The equation of the normal to the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$ at the point $(8,\;3\sqrt 3 )$ is

The locus of the centroid of the triangle formed by any point $\mathrm{P}$ on the hyperbola $16 \mathrm{x}^{2}-9 \mathrm{y}^{2}+$ $32 x+36 y-164=0$, and its foci is:

Let $\mathrm{A}\,(\sec \theta, 2 \tan \theta)$ and $\mathrm{B}\,(\sec \phi, 2 \tan \phi)$, where $\theta+\phi=\pi / 2$, be two points on the hyperbola $2 \mathrm{x}^{2}-\mathrm{y}^{2}=2$. If $(\alpha, \beta)$ is the point of the intersection of the normals to the hyperbola at $\mathrm{A}$ and $\mathrm{B}$, then $(2 \beta)^{2}$ is equal to ..... .

If a hyperbola passes through the point $\mathrm{P}(10,16)$ and it has vertices at $(\pm 6,0),$ then the equation of the normal to it at $P$ is

The eccentricity of a hyperbola passing through the points $(3, 0)$, $(3\sqrt 2 ,\;2)$ will be

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 :