Chapter 9.2, Problem 4BGP

To determine

To write: The equation for the parabola with focus $\left(5,6\right)$ and directrix $x=-2$ then plot the graph of the equation.

Answer to Problem 4BGP

The equation of the parabola is $x=\frac{1}{14}{\left(y-6\right)}^2+\frac{3}{2}$ and graph is shown in figure (1).

Explanation of Solution

Given:

The focus is $\left(5,6\right)$ and directrix is $x=-2$ .

Calculation:

The vertex is in the middle between the focus and directrix.

  $\begin{array}{c}\text{Vertex}=\left(\frac{5-2}{2},6\right)\\ =\left(\frac{3}{2},6\right)\\ =\left(h,k\right)\end{array}$

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

  $\begin{array}{c}-2=\frac{3}{2}-\frac{1}{4a}\\ \frac{1}{4a}=\frac{3}{2}+2\\ \frac{1}{4a}=2\\ a=\frac{1}{14}\end{array}$

Substitute $\frac{1}{14}$ for $a$ , $\frac{3}{2}$ for $h$ and 6 for $k$ in the equation $x=a{\left(y-k\right)}^2+h$ to find the equation of parabola.

  $x=\frac{1}{14}{\left(y-6\right)}^2+\frac{3}{2}$

Plot the graph of the parabola $x=\frac{1}{14}{\left(y-6\right)}^2+\frac{3}{2}$ as follows.

  

Figure (1)

Therefore, the equation of the parabola is $x=\frac{1}{14}{\left(y-6\right)}^2+\frac{3}{2}$ and graph is shown in figure (1).