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)

$\frac {\sqrt {3}} {2}$

Answer 2

The correct answer is D:$\frac{\sqrt{3}}{2}$
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 $=\pi \,a b$
Now, according to the question
$\pi a^{2}=2(\pi \,a b)$
$\Rightarrow a=2 b$
$\Rightarrow b=\frac{a}{2}\,\,\,\,\,\,...(i)$
$\therefore$Ecentricity (e)=$\sqrt{1-\frac{b^2}{a^2}}$
=$\sqrt{1-\frac{b^2}{4b^2}}$
=$\sqrt{1-\frac{1}{4}}$
e=$\frac{\sqrt{3}}{2}$