Chapter 9.1, Problem 76E

To determine

To find: The vertex, focus and directrix of the parabola $y^2-4x-4=0$ and verify with the help of graphing utility.

Answer to Problem 76E

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

Calculation:

Simplify the equation of the parabola.

  $\begin{array}{c}{y}^2-4x-4=0\\ y^2=4x+4\\ y^2=4\left(x+1\right)\end{array}$

Compare the given equation of parabola with general horizontal axis parabola equation ${\left(y-k\right)}^2=4p\left(x-h\right)$ . This gives the value of $h$ , $k$ , $p$ as $-1$ , $0$ , $1$ respectively.

The vertex of the parabola $y^2-4x-4=0$ is $\left(h,k\right)=\left(-1,0\right)$ .

The focus of the horizontal axis parabola $y^2-4x-4=0$ is:

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

The directrix of the parabola $y^2-4x-4=0$ is:

  $\begin{array}{c}x=h-p\\ =-1-1\\ =-2\end{array}$

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

The graph of the parabola $y^2-4x-4=0$ with vertex, focus and directrix is shown below.

  

Figure (1)

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