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 $\left| m _1\right|$ and $\left| m _2\right|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$ axis, then $\frac{(P Q)^2}{\alpha \beta}$ is equal to $............$.

The equation of the hyperbola in the standard form (with transverse axis along the $x$ -  axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ is

A rectangular hyperbola of latus rectum $2$ units passes through $(0, 0)$ and has $(1, 0)$ as its one focus. The other focus lies on the curve -

The normal to the rectangular hyperbola $xy = c^2$ at the point $'t_1'$ meets the curve again at the point $'t_2'$ . Then the value of $t_{1}^{3} t_{2}$ is

Let the foci of a hyperbola $\mathrm{H}$ coincide with the foci of the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$ and the eccentricity of the hyperbola $\mathrm{H}$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $\mathrm{H}$ is $\alpha$ and the length of its conjugate axis is $\beta$, then $3 \alpha^2+2 \beta^2$ is equal to :

The foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide. Then the value of $b^2$ is -