Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$, be 30 . If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2 t-t^2$, then $\left(a^2+b^2\right)$ is equal to

Let each of the two ellipses $\mathrm{E}_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$ and $\mathrm{E}_2: \frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1,(\mathrm{~A}<\mathrm{B})$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of $\mathrm{E}_1$ and $\mathrm{E}_2$ be $l_1$ and $l_2$, respectively, such that $2 l_1^2=9 l_2$. If the distance between the foci of $E_1$ is 8 , then the distance between the foci of $E_2$ is

If the points of intersection of the ellipses $x^2+2 y^2-6 x-12 y+23=0$ and

$4 x^2+2 y^2-20 x-12 y+35=0$ lie on a circle of radius $r$ and centre $(a, b)$, then the

value of $a b+18 r^2$ is :

Let the line $y-x=1$ intersect the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$ at the points A and B . Then the angle made by the line segment AB at the center of the ellipse is :