as coordinates of points, but where the law connecting them either is unknown, or is not known with sufficient definiteness to be expressed by an equation. For example, the speed of a motor boat is a function (amongst other things) of the rate of consumption of gasoline. The hourly rate of consumption can be taken as abscissas and the resulting speeds as ordinates, and a curve drawn which will express graphically this relation. The procedure in a case of this kind is first to plot the given points on a figure, adjusting the scales used, if necessary, so that the numbers expressing the abscissas and ordinates shall be of the same order of magnitude. The general shape of the curve can then be seen, and a form of equation adopted for trial which corresponds to the observed form of curve. The following examples illustrate the process. 1. Given the points whose coordinates are x = 0 0.5 1.0 1.5 2.0.2.5 3.0 3.5 4.0 y = 8.10 9.48 11.65 14.42 17.70 21.64 26.12 31.22 37.05 Find a good approximation to the equation of the curve on which they lie. The points are plotted in Fig. 109, the unit for x being ten times the unit for y. A convenient form of equation to use for a case of this kind, where in general y increases as x increases, is y = a + bx + CX2 + dx3 +.... (i) The solution of the problem consists in finding the values of a, b, c, etc., in (i), which will give the equation of a curve passing through a selected number of the given points, and as nearly as possible through the rest. An equation of the form (i) is chosen containing a definite number of terms on the right, and hence a definite number of undetermined coefficients, a, b, c, etc. An equal number of the given points must then be chosen, 198 ANALYTIC GEOMETRY [CHAP. XI] so that when their coordinates have been substituted for x and y there will be as many equations in a, b, c, etc., as there are quantities to be determined. In general the greater the number of terms taken in (i) the more accurate the result, but on the Y| I I I I * -XX,,,, *:< * - -- ^ --- ------------ ------ -~~~~~~~~~4 I I x FIG. 109 other hand the labor of computing the undetermined coefficients a, b, c, etc., increases rapidly with their number, so that practically it is desirable to use an equation with as few terms as will give a satisfactory result. We begin therefore with y = a + bx + CX2, (ii) the curve corresponding to which can be made to pass through any three points of the series, because the equation contains three undetermined coefficients, a, b, c. Taking the two end points of the series and the middle point (0, 8.10), (2.0, 17.70), (4.0, 37.05) and substituting in (ii) we have a = 8.10, a+2b + 4c = 17.70, a + 4b + 16c = 37.05, [ART. 110] EMPIRICAL EQUATIONS 199 from which a = 8.10, b = 2.362, c = 1.2188. Hence (ii) becomes y = 8.10 + 2.362x + 1.2188x2. (iii) If the ordinates to this curve (iii) be computed for the other given values of x they will be found to be x = 0.5 1.0 11.5 2.5 3.0 3.5 y = 9.59 11.68 14.39 21.62 26.16 31.30 Comparing these with the corresponding given values of y it will be seen that the differences are all small. The curve corresponding to (iii), therefore, represents fairly well the relation connecting the given pairs of values of x and y. Equation (iii) may therefore be used for computing values of y in this series corresponding to values of x lying between 0 and 4. An equation derived in this way should never be used, however, to extend the series in either direction beyond the limits fixed by the given values. Beyond these limits the curve corresponding to the assumed equation, and the curve which would correspond to the extended series may differ very widely. If it should be impossible to reach a satisfactory result by the use of equation (ii), an additional term from (i) should be added and the process repeated with the extended equation. 2. Find the equation of a curve which as nearly as possible shall pass through the points x = 1 2 3 14 5 6 7 8 y = 14 12.1 11.4 11.1 10.8 10.7 10.6 10.5 These points are plotted in Fig. 110, where one unit for x is the same as two units for y. Since the value of y tends to decrease as x increases, the equation y= a +- (i) x 200 ANALYTIC GEOMETRY [CHAP. XI] may be used. This equation contains two undetermined coefficients a and b, and therefore a curve with equation of this form can be determined so as to pass through any two of the eight given points. Using the first and last points of the series, Y *1;__ i I I I LI ILI_11_K O x FIG. 110 and substituting in (i), the results are a + b = 14, a + 8b = 10.5,.. b = 3.5, or b = 4, and hence a = 10. tion is therefore y = 10 +. X The resulting equa(ii) Substituting in (ii) the other six given values of x, the resulting values of y are x =2 3 4 5 l 6 7 y = 12 11.3 11.0 10.8 10.7 110.6 which agree closely with the given values of y. 111. Concluding remarks.-It is easily seen that each problem of the kind discussed in this chapter stands by itself. [ART. 111] EMPIRICAL EQUATIONS 201 Only broad general principles of procedure can be prescribed, as has been indicated in the examples given, and the details in each case must be adapted to the special requirements which it presents. The aim should be to produce an equation which expresses the law of variation of the data throughout the range for which the data are given, and within their limit of error. It may differ widely in form from the true theoretical function, and for this reason, in.using it to obtain values in addition to those given it should never be used beyond the limits within which the data lie, without careful investigation as to its reliability outside of these limits. As a rule it can safely be assumed to coincide closely with the true function only within this range. For example an equation for the curve traced through the points x = 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 y = 0.100 0.199 0.295 0.389 0.479 0.565 0.643 0.716 would be y = - 6x3 (i) which agrees closely with these values, and this equation could very properly be used to express the law connecting them, within the limits here given. The equation y = sin x, (ii) however, is also satisfied by the same numbers. Now if it be desired to extend the series of numbers beyond these limits it is found that when x = 1.0 1.2 1.5 2.0 y, from (i) = 0.833 0.912 0.937 0.667 etc. y, from (ii) = 0.842 0.932 0.997 0.909 so that the divergence of the two series of values of y increases rapidly. Hence, unless there is some means of determining which of the two equations, (i) or (ii) is the expression of the law upon which the numbers are based, the values thus derived, 202 ANALYTIC GEOMETRY [CHAP. XI] by extending the series beyond the range of the data, are of no value. If the results of a first try at a solution do not turn out satisfactorily another effort must be made either by extending the equation used in the first attempt by the addition of one or more terms, or by the use of some other form of equation appropriate to the arrangement of the given points. In addition to equations of the forms used in examples 1 and 2, Art. 110, others which will sometimes be found useful are y = cxn, where c and n are undetermined quantities; log y = a + bx, log y = cx", b y = a +X y=a+ -, where a, b, c and n are undetermined quantities. Other forms may also be found amongst the curves given for reference in the Appendix. EXERCISES ON CHAPTER XI In the following exercises the units for abscissas and ordinates should be so chosen that the resulting numbers used as the coordinates of points shall be of the same order of magnitude. See Ex. 1, p. 191. 1. In each of the following cases find an equation of the form (i), p. 197, which closely represents the data. (a) Tensile Tests of a Steel Wire Load Elongation Load Elongation Load Elongation 5,000 0.00000 25,000 0.00073 50,000 0.00170 10,000 0.00019 30,000 0.00090 55,000 0.00190 15,000 0.00037 40,000 0.00130 60,000 0.00210 20,000 0.00055 45,000 0.00150 65,000 0.00232 EMPIRICAL EQUATIONS 203 (b) Tensile Test of a Bronze Specimen Load Elongation Load Elongation Load Elongation 4,000 0.00013 12,000 0.00076 18,000 0.00158 6,000 0.00017 14,000 0.00098 20,000 0.00198 8,000 0.00042 16,000 0.00124 22,000 0.00243 10,000 0.00058 (c) Temperature Corrections for a Pyrometer T = temperature, C = correction. T = 800 900 1,000 1,100 1,200 1,300 1,400 1,500 C = 95 85 75 65 55 45 35 25 (d) Temperature Corrections for a Pyrometer T = temperature, C = correction. T = 700 800 900 1,000 1,100 1,200 1,300 1,400 C = 1 2 4 7 10 15 20 26 (e) Candle Power of Some Tungsten Lamps V = voltage, C = candle power. V =65.01 70.7 75.0 80.0 86.3 90.0 100.0 105.0 108.0 112.0 C =6.66 9.30 11.71 15.03 20.06 23.49 34.67 41.40 45.82 52.21 2. To determine the tensile strength of three lots of steel specimens were taken and tested with results as given below. Find for each specimen the equation of the straight line which best represents the data. Elongation in inches Load, Pounds Specimen a Specimen b Specimen c 5,000 0.0006 0.0011 0.0015 10,000 0.0010 0.0024 0.0031 15,000 0.0016 0.0039 0.0048 20,000 0.0020 0.0050 0.0064 25,000..... 0.0060 0.0079 30,000 0.0030 0.0072 0.0097 35,000..... 0.0087 0.0111 40,000 0.0042 0.0098 45,000 0.0047 0.0109 50,000 0.0050 55,000 0.0053 60,000 0.0059 204 ANALYTIC GEOMETRY [CHAP. XI] 3. Find the equation of the form (i), p. 197, which best fits each of the following sets of points. The values of x are the same for each set of points. x=0 1 2 13 4 5 6 7 8 y2 = 3.2 2.5 4.4 7.1 16.6 26.8 39.7 55.4 73.8 Y2 = 3.1 7.7 12.2 16.4 19.3 20.7 18.8 13.3 4.1 Y3 = 4.12 6.33 8.57 10.79 13.01 15.23 17.45 19.71 21.95 Y4 = 1.3 5.7 9.6 12.4 14.3 14.7 14.0 13.3 10.7 4. Determine the equation of the form aebd d ' which best represents the following data obtained from tests of wireless current received at varying distances from a constant source. d = distance, I = current, e = Napierian base. d = 300 400 500 600 800 1,000 1,200 1,500 2,000 I = 410 300 225 160 95 59 34 19 5' 5. For each of the sets of points below find a good approximation to the equation of the curve passing through them. In each case the proper form of equation to use is one of those given in Art. 111. () = 1.0 1.2 1.4 1.6 1.8 2.0 (a) y = 4.20 5.36 6.58 7.86 9.19 10.58 (b) x = 0.5 1.0 1.5 2.0 2.5 3.0 y) = 13.9 5.0 3.3 2.8 2.5 2.3 x = 1.2 1.6 2.0 2.4 2.8 3.2 y = 7.33 9.71 12.81 16.88 22.27 29.44 x = 0.2 0.4 0.6 0.8 1.0 1.2 1.4 (d) y = 0.96 0.86 0.69 0.52 0.38 0.24 0.14 6. For each of the sets of points below determine a good approximation to the equation of the curve which passes through them. The values of x are the same for each set of points. x = 0.5 1.0 1.5 2.0 2.5 3.0 y, = 4.72 3.10 2.57 2.28 2.13 2.04 Y2 = 0.33 2.49 8.42 20.12 39.0 67.6 y3 = 1.78 2.00 1.94 1.62 0.94 0.01 y4 = 0.83 1.14 1.58 2.31 3.48 5.38 CHAPTER XII EXTENSION OF COORDINATE GEOMETRY TO SOME SPACE PROBLEMS 112. Rectangular coordinates in space. THEOREM. The position of a point in space is determined when its distances from three planes, mutually perpendicular to each other are known. Let XOY, YOZ, ZOX be three mutually perpendicular planes, and P any point in space. From P draw a perpendicular to the plane XOY, intersecting this plane at M. The position of M in the plane XO Y is determined by the coordinates ON =x, NM = y. Hence these two distances together with the distance MP = z determine the position of P in space. Complete the rectangular parallelopiped which has 0 and P R 'I / 1f I II/ j 11 I X' i /I / X FIG M FIf. 111 for opposite vertices, and ON, NM, MP for its three dimensions. The edges QP = ON = x, and RP = NM = y are respectively the distances of P from the planes YOZ, ZOX. Hence the three distances QP, RP, MP determine the position of P. DEFINITIONS. I. The Coordinate System. The three planes, XOY, YOZ, ZOX, are called the coordinate planes, their common point of intersection, the origin, and the three lines X'X, Y'Y, Z'Z, in which they intersect, the coordinate axes. The distances of a point from the three coordinate planes are called 205 206 ANALYTIC GEOMETRY [CHAP. XII] the coordinates of the point, and are represented by the letters x, y, z, as indicated above, the x of a point being its distance from the plane YOZ, measured parallel to X'X, etc. II. Signs of the coordinates. Each coordinate of a point is measured from the corresponding coordinate plane, and hence is positive or negative according as it is measured in one direction or the opposite. As the figure is ordinarily drawn (see Fig. 111), the positive directions for x, y, z are respectively OX, OY, OZ. The three coordinates of a point are all positive in one, and only one, of the eight solid angles (octants) formed by the intersection of the three coordinate planes. III. Notation. In writing the coordinates of a point they are enclosed in parenthesis, and written in the order x, y, z. Thus the point (3, - 2, 4) is the point for which x = 3, y = - 2, z = 4. This point is found in the octant 0 - XY'Z, Fig. 111, since its y is negative, while its x and z are positive. IV. Drawing the Figure. In drawing figures for geometry of space a conventional perspective is used. See Fig 111. The axes X'X, Z'Z lie in the plane of the paper and are drawn at right angles. The axis Y'Y, which is perpendicular to both of the others, is drawn so as to make apparently equal angles with OX', OZ'. All parts of the figure which lie in the plane XOZ, or in planes parallel to XOZ, are drawn in their true shape, but when they lie in other planes they are more or less distorted. Distances parallel to OX or OZ are drawn in their true length, but those parallel to 0 Y must be foreshortened. 113. The distance between two points.-Let Pi(x1, y1, z1) and P2(x2, y2, z2) be any two points in space. Draw P1M1 and P2M2 perpendicular to the plane XOY. In the plane PIMIM2P2 draw P2Q parallel to M2M1, meeting M1P1 in Q. The triangle PlQP2 is right-angled at Q, hence PiP2 = 4pQ2 + QP2, [ART. 113] EXTENSION TO SPACE PROBLEMS 207 where P2Q = M2M1 = a(X -X2)2 + (y - y2)2, by (3), p. 6, and QP1 = MiP1 - M1Q = M1iP- M2P = z - z2. PP2 = 4(x- - X2)2 + (y1 - y2)2 + (1 - Z2)2. (1) P,2 O. I Q N, N, X FIG. 112 EXERCISES. 1. Draw a set of coordinate planes, and using any convenient unit of length, locate the points (3, 2, 4), (4, 3, 2), (- 3, 2, - 2), (0, 4, - 3), (0, 0, 5), (2, - 3, 5), (- 2, - 3, - 4). 2. What are the coordinates of a point on the X-axis at the distance 4 from the origin? 3. A point is on the Z-axis at the distance - 4 from the origin, what are its coordinates? 4. What is the locus of all points for which (a) z is zero, (b) x is zero, (c) x and y are both zero, (d) x is + 2, (e) x is 2 and y is 3? 5. What is the distance from the origin to each of the points (xl, yl, zi), (-x1, - Yl, Zi), (Xi, - Y1, zi), (xI, y1, - Zl), (Xi, - yi, - Z1), (- xi, - y1, - z1)? Which of these points are so situated that the straight line joining them passes through the origin? Which of them are symmetrically situated with respect to a coordinate plane? with respect to a coordinate axis? 6. Determine the lengths of the sides of the triangle whose vertices are the points (2, - 4, 7), (3, - 2, 0), (4; - 5, 4). Ans. 316, i14, 2-. 7. Find the lengths of the sides and diagonals of the quadrilateral whose vertices in order are the points (3, 2, 4), (2, - 3, 5), (1, - 2, - 3), (3, 2, - 1). Ans. Sides, 34, -6, 246, 5; Diagonals 6-9, 162. 208 ANALYTIC GEOMETRY [CHAP. XII] 114. Projections.-I. The orthogonal projection of a point on a line or plane is the foot of the perpendicular from the point to the line or plane. Let P be any point, RS any plane, and PM the perpendicular from P to the plane, then M is the II I K — FN. B' 1 FIG. 113 orthogonal projection of P on the plane RS. M is also the orthogonal projection of P on the line A'B' in the plane RS, passing through M. When the term projection is used in the following pages orthogonal projection will always be meant. II. The line or plane on which a projection falls is called the line or plane of projection. III. The projection of a line or curve on a plane is the locus of the projections on the plane of all the points on the line or curve. As proved in elementary geometry the projection of a straight line on a plane is in general a straight line. In Fig. 113 A'B' is the projection of AB on the plane RS. IV. The projection of a curve on a plane is in general another curve. Thus L'K' is the projection of LK. In particular, if the curve projected lies in a plane parallel to the plane of projection, the projection is identical in shape and size with the curve projected. 115. Direction angles and cosines. DEFINITION. The angles which a straight line makes with the three coordinate axes are called the direction angles of the line, [ART. 115] EXTENSION TO SPACE PROBLEMS 209 and the cosines of these angles are called the direction cosines of the line. If the line pass through the origin, as OP, Fig. 114, the direction angles are XOP = a, YOP = 1, ZOP = y. Since the angle between two non-intersecting lines in space is defined to be the angle formed by two lines drawn through a common point parallel respectively to the non-intersecting lines, the direction angles of a line which does not pass through the origin Z R P N Q / /P' M FIG. 114 are the same as those of a parallel line through the origin. NOTE. The direction angles of a line P'P passing through the origin may be taken as the angles which either OP, or OP' makes with the positive directions on the three axes. Either of these sets of angles determines completely the direction of the line in space. THEOREM I. The sum of the squares of the direction cosines of any line is equal to unity. Fram what has just been said it will be sufficient to prove the theorem for a line passing through the origin. Let OP, Fig. 114 be the line. From P let fall a perpendicular to OX. This will meet OX at N, where the perpendicular from M meets OX. The triangle OPN is right angled at N. Let OP = r. Therefore Similarly ON x OP r OQ y OR z cos =O - cos = =. OP r ' OP r (i) (ii) 15 210 ANALYTIC GEOMETRY [CHAP. XII] Squaring and adding, x2 + y2 + z2 Cos2 a + COS2 A + cos2._ -X - (iii) But OP is a diagonal of the rectangular parallelopiped whose three conterminous edges are ON = x, OQ = NM = y, OR =MP = z. Hence x2 + y2 + z2 = r2. from (iii) cOS2 a + COS2 P + COS2 = 1. (2) DEFINITION. The distance r of a point from the origin is called the radius vector of the point. From (i) and (ii) above x = r Cos a, y = rCOs, z = r cos. (3) THEOREM II. The direction cosines of the line joining two given points P1 (xl, yl, Z1), P2(x2, Y2, Z2) are XC X1 y2 - y COS ao-! COS -= d ' d (4) Z2 - Z cos 7 = d where d is the distance PIP2, Fig. 115. ` 1X FIG. 115 The proof is left to the student. [ART. 117] EXTENSION TO SPACE PROBLEMS 211 116. Cylindrical Coordinates.-The position of a point in space is determined by its distance from a fixed plane and the polar coordinates in this plane of z the projection of the point upon it. These coordinates are called cylin- z drical coordinates. Let P be the point, XOY the 0 -fixed plane, and OX the polar axis / \ x in this plane. Then the cylindrical r coordinates of P are FIG. 116 MP = z, OM1= r, XOM = 0. (5) NoTE. The letter r is used here to represent OM, instead of OP as in Arts. 115, 117. 117. Spherical coordinates.-Let ZOX be a fixed plane, and XOY, YOZ two planes perpendicular to XOZ and to each other. Let P be any point in space. Draw OP, let fall PM perpendicular to XOY, and draw OM. Then OP = r, XOM = 8, ZOP = (6) are the spherical coordinates of P. These coordinates determine the position of P because 0 deterZ p mines the plane through OZ which contains P, and (r, y) are the polar coordinates of P in this plane with reference to OZ as polar axis. 0 N X The rectangular coordinates of P may be expressed in terms of / M I r, 0, y, as follows. Draw MN perFrI. 117 pendicular to OX, then x = ON = OM cos 0, y = NM = OM sin. (i) 212 ANALYTIC GEOMETRY [CHAP. XII] In the right triangle OMP, right-angled at M, the angle MPO =, hence OM = r sin y, and z = MP = r cos y. (ii) Substituting the value of OM from (ii) in (i) the results are x = rcos 0 sin y, y = rsin 0 sin y, z r cos Y. (7) Conversely from equations (7), or from the figure r = 2+y +z2, 0 = tan-1 y, x (8) 7 = COS-1 4: +y2 + Z2 EXERCISES. 1. Find the direction cosines of the lines determined by the following pairs of points, taking the direction for each line from the first to the second given point. (a) (0, 0, 0), (1, 3, 7); (d) (0, 5, 0), (5, 0, 0); (b) (1, - 4, 3), (3, 2, 5); (e) (0, 0, 4), (5, 4, 0); (c) (- 1 2, - 4), (2, - 7, 3); (f) (1, 2, 3), (-1, -2, -3). Ans. (a) 5 9, 3 59, 7 59. 59 59 59 2. What are the direction cosines (a) of the positive direction of the X-axis, (b) of the negative direction of the Z-axis? 3. What are the direction cosines of the three lines joining the origin to the three points (1, 1, 0), (1, 1, 1), (- 1, - 1, - 1)? 4. Find the lengths of the sides and diagonals of the quadrilateral whose vertices in order are the points (1, 2, 4), (3, 2, - 1), (7, 5, - 3), (5, 5, 2). Prove that the figure is a parallelogram, by showing that the opposite sides are parallel. Ans. Sides all equal -9; diagonals, -94, I22. 5. Find the lengths of the projections of the line from (2, - 3, 3) to (- 2, 2, 4) on the X-, Y-, and Z-axes respectively. Ans. - 4, 5, 1. 6. Show that (2, - 3, 1) is the center of a sphere which passes through the points (8, 0, 3), (4, 3, 4), (0, - 9, - 2), (5, - 1, 7), (9, - 3, 1). [ART. 118] EXTENSION TO SPACE PROBLEMS 213 7. Two of the direction cosines of a line are - 2, and 1, what is the third? Ans.:t ]. 8. If a line makes angles of 60~ and 45~ respectively with the positive directions of the X- and Z-axes, what angle does it make with the Y-axis? Ans. 60~ or 120~. 9. Find the spherical coordinates of each of the following points (1, 2, 2), (1, 4, - 8), (- 4, - 3, 0), (4, - 4, 2). Ans. First point r = 3, 0 = tan-' 2, y = cos1 3. Second point r = 9, 0 = tan-' 4, 7 = cos-l. (- s). 10. Determine the rectangular coordinates of the point whose spherical coordinates are r = 4, 0 = 60~, y = 30~. Ans. (1, -3,'2/3). 118. Equations of the first degree containing one or two variables. THEOREM I. Equations of the form x = a, y = b, z = c, where a, b, c are constants, represent planes parallel to YOZ, ZOX, XOY respectively. The equation x = a defines a system of points all at the same distance a from YOZ. These points all lie in a plane parallel to YOZ and at the distance a from it. Hence x = a is z / the equation of this plane. Thus if OM = 4 units, the equation of the plane AMB is x = 4. D 7K Similarly the theorem is proved for y = b and z = c. M M / THEOREM II. Two simultane- X ous equations of the form x = a, N y = b, where a and b are con- Y A/H stants, represent a straight line F. FIG. ll8 parallel to OZ. Let OM = a, ON = b. By Theorem I, x = a represents the plane AMB, and y = b, the plane CND, hence the two equations taken simultaneously represent that system of points 214 ANALYTIC GEOMETRY [CHAP. XII] which is common to both planes. This is the straight line HK, parallel to OZ, in which the two planes intersect. Similarly the two pairs of equations y = b, z = c; x = a, z = c represent straight lines parallel respectively to OX and OY. The following important corollaries are easily deduced from Theorems I and II. The equation of YOZ is x = 0. (9) The equation of ZOX is y = 0. (10) The equation of XOYis z = 0. (11) The equations of OX are y = 0, z = 0. (12) The equations of OY are z = 0, x = 0. (13) The equations of OZ are x = 0, y = 0. (14) THEOREM III. The equation Ax + By + D = 0 represents a plane parallel to OZ. Let AB be the straight line in the plane XOY whose equation in that plane is Ax + By + D =. (i) Through AB pass a plane parallel to OZ, meeting XOZ in BD, and YOZ in AC. Let P be any point in this plane CABD, and C -N — through P draw QP parallel to OZ, N,<P - ^ meeting AB at Q. Since QP is parallel to OZ it is also parallel to B YOZ and to XOZ, so that every 0/ D, X-^ point on QP has the same x, and also Q B2 I the same y. Hence P has the same Y/A x and y as Q, and therefore the coorFIG. 119 dinates of P will satisfy equation (i) FIG. 119 since the coordinates of Q satisfy it. Equation (i) is therefore the equation of the plane CABD since it is satisfied by the coordinates of P, any point in it. [ART. 119] EXTENSION TO SPACE PROBLEMS 215 Similarly it may be shown that Ax + Cz + D = O, (ii) By + Cz + D = O, (iii) are the equations of planes parallel respectively to OY and OX. COROLLARY. Any two of the three equations (i), (ii), (iii), taken simultaneously, represent a straight line. The two equations taken simultaneously represent that system of points common to the loci of the equations considered separately, and since both loci are in this case planes the points common to both lie in the straight line in which they intersect. 119. Equations not of the first degree containing two variables. DEFINITIONS. A cylindrical surface is one which is generated by a straight line which moves so that it is always parallel to a fixed straight line and always intersects a given curve which does not lie in the same plane with the fixed line, nor in a plane parallel to it. Every position of the generating line of a cylindrical surface is called an element of the surface. THEOREM. Every equation, whose degree exceeds the first, which contains only two of the three variables, x, y, z, is the equation of a cylindrical surface, whose elements are parallel to that one of the coordinate axes which corresponds to the variable missing from the equation. The proof is essentially the same as that of III, Art. 118. Let f(x, y) = 0* (i) be any equation of higher degree than the first in x and y. In the plane XOY this equation represents some curve AB. * See Art. 17. 216 ANALYTIC GEOMETRY [CHAP. XII] Through every point of AB draw a straight line parallel to OZ. These lines will form a cylindrical surface. Take any point P in this surface. This point lies in an element of the surface PQ, and Q is the projection of P on XOY. Then as P and Q have the same x and the same y, and since Z the coordinates of Q satisfy (i) the coordinates of P will also satisfy it. Hence (i) is the equation of the cylindrical surface with elements parallel to OZ. v _ - _ _ _ _ _ X Similar conclusions follow with refB erence to the equations f(y, z) = 0, " -A f(z, x) = 0. Hence all equations of FIG. 120 the form fy, ) = 0, f(z, x) = 0, f(x, y) = 0, (15) represent cylindrical surfaces whose elements are parallel respectively to OX, OY, OZ. EXAMPLES. The equation (y - 3)2 + (z - 4)2 = 4, which as the equation of a plane locus represents a circle in the plane YOZ, with center at y = 3, z = 4, and radius = 2, represents as an equation in solid geometry a cylindrical surface with elements parallel to OX, all of whose cross-sections ~. parallel to YOZ are equal circles. -', Again the equation x2 + y2 = r2 is. X that of a cylindrical surface which in- tersects XO Y in the circle X2 + y2 = r2, and whose axis is OZ. FIG. 121 EXERCISES. 1. What surfaces do the following equations represent? (a) x =4, (b) y + 3 = O, (c) z = 2, (d) 3y +4 = 0, (e) 4z - 5 =0, () 2x +1 = 0, (g) x + y = 0, (h) x = y, (i) y = mx + b, [ART. 120] EXTENSION TO SPACE PROBLEMS 217 Ans. (a) A plane parallel to YOZ, 4 units on the positive side. (g) The plane through OZ which cuts XOY in the line x + y = 0. (i) The plane parallel to OZ, which cuts XOY in the line y = mx + b. 2. Write the equations of the following planes. (a) Parallel to XOY, and 4 units on the positive side of it. Ans. z = 4. (b) Parallel to YOZ, and two units on the negative side. (c) Perpendicular to OX, and + 4 units from the origin. (d) Parallel to ZOX, and passing through (4, 2, 7). Ans. y - 2 = 0. 3. What is the equation of the plane which is parallel to the Z-axis, and cuts the X- and Y-axes at + 2 and - 3 respectively? Ans. 3x - 2y- 6 = 0. 4. What is the equation of the plane, parallel to the Y-axis, which passes through the points (2, 0, 0) and (0, 0, - 2)? 5. What lines are represented by the following pairs of equations. (a) x = 3, y = 2. (d) 3x + 2 = 0, 4y - 7 = 0. (b) x = -2, z = 0. (e) 2x- 5 = 0, y = 3. (c) y + 2 = 0, z = 4. (f)4y - 3 = 0, z = - 2. Ans. (a) Parallel to OZ, and intersecting XOY at (3, 2). (b) Parallel to OY, and intersecting ZOX at (- 2, 0). 6. What locus is represented by the two simultaneous equations 3x + 4y -7 = 0, 4x -2y- 1 = 0. Construct the locus. 7. Describe the surfaces represented by the following equations, and make drawings of them. (a) y2 = 4x, (b) x2 + y2 = 25, (c) 42 + 9y = 36, (d) 4y2 = 3, (e) X2 - 4 =0, (f)4z2 - 9x2 = 36, (g) y = sin x, (h) x2 +2 + 4x - 6z- 12 = 0. Ans. (a) A parabolic cylinder with elements parallel to OZ. (b) A circular cylinder with elements parallel to OZ. 120, Equation of the first degree in three variables. THEOREM. Every equation of the first degree in three variables Ax + By + Cz + D = 0 (16) is the equation of a plane. 218 ANALYTIC GEOMETRY [CHAP. XII] Let Pi(xi, yi, Z1), P2(xo, y2, z2) be any two points on the locus represented by equation (16). Then Ax, + By1 + Cz1 + D = 0,(i) Ax2 + By2 + Cz2 + D =. (ii) Also let P3(x3, y3, z3) be any other point on the straight line through the points P1, P2 If it can be shown that P3 also lies in the locus whose equation is (16), then this locus must be a plane, because by elementary geometry a plane is a surface such that if a straight line be drawn through any two of its points then every point of this line lies in the surface. PROOF. Since by hypothesis P1, P2, P3 are in the same straight line the direction cosines of P1P3 and those of PiP2 must be equal, each to each. Let the length of P1P2 = d, and that of PiP3 = d'. Then by (4), p. 210, X3 - Xl X2 '- X1 d'- A = dX2-X1 or d(x3 - xl) = d'(x2 - x1), (iii) d' d and, similarly d(y3 - y1) = d'(y2 - y1), (iv) d(z3 - zx) = d'(z2 - z1). (v) Hence, multiplying (iii) by A, (iv) by B, and (v) by C Ad(x3 - xi) = Ad'(x2 - xi), 1 Bd(y3 - yi) = Bd'(y2 - y), (vi) Cd(z3 - zi) = Cd'(z2 - Z1). Adding the three equations (vi), d(Ax3 + By3 + Cz3) - d(Ax1 + Byi + Czl) = d'(Ax2 + By2 + Cz2) - d'(Ax + Byl + Czi). (vii) But from (i) and (ii) Axl + Byl + Czl = -D, Ax2 + By2 + Cz2 = -D, hence substituting these values, (vii) reduces to d(Ax3 + By3 + Cz3) + dD = d'(D - D) = 0 [ART. 122] EXTENSION TO SPACE PROBLEMS 219 or Ax3 + By3 + Cz3 + D = 0. That is, the coordinates (x3, Y3, z3) satisfy equation (16), hence P3 lies in the locus of which this is the equation, and that locus is therefore a plane. 121. The equation of the sphere. DEFINITION. A sphere is the locus of all points in space at a constant distance from a fixed point. The fixed point is called the center of the sphere, and the constant distance, the radius. The equation is obtained by translating the definition into algebraic language. Let (xo, yo, Zo) be the center, (x, y, z) any point on the locus, and r the radius. Then by (1), p. 207, the equation (x -Xo)2 +(y - yo))2+ (z -Z) = r2 (17) expresses that the distance between the points (x0, Yo, Zo) and (x, y, z) is always constant. Hence (17) z is the equation of a sphere of radius r, with center at (xo, yo, o). If the origin be taken at the center of the sphere, xo = o = zo = 0, and equation X (17) reduces to 2 + y2 + z = r. (18) FIG. 122 Fig. 122 represents one eighth of the sphere whose equation is (18). 122. Surfaces and curves.-In Arts. 120, 121 it has been shown that two well-known surfaces, the plane and sphere, are represented in analytic geometry by equations in three variables. Similarly other surfaces may be represented by other forms of equations in three variables. The relation between an equation in three variables and a surface is the same as that between an equation in two variables and a locus in a plane. 220 ANALYTIC GEOMETRY [CHAP. XII] The notation f(x, y, z), see Art. 17, is used to designate any expression containing the three variables x, y, z. Thus in general Ax, y, z) = 0 (19) is the equation of a surface. NOTE. The equation f(x, y, z) = 0 includes equations of the type x2 + y2 - a2 = 0, y - b = 0, etc., in which only one or two variables occur explicitly. These latter as the equations of loci in space are interpreted as equations in three variables in which the terms containing the missing variables have the coefficient zero. THEOREM. Two simultaneous equations in three variables, x, y, z, represent one or more lines, straight or curved. Let f(x, y, ) =, (x, yz) = 0, be the equations of two surfaces. The points whose coordinates satisfy both equations are those points which lie in both surfaces. The two equations taken simultaneously therefore represent the lines straight or curved in which the two surfaces intersect. 123. Interpretation of equations.-A method will now be described for determining the appearance and situation of a surface whose equation is given. A preliminary discussion will help to make clear the nature of the process. I. Let fl(x, y, z) = 0, f2(X, y, z) = 0 (i) be the equations of two surfaces. Let AB represent the curve of intersection of these surfaces. If one of the variables, for example z, be eliminated from the equations (i) the result will be an equation 4(x, y) = 0, (ii) which is satisfied by the coordinates of every point on the curve AB in which the surfaces (i) intersect. By Art. 119, equation (ii) [ART. 123] EXTENSION TO SPACE PROBLEMS 221 is the equation of a cylinder whose elements are parallel to the Z-axis, and hence it must be the equation of the cylinder, with elements P1Q1, P2Q2, P3Q3, etc., parallel to OZ, upon which Z B Q FIG. 123 lies the curve of intersection of the surfaces (i). It is therefore also the equation in the plane XOY of the curve CD, in which this cylinder intersects XOY, that is the equation of the projection of AB on XOY. Similar results follow if y or x be eliminated from the two equations (i). II. If one of the two surfaces given by equations (i) is a plane parallel to XOY, the curve of intersection AB, and its projection CD on XOY will be identical (see Art. 114, IV). The equation (ii) in this case may be taken either as the equation of CD in XOY, or as the equation of the curve of intersection itself in the cutting plane. Similar results follow for intersections of a surface by planes parallel to YOZ or XOZ. III. The form of a surface whose equation is given can often be outlined by determining its intersections with the coordinate planes, and with planes parallel to them, as indicated in II. For example, every plane parallel to XOY has for its equation z = k, a constant. If k be substituted for z in the equation of the surface, the resulting equation in x and y is the equation in the plane z = k of the curve in which this plane cuts the surface. This method will be illustrated by examples. 222 ANALYTIC GEOMETRY [CHAP. XII] 1. Construct the surface 4x + 3y + 5z = 12. By Art. 120 this is the equation of a plane. The intersection with the plane XOY is obtained by combining the given equation with z = 0, the equation of XOY. This gives 4x+ 3y = 12, which is therefore the equation in the plane XOY of the intersection, the line AB. Z CSimilarly, making y = 0, the result 4x + 5z = 12 is the equation of CA, the intersection of the given plane with o/ iA X ZOX, and making x = 0, the A X intersection BC of the given plane with YOZ is found to be '/fB 33y + 5z = 12. These three FIG. 124 intersections show the position of the plane. 2. Construct the surface x2 y2 z2 25 16 9 Cutting the surface by the plane XOY, by making z = 0 in the given equation, the result is the ellipse + y2 25 16 whose semi-axes are OA = 5, OB = 4. Similarly making y = 0, the section formed by the plane ZOX is the ellipse x2 z2 _+ = 25 9 whose semi-axes are OA = 5, OC = 3. [ART. 123] EXTENSION TO SPACE PROBLEMS 223 Finally, making x = 0, the section formed by the plane YOZ is the ellipse y2 z2 16 91, whose semi-axes are OB = 4, OC = 3. Only one eighth of the figure z E is drawn, as it is evident that the surface is symmetrical with respect to the origin. / / Sections of the surface formed / by planes parallel to YOZ are obtained by combining the given B D D equation with the equations of I. 125 such planes, which all have the form x = k, where k is a constant. Substituting x = k the given equation becomes y2 z2 I2 y2+ = 16 9 25' or y2 + 1. 16( ) (2 9 # )k This is the equation of a real ellipse in the plane x = k for all values of k2 < 25, but if k2 > 25 the curve is imaginary. Hence all sections of the surface parallel to YOZ are ellipses, and the surface does not extend beyond A, where k = 5. DiE1, D2E2, etc., represent quadrants of some of these ellipses. Similarly by assigning constant values to y and z in turn it is shown that sections of the surface parallel to ZOX and XOY respectively, are ellipses, and that the surface does not extend beyond B, where OB = 4 in the direction OY, nor beyond C, where OC = 3 in the direction OZ. 224 ANALYTIC GEOMETRY [CHAP. XII] Similar results may be obtained for any surface whose equation is of the form — 2 ~+ 2 z2 (20) or for any surface whose equation can be reduced to that form, such as Ax2 + By2 + C2 = ZD, (21) where A, B, C, D are positive constants. This surface is called an ellipsoid, and the three distances OA = a, OB = b, OC = c, Fig. 125, are called the semi-axes of the ellipsoid. If the three semi-axes, a, b, c, of an ellipsoid are equal the surface is a sphere. See equation (18), p. 219. If two of the semi-axes, as b and c are equal, equation (20) takes the form X2 y2 2 a2 + b2 + = 1, (22) and the series of sections of this surface determined by the planes x = k have equations of the form y2 +z2 b2 1 -, which are circles. The surface represented by equation (22) is therefore a surface of revolution. It may be conceived as being generated by the rotation of the ellipse (x2/a2) + (y2/b2) = 1 about its major-axis. An ellipsoid of revolution is usually called a spheroid. If the two equal axes are shorter than the third axis, that is if b < a in (22) the surface is called a prolate spheroid, but if the two equal axes are longer than the third, or b > a in (22), the surface is an oblate spheroid. [ART. 123] EXTENSION TO SPACE PROBLEMS 225 3. Construct the surface x2 y2 z2 25 16 9 The analysis of this equation is made as in Ex. 2. The section by the plane XOY, or z = 0, is the ellipse AB x2 y2 25+ 16 whose semi-axes are OA = 5, OB = 4. The section by the plane ZOX, or y = 0, is the hyperbola AC X2 Z2 25 9 which cuts OX at A, where OA = 5. The section by the plane YOZ, or x = 0, is the hyperbola BD 2 2 1 16 9 which cuts OY at B, where OB = 4. Only one fourth of each of these sections is shown in the figure. Sections parallel to XOY, obtained Z by giving to z a series of constant values, positive or negative, are al- D ~A3 ways real ellipses. Thus if z = + k \ wy./j '-j/A1 X the equations of these sections in their respective planes are B x2;y2 k2 -x+ = 1+ - FIG. 126 25 16 9 Quadrants of these elliptic sections are shown at A1B1, A2B2, etc. Sections parallel to YOZ, and to ZOX, obtained by making x = k and y = k, respectively are in general hyperbolas, as the 16 226 ANALYTIC GEOMETRY [CHAP. XII] student can show, but there are important exceptions to this general statement. Thus if the surface be cut by a plane through B, by making y = 4, the result is T-X2 - - 2 = 0, or X ~ 3Z =0, which represents in the plane y = 4 a pair of straight lines meeting at B. Similarly, making x = 5, it is seen that this plane cuts the surface in a pair of straight lines meeting at A, whose equations in the plane x = 5 are y -~ z = 0. Similar results are given by the two planes y = - 4, and x = - 5. Similar results may be obtained for any surface whose equation has the form X2 y2 X2 a2 + b-c2 =2 1) (23) or, more generally, whose equation has the form Ax2 + By2 + Cz2 = D (24) where one of the three letters A, B, C stands for a negative number and the other two, with D, for positive numbers. This surface is called a hyperboloid of one sheet. If a = b the sections A1B1, A2B2, etc., are circles, and equation (23) then represents a hyperboloid of revolution of one sheet. It can be produced by rotating the hyperbola (x2/a2) - (2/c2) = 1 about its conjugate axis. 4. Construct the surface x2 y2 z2 9 16 25 If this surface be cut by the plane XOY by substituting z = 0 in the given equation, the section is the hyperbola x2 y2 _9- = 16 [ART. 123] EXTENSION TO SPACE PROBLEMS 227 AB, A'B' in the figure, in which OA = 3. Similarly the section by the plane ZOX is the hyperbola AC, A'C', with the same transverse axis. If x = 0 the equation z is - (y2/16) - (z2/25) = 1, showing that the section by the plane YOZ is im- x' -A 7 A Iaginary.' If, however, sections parallel to YOZ be FIG. 127 FIG. 127 formed by making x = k, a constant, the resulting equation is y2 z2 16 ( -1 25 (9-1) which represents a real ellipse in the plane x = k if k > 3, or < - 3. Hence all sections parallel to YOZ are real ellipses if the cutting plane is more than three units from YOZ on either side. Similar results may be obtained for any surface whose equation has the form x2 y2 z2 - - 2- 2 = 1' (25) a2 b2 c2 or, Ax2 + By2 + Cz2 = D (26) where two of the three letters A, B, C stand for negative numbers, and the other, with D, for positive numbers. This surface is called the hyperboloid of two sheets. If b = c in equation (25) the sections parallel to YOZ are circles, and the surface is a hyperboloid of revolution of two sheets. It may be produced by rotating the hyperbola (x2/a2) - (y2/b2) = 1 about its transverse axis. 228 ANALYTIC GEOMETRY [CHAP. XII] 5. Construct the surface x2 y2 z 25 9 4' Cutting the surface by XOY by making z = 0, the result is I x2 + oy2 = 0. z A 4 M I MN1 X FIG. 128 This equation is satisfied by only one pair of real values of x and y, x = 0, y = 0. Hence the surface has only one real point, the origin, in the plane XOY. Making y = 0, the section by the plane ZOX is found to be X2 = -2 Z, a parabola OA tangent to OX at the vertex O. Similarly, making x = 0, the section by the plane YOZ is y2 =_ z, a parabola OB tangent to O Y at the vertex O. To cut the surface by planes parallel to XOY give to z a series of constant values z = k. Thus we have 2 y2 k 25 9 4' which is always an ellipse in the plane z = k, real if k is positive, and imaginary if k is negative. M1N,, M2N2, etc., represent quadrants of these elliptic sections. The student may show that sections of this surface parallel to YOZ, or to ZOX, are always parabolas. The general equation of this type is a+2 b2c (27) ~ ---b2 [ART. 123] EXTENSION TO SPACE PROBLEMS 229 or Ax2 + By2 = Cz, (28) where A and B are positive. This surface is called the elliptic paraboloid. If a = b in (27) it becomes a paraboloid of revolution, that is the surface produced by revolving a parabola about its axis. 6. Construct the surface x2 y2 z 25 16 3 Making z = 0, the section formed by the plane XOY is seen to be a pair of straight lines (x2/25)- G (y2/16) = 0, or (x/5) a= (y/4) = 0, I one of which, OB, is drawn in the F figure. HI Making y = 0, the section formed K - by the plane ZOX is found to be NK /Kthe parabola OC, - X2 = 2 z. FIG. 129 Making x = 0 the section by the plane YOZ is the parabola OD, 2= - - z. The series of planes z = k, parallel to XOY, form sections which are always hyperbolas x2 y2 k 25 16 3 The transverse axes of these hyperbolas are parallel to OX if k is positive, and parallel to OY if k is negative. The hyperbolas GH and EF, represent portions of two of these sections for 230 ANALYTIC GEOMETRY [CHAP. XII] positive values of k, and KL, K'L' represent portions of these hyperbolic sections for negative values of k. For all planes parallel to XOZ, given by y = k, the sections are parabolas with axes extending upward, and for all planes parallel to YOZ, given by x = k, the sections are parabolas with axes extending downward. The general equation of this type is X2 y2 z x b2 S (29) or Ax2 + By2 = Cz, (30) where A and B represent numbers having opposite signs. This surface is called a hyperbolic paraboloid. It has quite important uses and applications. Its general appearance is best studied with the aid of a solid model. 7. Construct the surface Z B _X 2 y2 z2 C - 0. 16 9 25.~;-'A - - By methods fully illustrated in the 'i-U r-preceding examples it can be shown that __So the plane XOY cuts the surface in a I x point (the origin), and that ZOX, and / l-i-X YOZ cut it respectively in the pairs of Y/ ---i tPV straight lines (x/4) ~ (z/5) = 0, or OC, c Yi,\ od' 2OD, and (y/3) - (z/5) = 0, or OA, OB. FIG. 130 The sections parallel to XOY, found by making z = k, are ellipses for all values of k, positive and negative. Let the surface be intersected by any plane y = mx passing through OZ. Eliminating y between this and the given equation the result is X2(.~ + 1m2) - -z2 = 0. [ART. 123] EXTENSION TO SPACE PROBLEMS 231 By I, p. 220, this equation is that of the projection of the intersection of y = mx with the surface on the plane XOZ. From the form of the equation it is seen that this projection is a pair of straight lines meeting at 0. If x be eliminated between y = mx and the given equation the result is Y2( 1 + 1 - _ Z2 = 0, showing that the projection of the intersection on YOZ is also a pair of straight lines meeting at the origin. Since both of these projections on perpendicular planes consist of pairs of straight lines the intersection itself must be a pair of straight lines, for all values of m. That is, every plane through OZ cuts the surface in a pair of straight lines meeting at 0. The surface is therefore a cone, with vertex at 0, and axis coincident with OZ. NOTE. A conical surface, or a cone, is a surface generated by a straight line moving in space so that it always passes through a fixed point called the vertex, and always intersects a given curve which does not lie in the same plane with the vertex. Every position of the moving line is called an element of the cone. The general equation of this type is X2 y2 2,+t - 2 0+ (31) or Ax2 + By2 + z2 = 0, (32) where A, B, and C do not all have the same sign. The cone represented by (32) will have its axis coincident with OX, OY, or OZ, according as the sign of A, B or C differs from that of the other two of these letters. If in equation (31) a = b the surface is a right circular cone, or a cone of revolution. 232 ANALYTIC GEOMETRY [CHAP. XII] 124. Concluding remarks.-This chapter is intended to give only the briefest possible introduction to some of the principles of coordinate geometry of space, especially those which will be needed when, in the study of calculus, the student comes to applications in which solids and the surfaces of solids are treated. A fuller introduction to this interesting, important, and extensive branch of mathematics may be obtained from the special elementary treatises on the subject. EXERCISES ON CHAPTER XII Normal Exercises 1. Locate on a figure the points whose rectangular coordinates are (1, 3, 2) and (- 4, 2, - 3). 2. Locate the points whose spherical coordinates are r = 4, 0 = 60~, = 30~, and r = 2, 0 = 225~, y = 150~. 3. Find the lengths of the sides of the triangle whose vertices are the points (4, 2, - 3), (2, 1, - 1), (- 2, - 3, 1). Ans. 3, 6, /67. 4. Find the lengths of the projections on the three coordinate axes of the line joining (- 2, 4, 5) and (3, - 8, 5). 5. Find the direction cosines of the line joining (a) the origin to (1, -4, 4), (b) (- 2, 4, 5) to (3, - 8, 5). Ans. (a) I: ~-33, -4: 133, 4: -3-3. (b) 5, -TX, 0. 6. A line joining the origin and a point whose coordinates are all positive makes with the X-axis an angle of 40~, and with the Z-axis an angle of 65~. What angle does it make with the Y-axis? Ans. 61~ 2'. 7. Determine the spherical coordinates of the point whose rectangular coordinates are (4, - 4, - 2). Ans. r = 6, 0 = 315~, y = cos-1 (- ). 8. If the spherical coordinates of a point are r = 3, 0 = 120~, y = 135~, what are its rectangular coordinates? Ans. (-4i2, i, - I2 ). EXTENSION TO SPACE PROBLEMS 233 9. Determine the surfaces represented by the following equations (a) x = 4, (b) 2 -y2 = 0, (c) z - 3x + 5 = 0, (d) y = -2, (e) 4x + y = 0, (f)4x + 3y - z - 24 = 0, (g) x2 + y2 = 4, (h) 3x-4y + 2z- 5 = 0, (i) x2 + y2 z2-25 =0, (j) 4x2 + 4y2 + 4z2 - 8x + 12z - 3 = 0. 10. Find the equations of the following spheres (a) center at the origin, radius = 5, (b) center at (- 1, 2, - 3), radius = 4. 11. Construct the loci represented by the following pairs of equations (a) x = 2, y = 4. (d) x + 3y - 7 = 0, y = 4. (b) x + 3 = 0, z = 4. (e) x + 3y - 7 = 0, z = 4. (c) x2 + y2 = 16, z =4. (f) 2 -4y2 =9, = 0. 12. Discuss and sketch the following surfaces (a) 42 + 9y2 + 2 - 36 = 0, (d) 4z2 + 9y2-x =0, (b) 4x2 + 4y2 z2 - 16 = 0, (e) y2 - 4X2 - 4z = 0, (c) 2 - 4x2 -y2 - 64 = 0, (f) 42 - 9y2 + 9z2 = 0. General Exercises 13. In the case of the following planes find their intercepts on the coordinate axes, and the equations of the lines in which they intersect the coordinate planes. (a) 4x + 3y + 7z- 12 = 0, (d) x + y + z + 1 = 0, (b) 2x - y + z- 16 = 0, (e) x + y + z = 0, (c) 8x +3y - 4z = 0, (f)3x +4y -2- 2 12 = 0. Ans. (a) Intercepts are 3, 4, —; intersections with YOZ, ZOX, XOY, respectively are 3y + 7z - 12 = 0, 4x + 7z - 12 = 0, 4x + 3y -12 = 0. 14. Find the equation of the plane which passes through each of the following sets of points (a) (1, 3, 4), (2, 4, 3), (4, 1, 1). (c) (5, 0, 0), (0, 5, 0), (0, 0, 5). (b) (- 2, 3, 5), (1, 0, - 3), (3, 2, 0). (d) (0, 0, 0), (0, 0, 1), (0, 5, 0). Ans. (a) x + z - 5 = 0, (b) 7x - 25y + 12z + 29 = 0. 234 ANALYTIC GEOMETRY [CHAP. XII] 15. Find the equations of the planes determined by the following sets of conditions. (a) Passing through the X-axis, and the point (4, 5, 6). Ans. 6y - 5z = 0. (b) Passing through the points (2, - 1, 3), (3, 1, - 2), and parallel to the Z-axis. Ans. 2x - y - 5 = 0. (c) Passing through the line x = 2, y = 3, and the point (3, 1, 7). Ans. 2x + y - 7 = 0. (d) Passing through the line 3x + 4y - 7 = 0, z = 0, and the point (5, - 2, 3). Ans. 3x + 4y - 7 = 0. (e) Passing through the line 3x + 4y - 5z - 12 = 0, 4x + 2y + 5z + 7 = 0, and parallel to the Z-axis. Ans. 7x + 6y - 5 = 0. 16. Find the equation of the sphere (a) whose center is at (2, 1, 3), and radius 4, (b) whose center is at (0, 0, 0), and radius 5, (c) with center at (1, 2, 3), and passing through (3, 1, 1), (d) with center at (4, 3, 5), and tangent to XOY. Ans. (a) x2 + y2 + z2 - 4x -2y - 6z -2 = 0, (b) x2 + y2 + z2- 2x -4y - 6z + 5 = 0. 17. Determine the centers and radii of the following spheres. (a) X2 + y2 + 2 - 4x + 2y - 8z - 4 = 0, (b) 4x2 + 4y2 + 42 + 8x - 4y + 12z + 65 = 0. Ans. (a) Center, (2, - 1, 4), r = 5. (b) Imaginary sphere. 18. Find the equation of the spheres determined by the following sets of conditions. (a) Tangent to the XY plane, center at (4, 5, 6). Ans. 2 + y2 + z2 - 8x - lOy - 12z + 41 = 0. (b) Tangent to the X-axis, center at (2, 3, 4). Ans. x2 + y2 + 2 - 4x - 6y - 8z + 4 = 0. (c) Center at (2, 3, 4), and tangent to the plane x = 5. (d) Passing through the four points (2, 2, - 4), (- 2, 6, 0), (6, - 2, 4), (4, 6, 6). Ans. 2 + y2 + 2-4x -4y - 4z - 24 = 0. EXTENSION TO SPACE PROBLEMS 235 19. Discuss and sketch the following surfaces (a) 4x2 + 9y2 + z2-36 = 0, (i) x2 + y2- 4z = 4, (b) x2 + y2 -_ 2 = 36, (j) z2 - 9y2 - 16x2 = - 36, (c) 4x2 + y2 -9z2 =36, (k) 4x2 + 9y2 + 2 + 4 =0, (d) 42 + 4y2-9z = 0, (1) z2- 4y2-4x = 0, (e) 3x + 4y + z = 12, (m) 4y2 + 9x2 + 36z + 36 = 0, (f) y2 - 4x2- 2522 = 4, (n) 2 - y2 + 22 = 0, (g) 9x2 - 36y2 + 25z2 = 225, (o) 4x2 + 4y2- 25z2 = 0, (h) 42 + 4y2+ 4z2-25 = 0, (p) 2 + 3y2 -27x2 = 0. Ans. (a) Ellipsoid, a = 3, b = 4, c = 6. (b) Hyperboloid of revolution of one sheet, a = b = c = 6. (c) Hyperboloid of one sheet, a = 3, b = 6, c = 2. (d) Paraboloid of revolution, with axis in OZ. 20. Discuss and sketch the following surfaces. Note the changes produced by changes in value and sign of the coefficients. (a) 4x2 + 9y2 + z2-36 = 0, (e) 9x2 - 4y2 - 2 = 0, (b) 9x2 + 4y2 + 2- 36 = 0, (f) 9x2 + 4y2 - z2 = 0, (c) 9x2 + 4y2 + z2 + 36 = 0, (g) 9x2 + 4y2 - z = 0, (d) 9x2 + 4y2- z2-36 = 0, (h) 92 + 4y2- - 36 = 0, 21. If the direction cosines of a line are to each other as I: m: n, what are their values? 22. Find the coordinates of the middle point of the line joining (3, - 4, 6) and (7, 2, - 2). Ans. (5, - 1, 2). 23. Derive formulas for the coordinates of the middle point of the line joining (xi, y1, zi) and (X2, Y2, Z2). 24. Find the equation of the surface every point of which is equally distant from (4, 3, 7) and (2, 4, 3). Describe and sketch the surface. Ans. 4x - 2y + 8z - 45 = 0. 25. Find the equation of the locus of a point whose distance from (4, 0, 0) is always twice its distance from (1, 0, 0). Describe and sketch the surface. Ans. x2 + y2 + z2 = 4. 26. A point moves in space so that the sum of its distances from (2, 0, 0) and (- 2, 0, 0) is always equal to 8. Describe and sketch its locus. Ans. 3x2 + 4y2 + 4z2 = 48. APPENDIX CURVES FOR REFERENCE A number of curves, not referred to in the preceding pages, which are either important practically, or are of interest historically, are gathered here for convenience of reference. In the equations which follow a and b always stand for given arbitrary lengths, k, for a numerical constant. When an equation contains one arbitrary length a or b, a change in its value simply alters the scale of the drawing. A change in the value of a numerical constant, however, or a change in the relative values of two arbitrary lengths, may change completely the appearance of the curve. The cubical parabola, y = x3 or The semi-cubical parabola, a2y = x3. y2 = x3, or ay2 =. FIG. 131 FIG. 132 236 The cissoid. /2 = 2 x 3 y 2a -x' Yl CURVES FOR REFERENCE The parabola. 2a sin2 0 2a sin x1/2 + yl2 = a/2. cos 0 237 Y (1 ko,a), -— r, FIG. 133 The witch. 8a3 Y = 4a2 + x2 Y (o,2a) 0 X 01 ta,u) X FIG. 134 The probability curve. y = ae-k22 Y 0 X FIG. The conchoid. 135 (X2 + y2)(y - a)2 FIG. 136 = b2y2. r = a csc 0 + b. (o, a+b) ____/___ (0o, a) 0 X a> b x FIG. 137 o, a-b) a <b 238 APPENDIX The limagon. r = a - b cos 0. (a+b,ir) -(a-b, o) a B FIG. 138 a<b The cardioid. r = a(1 - cos 0). (2a,r) A B FIG. 139 The hyperbolic spiral, or reciprocal spiral. rO = a. The catenary. y = 2a(exla + e-/a). Y (o,a) 01 FIG. 140 The logarithmic spiral. r = at. K FIG. 141 FIG. 141 (al) BG4 FIG. 142 CURVES FOR REFERENCE 239 The tangent curve. y = tan x. )Y )1 The secant curve. y = sec x. I w -2?f FIG. 143 FIG. 143 2T 31r X Jr 2A 1 0 1 s, FI. FIG. 144 x r = a sin 20. r = a tan 30. FIG. 145 FIG. 146 The trochoid. x = a - b sin 0. y = a - b cos 0. FIG. 147 a < b