Answer
$x^2=2y$, $(-1,\frac{1}{2}),(1,\frac{1}{2})$. See graph.
Work Step by Step
1. Given directrix $y=-\frac{1}{2}$ and vertex $(0,0)$, we have $p=\frac{1}{2}$. The general equation is $4py =x^2$ or $x^2=2y$
2. Let $y=\frac{1}{2}$, we have $x^2=1$ and $x=\pm1$, thus we have the two points that define the latus rectum $(-1,\frac{1}{2}),(1,\frac{1}{2})$
3. See graph.