Chapter 9.1, Problem 67E

To determine

To find: The vertex, focus and directrix of the parabola ${\left(x+1\right)}^2-8\left(y+2\right)=0$ and verify with the help of graphing utility.

Answer to Problem 67E

The vertex, focus and directrix of the parabola ${\left(x+1\right)}^2-8\left(y+2\right)=0$ are $\left(-1,-2\right)$ , $\left(-1,0\right)$ , $y=-4$ respectively. The graph of the parabola with vertex, focus and directrix is shown in Figure(1).

Explanation of Solution

Given information:

The equation of the parabola is ${\left(x+1\right)}^2-8\left(y+2\right)=0$ .

Calculation:

Simplify the equation of the parabola.

  $\begin{array}{c}{\left(x+1\right)}^2-8\left(y+2\right)=0\\ {\left(x+1\right)}^2=8\left(y+2\right)\end{array}$

Compare the given equation of parabola with general vertical axis parabola equation ${\left(x-h\right)}^2=4p\left(y-k\right)$ . This gives the value of $h$ , $k$ , $p$ as $-1$ , $-2$ , $2$ respectively.

The vertex of the parabola ${\left(x+1\right)}^2-8\left(y+2\right)=0$ is $\left(h,k\right)=\left(-1,-2\right)$ .

The focus of the horizontal axis parabola ${\left(x+1\right)}^2-8\left(y+2\right)=0$ is:

  $\begin{array}{c}\left(h,k+p\right)=\left(-1,-2+2\right)\\ =\left(-1,0\right)\end{array}$

The directrix of the parabola ${\left(x+1\right)}^2-8\left(y+2\right)=0$ is:

  $\begin{array}{c}y=k-p\\ =-2-2\\ =-4\end{array}$

Therefore, The vertex, focus and directrix of the parabola ${\left(x+1\right)}^2-8\left(y+2\right)=0$ are $\left(-1,-2\right)$ , $\left(-1,0\right)$ , $y=-4$ respectively.

The graph of the parabola ${\left(x+1\right)}^2-8\left(y+2\right)=0$ with vertex, focus and directrix is shown below.

  

Figure (1)

Here we can verify the vertex, focus, directrix of the parabola ${\left(x+1\right)}^2-8\left(y+2\right)=0$ are $\left(-1,-2\right)$ , $\left(-1,0\right)$ , $y=-4$ respectively.