If the point $P$ on the curve $4x^{2} + 5y^{2} = 20$ is farthest from the point $Q(0, -4)$,then $PQ^{2}$ is equal to:

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 $E_2$?

The equation of the ellipse referred to its axes as the axes of coordinates with latus rectum of length $4$ and distance between foci $4 \sqrt{2}$ 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

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

$P$ is a variable point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with foci $F_1$ and $F_2$. If $A$ is the area of the triangle $P F_1 F_2$,then the maximum value of $A$ is