Question

If a latus-rectum of an ellipse subtends a right angle at the centre of the ellipse, then write the eccentricity of the ellipse.

Answer 1

The equation of ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where a>b
Now,
$△PO{P}^{\prime }$ is right angle triangle
$\therefore (P^{\prime }P{)}^2=(OP{)}^2+(O{P}^{\prime }{)}^2$
$\Rightarrow (ae-ae{)}^2+(\frac{b^2}{a}+\frac{b^2}{a})$
$=\left[\right(ae-0{)}^2+{(\frac{b^2}{a}-0)}^2]+\left[\right(ae-0{)}^2+{(0-\frac{b^2}{a})}^2]$
$\Rightarrow 0+{\left(\frac{2b^2}{a}\right)}^2=(ae{)}^2+{\left(\frac{b^2}{a}\right)}^2+(ae{)}^2+{\left(\frac{-b^2}{a}\right)}^2$
$\Rightarrow \frac{4b^4}{a^2}=2(ae{)}^2+\frac{2b^4}{a^2}$
$\Rightarrow \frac{4b^4}{a^2}-\frac{2b^4}{a^2}=2a^2{e}^2$
$\Rightarrow \frac{2b^4}{a^2}=2a^2{e}^2$
$\Rightarrow \frac{b^4}{a^4}=e^2$
$\Rightarrow e=\frac{b^2}{a^2}$
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow \frac{b^2}{a^2}=1-e^2$
$\Rightarrow e=1-e^2$
$\Rightarrow e^2+e-1=0$
$\Rightarrow e=\frac{-\left(1\right)\pm \sqrt{(1{)}^2-4\times 1\times (-1)}}{2\times 1}$
$\Rightarrow e=\frac{-1\pm \sqrt{1+4}}{2}$
$\Rightarrow e=\frac{-1\pm \sqrt{5}}{2}$
$\Rightarrow e=\frac{\sqrt{5}-1}{2}$
$\Rightarrow e=\frac{\sqrt{5}-1}{2}$
Hence, $e=\frac{\sqrt{5}-1}{2}$