inclination to the axis a; then xp cos.0, y=p sin. 0, and by substitution in equation (a) we have b2 a2 Hence p2 = :: 51. Now it is well known that a2 — b2 = (dist. between focus and centre) b2 e2 cos.20 ab a2- a2e2 cos.2 0 a2b1 a2 — (a2 — b3) cos.20 which gives x = y3 = ax2 + x3. Let x = 0, and each = 0. Hence, putting AB (Fig. 37), = pass through A and B. taking Am = = x + To trace the curve whose equation is 3x3 (a + x) p2 cos: 0) Again, let r± ∞o. Then yoo, or we have two infinite branches AQ, Bq lying on different sides of the axis Cc. To find the greatest ordinate of the branch AMB, we have dy 2a+3x dx = 2 3 a. Then the corresponding values of y are a, the curve will the equation required. 2a 2, and ..y = (ax2+23)+ 3 y = x. (1 + ax−13 = x (1 + 2% ~ x-1 3 AB and drawing mM = at right angles to Cc, we obtain the position and magnitude of the greatest ordinate required. To find the asymptotes, we have = 3 a2 x−2 + &c.) Hence, Hence, the equation to the rectilinear-asymptote, (see Francœur, Mathemat. Pures, p. 323, or Stirling, Linea Tertü Ordinis Newtonianæ, p. 48,) is a y' = x' + —, x' being measured from A. Let y' = 0. Then x = 52. a Let x' = 0. =AR. 3 Therefore DR being joined and produced indefinitely will give the asymptotes DE, De to the respective branches AQ, Bq. For the common method of finding the asymptotes to this curve, see Vince's Fluxions, p. 52. Then y = dy 2a+3x Since tan. = being the inclination of the dx зx(a + x) tangent to the axis. By substituting the singular values of x,viz.,0,-a, the corresponding values of tan. 0, are oo, oo and 0. 2a and 3 y = ±x a 3 Consequently the tangents at A and B are to the axis, and that at M is parallel to it. To trace the curve whose equation is √ a2 x2 = AD (Fig. 37). a2 + x2 Let y = 0; then x 0 ora. Take therefore AB = + a, Ab= -α a (Fig. 1), in the same straight line, and the curve passes through the points A, B, b. Again, B and b are the limits of x, since y is imaginary for every value of x> a. Again, putting x = 0 and a, in a4 2a2x2 x4 dy = we get x2 Stan. 4507 ten. 135° Jand (and tan. 90°. .. The point A is double, its two tangents being inclined to the axis at the angles 45° and 135° respectively, and the tangents at Bb are each axis. The greatest ordinates PM, pm are given by putting a*—2a2x2-x10, and thence obtaining xa √2-1= AP or Ap; which being substituted in the given equation afford y = ± a 1 for the pairs of maximum ordinates at √2 P and p, viz, PM, PM' and pm, pm' respectively. a2 x2 Sydz = fxdx a2 + x2 But C = 2a2 + 1 √a+ x4 AMB= u2. Again let u, and we finally get 2a2 du S 2a2 + 1 + √ a1 - x4 2a+1 + a2 2a2 + 1 √ a1 — x1 — a2 2a+1 as 2a2+ 1 + + 2a2 45° 2a2 + 1 2a2 2a2 + 1 the general value of the area A M P. Let Ia; then the area as 2a2+ 1 + and putting a2 + x2 = u2, and 2a2 2a2 + 1 + 2a2 2a2 + 1' √2a2 2a9 2a+1 sin. (90° u2 du √ 2 六 9 น ao √ a2 + x2 √2 (sin.-" √ a2 + x2 √2 45°) + C 180° × a 45°) locity = TT = x + NT' = x and by the question TT' s = PP' = .. ns = x Let 53. To find the equation to the curve of pursuit. velocity Let T, moving uniformly along the straight line TM with a v, be pursued by P moving also uniformly with a ve; and let PT, be that path of P which is TM. av n dx dy Then Let also PT be any other contemporaneous positions of P,T. Then, since P'T is evidently a tangent at P' and PN is decreasing, we have -- = P. -ny n ydx dy Hence, making dy constant, we have d2x n. √ dx2+dy2 = dx — dx dy - n - dp dy + = 2 2a+1 2a2+ 1 ydx dy - y. x (area of O, rad. a.) since they move uniformly. = √1+ p2 = dex dy y dy dp y √1+ p2 and l.cly" l. (p + √ 1 + po). 0; then y = a dx cot. dy (—)* − y and squaring both sides, we get 2 a*p = -y"; ya 2n 2a"dx a""y-"dy-y"dy, and integrating, on the supposition that dy is negative, we obtain p = √1+ p2 n cot. 90° = 0, a2" 1-n equation of the Curve of Pursuit. Let u = and y = 0. Then TM = 3α a.} = the distance described by T before 1-4 the general P overtakes it. The problem admits of being further generalized. T might be made to move in a curve, and with a velocity varying according to a given law, P following it likewise with a variable velocity. The determination of the problem under these circumstances we recommend, as an useful exercise. 54. To transform the equation to the Lemniscata, (x2 + y2)2= ao. (x2 - y3). from rectangular to polar co-ordinates, Make A, (Fig. 1) the origin of the rectangular, the pole of the polar co-ordinates, and put AM = p, and MAB. Then, since y AM. sin. = P. sin. 0, and x AP cos. 0, by substituting in the given equation, we get |