Find the length of the latus rectum of the ellipse $\frac{x^2}{36} + \frac{y^2}{49} = 1$.

The eccentricity of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$ is

The distance of the point $\theta$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ from a focus is

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 $\sqrt{a^2 + 6ab + b^2} =$

Prove that the product of the lengths of the perpendiculars drawn from the points $(\sqrt{a^{2}-b^{2}}, 0)$ and $(-\sqrt{a^{2}-b^{2}}, 0)$ to the line $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$ is $b^{2}$.

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 $QR = \left| \frac{a^2 \sin^2 \theta + b^2 \cos^2 \theta}{a \cos \theta} \right|$.