Explanation

(c)
The equation of the tangent at $\left( 4\cos\phi,\ \frac{16}{\sqrt{11}}\sin\phi \right)$ to the ellipse ${16x}^{2} + {11y}^{2} = 256$ is
$16\left( 4\cos\phi \right)x + 11\left( \frac{16}{\sqrt{11}}\sin\phi \right)y = 256$
$\Rightarrow 4x\cos\phi + \sqrt{11}y\sin\phi = 16$
This touches the circle $({x - 1)}^{2} + y^{2} = 4^{2}$
$\therefore\left| \frac{4\cos\phi - 16}{\sqrt{16\cos^{2}\phi + 11\sin^{2}\phi}} \right| = 4$
$\Rightarrow {(\cos\phi - 4)}^{2} = 16\cos^{2}\phi + 11\sin^{2}\phi$
$\Rightarrow 4\cos^{2}\phi + 8\cos\phi - 5 = 0$
$\Rightarrow \left( 2\cos\phi - 1 \right)\left( 2\cos\phi + 5 \right) = 0$
$\Rightarrow \cos\phi = \frac{1}{2} \Rightarrow \phi = \pm \frac{\pi}{3}$ $\left( \because\cos\phi \neq \frac{5}{2} \right)$