$A$ 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:

For an ellipse with eccentricity $e = \frac{1}{2}$,the centre is at the origin. If one of its directrices is $x = 4$,then the equation of the ellipse is

Let $S \equiv \frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$ and $S^{\prime} \equiv \frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}-1=0$ be two intersecting ellipses. If $P(a \cos \theta, b \sin \theta)$ and $Q\left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right)$ are their points of intersection,then $\frac{1}{2}\left(a^2 \beta^2+b^2 \alpha^2\right)=$

The eccentricity of the ellipse $25x^2 + 16y^2 = 100$ 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

If $A_1, A_2, A_3$ are the areas of the ellipse $x^2+4y^2-4=0$, its director circle, and its auxiliary circle respectively, then $A_2+A_3-A_1=$ (in $\pi$)