Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at the point $P$ intersects the $x$-axis at $(9,0)$, then the eccentricity of the hyperbola is

Eccentricity of the hyperbola conjugate to the hyperbola $\frac{{{x^2}}}{4} – \frac{{{y^2}}}{{12}} = 1$ is

Tangents are drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$, parallel to the straight line $2 x-y=1$. The points of contacts of the tangents on the hyperbola are

$(A)$ $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ $(B)$ $\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

$(C)$ $(3 \sqrt{3},-2 \sqrt{2})$ $(D)$ $(-3 \sqrt{3}, 2 \sqrt{2})$

If transverse and conjugate axes of a hyperbola are equal, then its eccentricity is

Let $H : \frac{ x ^{2}}{ a ^{2}}-\frac{y^{2}}{ b ^{2}}=1$, a $>0, b >0$, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4(2 \sqrt{2}+\sqrt{14})$. If the eccentricity $H$ is $\frac{\sqrt{11}}{2}$, then value of $a^{2}+b^{2}$ is equal to