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 $\alpha + \beta$.

The equation of the asymptotes of the hyperbola $2 x^2+5 x y+2 y^2-11 x-7 y-4=0$ is

Let the points $P_1\left(\frac{\pi}{4}\right), P_2\left(\frac{3 \pi}{4}\right), P_3\left(\frac{5 \pi}{4}\right)$ and $P_4\left(\frac{7 \pi}{4}\right)$ given in parametric form,lie on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$. Then these four points in that order form

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $2a$ and $2b$,respectively,and one focus and the corresponding directrix of this hyperbola be $(-5, 0)$ and $5x + 9 = 0$,respectively. If the product of the focal distances of a point $(\alpha, 2\sqrt{5})$ on the hyperbola is $p$,then $4p$ is equal to . . . . . . .

$A$ normal to the hyperbola $4x^2 - 9y^2 = 36$ meets the coordinate axes $x$ and $y$ at $A$ and $B$,respectively. If the parallelogram $OABP$ ($O$ being the origin) is formed,then the locus of $P$ is

Find the equation of the hyperbola satisfying the given conditions: Vertices $(0, \pm 5)$,foci $(0, \pm 8)$.