Explanation

(a)
$Given,\ \sqrt{3}\sin{\theta + 2\cos{\theta = \frac{4}{r}}\ \ \ldots(i)}$
Any line perpendicular to Eq.(i) is
$\Rightarrow \ \sqrt{3}\cos{\theta - 2\sin{\theta = \frac{k}{r}}\ \ }$
$It\ passes\ through\ \left( - 1,\frac{\pi}{2} \right),\ then\$
$\sqrt{3}\cos{\frac{\pi}{2} - 2\sin{\frac{\pi}{2} = \frac{k}{- 1}}}$
$- 2 = \frac{k}{- 1}\ \ \Rightarrow \ k = 2$
Thus, the equation is
$\sqrt{3}\cos{\theta - 2\sin{\theta = \frac{2}{r}}}$
$\Rightarrow \ \sqrt{3}r\cos{\theta - 2r\sin{\theta = 2}}$