University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{2}{2+\sin \theta}$; an ellipse
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: $e=\dfrac{1}{2},k=2$ Then $y=2$ Thus, equation (1), becomes $r=\dfrac{1}{1+(\dfrac{1}{2})\sin \theta}=\dfrac{2}{2+\sin \theta}$