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

Let the hyperbola $H : \frac{ x ^{2}}{ a ^{2}}- y ^{2}=1$ and the ellipse $E: 3 x^{2}+4 y^{2}=12$ be such that the length of latus rectum of $H$ is equal to the length of latus rectum of $E$. If $e_{ H }$ and $e_{ E }$ are the eccentricities of $H$ and $E$ respectively, then the value of $12\left( e _{ H }^{2}+ e _{ E }^{2}\right)$ is equal to

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

For the hyperbola $\frac{{{x^2}}}{9} – \frac{{{y^2}}}{3} = 1$  the incorrect statement is :

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 ………………..