If a hyperbola passes through the point $P(10, 16)$ and it has vertices at $(\pm 6, 0)$,then the equation of the normal to it at $P$ is

$A$ 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

Let the eccentricity of the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $\sqrt{\frac{5}{2}}$ and the length of its latus rectum be $6\sqrt{2}$. If $y = 2x + c$ is a tangent to the hyperbola $H$,then the value of $c^2$ is equal to

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

The eccentricity of the hyperbola conjugate to ${x^2} - 3{y^2} = 2x + 8$ is

The equation of the normal to the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ at $(-4, 0)$ is