## Precalculus (10th Edition)

$(x-3)^2=-8y$ Latus rectum points: $(-1,-2),(7,-2)$ See graph
We are given the parabola: Focus: $(3,-2)$ Vertex: $(3,0)$ Because the vertex and the focus have the same $x$-coordinate, the parabola is vertical. Its standard equation is: $(x-h)^2=4p(y-k)$ Use the coordinates of the vertex to determine $h,k$: $(h,k)=(3,0)$ $h=3$ $k=0$ Determine $p$ using the coordinates of the focus: $(h,k+p)=(3,-2)$ $(3,0+p)=(3,-2)$ $p=-2$ Determine the parabola's equation: $(x-3)^2=4(-2)(y-0)$ $(x-3)^2=-8y$ Determine the two points defining the latus rectum: $y=-2$ $(x-3)^2=-8(-2)$ $(x-3)^2=16$ $x-3=\pm 4$ $x-3=-4\Rightarrow x_1=-1$ $x-3=4\Rightarrow x_2=7$ $\Rightarrow (-1,-2),(7,-2)$ Plot the points and graph the parabola: