Chapter 10 - Section 10.6 - Conic Sections in Polar Coordinates - 10.6 Exercises - Page 718: 16

(a) $\frac{1}{2}$ (b) Ellipse (c) $y=1$ (d) See graph

Part (a) $r=\frac{1}{2+\sin\theta}$ $\frac{ed}{1+e\sin\theta}=\frac{\frac{1}{2}}{1+\frac{1}{2}\sin\theta}$ $e=\frac{1}{2}$ and $ed=\frac{1}{2}$ $e=\frac{1}{2}$ and $d=1$ So, the eccentricity is $\frac{1}{2}$. Part (b) Since $e=\frac{1}{2}<1$, the conic is an ellipse. Part (c) We have obtained $d=1$. Using Theorem 6, and part (c) of Figure 2, the directrix has the equation $y=1$. Part (d) See graph