Question

If the ratio of area of triangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to that of triangle formed by the corresponding points on the auxiliary circle is $\frac{1}{2}$, then the eccentricity of the ellipse is

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

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}$

The ratio of area of any triangle inscribed in an ellipse to the area of triangle formed by corresponding points on the auxiliary circle is equal to the ratio of semi-minor axis to semi major axis.
For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Assume $a>b.$ So,the given ratio is $\frac{b}{a}=\frac{1}{2}$
Now $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{4}}$
$\Rightarrow e=\frac{\sqrt{3}}{2}$