If the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has eccentricity $e = \frac{5}{4}$ and the length of the latus rectum equal to $9$,then $ab$ is equal to

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 $y^2-x^2=$

The equation of the normal to the curve $3x^2 - y^2 = 8$,which is parallel to the line $x + 3y = 10$,is

The equation of the hyperbola whose foci are $(-2, 0)$ and $(2, 0)$ and eccentricity is $2$ is given by :-

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ be two points such that $\theta+\phi=\frac{\pi}{2}$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$,then $k=$

If the latus rectum of a hyperbola subtends a right angle at the center,then its eccentricity is: