Question

# Find the eccentricity of an ellipse whose latus rectum is (i) half of its minor axis (ii) half of its major axis.

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Solution

## $\text{(i) According to the question, the latus rectum is half its minor axis.}\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.\frac{2{b}^{2}}{a}=\frac{1}{2}×\left(2b\right)\phantom{\rule{0ex}{0ex}}⇒2{b}^{2}=ab\phantom{\rule{0ex}{0ex}}⇒2b=a\phantom{\rule{0ex}{0ex}}\mathrm{Now},e=\sqrt{1-\frac{{b}^{2}}{4{b}^{2}}}\phantom{\rule{0ex}{0ex}}⇒e=\sqrt{1-\frac{1}{4}}\phantom{\rule{0ex}{0ex}}⇒e=\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}2\right)\text{According to the question, the latus rectum is half its minor axis.}\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.\frac{2{b}^{2}}{a}=\frac{1}{2}×\left(2a\right)\phantom{\rule{0ex}{0ex}}⇒2{b}^{2}={a}^{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=2{b}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Now},e=\sqrt{1-\frac{{b}^{2}}{2{b}^{2}}}\phantom{\rule{0ex}{0ex}}⇒e=\sqrt{1-\frac{1}{2}}\phantom{\rule{0ex}{0ex}}⇒e=\frac{1}{\sqrt{2}}$

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