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(D) The equation of the normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at point $(x_{1}, y_{1})$ is given by $\frac{a^{2}x}{x_{1}}-

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

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(D) The equation of the normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at point $(x_{1}, y_{1})$ is given by $\frac{a^{2}x}{x_{1}}-\frac{b^{2}y}{y_{1}}=a^{2}-b^{2}=a^{2}e^{2}$.The coordinates of an end of the latus rectum are $(ae, \frac{b^{2}}{a})$.Substituting these into the normal equation:$\frac{a^{2}x}{ae}-\frac{b^{2}y}{b^{2}/a} = a^{2}e^{2}$$\frac{ax}{e}-ay = a^{2}e^{2} \Rightarrow \frac{x}{e}-y = ae^{2}$.Since this normal passes through an extremity of the minor axis $(0, b)$ or $(0, -b)$,we substitute $(0, b)$:$0 - b = ae^{2} \Rightarrow b = -ae^{2}$.Since $b$ and $a$ are lengths,we consider the magnitude $b = ae^{2}$,so $b^{2} = a^{2}e^{4}$.Using $b^{2} = a^{2}(1-e^{2})$:$a^{2}(1-e^{2}) = a^{2}e^{4}$$1-e^{2} = e^{4} \Rightarrow e^{4}+e^{2}-1=0$.

If the normal at an end of a latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through an extremity of the minor axis,then the eccentricity $e$ of the ellipse satisfies:

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