Let $E$ be the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. For any three distinct points $P, Q$ and $Q^{\prime}$ on $E$, let $M(P, Q)$ be the mid-point of the line segment joining $P$ and $Q$, and $M \left( P , Q ^{\prime}\right)$ be the mid-point of the line segment joining $P$ and $Q ^{\prime}$. Then the maximum possible value of the distance between $M ( P , Q )$ and $M \left( P , Q ^{\prime}\right)$, as $P, Q$ and $Q^{\prime}$ vary on $E$, is. . . . .

Area of the quadrilaterals formed by drawing tangents at the ends of latus recta of $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ is

The equation of the tangent at the point $(1/4, 1/4)$ of the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{12}} = 1$ is

A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at the points $P$ and $Q$. Let the tangents to the ellipse at $P$ and $Q$ meet at the point $R$. If $\Delta(h)=$ area of the triangle $P Q R, \Delta_1=\max _{1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta_2=\min _{1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta_1-8 \Delta_2=$

Let the tangent and normal at the point $(3 \sqrt{3}, 1)$ on the ellipse $\frac{x^2}{36}+\frac{y^2}{4}=1$ meet the $y$-axis at the points $A$ and $B$ respectively. Let the circle $C$ be drawn taking $A B$ as a diameter and the line $x =2 \sqrt{5}$ intersect $C$ at the points $P$ and $Q$. If the tangents at the points $P$ and $Q$ on the circle intersect at the point $(\alpha, \beta)$, then $\alpha^2-\beta^2$ is equal to

For some $\theta \in\left(0, \frac{\pi}{2}\right),$ if the eccentricity of the hyperbola, $x^{2}-y^{2} \sec ^{2} \theta=10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^{2} \sec ^{2} \theta+y^{2}=5,$ then the length of the latus rectum of the ellipse is