Let $S$ be the focus of the hyperbola $\frac{x^2}{3}-\frac{y^2}{5}=1$, on the positive $\mathrm{x}$-axis. Let $\mathrm{C}$ be the circle with its centre at $\mathrm{A}(\sqrt{6}, \sqrt{5})$ and passing through the point $\mathrm{S}$. if $\mathrm{O}$ is the origin and $\mathrm{SAB}$ is a diameter of $\mathrm{C}$ then the square of the area of the triangle $OSB$ is equal to ....................

The minimum value of ${\left( {{x_1} - {x_2}} \right)^2} + {\left( {\sqrt {2 - x_1^2}  - \frac{9}{{{x_2}}}} \right)^2}$ where ${x_1} \in \left( {0,\sqrt 2 } \right)$ and ${x_2} \in {R^ + }$.

Foci of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{{(y - 2)}^2}}}{9} = 1$ are

Circles are drawn on chords of the rectangular hyperbola $ xy = c^2$  parallel to the line $ y = x $ as diameters. All such circles pass through two fixed points whose co-ordinates are :

Let the tangent drawn to the parabola $y ^{2}=24 x$ at the point $(\alpha, \beta)$ is perpendicular to the line $2 x$ $+2 y=5$. Then the normal to the hyperbola $\frac{x^{2}}{\alpha^{2}}-\frac{y^{2}}{\beta^{2}}=1$ at the point $(\alpha+4, \beta+4)$ does $NOT$ pass through the point.

 A hyperbola passes through the point $P\left( {\sqrt 2 ,\sqrt 3 } \right)$ has foci at $\left( { \pm 2,0} \right)$. Then the tangent to this hyperbola at  $P$ also passes through the point