The distance between the foci of an ellipse is $16$ and its eccentricity is $\frac{1}{2}$. The length of the major axis of the ellipse is:

$A$ line passing through the point $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at $A$ and $B$ such that $(PA) \cdot (PB)$ is maximum. Then $5(PA^2 + PB^2)$ is equal to:

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 equation of the normal to the curve $4x^2 + 9y^2 = 36$ at the point where the parametric angle is $\theta = \frac{7\pi}{4}$ is

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6, 0)$,foci $(\pm 4, 0)$.

Consider an ellipse,whose centre is at the origin and its major axis is along the $x-$ axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is $6$,then the area (in sq. units) of the quadrilateral inscribed in the ellipse,with the vertices as the vertices of the ellipse,is