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