Answer
$x^2=\dfrac{4}{3}y$
Latus rectum points: $\left(-\dfrac{2}{3},\dfrac{1}{3}\right),\left(\dfrac{2}{3},\dfrac{1}{3}\right)$
See graph
Work Step by Step
We are given the parabola:
Vertex: $(0,0)$
Axis of symmetry:: $x=0$
Point on the graph: $(2,3)$
Because the axis of symmetry is vertical, the parabola is vertical. Its standard equation is:
$(x-h)^2=4p(y-k)$
Use the coordinates of the vertex to determine $h,k$:
$(h,k)=(0,0)$
$h=0$
$k=0$
Determine $p$ using the point $(2,3)$:
$(2-0)^2=4p(3-0)$
$4=12p$
$p=\dfrac{4}{12}=\dfrac{1}{3}$
Determine the parabola's equation:
$x^2=4\left(\dfrac{1}{3}\right)y$
$x^2=\dfrac{4}{3}y$
The focus is:
$(h,k+p)=\left(0,0+\dfrac{1}{3}\right)=\left(0,\dfrac{1}{3}\right)$
Determine the two points defining the latus rectum:
$y=\dfrac{1}{3}$
$x^2=\dfrac{4}{3}\left(\dfrac{1}{3}\right)$
$x^2=\dfrac{4}{9}$
$x=\pm \dfrac{2}{3}$
$x=-\dfrac{2}{3}\Rightarrow x_1=-\dfrac{2}{3}$
$x=\dfrac{2}{3}\Rightarrow x_2=\dfrac{2}{3}$
$\Rightarrow \left(-\dfrac{2}{3},\dfrac{1}{3}\right),\left(\dfrac{2}{3},\dfrac{1}{3}\right)$
Plot the points and graph the parabola:
