Answer
$x^2=-4y$
Latus rectum points: $(-2,-1),(2,-1)$
See graph
Work Step by Step
We are given the parabola:
Focus: $(0,-1)$
Directrix: $y=1$
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 $h,k,p$:
$(h,k+p)=(0,-1)$
$y=k-p=1$
$\begin{cases}
h=0\\
k+p=-1\\
k-p=1
\end{cases}$
$\begin{cases}
h=0\\
k+p=-1\\
k-p+k+p=1-1
\end{cases}$
$\begin{cases}
h=0\\
k+p=-1\\
2k=0
\end{cases}$
$\begin{cases}
h=0\\
k+p=-1\\
k=0
\end{cases}$
$k+p=-1$
$0+p=-1$
$p=-1$
Determine the parabola's equation:
$(x-0)^2=4(-1)(y-0)$
$x^2=-4y$
Determine the two points defining the latus rectum:
$y=-1$
$x^2=-4(-1)$
$x^2=4$
$x=\pm 2$
$x=-2\Rightarrow x_1=-2$
$x=2\Rightarrow x_2=2$
$\Rightarrow (-2,-1),(2,-1)$
Plot the points and graph the parabola:
