Foci of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{{(y - 2)}^2}}}{9} = 1$ are

If the foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ coincide with the foci of the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}},$ then $b^2$ is equal to

Consider the hyperbola

$\frac{x^2}{100}-\frac{y^2}{64}=1$

with foci at $S$ and $S_1$, where $S$ lies on the positive $x$-axis. Let $P$ be a point on the hyperbola, in the first quadrant. Let $\angle SPS _1=\alpha$, with $\alpha<\frac{\pi}{2}$. The straight line passing through the point $S$ and having the same slope as that of the tangent at $P$ to the hyperbola, intersects the straight line $S_1 P$ at $P_1$. Let $\delta$ be the distance of $P$ from the straight line $SP _1$, and $\beta= S _1 P$. Then the greatest integer less than or equal to $\frac{\beta \delta}{9} \sin \frac{\alpha}{2}$ is. . . . . . . 

Let the foci of a hyperbola $\mathrm{H}$ coincide with the foci of the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$ and the eccentricity of the hyperbola $\mathrm{H}$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $\mathrm{H}$ is $\alpha$ and the length of its conjugate axis is $\beta$, then $3 \alpha^2+2 \beta^2$ is equal to :

The foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide. Then the value of $b^2$ is -

The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2 -2y^2 -2=0$  to its asymptotes is