Answer
$x^{2}=\displaystyle \frac{4}{3}y$
The latus rectum has endpoints $(-\displaystyle \frac{2}{3},\frac{1}{3})$ and $(\displaystyle \frac{2}{3},\frac{1}{3})$
Work Step by Step
1.$\quad $The axis of symmetry is vertical, the point (2,3) is above the vertex $\quad \Rightarrow\quad $opens up.
2.$\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^{2}=4ay\quad $
3.$\quad $Find $a$ by using the fact that (2,3) lies on the parabola:
$2^{2}=4a(3)$
$a=\displaystyle \frac{1}{3}\quad \Rightarrow\quad $Focus : $(h,k+a)$ = $(0,\displaystyle \frac{1}{3})$, directrix: $y=-\displaystyle \frac{1}{3}$
4.$\quad $Equation:$\displaystyle \quad x^{2}=\frac{4}{3}y$
5. $\quad $For $y=\displaystyle \frac{1}{3}, \quad x^{2}=\displaystyle \frac{4}{9}\quad \Rightarrow\quad x=\pm\frac{2}{3}$
The latus rectum (line segment parallel to the directrix, containing the focus)
has endpoints $(-\displaystyle \frac{2}{3},\frac{1}{3})$ and $(\displaystyle \frac{2}{3},\frac{1}{3})$
We have enough details for the graph.
