Question

If the normal at the end of latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through $(0,-b)$, then $e^4+e^2$ (where $E$ is eccentricity) equals

• A. eccentricity of the ellipse is $\sqrt{\frac{\sqrt{5}-1}{2}}$
• B. ratio of the minor and major axes is $\frac{\sqrt{5}+1}{2}$
• C. square of the eccentricity is equal to the ratio of the minor and major axes
• D. $noneofthese$

Answer 1

Answer: (A)

eccentricity of the ellipse is $\sqrt{\frac{\sqrt{5}-1}{2}}$

Normal at the extremity of latus rectum in the first quadrant $(ae,\frac{b^2}{a})$ is
$\frac{x-ae}{ae/a^2}=\frac{y-b^2/a}{b^2/a{b}^2}$
As it passes through $(0,-b)$
$\frac{-ae}{ae/a^2}=\frac{-b-b^2/a}{1/a}$
$\Rightarrow -a^2=-ab-b^2$
$\Rightarrow a^2-b^2=ab$
$\Rightarrow a^2{e}^2=ab$
$e^2=b/a$
Therefore, $e^4=\frac{b^2}{a^2}=1-e^2$
$\Rightarrow e^2+e^2=1$