WBJEE · Maths · Ellipse
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 \(\frac{\sqrt{3}}{2}\)
- B \(\frac{1}{2}\)
- C \(\frac{\sqrt{7}}{3}\)
- D \(\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
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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