Answer
$x^2=\frac{4}{3}y$, $(-\frac{2}{3},\frac{1}{3}),(\frac{2}{3},\frac{1}{3})$. See graph.
Work Step by Step
1. Given vertex $(0,0)$ and y-axis as the symmetry axis, we have the general equation as $y=kx^2$.
2. Use point $(2,3)$, we have $3=2^2k$, thus $k=\frac{3}{4}$ and $y=\frac{3}{4}x^2$ or $x^2=\frac{4}{3}y$, thus $p=\frac{1}{3}$.
3. Let $y=\frac{1}{3}$, we have $x^2=\frac{4}{9}$ and $x=\pm\frac{2}{3}$, thus we have the two points that define the latus rectum $(-\frac{2}{3},\frac{1}{3}),(\frac{2}{3},\frac{1}{3})$
4. See graph.