Let a focus of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $S(4, 0)$ and its eccentricity be $\frac{4}{5}$. If the point $P(3, \alpha)$ lies on $E$ and $O$ is the origin, then the area of $\triangle POS$ is equal to: (in $/ 5$)

The equation of the ellipse in the standard form whose length of the latus rectum is $4$ and whose distance between the foci is $4 \sqrt{2}$,is

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

$P$ is a variable point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with foci $F_1$ and $F_2$. If $A$ is the area of the triangle $P F_1 F_2$,then the maximum value of $A$ is

Let each of the two ellipses $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, (A < B)$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of $E_1$ and $E_2$ be $\ell_1$ and $\ell_2$,respectively,such that $2\ell_1^2 = 9\ell_2$. If the distance between the foci of $E_1$ is $8$,then the distance between the foci of $E_2$ is:

If the area of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (where $a > b$) is twice the area of the ellipse,then the eccentricity of the ellipse is