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
$15$
$14$
$17$
$16$
The equation of the normal at the point $(6, 4)$ on the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 3$, is
The length of the transverse axis of a hyperbola is $7$ and it passes through the point $(5, -2)$. The equation of the hyperbola is
The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and $\mathrm{x}= \pm \frac{4}{\sqrt{3}}$, respectively. Let the line $y-\sqrt{3} \mathrm{x}+\sqrt{3}=0$ touch this hyperbola at $\left(\mathrm{x}_0, \mathrm{y}_0\right)$. If $\mathrm{m}$ is the product of the focal distances of the point $\left(\mathrm{x}_0, \mathrm{y}_0\right)$, then $4 \mathrm{e}^2+\mathrm{m}$ is equal to ...........
Area of the triangle formed by the lines $x -y = 0, x + y = 0$ and any tangent to the hyperbola $x^2 -y^2 = a^2$ is :-
The point $\mathrm{P}(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having eccentricity $\frac{\sqrt{5}}{2} .$ If the tangent and normal at $\mathrm{P}$ to the hyperbola intersect its conjugate axis at the point $\mathrm{Q}$ and $\mathrm{R}$ respectively, then $QR$ is equal to :