## Precalculus (10th Edition)

We are given the equation in polar coordinates: $r=\dfrac{6}{8+2\sin\theta}$ Rewrite the equation: $r=\dfrac{\dfrac{6}{8}}{\dfrac{8+2\sin\theta}{8}}$ $r=\dfrac{\dfrac{3}{4}}{1+\dfrac{1}{4}\sin\theta}$ The equation is in the form: $r=\dfrac{ep}{1+e\sin\theta}$ Identify $e$ from the denominator, then $p$ from the numerator: $e=\dfrac{1}{4}$ $ep=\dfrac{3}{4}\Rightarrow \dfrac{1}{4}p=\dfrac{3}{4}\Rightarrow p=3$ Because $e<1$, the conic is an ellipse. The directrix is parallel to the polar axis at a distance of 3 units above the pole.