The area (in square units) of the quadrilateral formed by joining the foci of the two ellipses $\frac{x^2}{9}+\frac{y^2}{5}=1$ and $\frac{x^2}{5}+\frac{y^2}{9}=1$ is

Let a line $L$ pass through the point of intersection of the lines $bx + 10y - 8 = 0$ and $2x - 3y = 0$,where $b \in R - \{\frac{4}{3}\}$. If the line $L$ also passes through the point $(1, 1)$ and touches the circle $17(x^2 + y^2) = 16$,then the eccentricity of the ellipse $\frac{x^2}{5} + \frac{y^2}{b^2} = 1$ is:

If the line $y = mx + c$ touches the ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,then $c = $

The area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{\pi}{6}$ sq. units. Which of the following is a possible equation of the ellipse?

Assertion $(A)$: The image of $\frac{x^2}{25}+\frac{y^2}{16}=1$ in the line $x+y=10$ is $\frac{(x-10)^2}{16}+\frac{(y-10)^2}{25}=1$.
Reason $(R)$: The image of a curve '$C$' in a line $L$ is the locus of the image of every point of $C$ with respect to the line $L$.
The correct option among the following is:

The eccentricity of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$ is