Answer
$y^{2}=-16x$
The latus rectum has endpoints $(-4,-8)$ and $(-4,8)$
Work Step by Step
1. The focus and the vertex lie on the horizontal line $\quad y=0.\quad a=4.$
2. The focus $(-4,0)$ is left of the vertex $(0,0)\Rightarrow$the parabola opens left.
3. By table 2, the equation is$\quad (y-k)^{2}=-4a(x-h)$
that is, $\quad y^{2}=-16x$
4. For $x=-4, \quad y^{2}=64\quad \Rightarrow\quad y=\pm 8$
The latus rectum (line segment parallel to the directrix, containing the focus)
has endpoints $(-4,-8)$ and $(-4,8)$
5. Directrix: $\quad \quad x=h+a\quad \quad \Rightarrow\quad x=4$
We have enough details for the graph.
