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

The Correct Option is D

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=\(\frac{\sqrt{3}}{2}\)