An ellipse of eccentricity (2 √2/3) is inscribed in a circle. A point is chosen inside the circle at random. The probability that the point lies outside the ellipse is

Question

An ellipse of eccentricity (2 √2/3) is inscribed in a circle. A point is chosen inside the circle at random. The probability that the point lies outside the ellipse is (A) (1/3) (B) (2/3) (C) (1/9) (D) (2/9)

Tardigrade Admin · Accepted Answer

Given, e=(2 √2/3) e2=1-(b2/a2) ⇒ (b2/a2)=1-(8/9) ...(i) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ed1b39aa8949e0d00268b1405ffc7494-.png" /> P(x) =(π a2-π a b/π a2) =1-(b/a) =1-(1/3) [using Eq. (i)] P(x) =(2/3)

Q. An ellipse of eccentricity $\frac{2 2}{3}$ is inscribed in a circle. A point is chosen inside the circle at random. The probability that the point lies outside the ellipse is

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{1}{9}$

$\frac{2}{9}$

$\frac{1}{27}$

Given, $e = \frac{2 2}{3}$
$e^{2} = 1 - \frac{b ^{2}}{a ^{2}}$
$\Rightarrow \frac{b ^{2}}{a ^{2}} = 1 - \frac{8}{9} ... (i)$

$P (x) = \frac{π a ^{2} - πab}{π a ^{2}}$
$= 1 - \frac{b}{a}$
$= 1 - \frac{1}{3}$ [using Eq. (i)]
$P (x) = \frac{2}{3}$

Q. An ellipse of eccentricity 322​​ is inscribed in a circle. A point is chosen inside the circle at random. The probability that the point lies outside the ellipse is

31​

32​

91​

92​

271​

Given, e=322​​ e2=1−a2b2​ ⇒a2b2​=1−98​...(i) P(x)=πa2πa2−πab​ =1−ab​ =1−31​ [using Eq. (i)] P(x)=32​