Answer
$y^2=\dfrac{9}{2}x$
Latus rectum points: $\left(\dfrac{9}{8},-\dfrac{9}{4}\right),\left(\dfrac{9}{8},\dfrac{9}{4}\right)$
See graph
Work Step by Step
We are given the parabola:
Vertex: $(0,0)$
Axis of symmetry: $y=0$
Point on the graph: $(2,3)$
Because the axis of symmetry is horizontal, the parabola is horizontal. Its standard equation is:
$(y-k)^2=4p(x-h)$
Use the coordinates of the vertex to determine $h,k$:
$(h,k)=(0,0)$
$h=0$
$k=0$
Determine $p$ using the point $(2,3)$:
$(3-0)^2=4p(2-0)$
$9=8p$
$p=\dfrac{9}{8}$
Determine the parabola's equation:
$y^2=4\left(\dfrac{9}{8}\right)x$
$y^2=\dfrac{9}{2}x$
The focus is:
$(h+p,k)=\left(0+\dfrac{9}{8},0\right)=\left(\dfrac{9}{8},0\right)$
Determine the two points defining the latus rectum:
$x=\dfrac{9}{8}$
$y^2=\dfrac{9}{2}\left(\dfrac{9}{8}\right)$
$y^2=\dfrac{81}{16}$
$y=\pm \dfrac{9}{4}$
$y=-\dfrac{9}{4}\Rightarrow y_1=-\dfrac{9}{4}$
$y=\dfrac{9}{4}\Rightarrow y_2=\dfrac{9}{4}$
$\Rightarrow \left(\dfrac{9}{8},-\dfrac{9}{4}\right),\left(\dfrac{9}{8},\dfrac{9}{4}\right)$
Plot the points and graph the parabola:
