Explanation

(a)
Normal at the extremity of latusrectum in the first quadrant $(ae,\ b^{2}/a)$ is
$\frac{x - ae}{\frac{ae}{a^{2}}} = \frac{y - \frac{b^{2}}{a}}{\frac{b^{2}}{{ab}^{2}}}$
As it passes through $(0,\ - b)$
$\frac{- ae}{\frac{ae}{a^{2}}} = \frac{- b - \frac{b^{2}}{a}}{\frac{1}{a}}$
$\Rightarrow {- a}^{2} = - ab - b^{2}$
$\Rightarrow a^{2} - b^{2} = ab$
$\Rightarrow a^{2}e^{2} = ab$
Or $e^{2} = \frac{b}{a}$
$\therefore e^{4} = \frac{b^{2}}{a^{2}} = 1 - e^{2}$
$\Rightarrow e^{4} + e^{2} = 1$