If the area of the auxiliary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (where $a > b$) is twice the area of the ellipse,then the eccentricity of the ellipse is

An ellipse having foci at $(3, 3)$ and $(-4, 4)$ and passing through the origin has eccentricity equal to

The length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is $\frac{8}{3}$. If the distance from the centre of the ellipse to its focus is $\sqrt{5}$,then $\sqrt{a^2 + 6ab + b^2} =$

The eccentricity of the ellipse $4x^2 + 25y^2 = 100$ 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:

If $\tan \theta_1 \times \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining $2$ points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at