Explanation

(a)
Any line which is perpendicular to $\sqrt{3}\sin\theta + 2\cos{\theta =}\frac{4}{r}$ is
$\sqrt{3}\sin\left( \frac{\pi}{2} + \theta \right) + 2\cos\left( \frac{\pi}{2} + \theta \right) = \frac{k}{r}$ …(i)
Since, it is passing through $\left( - 1,\frac{\pi}{2} \right)$
$\therefore\ \sqrt{3}\sin\pi + 2\cos\pi = \frac{k}{- 1} \Rightarrow k = 2$
On putting $k = 2$ n Eq. (i), we get
$\sqrt{3}\cos\theta - 2\sin\theta = \frac{2}{r}$
$\Rightarrow 2 = \sqrt{3}r\cos\theta - 2r\sin\theta$