The eccentricity of the conic ${x^2} - 4{y^2} = 1$ is

The length of the latus rectum of the curve $xy = 7x + 5y$ is

If $e_1$ is the eccentricity of the hyperbola $x = \sec \theta, y = \sqrt{2} \tan \theta$ and $e_2$ is the eccentricity of the hyperbola $x = \sqrt{2} \sec \theta, y = \tan \theta$,then $\frac{e_2^2}{e_1^2} = $

$PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle,where $O$ is the centre of the hyperbola. Then the eccentricity $e$ of the hyperbola satisfies:

Let the hyperbola be $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$. Let $A$ be a vertex of the hyperbola. Let $B$ be an endpoint of the latus rectum near $A$. If $C$ is the focus of the hyperbola nearest to $A$,find the area of triangle $ABC$.

$A$ hyperbola,having the transverse axis of length $2 \sin \theta$,is confocal with the ellipse $3 x^{2}+4 y^{2}=12$. Its equation is