Area of the quadrilateral formed with the foci of the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ and $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} =  – 1$ is

An ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the vertices of the hyperbola $H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1$. Let the major and minor axes of the ellipse $E$ coincide with the transverse and conjugate axes of the hyperbola $H$. Let the product of the eccentricities of $E$ and $H$ be $\frac{1}{2}$. If $l$ is the length of the latus rectum of the ellipse $E$, then the value of $113\,l$ is equal to $….$

Let $H : \frac{ x ^{2}}{ a ^{2}}-\frac{y^{2}}{ b ^{2}}=1$, a $>0, b >0$, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4(2 \sqrt{2}+\sqrt{14})$. If the eccentricity $H$ is $\frac{\sqrt{11}}{2}$, then value of $a^{2}+b^{2}$ is equal to

Let $P \left( x _0, y _0\right)$ be the point on the hyperbola $3 x ^2-4 y ^2$ $=36$, which is nearest to the line $3 x+2 y=1$. Then $\sqrt{2}\left( y _0- x _0\right)$ is equal to :

The equation of the normal at the point $(6, 4)$ on the hyperbola $\frac{{{x^2}}}{9} – \frac{{{y^2}}}{{16}} = 3$, is