Question 11.2.3: Graphing a Parabola Find the vertex, focus, and directrix of...
Graphing a Parabola
Find the vertex, focus, and directrix of the parabola 2y² – 8y – x + 7 = 0. Sketch the graph of the parabola.
First, complete the square on y. Then you can compare the resulting equation with one of the standard forms of the equation of a parabola.
\begin{aligned}2 y^{2}-8 y-x+7 &=0 & & \text { The given equation } \\2 y^{2}-8 y &=x-7 & & \text { Isolate terms containing } y \\2\left(y^{2}-4 y\right) &=x-7 & & \text { Factor out the common factor, } 2 .\end{aligned}
2\left(y^{2}-4 y+4\right)=x-7+8 \quad \operatorname{Add} 2\left(-\frac{4}{2}\right)^{2}=2(-2)^{2}=2 \cdot 4=8 \text { to both sides to complete the square.}
2(y-2)^{2}=x+1 \quad y^{2}-4 y+4=(y-2)^{2} simplify.
(y-2)^{2}=\frac{1}{2}(x+1) Divide both sides by 2.
The last equation is in one of the standard forms of the parabola. Comparing it with the form (y-k)^{2}=4 a(x-h) \text {, we have } h=-1 ; k=2 ; \text { and } 4 a=\frac{1}{2} \text {, or } a=\frac{1}{8}.
The parabola opens to the right, the vertex is (h, k)=(-1,2), the focus is (h+a, k)=\left(-1+\frac{1}{8}, 2\right)=\left(-\frac{7}{8}, 2\right), the axis is the horizontal line y = 2, and the directrix is the vertical line x=h-a=-1-\frac{1}{8}=-\frac{9}{8} . The graph is shown in Figure 10.
