Chapter 11, Problem 5CLR

To determine

To calculate:

The equation of the parabola with graph of following equation:

Vertex: $\left(4,0\right)$

Directrix: $x=0$

Answer to Problem 5CLR

The equation of the parabola is $x=\frac{1}{16}{y}^2+4$ .

  

Graph: $x=\frac{1}{16}{y}^2+4$

Explanation of Solution

Given information:

Vertex: $\left(4,0\right)$

Directrix: $x=0$

Calculation:

Consider the formula of parabola:

  $x-h=\frac{1}{4\times p}{\left(y-k\right)}^2\mathrm{...}\left(1\right)$

Then put the value of $k$ and $h$ :

  $x-4=\frac{1}{4p}{\left(y-0\right)}^2$

The directrix equal to $0$ and solve for $p$ :

  $\begin{array}{l}4-p=0\\ p=4\end{array}$

Put the value of $p$ in equation (1):

  $\begin{array}{l}x-4=\frac{1}{4\times 4}{\left(y-0\right)}^2\\ x=\frac{1}{16}{y}^2+4\end{array}$

  

Graph: $x=\frac{1}{16}{y}^2+4$