In an ellipse,two vertices are $(5,0)$ and $(0,-4)$. Then the equation of the ellipse is

Let an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a < b$, pass through the point $(4, 3)$ and have eccentricity $\frac{\sqrt{5}}{3}$. Then the length of its latus rectum is:

The eccentricity of the ellipse $x^2+4 y^2+2 x+16 y+13=0$ is

If $c \in \mathbb{R}$ be such that the line $4x - y + c = 0$ touches the ellipse $x^2 + 4y^2 = 4$,then an equation having all such values of $c$ among its roots is

$L_1^{\prime}$ 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 $a =$

The line $x = 8$ is the directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the corresponding focus $(2, 0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0, 4\sqrt{3})$ and intersects the $x$-axis at $Q$,then $(3PQ)^2$ is equal to $........$