## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{2}{1+\sin \theta}$
The polar equation of a conic with eccentricity $e$ and directrix $y=k$ is defined as: $r=\dfrac{ke}{1+e \sin \theta}$ ...(1) We are given that the vertices are: $e=1,k=2$ Then $x=2$ Thus, equation (1), becomes $r=\dfrac{(2)(1)}{1+(1) \sin \theta}=\dfrac{2}{1+\sin \theta}$