The line $y = mx + c$ touches the curve $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, if

A square $ABCD$ has all its vertices on the curve $x ^{2} y ^{2}=1$. The midpoints of its sides also lie on the same curve. Then, the square of area of $ABCD$ is

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:

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 :

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 ....................

Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 4,\,0),$ the latus rectum is of length $12$