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KCET · Maths · Ellipse

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

  1. A \(\frac{1}{\sqrt{3}}\)
  2. B \(\frac{1}{2}\)
  3. C \(\frac{1}{\sqrt{2}}\)
  4. D \(\frac{\sqrt{3}}{2}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

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
\(\begin{aligned}
\pi a^{2}=2(\pi a b) \\
\Rightarrow \quad a b &=2 b \Rightarrow b=\frac{a}{2}...(i)
\end{aligned}\)
\(\therefore\) Eccentricity of an ellipse
\(\begin{aligned}
&=\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{\frac{a^{2}-\frac{a^{2}}{4}}{a^{2}}} \quad \text { [from Eq. (i)] } \\
&=\sqrt{\frac{3 a^{2}}{4 a^{2}}}=\frac{\sqrt{3}}{2}
\end{aligned}\)