Explanation

(b)
Equation of tangent at $\left( 3\sqrt{3}\cos\theta,\ \sin\theta \right)$ to the ellipse $\frac{x^{2}}{27} + y^{2} = 1$ is $\frac{x\cos\theta}{3\sqrt{3}} + y\sin\theta = 1$
It cuts intercepts on the coordinate axes.
$\therefore$ Sum of intercepts on axes is
$3\sqrt{3}\sec\theta + cosec\theta = f(\theta)$ (say)
On differentiating $w.r.t.$ $\theta$
$f^{'}(\theta) = \frac{3\sqrt{3}\sin^{3}\theta - \cos^{3}\theta}{\sin^{2}\theta\cos^{2}\theta}$
For maxima and minima, put $f^{'}(\theta) = 0$
$\Rightarrow 3\sqrt{3}\sin^{3}\theta - \cos^{3}\theta = 0$
$\Rightarrow \tan^{3}\theta = \frac{1}{3\sqrt{3}}$
$\Rightarrow \tan\theta = \frac{1}{\sqrt{3}} \Rightarrow \theta\ = \frac{\pi}{6}$
At $\theta = \frac{\pi}{3},\ f^{''}(\theta) > 0$
$\therefore f(\theta)$ is minimum at $\theta = \frac{\pi}{6}$