Consider the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having one of its foci at $P(-3,0)$. If the latus rectum through its other focus subtends a right angle at $P$ and $a^2b^2 = \alpha\sqrt{2} - \beta$,where $\alpha, \beta \in N$,then find the value of $\alpha + \beta$.

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 two tangents are drawn to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that the product of their gradients is $c^2$,then they intersect on the curve:

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 $13[{(x - 1)^2} + {(y - 2)^2}] = 3{(2x + 3y - 2)^2}$ represents

If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ of eccentricity $e = \sqrt{3}$ from its asymptotes is equal to $6$,then the length of the transverse axis of the hyperbola is