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

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:

$A$ hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$. Then the tangent to this hyperbola at $P$ also passes through the point:

The area (in sq units) of the triangle formed by the tangent at $(\sqrt{3}, 0)$ to the hyperbola $x^2-3y^2=3$ with the pair of asymptotes of the hyperbola is

For all real values of $m$,the straight line $y = mx + \sqrt{9m^2 - 4}$ is a tangent to the curve:

$A$ hyperbola with centre at $(0,0)$ has its transverse axis along the $X$-axis,and its length is $12$. If $(8,2)$ is a point on the hyperbola,then its eccentricity is