Q.If the normal at one end of a latus rectum of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ passes through one end of the minor axis, then show that $e^4 + e^2 = 1$ (e is the eccentricity of the ellipse).

Write the normal at the latus-rectum end $(ae,b^2/a)$, require it to pass through the minor-axis end $(0,-b)$, and simplify using $b^2=a^2(1-e^2)$.

For the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, the normal at a point $(x_1,y_1)$ on it is

$\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2.$

One end of the latus rectum is $\left(ae,\dfrac{b^2}{a}\right)$. Substituting $x_1=ae,\ y_1=\dfrac{b^2}{a}$:

$\frac{a^2x}{ae}-\frac{b^2y}{b^2/a}=a^2-b^2 \implies \frac{ax}{e}-ay=a^2-b^2.$

Since $a^2-b^2=a^2e^2$, dividing by $a$:

$\frac{x}{e}-y=ae^2. \quad (\ast)$

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