If $ PN$  is the perpendicular from a point on a rectangular hyperbola $x^2 - y^2 = a^2 $ on any of its asymptotes, then the locus of the mid point of $PN$  is :

If $\mathrm{e}_{1}$ and $\mathrm{e}_{2}$ are the eccentricities of the ellipse, $\frac{\mathrm{x}^{2}}{18}+\frac{\mathrm{y}^{2}}{4}=1$ and the hyperbola, $\frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1$ respectively and $\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)$ is a point on the ellipse, $15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k},$ then $\mathrm{k}$ is equal to 

If line $ax$ + $by$ = $1$ is normal to the hyperbola $\frac{{{x^2}}}{{{p^2}}} - \frac{{{y^2}}}{{{q^2}}} = 1$ then $\frac{{{p^2}}}{{{a^2}}} - \frac{{{q^2}}}{{{b^2}}} = 1$ is equal to (where $a$,$b$,$p$, $q \in {R^ + })$-

The length of the latus rectum of the hyperbola $25x^2 -16y^2 = 400$ is -

The equation to the hyperbola having its eccentricity $2$ and the distance between its foci is $8$

Let $H : \frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$, where $a > b >0$, be $a$ hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4 \sqrt{3}$..

The correct option is: