Question

The latus rectum subtends a right angle at the centre of the ellipse, then its eccentricity is

• A. $2\sin {18}^{\circ }$
• B. $2\cos {18}^{\circ }$
• C. $2\sin {54}^{\circ }$
• D. $2\cos {54}^{\circ }$

Answer 1

The correct option is A $2\sin {18}^{\circ }$

Let the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$

end points of latus rectum are :$L(ae,\frac{b^2}{a})$ and $L^{\prime }(ae,\frac{-b^2}{a})$
As $L{L}^{\prime }$ subtend right angle at center $O(0,0)$, so
Slope of $OL\times O{L}^{\prime }=-1$
$\Rightarrow \frac{\frac{b^2}{a}}{ae}\times \frac{\frac{-b^2}{a}}{ae}=-1\Rightarrow e^2=\frac{b^4}{a^4}\Rightarrow e^2=(1-e^2{)}^2\Rightarrow e^2+e-1=0\text{or}{e}^2-e-1=0\Rightarrow e=\frac{-1+\sqrt{5}}{2}\text{or}e=\frac{1+\sqrt{5}}{2}(\because 0<e<1\text{for ellipse)}\therefore e=2\sin {18}^{\circ }$