Find the locus of the midpoint of the portion of any tangent to the ellipse $x^{2} + 2y^{2} = 2$ intercepted between the axes.

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:

Let $P$ be any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. If $S_1$ and $S_2$ are its foci,then the maximum area of $\Delta PS_1S_2$ is (in square units):

The centre of the ellipse $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$ is

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is

If the curves $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut each other orthogonally, then $a^2-b^2$ equals to