Search any question & find its solution
Question:
Answered & Verified by Expert
If the area of the auxillary 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
Options:
Solution:
2973 Upvotes
Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{2}$
Equation of ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Then, equation of auxillary circle will be
$$
x^{2}+y^{2}=a^{2} \quad(a>b)
$$
Now, it is given that
Area of auxillary circle $=2 \times$ Area of ellipse
$$
\begin{array}{ll}
\Rightarrow & \pi a^{2}=2 \times \pi a b \\
\Rightarrow & a=2 b
\end{array}
$$
Now, cocentricity of the ellipse,
$$
\begin{aligned}
e &=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{b^{2}}{(2 b)^{2}}} \\
&=\sqrt{1-\frac{b^{2}}{4 b^{2}}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}
\end{aligned}
$$
Then, equation of auxillary circle will be
$$
x^{2}+y^{2}=a^{2} \quad(a>b)
$$
Now, it is given that
Area of auxillary circle $=2 \times$ Area of ellipse
$$
\begin{array}{ll}
\Rightarrow & \pi a^{2}=2 \times \pi a b \\
\Rightarrow & a=2 b
\end{array}
$$
Now, cocentricity of the ellipse,
$$
\begin{aligned}
e &=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{b^{2}}{(2 b)^{2}}} \\
&=\sqrt{1-\frac{b^{2}}{4 b^{2}}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.