Answer
(a) $\frac{3}{2}$
(b) Hyperbola
(c) $x=\frac{4}{3}$
(d) See graph
Work Step by Step
Part (a)
$r=\frac{4}{2+3\cos\theta}$
$\frac{ed}{1+e\cos\theta}=\frac{2}{1+\frac{3}{2}\cos\theta}$
$e=\frac{3}{2}$ and $ed=2$
$e=\frac{3}{2}$ and $d=\frac{4}{3}$
So, the eccentricity is $\frac{3}{2}$.
Part (b)
Since $e=\frac{3}{2}>1$, the conic is a hyperbola.
Part (c)
We have obtained $d=\frac{4}{3}$.
Using Theorem 6, and part (a) of Figure 2, the directrix has the equation $x=\frac{4}{3}$.
Part (d)
We draw the graph.
