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