If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies:

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$e^{2} + e - 1 = 0$

$e^{2} + 2 e - 1 = 0$

$e^{4} + e^{2} - 1 = 0$

$e^{4} + 2 e^{2} - 1 = 0$

answer is C.

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Detailed Solution

$Equation of the ellipse is \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$One end of the latursectum is L (a e, \frac{b^{2}}{a})$

Equation of the normal at the point L is

$\begin{matrix} \frac{a^{2} x}{x_{1}} - \frac{b^{2} y}{y_{1}} = a^{2} - b^{2} \\ \frac{a^{2} x}{a e} - \frac{b^{2} y}{(b^{2} / a)} = a^{2} e^{2} \end{matrix}$

$The above normal is passing through the point (0, - b), one end of the minor axis$

$\begin{matrix} \Rightarrow 0 + a b = a^{2} e^{2} \\ \Rightarrow \frac{b}{a} = e^{2} \\ \Rightarrow \frac{b^{2}}{a^{2}} = e^{4} \Rightarrow 1 - e^{2} = e^{4} \\ \Rightarrow e^{4} + e^{2} - 1 = 0 \end{matrix}$