The distance between the directrices of a rectangular hyperbola is $10$ units, then distance between its foci is

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

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 director circle of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{4} = 1$ is given by

The straight line $x + y = \sqrt 2 p$ will touch the hyperbola $4{x^2} - 9{y^2} = 36$, if

The equation of the tangent parallel to $y - x + 5 = 0$ drawn to $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$ is