The equation of a hyperbola,whose foci are $(5, 0)$ and $(-5, 0)$ and the length of whose conjugate axis is $8$,is

The asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,with any tangent to the hyperbola form a triangle whose area is $a^2 \tan (\alpha)$. Then its eccentricity equals

The equation of the tangent parallel to $y-x+5=0$ drawn to $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$ is

The point $P(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having eccentricity $e = \frac{\sqrt{5}}{2}$. If the tangent and normal at $P$ to the hyperbola intersect its conjugate axis at the points $Q$ and $R$ respectively,then $QR$ is equal to :

The length of the latus rectum and directrices of a hyperbola with eccentricity $e$ are $9$ and $x = \pm \frac{4}{\sqrt{13}}$,respectively. Let the line $y - \sqrt{3}x + \sqrt{3} = 0$ touch this hyperbola at $(x_0, y_0)$. If $m$ is the product of the focal distances of the point $(x_0, y_0)$,then $4e^2 + m$ is equal to ...........

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 $\sqrt{a^2+e^2}=$