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