Answer
$(x+3)^2=4(y-3)$
Latus rectum points: $(-5,4),(-1,4)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(-3,4)$
Directrix: $y=2$
Because the directrix is in the form $y=a$, the parabola is vertical. Its standard equation is:
$(x-h)^2=4p(y-k)$
Use the coordinates of the focus and the directrix to determine $h,k,p$:
$(h,k+p)=(-3,4)$
$y=k-p=2$
$\begin{cases}
h=-3\\
k+p=4\\
k-p=2
\end{cases}$
$\begin{cases}
h=-3\\
k+p=4\\
k-p+k+p=2+4
\end{cases}$
$\begin{cases}
h=-3\\
k+p=4\\
2k=6
\end{cases}$
$\begin{cases}
h=-3\\
k+p=4\\
k=3
\end{cases}$
$k+p=4$
$3+p=4$
$p=1$
Determine the parabola's equation:
$(x+3)^2=4(1)(y-3)$
$(x+3)^2=4(y-3)$
Determine the two points defining the latus rectum:
$y=4$
$(x+3)^2=4(4-3)$
$(x+3)^2=4$
$x+3=\pm 2$
$x+3=-2\Rightarrow x_1=-5$
$x+3=2\Rightarrow x_2=-1$
$\Rightarrow (-5,4),(-1,4)$
Plot the points and graph the parabola:
