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 :

Equation of normal at $\left( {ae,{{{b^2}} \over a}} \right)$ <br><br>${{{a^2}x} \over {ae}} - {{{b^2}y} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}$ <br><br>It passes through (0,–b) <br><br>$\therefore$ $0 - {{{b^2}\left( { - b} \right)} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}$ <br><br>$\Rightarrow$ $a$b = ${a^2} - {b^2}$ <br><br>$\Rightarrow$ $a$b = ${a^2}{e^2}$ [as b² = ${a^2}\left( {1 - {e^2}} \right)$] <br><br>$\Rightarrow$ $a$²b² = ${a^4}{e^4}$ <br><br>$\Rightarrow$ ${{{{b^2}} \over {{a^2}}}}$ = e⁴ <br><br>$\Rightarrow$ ${1 - {e^2}}$ = e⁴ <br><br>$\Rightarrow$ e⁴ + e² – 1 = 0