Question

If the normal at an end of a latus-rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through one extremity of the minor axis, the eccentricity of the ellipse is given by:

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

Answer 1

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

Let $a>b$, then one of latus rectum of the ellipse is $(ae,\frac{b^2}{a})$

Thus equation of normal at this point is given by,

$\frac{a^{2}x}{ae}-\frac{b^{2}y}{b^2/a}=a^2{e}^2$
Given it passes through one of minor axis ,which is $(0,-b)$
$\Rightarrow \frac{a^2\left(0\right)}{ae}-\frac{b^2(-b)}{b^2/a}=a^2{e}^2$
$\Rightarrow e^2=\frac{b}{a}$

Now using $e^2=1-\frac{b^2}{a^2}$
we get, $e^4+e^2-1=0$
$e^2=\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$(not possible)

$\therefore e^2=\frac{-1+\sqrt{5}}{2}$