Answer
(a) $e=3$: hyperbola
(b) Vertices: $(1,0)$ and $(-2,\pi)$
Work Step by Step
The polar equation:
$r=\frac{4}{1+3~cos~θ}$
corresponds to:
$r=\frac{ed}{1+e~cos~θ}~~$ (Page 826)
So:
$e=3\gt1~~$ (hyperbola)
(b) Main points:
$θ=0$
$r=\frac{4}{1+3~cos~0}=\frac{4}{4}=1$
$θ=\frac{\pi}{2}$
$r=\frac{4}{1+3~cos~\frac{\pi}{2}}=\frac{4}{1}=4$
$θ=\pi$
$r=\frac{4}{1+3~cos~0}=\frac{4}{-2}=-2$
$θ=\frac{3\pi}{2}$
$r=\frac{4}{1+3~cos~\frac{3\pi}{2}}=\frac{4}{1}=4$
