Answer
$y^{2}=-8x$
The latus rectum has endpoints $(-2,-4)$ and $(-2,4)$
Work Step by Step
1. The directrix is vertical$\quad \Rightarrow\quad $the axis of symmetry is horizontal.
2.$\quad $The focus lies on the line $y=0$. So does the vertex.
3.$\quad $The focus is left of the directrix, the parabola opens left. $a=2$
4.$\quad $From Table 2,
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h-a,k)}&{x=h+a}&{(y-k)^{2}=-4a(x-h)} &left \\ \hline \end{array}$
$(h-a,k)=(-2,0)\quad \Rightarrow$Vertex: $(0,0)$
The equation is $\quad (y-0)^{2}=-4(2)(x-0)$
that is, $\quad y^{2}=-8x$
5. For $x=-2, \quad y^{2}=16\quad \Rightarrow\quad y=\pm 4$
The latus rectum (line segment parallel to the directrix, containing the focus)
has endpoints $(-2,-4)$ and $(-2,4)$
We have enough details for the graph.
