Question

Let $S,S^{\prime }$ be the foci and $B,B^{\prime }$ be the extremities of minor axis of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. If a point $P$ is selected at random inside the ellipse, then the probability that $P$ lies inside the quadrilateral $BS{B}^{\prime }{S}^{\prime }$ is

• A. $\frac{\sqrt{7}}{4\pi }$
• B. $\frac{\sqrt{7}}{8\pi }$
• C. $\frac{\sqrt{7}}{2\pi }$
• D. $\frac{\sqrt{7}}{3\pi }$

Answer 1

The correct option is C $\frac{\sqrt{7}}{2\pi }$

$\frac{x^2}{16}+\frac{y^2}{9}=1$
$a=4,b=3$
$e=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}$
$\therefore$ Coordinates of foci are $(\pm ae,0)$
$S(\sqrt{7},0)$ and $S^{\prime }(-\sqrt{7},0)$
Coordinates of the ends of minor axis are $(0,\pm b)$
$B(0,3)$ and $B^{\prime }(0,-3)$

Area of the ellipse $=\pi ab=12\pi$
Area of the quadrilateral $BS{B}^{\prime }{S}^{\prime }$ $=2aeb=6\sqrt{7}$
$\therefore$ Required probability $=\frac{6\sqrt{7}}{12\pi }=\frac{\sqrt{7}}{2\pi }$