The normal drawn at the point $\left(\sqrt{9} \cos \frac{\pi}{4}, \sqrt{7} \sin \frac{\pi}{4}\right)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{7} = 1$ intersects its major axis at the point

An ellipse has $OB$ as semi-minor axis,$F$ and $F'$ as its foci,and the angle $\angle FBF'$ is a right angle. Then the eccentricity of the ellipse 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:

The coordinates of any point,in the parametric form,on the ellipse whose foci are $(-2,0)$ and $(8,0)$ and eccentricity is $\frac{1}{\sqrt{2}}$,is

An ellipse is drawn by taking a diameter of the circle $(x - 1)^2 + y^2 = 1$ as its semi-minor axis and a diameter of the circle $x^2 + (y - 2)^2 = 4$ as its semi-major axis. If the center of the ellipse is at the origin and its axes are the coordinate axes,then the equation of the ellipse is:

If the normal at one end of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through one end of the major axis,then: