An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ and the parabola $x^2=4(y+b)$ are such that the two foci of the ellipse and the end points of the latusrectum of parabola are the vertices of a square. The eccentricity of the ellipse is

Area of the quadrilaterals formed by drawing tangents at the ends of latus recta of $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ is

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:

Minimum distance between two points $P$ and $Q$ on the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{4} = 1$ , if difference between eccentric angles of $P$ and $Q$ is $\frac{{3\pi }}{2}$ , is

Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$, be $\frac{1}{4}$. If this ellipse passes through the point $\left(-4 \sqrt{\frac{2}{5}}, 3\right)$, then $a^{2}+b^{2}$ is equal to