The equation of the hyperbola whose foci are $(6, 5), (- 4, 5)$ and eccentricity $\frac{5}{4}$ is:

\frac{(x - 1)^{2}}{16} - \frac{(y - 5)^{2}}{9} = 1

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\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1

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\frac{(x - 1)^{2}}{16} - \frac{(y - 5)^{2}}{9} = - 1

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\frac{(x - 1)^{2}}{4} - \frac{(y - 5)^{2}}{9} = 1

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Solution

The correct option is A $\frac{(x - 1)^{2}}{16} - \frac{(y - 5)^{2}}{9} = 1$
Centre of the ellipse $=$ mid point of foci $= (1, 5)$

Distance between foci $= \sqrt{(6 - (- 4))^{2} + (5 - 5^{2})}$ $= 10$

$2 a e = 6 - (- 4) = 10 \Rightarrow a = 5 / e = 4$
$\Rightarrow b^{2} = a^{2} (e^{2} - 1) = 9$
Hence required hyperbola is $\frac{(x - 1)^{2}}{16} - \frac{(y - 5)^{2}}{9} = 1$

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Equation of the hyperbola with foci (0,± 5) and e = $\frac{5}{3}$ is

\frac{3}{x + y} + \frac{2}{x - y} = 2 and \frac{9}{x + y} - \frac{4}{x - y} = 1

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