Answer
(a) $\frac{1}{3}$
(b) Ellipse
(c) $x=\frac{9}{2}$
(d) See graph
Work Step by Step
Part (a)
$r=\frac{9}{6+2\cos\theta}$
$\frac{ed}{1+e\cos\theta}=\frac{\frac{9}{6}}{1+\frac{2}{6}\cos\theta}$
$\frac{ed}{1+e\cos\theta}=\frac{\frac{3}{2}}{1+\frac{1}{3}\cos\theta}$
$e=\frac{1}{3}$ and $ed=\frac{3}{2}$
$e=\frac{1}{3}$ and $d=\frac{9}{2}$
So, the eccentricity is $\frac{1}{3}$.
Part (b)
Since $e=\frac{1}{3}<1$, the conic is an ellipse.
Part (c)
We have obtained $d=\frac{9}{2}$.
Using Theorem 6, and part (a) of Figure 2, the directrix has the equation $x=\frac{9}{2}$.
Part (d)
We graph the conic.
