The magnitude of the gradient of the tangent at an extremity of the latera recta of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is equal to (where $e$ is the eccentricity of the hyperbola).

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 . . . . . . .

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

Let $\lambda x - 2y = \mu$ be a tangent to the hyperbola $a^{2}x^{2} - y^{2} = b^{2}$. Then $\left(\frac{\lambda}{a}\right)^{2} - \left(\frac{\mu}{b}\right)^{2}$ is equal to

If $e_1$ is the eccentricity of the hyperbola $x = \sec \theta, y = \sqrt{2} \tan \theta$ and $e_2$ is the eccentricity of the hyperbola $x = \sqrt{2} \sec \theta, y = \tan \theta$,then $\frac{e_2^2}{e_1^2} = $

Find the equation of the hyperbola satisfying the given conditions: Vertices $(0, \pm 3)$,foci $(0, \pm 5)$.