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If the latus rectum of an ellipse is equal to half of the minor axis, then what is its eccentricity?
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Verified Answer
The correct answer is:
$\frac{\sqrt{3}}{2}$
Length of minor axis $=2 \mathrm{~b}$ and latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$
According to given condition $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\mathrm{b}$
$\Rightarrow 2 \mathrm{~b}=\mathrm{a}$
Now, $\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\sqrt{1-\frac{\mathrm{b}^{2}}{4 \mathrm{~b}^{2}}}=\sqrt{\frac{3}{4}}$
$\Rightarrow \mathrm{e}=\frac{\sqrt{3}}{2}$
According to given condition $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\mathrm{b}$
$\Rightarrow 2 \mathrm{~b}=\mathrm{a}$
Now, $\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\sqrt{1-\frac{\mathrm{b}^{2}}{4 \mathrm{~b}^{2}}}=\sqrt{\frac{3}{4}}$
$\Rightarrow \mathrm{e}=\frac{\sqrt{3}}{2}$
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