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

If $(8,2)$ is a point on the hyperbola whose length of the transverse axis is $12$ and conjugate axis is $x=0$,then the eccentricity of that hyperbola is

$PQ$ 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:

$A$ tangent to the curve $9 b^2 x^2 - 4 a^2 y^2 = 36 a^2 b^2$ makes intercepts of unit length on each of the coordinate axes. Then the point $(a, b)$ lies on:

Let $x$ be the eccentricity of a hyperbola whose transverse axis is twice its conjugate axis. Let $y$ be the eccentricity of another hyperbola for which the distance between the foci is $3$ times the distance between its directrices. Then $y^2-x^2=$

If the angle between the asymptotes of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $2 \tan^{-1}\left(\frac{b}{a}\right) = 2 \tan^{-1}\left(\frac{2}{3}\right)$ and $a^2-b^2=45$,then $ab=$