Answer
$r \cos (\theta +\dfrac{\pi}{6})=\dfrac{1}{2}$
Work Step by Step
Conversion of polar coordinates and Cartesian coordinates are as follows:
a) $r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$
b) $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$
c) $x=r \cos \theta$
d) $y=r \sin \theta$
We are given that $\sqrt 3 x-\sqrt y=1$
This implies that $\cos (\dfrac{\pi}{6}) (r \cos \theta)+\sin (\dfrac{\pi}{6}) (r \sin \theta)=\dfrac{1}{2}$
or, $r[\cos (\dfrac{\pi}{6}) (\cos \theta)+\sin (\dfrac{\pi}{6}) (\sin \theta)]=\dfrac{1}{2}$
Thus, $r \cos (\theta +\dfrac{\pi}{6})=\dfrac{1}{2}$
