What is the length of the latus rectum of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$?

Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$,be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$,then $\beta$ is equal to

If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ and $e_2$ is the eccentricity of a hyperbola passing through the foci of the given ellipse and $e_1 e_2=1$,then the equation of such a hyperbola among the following is

Let $H$ be the hyperbola,whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $e = \sqrt{2}$. Then the length of its latus rectum is

The midpoint of the chord $4x - 3y = 5$ of the hyperbola $2x^2 - 3y^2 = 12$ is

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