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} $
$\Rightarrow b = ae^{2}$
$\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} $
$\Rightarrow e^{4}+e^{2}=1$