Let $P (10, 2 \sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ whose foci are $S$ and $S'$. If the length of its latus rectum is $8$, then the square of the area of $\Delta PSS'$ is equal to:

Tangents are drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$,parallel to the straight line $2x-y=1$. The points of contact of the tangents on the hyperbola are:
$(A) \left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$(B) \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$
$(C) (3\sqrt{3}, -2\sqrt{2})$
$(D) (-3\sqrt{3}, 2\sqrt{2})$

$A$ square $ABCD$ has all its vertices on the curve $x^{2}y^{2}=1$. The midpoints of its sides also lie on the same curve. Then,the square of the area of $ABCD$ is

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

The values of $m$ for which the line $y=mx+2$ is a tangent to the hyperbola $4x^2-9y^2=36$ are

If the area of the region bounded by $16x^2 - 9y^2 = 144$ and $8x - 3y = 24$ is $A$, then $3(A + 6 \ln(3))$ is equal to . . . . . . .