Explanation

(b)
Equation of ellipse $\frac{x^{2}}{14} + \frac{y^{2}}{5} = 1$
Any point on the ellipse is $(\sqrt{14}\cos\theta,\ \sqrt{5}\sin\theta)$
$\therefore$ Equation of normal at $(\sqrt{14}\cos\theta,\ \sqrt{5}\sin\theta)$ is
$\sqrt{14}x\sec\theta - \sqrt{5}y\ cosec\theta = 9$
It passes through $\left( a\cos{2\theta},\ b\sin{2\theta} \right)$
$\Rightarrow \sqrt{14}\sqrt{14}{\cos{2\theta}\sec}\theta - \sqrt{5}\sqrt{5}\ \sin{2\theta}cosec\ \theta$=9
$\Rightarrow 14\frac{\cos{2\theta}}{\cos\theta} - 5\frac{\sin{2\theta}}{\sin\theta} = 9$
$\Rightarrow 14\left( 2\cos^{2}\theta - 1 \right) - 10\cos^{2}\theta = 9\cos\theta$
$\Rightarrow 18\cos^{2}\theta - 9\cos\theta - 14 = 0$
$\Rightarrow 18\cos^{2}\theta - 21\cos\theta + 12\cos\theta - 14 = 0$
$\Rightarrow \left( 3\cos\theta + 2 \right)\left( 6\cos\theta - 7 \right) = 0$
$\Rightarrow \cos\theta = - \frac{2}{3},\ \cos\theta \neq - \frac{7}{6}$