## College Algebra (10th Edition)

$x^{2}=8y$ The latus rectum has endpoints $(-4,2)$ and $(4,2)$
Given: $a=2,\ (h,k)=(0,0)$ 1. The focus and the vertex lie on the vertical line $\quad x=0.$ 2. The focus $(0,2)$ is above the vertex $(0,0)$ So the parabola opens up. 3. By table 2, the equation is$\quad (x-h)^{2}=4a(y-k)$ that is, $\quad x^{2}=8y$ 4. For $y=2, \quad x^{2}=16\Rightarrow x=\pm 4$ The latus rectum (line segment parallel to the directrix, containing the focus) has endpoints $(-4,2)$ and $(4,2)$ 5. Directrix: $\quad \quad y=k-a\quad \quad \Rightarrow\quad y=-2$ We have enough details for the graph.