If the latus rectum of an ellipse is equal
Question:

If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.

Solution:

Equation of an ellipse $=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Whereas

Length of latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$

Length of minor axis $=2 b$

$\mathrm{SO}$

$\frac{2 b}{2}=\frac{2 b^{2}}{a}$ (Given)

$A b=2 b^{2}$

$2 b^{2}-a b=0$

$b(2 b-a)=0$

So, $b=0$ or $a=2 b$

$b^{2}=a^{2}\left(1-e^{2}\right)$

$\mathrm{b}^{2}=(2 \mathrm{~b})^{2}\left(1-\mathrm{e}^{2}\right)$

$b^{2}=4 b^{2}\left(1-e^{2}\right)$

$1-e^{2}=\frac{1}{4}$

$\mathrm{e}^{2}=\frac{3}{4}$

$e=\frac{\sqrt{3}}{2}$

Hence, the eccentricity is $\frac{\sqrt{3}}{2}$

$\mathrm{Ab}=2 \mathrm{~b}^{2}$

$2 b^{2}-a b=0$

$b(2 b-a)=0$

So, $b=0$ or $a=2 b$

$b^{2}=a^{2}\left(1-e^{2}\right)$

$b^{2}=(2 b)^{2}\left(1-e^{2}\right)$

$b^{2}=4 b^{2}\left(1-e^{2}\right)$

On rearranging we get

$1-e^{2}=\frac{1}{4}$

$e^{2}=\frac{3}{4}$

$e=\frac{\sqrt{3}}{2}$

Hence, the eccentricity is $\frac{\sqrt{3}}{2}$

 

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