Answer
$(x-2)^{2}=-8(y+3)$
The latus rectum has endpoints $(-2,-5)$ and $(6,-5).$
Work Step by Step
1.$\quad $The vertex and focus lie on the line $x=2\quad \Rightarrow\quad $parabola opens up or down.
The focus is below the vertex$\quad \Rightarrow\quad $opens down.
2.$\quad $From Table 2,
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h,k-a)}&{y=k+a}&{(x-h)^{2}=-4a(y-k)}& down \\ \hline \end{array}$
$(h,k)=(2,-3)$,
3.$\quad $The equation is $\quad (x-2)^{2}=-4a(y+3)\quad $
Since the focus is at $\quad (h,k-a)=(2,-3-a)$,
given the focus is $\quad (2,-5)$, we find $a$
$-3-a=-5$
$-3+5=a$
$a=2$
So, the equation is$\quad (x-2)^{2}=-8(y+3)$
4. $\quad $For the latus rectum, set $y=-5, \quad $
$(x-2)^{2}=-8(-5+3)$
$(x-2)^{2}=16$
$x-2=\pm 4$
$x=2\pm 4$
The latus rectum (line segment parallel to the directrix, containing the focus)
has endpoints $(-2,-5)$ and $(6,-5).$
5. $\quad $We have enough details for the graph.
Graph the directrix, $\quad y=-1$
