Q86 / 155 · WBJEE 2023 · Mathematics

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

  • A
    \(\frac{\sqrt{3}}{2}\)
  • Correct answer
    B
    \(\frac{1}{2}\)
  • C
    \(\frac{\sqrt{7}}{3}\)
  • D
    \(\frac{1}{\sqrt{3}}\)

Solution

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

Hence, the correct answer is option: 'B'.

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