If the lengths of the transverse axis and the latus rectum of a hyperbola are $6$ and $\frac{8}{3}$ respectively,then the equation of the hyperbola is $ . . . . . . $

The eccentricity of the hyperbola $5x^2 - 4y^2 + 20x + 8y = 4$ is

Let $H_1: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $H_2:-\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ be two hyperbolas having lengths of latus rectums $15 \sqrt{2}$ and $12 \sqrt{5}$ respectively. Let their eccentricities be $e_1=\sqrt{\frac{5}{2}}$ and $e_2$ respectively. If the product of the lengths of their transverse axes is $100 \sqrt{10}$,then $25 e_2^2$ is equal to . . . . . . .

$A$ hyperbola having its centre at the origin passes through the point $(5, 2)$ and has a transverse axis of length $8$ along the $X$-axis. What is the eccentricity of its conjugate hyperbola?

If the normal drawn to the hyperbola $xy=16$ at $(8,2)$ meets the hyperbola again at a point $(\alpha, \beta)$,then $|\beta|+\frac{1}{|\alpha|}=$

If $\theta$ is the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$ and $\cos \theta=\frac{5}{13}$,then $a^2=$