Answer
$x^{2}=2y$
The latus rectum has endpoints $(-1,\displaystyle \frac{1}{2})$ and $(1,\displaystyle \frac{1}{2}).$
Work Step by Step
1. The directrix is horizontal$\quad \Rightarrow\quad $the axis of symmetry is vertical.
The vertex lies on the line $x=0$. So does the focus.
2.$\quad $The vertex $(0,0)$ is 0.5 units above the directrix, the parabola opens up.
3.$\quad $The focus is another 0.5 units above the vertex $F=(0,0.5),\quad a=0.5$
4.$\quad $From Table 2,
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h,k+a)}&{y=k-a}&{(x-h)^{2}=4a(y-k)}&up\\ \hline \end{array}$
The equation is $\quad (x-0)^{2}=4(0.5)(y-0)$
that is, $\quad x^{2}=2y$
5. For $y=\displaystyle \frac{1}{2}, \quad x^{2}=1\quad \Rightarrow\quad x=\pm 1$
The latus rectum (line segment parallel to the directrix, containing the focus)
has endpoints $(-1,0.5)$ and $(1,0.5)$
We have enough details for the graph.
