If the centre,one of the foci,and the semi-major axis of an ellipse are $(0, 0)$,$(0, 3)$,and $5$ respectively,then its equation is:

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1$,where $a > 2$,having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the $y$-axis,be $6 \sqrt{3}$. Then the eccentricity of the ellipse is

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

If an ellipse with its axes as coordinate axes,$2a$ and $2b$ as the lengths of its major and minor axes respectively,passes through the points $(2,2)$ and $(3,1)$,then $3a^2+5b^2=$

$L_1^{\prime}$ is the end of a latus rectum of the ellipse $3x^2 + 4y^2 = 12$ which is lying in the third quadrant. If the normal drawn at $L_1^{\prime}$ to this ellipse intersects the ellipse again at the point $P(a, b)$,then $a =$