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If the normal at one end of the latus rectum of an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ passes through one end of the minor axis then :
$e^4 -e^2 + 1 = 0$
$e^2 -e -1 = 0$
$e^2 + e + 1 = 0$
$e^4 + e^2 -1 = 0$
Solution
Ends of latus rectum ($ \pm ae$, $\pm \mathrm{b}^{2} / \mathrm{a}$ )
equation of normal at Pt. $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ on ellipse is
$\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}=a^{2}-b^{2}$
So, equation of normal at $\left( {ae,{b^2}/a} \right)$ is
$x-e y=a e^{3}$
Normal passes through $(0,-b)$ then
$0+\mathrm{eb}=\mathrm{ae}^{3}$
$\Rightarrow \mathrm{b}=\mathrm{ae}^{2}$
$\Rightarrow \mathrm{b}^{2}=\mathrm{a}^{2} \mathrm{e}^{4} \Rightarrow \mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)=\mathrm{a}^{2} \mathrm{e}^{4}$
$\Rightarrow e^{4}+e^{2}-1=0$
