Let an ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$, passes through $\left(\sqrt{\frac{3}{2}}, 1\right)$ and has ecentricity $\frac{1}{\sqrt{3}} .$ If a circle, centered at focus $\mathrm{F}(\alpha, 0), \alpha>0$, of $\mathrm{E}$ and radius $\frac{2}{\sqrt{3}}$, intersects $\mathrm{E}$ at two points $\mathrm{P}$ and $\mathrm{Q}$, then $\mathrm{PQ}^{2}$ is equal to:

The pole of the straight line $x + 4y = 4$ with respect to ellipse ${x^2} + 4{y^2} = 4$ is

Find the equation for the ellipse that satisfies the given conditions: $b=3,\,\, c=4,$ centre at the origin; foci on the $x$ axis.

If tangents are drawn from the point ($2 + 13cos\theta , 3 + 13sin\theta $) to the ellipse $\frac{(x-2)^2}{25} + \frac{(y-3)^2}{144} = 1,$ then angle between them, is

The line $x =8$ is the directrix of the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with the corresponding focus $(2,0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0,4 \sqrt{3})$ and intersects the $x$-axis at $Q$, then $(3PQ)^2$ is equal to $........$

The equation of an ellipse, whose vertices are $(2, -2), (2, 4)$ and eccentricity $\frac{1}{3}$, is