Let the length of a latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $10$. If its eccentricity is the minimum value of the function $f(t) = t^2 + t + \frac{11}{12}$,$t \in R$,then $a^2 + b^2$ is equal to:

If $P$ is any point on the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$ and $S$ and $S^{\prime}$ are the foci,then $PS + PS^{\prime}$ 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 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

The tangent to the ellipse $9x^2 + 16y^2 = 288$ making equal intercepts on the coordinate axes intersects the $X$-axis and the $Y$-axis at points $A$ and $B$ respectively. Then,the area of $\triangle OAB$ (where $O$ is the origin) is:

The equation of the ellipse having a vertex at $(6,1)$,a focus at $(4,1)$ and the eccentricity $e = \frac{3}{5}$ is