Chapter 9.2, Problem 30PPS

To determine

To write: The equation for the parabola with vertex $\left(-6,0\right)$ and directrix $x=2$ then plot the graph of the equation.

Answer to Problem 30PPS

The equation of the parabola is $x=-\frac{1}{32}{y}^2-6$ and graph is shown in figure (1).

Explanation of Solution

Given:

The vertex is $\left(-6,0\right)$ and directrix is $x=2$ .

Calculation:

The vertex is $\left(-6,0\right)$ . So, $h=-6$ and $k=0$ .

Use the equation of directrix $x=h-\frac{1}{4a}$ to find the value of $a$ .

  $\begin{array}{c}2=-6-\frac{1}{4a}\\ \frac{1}{4a}=-6-2\\ \frac{1}{4a}=-8\\ a=-\frac{1}{32}\end{array}$

Substitute $-\frac{1}{32}$ for $a$ , $-6$ for $h$ and 0 for $k$ in the equation $x=a{\left(y-k\right)}^2+h$ to find the equation of parabola.

  $\begin{array}{c}x=-\frac{1}{32}{\left(y-0\right)}^2-6\\ =-\frac{1}{32}{y}^2-6\end{array}$

Plot the graph of the parabola $x=-\frac{1}{32}{y}^2-6$ as follows.

  

Figure (1)

Therefore, the equation of the parabola is $x=-\frac{1}{32}{y}^2-6$ and graph is shown in figure (1).