The ellipse $E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ circumscribes the rectangle $R$ and passes through the point $(0, 4)$. What is the eccentricity of the ellipse $E_2$?

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

The centre of an ellipse is $C$,$PN$ is any ordinate,and $A$,$A'$ are the end points of the major axis. Then the value of $\frac{PN^2}{AN \cdot A'N}$ is

Let $P \left(\frac{2 \sqrt{3}}{\sqrt{7}}, \frac{6}{\sqrt{7}}\right), Q, R$ and $S$ be four points on the ellipse $9x^2 + 4y^2 = 36$. Let $PQ$ and $RS$ be mutually perpendicular chords passing through the origin. If $\frac{1}{(PQ)^2} + \frac{1}{(RS)^2} = \frac{p}{q}$,where $p$ and $q$ are coprime,then $p+q$ is equal to $.........$.

The product of the perpendiculars from the two foci of the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$ on the tangent at any point on the ellipse is:

The minimum area of a triangle formed by any tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{81} = 1$ and the coordinate axes is