Let for two distinct values of p the lines $y=x+\mathrm{p}$ touch the ellipse $\mathrm{E}: \frac{x^2}{4^2}+\frac{y^2}{3^2}=1$ at the points A and B . Let the line $y=x$ intersect E at the points C and D . Then the area of the quadrilateral $A B C D$ is equal to :

The centre of a circle C is at the centre of the ellipse $\mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$. Let C pass through the foci $F_1$ and $F_2$ of E such that the circle $C$ and the ellipse $E$ intersect at four points. Let P be one of these four points. If the area of the triangle $\mathrm{PF}_1 \mathrm{~F}_2$ is 30 and the length of the major axis of $E$ is 17 , then the distance between the foci of $E$ is :

A line passing through the point $\mathrm{A}(-2,0)$, touches the parabola $\mathrm{P}: y^2=x-2$ at the point $B$ in the first quadrant. The area, of the region bounded by the line $A B$, parabola $P$ and the $x$-axis, is :

Let the product of $\omega_1=(8+i) \sin \theta+(7+4 i) \cos \theta$ and $\omega_2=(1+8 i) \sin \theta+(4+7 i) \cos \theta$ be $\alpha+i \beta$, $i=\sqrt{-1}$. Let p and q be the maximum and the minimum values of $\alpha+\beta$ respectively. Then $\mathrm{p}+\mathrm{q}$ is equal to :