Answer
Graph:
.
Work Step by Step
The polar equation for a conic with eccentricity $e$ is
$r=\displaystyle \frac{ke}{1+e\cos\theta},$
where $x=k \gt 0$ is the vertical directrix, and
$\left\{\begin{array}{ll}
e=1 & \Rightarrow\text{parabola}\\
e \lt 1 & \Rightarrow\text{ellipse}\\
e \gt 1 & \Rightarrow\text{hyperbola}
\end{array}\right.$
Here,
$e=1 \Rightarrow$ this is a parabola
$k=1, \Rightarrow$ directrix: $x=1$ (opens left, since the focus is at the origin)
The vertex is at $\theta=0\quad $(the x-axis).
$r=\displaystyle \frac{1}{1+1}=\frac{1}{2}$
The polar coordinates of the vertex are: $(\displaystyle \frac{1}{2},0)$
