Let the ellipse $E: \frac{x^{2}}{144}+\frac{y^{2}}{169}=1$ and the hyperbola $H: \frac{x^{2}}{16}-\frac{y^{2}}{\lambda^{2}}=-1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$, then the value of $24(e+L)$ is:

At which point does the line $2x + \sqrt{6}y = 2$ touch the hyperbola $x^2 - 2y^2 = 4$?

If $3x + 2\sqrt{2}y + k = 0$ is a normal to the hyperbola $4x^2 - 9y^2 - 36 = 0$ making positive intercepts on both the axes,then $k=$ (in $\sqrt{2}$)

The foci of a hyperbola are $(\pm 2, 0)$ and its eccentricity is $\frac{3}{2}$. $A$ tangent,perpendicular to the line $2x + 3y = 6$,is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $x$- and $y$-axes are $a$ and $b$ respectively,then $|6a| + |5b|$ is equal to $..........$.

The asymptotes of a hyperbola are parallel to $2x + 3y = 0$ and $3x + 2y = 0$. The equation of the hyperbola whose center is at $(1, 2)$ and which passes through $(5, 3)$ is:

The tangent drawn at an extremity (in the first quadrant) of the latus rectum of the hyperbola $\frac{x^2}{4}-\frac{y^2}{5}=1$ meets the $x$-axis and $y$-axis at $A$ and $B$ respectively. If $O$ is the origin,then $(OA)^2-(OB)^2=$