Question

If the area of the auxiliary circle of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ is twice the area of the ellipse, then the eccentricity of the ellipse is

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

Answer 1

Answer: (D)

Equation of auxiliary circle of the ellipse,
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ is }{x}^2+y^2=a^2$
$\therefore$ Area of auxiliary circle $=\pi a^2$
and area of an ellipse $=\pi ab$
Now, according to the question
$\pi a^2=2(\pi ab)$
$\Rightarrow a=2b$
$\Rightarrow b=\frac{a}{2}...(i)$
$\therefore$ Eccentricity of an ellipse
$=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{a^2-\frac{a^2}{4}}{a^2}}$ [from Eq.(i)]
$=\sqrt{\frac{3a^2}{4a^2}}=\frac{\sqrt{3}}{2}$