$A$ line passing through the point $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at $A$ and $B$ such that $(PA) \cdot (PB)$ is maximum. Then $5(PA^2 + PB^2)$ is equal to:

Let $A_1$ be the area of the given ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let $A_2$ be the area of the region bounded by the curve which is the locus of the midpoint of the line segment joining the focus of the ellipse and a point $P$ on the given ellipse. Then $A_1 : A_2$ is equal to:

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 midpoint of a chord of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is $(\sqrt{2}, 4/3)$,and the length of the chord is $\frac{2 \sqrt{\alpha}}{3}$,then $\alpha$ is :

The angle between the lines joining the foci of an ellipse to one particular extremity of the minor axis is $90^{\circ}$. The eccentricity of the ellipse is

$A$ point on the ellipse $4x^2 + 9y^2 = 36$,where the normal is parallel to the line $4x - 2y - 5 = 0$,is