If the product of the lengths of the perpendiculars from any point on the hyperbola $16x^2 - 25y^2 = 400$ to its asymptotes is $p$ and the angle between the two asymptotes is $\theta$,then $p \tan \frac{\theta}{2} =$

If the line $5x - 2y - 6 = 0$ is a tangent to the hyperbola $5x^2 - ky^2 = 12$,then the equation of the normal to this hyperbola at the point $(\sqrt{6}, p)$ where $p < 0$ 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 length of the latus rectum of the hyperbola $25x^2 - 16y^2 = 400$ is -

The vertices of a hyperbola $H$ are $(\pm 6, 0)$ and its eccentricity is $\frac{\sqrt{5}}{2}$. Let $N$ be the normal to $H$ at a point in the first quadrant and parallel to the line $\sqrt{2} x + y = 2 \sqrt{2}$. If $d$ is the length of the line segment of $N$ between $H$ and the $y$-axis,then $d^2$ 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:
$(A) \left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$(B) \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$
$(C) (3\sqrt{3}, -2\sqrt{2})$
$(D) (-3\sqrt{3}, 2\sqrt{2})$