Answer
$y^{2}=\displaystyle \frac{9}{2}x$
The latus rectum has endpoints $ (\displaystyle \frac{9}{8},-\frac{9}{4})$ and $(\displaystyle \frac{9}{8},\frac{9}{4})$
Work Step by Step
1.$\quad $The axis of symmetry is horizontal, the point (2,3) is right of the vertex $\quad \Rightarrow\quad $opens right.
2.$\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^{2}=4ax\quad $
3.$\quad $Find $a$ by using the fact that (2,3) lies on the parabola:
$3^{2}=4a(2)$
$a=\displaystyle \frac{9}{8}\quad \Rightarrow\quad $Focus : $(h+a, k)$= $(\displaystyle \frac{9}{8},0)$, directrix: $y=-\displaystyle \frac{9}{8}$
4.$\quad $Equation:$\displaystyle \quad y^{2}=\frac{9}{2}x$
5. $\quad $For $x=\displaystyle \frac{9}{8}, \quad y^{2}=\displaystyle \frac{81}{16}\quad \Rightarrow\quad y=\pm\frac{9}{4}$
The latus rectum (line segment parallel to the directrix, containing the focus)
has endpoints $(\displaystyle \frac{9}{8},-\frac{9}{4})$ and $(\displaystyle \frac{9}{8},\frac{9}{4})$
We have enough details for the graph.