One of the latus recta of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends an angle $2 \operatorname{Tan}^{-1}\left(\frac{3}{2}\right)$ at the centre of the hyperbola. If $b^2=36$ and $e$ is the eccentricity of the given hyperbola,then $\sqrt{a^2+e^2}=$

The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. The equation of the hyperbola with eccentricity $e = 2$ is

The normal to the rectangular hyperbola $xy = c^2$ at the point $t_1$ meets the curve again at the point $t_2$. Then the value of $t_1^3 t_2$ is

The point from which two distinct tangents can be drawn to two different branches of the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$,but no two different tangents can be drawn to the circle $x^2 + y^2 = 36$,is:

Let $H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ be the hyperbola,whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6)$,where $\alpha > 0$,lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$,then $\alpha^2+\beta$ is equal to:

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: