Answer
Graph:
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Work Step by Step
The polar equation for a conic with eccentricity $e$ is $\displaystyle \quad r=\frac{ke}{1+e\cos\theta},$
where
$x=k \gt 0$ is the vertical directrix, and $\left\{\begin{array}{ll}
e=1 & \Rightarrow\text{parabola}\\
e \lt 1 & \Rightarrow\text{ellipse}\\
e \gt 1 & \Rightarrow\text{hyperbola}
\end{array}\right.$
$\displaystyle \frac{6\div 2}{(2+\cos\theta)\div 2}=\frac{3}{1+(\frac{1}{2})\cos\theta}=\frac{6(\frac{1}{2})}{1+(\frac{1}{2})\cos\theta}$
So, $e=\displaystyle \frac{1}{2},\qquad \Rightarrow\text{ellipse}$
and $\quad k=6 \Rightarrow x=6$ is the right directrix.
$ r=\displaystyle \frac{a(1-e^{2})}{1+e\cos\theta}\quad$ is the polar equation of an ellipse,
$a(1-e^{2})=3$
$a(1-\displaystyle \frac{1}{4})=3$
$a\displaystyle \cdot\frac{3}{4}=3$
$a=4$
$c=ea=2$
The center of the ellipse is (in polar coordinates) at $(2,\pi)$ (2 units left of the focus at the origin).
The vertices are at $4$ units left/right of the center;
in polar coordinates,
$(6,\pi)$ and
$(-2,\pi)$ or $(2,0)$
