Answer
$y^2=-16x$, $(-4,-8),(-4,8)$,
See graph.
Work Step by Step
1. Given the focus at $(-4,0)$, vertex at $(0,0)$, we can write the parabola equation in a standard form $4px=-y^2$. With $p=4$, we get $y^2=-16x$
2. Let $x=-4$, we have $y^2=64$ and $y=\pm8$, thus the two points that define the latus rectum are $(-4,-8),(-4,8)$
3. See graph.