Find the equation of the tangent to the hyperbola $x^2 - 4y^2 = 36$ which is perpendicular to the line $x - y + 4 = 0$.

Statement $I$: The eccentricity of the hyperbola $9x^2-16y^2-72x+96y-144=0$ is $5/4$.
Statement $II$: The eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\sqrt{1+\frac{b^2}{a^2}}$.

The equation of the hyperbola whose foci are the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and the eccentricity is $2$,is

The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$. The equation of the hyperbola is:

On a rectangular hyperbola $x^2-y^2=a^2, a > 0$,three points $A, B, C$ are taken as follows: $A=(-a, 0)$; $B$ and $C$ are placed symmetrically with respect to the $X$-axis on the branch of the hyperbola not containing $A$. Suppose that the $\triangle ABC$ is equilateral. If the side length of the $\triangle ABC$ is $ka$,then $k$ lies in the interval

$A$ tangent to the hyperbola $\frac{x^2}{4} - \frac{y^2}{2} = 1$ meets the $x-$axis at $P$ and the $y-$axis at $Q$. Lines $PR$ and $QR$ are drawn such that $OPRQ$ is a rectangle (where $O$ is the origin). Then $R$ lies on