Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be reciprocal to that of the ellipse $x^{2}+9y^{2}=9$. Then the ratio $a^{2}:b^{2}$ equals:

The eccentricity of the curve $x^2 - y^2 = a^2$ is

Let $P$ be the foot of the perpendicular from the focus $S$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ on the line $bx-ay=0$ and let $C$ be the centre of the hyperbola. Then, the area of the rectangle whose sides are equal to $SP$ and $CP$ is

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

If $e_{1}$ and $e_{2}$ represent the eccentricity of the curves $16x^{2}-9y^{2}=144$ and $9x^{2}-16y^{2}=144$ respectively,then $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}$ is equal to

Find the equation of the hyperbola satisfying the given conditions: Foci $(0, \pm \sqrt{10})$,passing through $(2, 3)$.