If the lines joining the focii of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a b$, and an extremity of its minor axis is inclined at an angle $60^{\circ}$, then the eccentricity of the ellipse is

Question

If the lines joining the focii of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a > b$, and an extremity of its minor axis is inclined at an angle $60^{\circ}$, then the eccentricity of the ellipse is, $\frac{\sqrt{3}}{2}$, $\frac{1}{2}$, $\frac{\sqrt{7}}{3}$, $\frac{1}{\sqrt{3}}$

MARKS App · Accepted Answer

Hint: $\frac{b}{a e}=\tan 60^{\circ} \Rightarrow \frac{b}{a}=e \sqrt{3}$ $\begin{aligned} & \because e^2=1-\frac{b^2}{a^2}=1-3 e^2 \ & \Rightarrow e=+\frac{1}{2} \quad\left(e \neq-\frac{1}{2}\right) \end{aligned}$

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