The length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is $\frac{8}{3}$. If the distance from the centre of the ellipse to its focus is $\sqrt{5}$,then $\sqrt{a^2 + 6ab + b^2} =$

The ellipse $4x^2 + 9y^2 = 1$ intersects the positive $Y$-axis at point $A$. If $S$ and $S'$ are its foci, then the area (in sq. units) of $\Delta SAS'$ is

Let $C$ be the centre of an ellipse and $PQ$ be a chord of it with $\angle PCQ = 90^{\circ}$. If $R$ is the point of intersection of the tangents to the ellipse at $P$ and $Q$,then $R$ lies on

Assertion $(A)$: If the tangent and normal to the ellipse $9x^2 + 16y^2 = 144$ at the point $P(\frac{\pi}{3})$ on it meet the major axis in $Q$ and $R$ respectively,then $QR = \frac{57}{8}$.
Reason $(R)$: If the tangent and normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $P(\theta)$ on it meet the major axis in $Q$ and $R$ respectively,then $QR = \left| \frac{a^2 \sin^2 \theta + b^2 \cos^2 \theta}{a \cos \theta} \right|$.

Find the length of the latus rectum of the ellipse $4x^2 + 9y^2 - 36y + 4 = 0$.

If $\frac{\pi}{3}$ and $\theta$ are the eccentric angles of the ends of a focal chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$,then $\tan \theta=$