The eccentricity of the hyperbola conjugate to the hyperbola $\frac{x^2}{4} - \frac{y^2}{12} = 1$ is

The foci of the hyperbola $2x^2 - 3y^2 = 5$ are

The eccentricity of the hyperbola $25x^2 - 9y^2 = 225$ is...

Let a line $L_{1}$ be tangent to the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ and let $L_{2}$ be the line passing through the origin and perpendicular to $L_{1}$. If the locus of the point of intersection of $L_{1}$ and $L_{2}$ is $(x^{2}+y^{2})^{2} = \alpha x^{2}+\beta y^{2}$,then $\alpha+\beta$ is equal to

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 . . . . . . .

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be $\frac{5}{4}$. If the equation of the normal at the point $\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ on the hyperbola is $8 \sqrt{5} x + \beta y = \lambda$,then $\lambda - \beta$ is equal to