The equation of the hyperbola,whose eccentricity is $\sqrt{2}$ and whose foci are $16$ units apart,is

Find the equation of the hyperbola with foci $(0, \pm 3)$ and vertices $(0, \pm \frac{\sqrt{11}}{2})$.

The line $y=x$ intersects the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ at the points $P$ and $Q$. The eccentricity of the ellipse with $PQ$ as the major axis and a minor axis of length $\frac{5}{\sqrt{2}}$ is

The vertices of a hyperbola $H$ are $(\pm 6, 0)$ and its eccentricity is $\frac{\sqrt{5}}{2}$. Let $N$ be the normal to $H$ at a point in the first quadrant and parallel to the line $\sqrt{2} x + y = 2 \sqrt{2}$. If $d$ is the length of the line segment of $N$ between $H$ and the $y$-axis,then $d^2$ is equal to $............$.

If the latus rectum of a hyperbola is $8$ and the eccentricity is $3/\sqrt{5}$,then the equation of the hyperbola is:

Let $m_1$ and $m_2$ be the slopes of the tangents drawn from the point $P(4, 1)$ to the hyperbola $H: \frac{y^2}{25} - \frac{x^2}{16} = 1$. If $Q$ is the point from which the tangents drawn to $H$ have slopes $|m_1|$ and $|m_2|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$-axis,then $\frac{(PQ)^2}{\alpha \beta}$ is equal to $............$.