Answer
$\dfrac{6}{3-\sin \theta}$
Work Step by Step
The polar equation of a conic with eccentricity $e$ and directrix $y=-k$ is:
$r=\dfrac{ke}{1-e \sin \theta}$ ...(1)
We are given that the vertices are: $e=\dfrac{1}{3},k=6$
Then $y=-6$
Thus, equation (1), becomes
$r=\dfrac{2}{1-(\dfrac{1}{3})\sin \theta}=\dfrac{6}{3-\sin \theta}$
