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Question
If the latus rectum of an ellipse is one half of its minor axis, then its eccentricity is
Options
- \[\frac{1}{2}\]
- \[\frac{1}{\sqrt{2}}\]
- \[\frac{\sqrt{3}}{2}\]
- \[\frac{\sqrt{3}}{4}\]
Solution
\[\frac{\sqrt{3}}{2}\]
According to the question, the latus rectum of an ellipse is half its minor axis.
\[i . e . \frac{2 b^2}{a} = \frac{1}{2} \times 2b\]
\[ \Rightarrow 2 b^2 = ab\]
\[ \Rightarrow a = 2b\]
\[\text{ Now, }e = \sqrt{1 - \frac{b^2}{a^2}}\]
\[ \Rightarrow e = \sqrt{1 - \frac{b^2}{4 b^2}}\]
\[ \Rightarrow e = \sqrt{1 - \frac{1}{4}}\]
\[ \Rightarrow e = \sqrt{\frac{3}{4}}\]
\[ \Rightarrow e = \frac{\sqrt{3}}{2}\]
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