Question

An ellipse of eccentricity $\frac{\sqrt{5}}{3}$ is inscribed in a circle and a point with in the circle is selected at random. The probability that the point lies out side the ellipse is

• A. $\frac{4}{5}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{5}$

Answer 1

Answer: (C)

$\frac{1}{3}$

The standard equation of an ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$(a>b)$ and equation of circle be $x^2+y^2=a^2$
$\therefore$ Area of circle $=\pi a^2$ and Area of ellipse $\pi ab$
Now point lies outside the ellipse with in the circle
$\therefore$ Area of region in which point lies $=\pi a^2-\pi ab$
$\therefore$ Required probability
$=\frac{\pi a^2-\pi ab}{\pi a^2}\left(\frac{Favourablearea}{Totalarea}\right)$
$=1-\frac{b}{a}=1-\sqrt{\frac{b^2}{a^2}}=1-\sqrt{1-e^2}$
$1-\sqrt{1-\frac{5}{9}}(\because e=\frac{\sqrt{5}}{3})=1-\frac{2}{3}=\frac{1}{3}$