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