Question

If the normal at an end point of a latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ passes through one extremity of the minor axis. Then the eccentricity of the ellipse is equal to

• A. $\frac{\sqrt{5}-1}{2}$
• B. $\frac{1}{5^{\frac{1}{4}}}$
• C. $\frac{1}{2}$
• D. $\sqrt{\frac{\sqrt{5}-1}{2}}$

Answer 1

The correct option is D $\sqrt{\frac{\sqrt{5}-1}{2}}$

The coordinates of an end point $P$ of a latus-rectum are $(ae,\frac{b^2}{a})$

Equation of normal at $P$ is
$\frac{a^{2}x}{ae}-\frac{b^{2}y}{\left(b^2/a\right)}=a^2-b^2$
$\Rightarrow \frac{ax}{e}-ay=a^2{e}^2$


If it passes through one extremity of the minor axis, whose coordinates are $(0,\pm b),$ but for $(0,b)$
$0-ab=a^2{e}^2$ not possible as $a,b,e>0$
Now using $(0,-b)$
$\Rightarrow 0+ab=a^2{e}^2$
$\Rightarrow \frac{b}{a}=e^2$
$\Rightarrow \frac{b^2}{a^2}=e^4$
$\Rightarrow 1-e^2=e^4$
$\Rightarrow e^4+e^2-1=0$
$\Rightarrow (e^2{)}^2+e^2-1=0$
$\Rightarrow e^2=\frac{-1\pm \sqrt{1+4}}{2}$
$\Rightarrow e=\sqrt{\frac{\sqrt{5}-1}{2}}$
(taking positive sign as $e>0$)