Find the eccentricity, foci, centre, length of latus rectum vertices and the equation to the directrices of the hyperbola.
(a) $9 x^{2} - 16 y^{2} + 72 x - 32 y - 16 = 0$
(b) $4 x^{2} - 5 y^{2} - 16 x + 10 y + 31 = 0$
(c) $5 x^{2} - 4 y^{2} + 20 x + 8 y - 4 = 0$
(d) $4 (y + 3)^{2} - 9 (x - 2)^{2} = 1$
(e) $4 x^{2} - 9 y^{2} - 8 x - 32 = 0$ .

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Solution

(a) $e = \frac{5}{4}$ , latus rectum $= \frac{9}{2}, c = (- 4, - 1), f o c i (1, - 1), (- 9, - 1), V e r t i c e s = (0, 1) (- 8, - 5) D i r e c t r i c e s 5 x + 36 = 0, 5 x + 4 = 0$ .

(b) $e = \frac{3}{2}, l a t u s r e c t r u m = 5, c = (2, 1), f o c i = (2, 4), (2, - 2), v e r t i c e s (2, 3), (2, - 1), d i r e c t r i c e s 3 y - 7 = 0, 3 y + 1 = 0$ .

(c) $e = \frac{3}{2}, l a t u s r e c t u m = 5, f o c i (1, 1) (- 5, 1), c e n t r e = (- 2, 1), v e r t i c e s (0, 1), (- 4, 1), d i r e c t r i c e s 3 x + 2 = 0, 3 x + 10 = 0$ .

(d) $e = \frac{\sqrt{13}}{3}, f o c i = (2, - 3 \pm \frac{1}{2} \frac{\sqrt{13}}{3}), l a t u s r e c t u m = \frac{4}{9}, v e r t i c e s (2, - 3, \pm \frac{1}{2}), d i r e c t r i c e s y = - 3 \pm \frac{2}{2 \sqrt{3}}$ .

(e) $e = \frac{\sqrt{13}}{3}, c e n t r e = (1, 0), f o c i = (1 \pm \sqrt{13}, 0), l a t u s r e c t u m = \frac{8}{3}, v e r t i c e s (4, 0), (- 2, 0), d i r e c t r i c e s x = 1 \pm \frac{9}{\sqrt{13}}$ .

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Find the eccentricity, foci, centre, length of latus rectum vertices and the equation to the directrices of the hyperbola.(a) 9x2−16y2+72x−32y−16=0(b) 4x2−5y2−16x+10y+31=0(c) 5x2−4y2+20x+8y−4=0(d) 4(y+3)2−9(x−2)2=1(e) 4x2−9y2−8x−32=0.