Find the equation of the hyperbola whose foci are $(-2, 0)$ and $(2, 0)$ and eccentricity is $2$.

If the line $5x - 2y - 6 = 0$ is a tangent to the hyperbola $5x^2 - ky^2 = 12$,then the equation of the normal to this hyperbola at the point $(\sqrt{6}, p)$ where $p < 0$ is:

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

Let $PQ$ be a chord of the hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$, perpendicular to the $x$-axis such that $OPQ$ is an equilateral triangle, where $O$ is the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$, then the area of the triangle $OPQ$ is:

The line $3x - 4y = 5$ is a tangent to the hyperbola $x^2 - 4y^2 = 5$. The point of contact is

If $P(\theta) = (x_1, \frac{3 \sqrt{5}}{2})$,$0 < \theta < \frac{\pi}{2}$ is a point on the hyperbola $\frac{x^2}{25} - \frac{y^2}{9} = 1$,where $\theta$ is the parameter in its parametric form,then $2 x_1 + 9 \sin^2 \theta = $