Question

If the normal at one 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, then the eccentricity of the ellipse is given by the equation

• A. $e^2+e-1=0$
• B. $e^2+e+1=0$
• C. $e^4+e^2+1=0$
• D. $e^4+e^2-1=0$

Answer 1

Answer: (D)

Normal at $\left(ae,\frac{b^2}{a}\right)$ of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$\frac{x-ae}{ae/a^2}=\frac{y-b^2/a}{\frac{b^2}{a}/b^2}\left(\frac{x-x_1}{x_1/a^2}=\frac{y-y_1}{y_1/b^2}\right)$

It passes thro’ $\left(0,-b\right)$ if $\frac{0-ae}{ae/a^2}=\frac{-b-b^2/a}{1/a}$
$\Rightarrow -a^2=-a\left(b+\frac{b^2}{a}\right)$
$\Rightarrow a=\frac{ab+b^2}{a}$
$\Rightarrow a^2=ab+b^2=ab+a^2-a^2{e}^2$
$\Rightarrow ab=a^2{e}^2$
$\Rightarrow b=a{e}^2$
$\Rightarrow b^2=a^2{e}^4$
$\Rightarrow a^2\left(1-e^2\right)=a^2{e}^4$
$\Rightarrow 1-e^2=e^4$
$\Rightarrow e^4+e^2=1$