If $(0,\; \pm 4)$ and $(0,\; \pm 2)$ be the foci and vertices of a hyperbola, then its equation is

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

$C$ the centre of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$. The tangents at any point $P$ on this hyperbola meets the straight lines $bx - ay = 0$ and $bx + ay = 0$ in the points $Q$ and $R$ respectively. Then $CQ\;.\;CR = $

Let a line $L: 2 x+y=k, k\,>\,0$ be a tangent to the hyperbola $x^{2}-y^{2}=3 .$ If $L$ is also a tangent to the parabola $y^{2}=\alpha x$, then $\alpha$ is equal to :

The equation of the director circle of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{4} = 1$ is given by

Area of the quadrilateral formed with the foci of the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ and $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} =  - 1$ is