Question

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

Answer 1

Let the equation of ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (where $a>b$)

Given, latus rectum$=\frac{1}{2}\times$(minor axis)

$\Rightarrow \frac{2b^2}{a}=\frac{1}{2}\left(2b\right)$

$\left[\right(\because latusrectumoftheellipseis\frac{2b^2}{a}andminoraxisis2b\left)\right]$

$\Rightarrow 2b=a$

$\Rightarrow 4b^2=a^2$

[ on squaring on both sides]

$\Rightarrow 4a^2(1-e^2)=a^2$

$[\because b^2=a^2(1-e^2\left)\right]$

$\Rightarrow 4-4e^2=1$

$\Rightarrow 4e^2=3$

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

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

$[\because$ eccentricity is always positive]

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