question_answer

An ellipse of eccentricity \[\frac{2\sqrt{2}}{3}\] is inscribed in a circle and a point with in the circle is chosen at random. The probability that this point lies outside the ellipse is:

A) \[\frac{1}{9}\]

B) \[\frac{4}{9}\]

C) \[\frac{1}{3}\]

D) \[\frac{2}{3}\]

Correct Answer: D

Solution :

Let the radius of the circle be a, then the major axis of the inscribed ellipse is of length\[2a\].

The required probability

\[=\frac{\pi {{a}^{2}}-\pi {{a}^{2}}\sqrt{1-{{e}^{2}}}}{\pi {{e}^{2}}}\]

\[=1-\sqrt{1-{{e}^{2}}}=-\sqrt{1-\frac{8}{9}}=\frac{2}{3}\]

[Area of ellipse\[=\pi ab=\pi a.a\sqrt{1-{{e}^{2}}},'e'\] being eccentricity.]