The eccentricity of the hyperbola can never be equal to

For $0<\theta<\pi / 2$, if the eccentricity of the hyperbola $\mathrm{x}^2-\mathrm{y}^2 \operatorname{cosec}^2 \theta=5$ is $\sqrt{7}$ times eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta+y^2=5$, then the value of $\theta$ is :

If a directrix of a hyperbola centered at the origin and passing through the point $(4, -2\sqrt 3)$ is $5x = 4\sqrt 5$ and its eccentricity is $e$, then

Let $H : \frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$, where $a > b >0$, be $a$ hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4 \sqrt{3}$..

The correct option 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 ………………..