Explanation

(a)
$\because e^{2} = 1 - \frac{b^{2}}{a^{2}}$
$\therefore a^{2} - b^{2} = a^{2}e^{2}$
So, the points on the minor axis at a distance $\sqrt{a^{2} - b^{2}}$ from the centre $(0,\ 0)$ of the ellipse are $(0, \pm ae)$
The equation of tangent at any point $(a\cos\theta,\ b\sin\theta)$ on the ellipse is
$\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$
$\therefore$ Required sum
$= \left\lbrack \frac{\frac{ae\sin\theta}{b} - 1}{\sqrt{\frac{\cos^{2}\theta}{a^{2}} + \frac{\sin^{2}\theta}{b^{2}}}} \right\rbrack^{2} + \left\lbrack \frac{\frac{- ae\sin\theta}{b} - 1}{\sqrt{\frac{\cos^{2}\theta}{a^{2}} + \frac{\sin^{2}\theta}{b^{2}}}} \right\rbrack^{2}$
$= \frac{\left( ae\sin\theta - b \right)^{2} + \left( ae\sin\theta + b \right)^{2}}{(b^{2}\cos^{2}\theta + a^{2}\sin^{2}\theta)} \times a^{2}$
$= \frac{2a^{2}\lbrack\left( a^{2} - b^{2} \right)\sin^{2}\theta + b^{2}\rbrack}{b^{2}\cos^{2}\theta + a^{2}\sin^{2}\theta} = 2a^{2}$