Correct Answer - A
Perpendicular to `sqrt(3) sin theta +2 cos theta = (4)/(r)` is
`sqrt(3) sin ((pi)/(2) +theta) +2 cos ((pi)/(2)+theta) = (k)/(r )`
It is passing through `(-1,pi//2)`
`:. sqrt(3) sin pi +2 cos pi =(k)/(-1) rArr k = 2`
`:. sqrt(3) cos theta - 2 sin theta = (2)/(r) rArr 2 = sqrt(3)r cos theta - 2r sin theta = 2`.