Answer
$y^2=-8x$
Latus rectum points: $(-2,-4),(-2,4)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(-2,0)$
Directrix: $x=2$
Because the directrix is in the form $x=a$, the parabola is horizontal. Its standard equation is:
$(y-k)^2=4p(x-h)$
Use the coordinates of the focus and the directrix to determine $h,k,p$:
$(h+p,k)=(-2,0)$
$x=h-p=2$
$\begin{cases}
h+p=-2\\
k=0\\
h-p=2
\end{cases}$
$\begin{cases}
h+p=-2\\
k=0\\
h-p+h+p=2-2
\end{cases}$
$\begin{cases}
h+p=-2\\
k=0\\
2h=0
\end{cases}$
$\begin{cases}
h+p=-2\\
k=0\\
h=0
\end{cases}$
$h+p=-2$
$0+p=-2$
$p=-2$
Determine the parabola's equation:
$(y-0)^2=4(-2)(x-0)$
$y^2=-8x$
Determine the two points defining the latus rectum:
$x=-2$
$y^2=-8(-2)$
$y^2=16$
$y=\pm 4$
$y=-4\Rightarrow y_1=-4$
$y=4\Rightarrow y_2=4$
$\Rightarrow (-2,-4),(-2,4)$
Plot the points and graph the parabola:
