Question

If the distance between the foci of an ellipse is equal to the length of the latus-rectum, write the eccentricity of ellipse.

Answer 1

Let the equation of ellipse be
$\frac{x^2}{a^2}=1,$ where a>b
According to equation
$\Rightarrow 2ae=\frac{2b^2}{a}$
$\Rightarrow a^2=b^2$
$\Rightarrow e=\frac{b^2}{a^2}$
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow \frac{b^2}{a^2}=(1-e^2)$
$\Rightarrow e=1-e^2$
$\Rightarrow e^2+e-1=0$
$\Rightarrow e=\frac{-(-1)\pm \sqrt{(1{)}^2-4\times 1\times (-1)}}{2\times 1}$
$\Rightarrow e=\frac{-1\pm \sqrt{1+4}}{1}$
$\Rightarrow e=\frac{-1\pm \sqrt{5}}{2}$
$\Rightarrow e=\frac{\sqrt{5}-1}{2}$ $[\because e\ne \frac{\sqrt{5}-1}{2}]$
$\Rightarrow e=\frac{\sqrt{5}-1}{2}$