Question

A circle is inscribed in an ellipse. If ‘P’ is the probability that a point within the ellipse chosen at random lies outside the circle, then the

eccentricity of the ellipse is​

Answer 1

Answer:

$\sqrt{p(1-p)}$

Step-by-step explanation:

1) Let the lengths of  major and minor axis be 'a'  and 'b' respectively.

Since, circle is inscribed in an ellipse.

=> radius of circle =b.

2) Now,

Probability that random selected point in ellipse will lie outside the circle is p.

That is,

$p=\frac{Area\;of\;Ellipse\:-Area\:of\:Circle}{Area\;of\;Ellipse}\\ \\=>p=\frac{\pi*a*b-\pi*b^2}{\pi*a*b} =\frac{a-b}{a}\\ \\=>p=1-\frac{b}{a}\\ \\=>\frac{b}{a}=1-p$

3) Now, eccentricity of ellipse

$e=\sqrt{1-\frac{b^2}{a^2}}\\ \\=\sqrt{1-(1-p)^2}\\ \\=\sqrt{p(2-p)}$

Hence, we got our required eccentricity.