Chapter 9.3, Problem 47E

To determine

To find: The standard form of the equation of hyperbola with the given characteristics.

Answer to Problem 47E

The standard form of the equation of the hyperbola is $\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-2\right)}^2}{4}=1$ .

Explanation of Solution

Given information:

The characteristics are vertices: $\left(0,2\right),\left(6,2\right)$ ; asymptotes: $y=\frac{2}{3}x,y=4-\frac{2}{3}x$ .

Calculation:

The center is the midpoint of the vertices.

  $\begin{array}{c}\text{Center}=\left(\frac{0+6}{2},\frac{2+2}{2}\right)\\ =\left(3,2\right)\end{array}$

The distance between one of the vertices and the center is $a=3$ .

The vertices lie on a horizontal line. So, the hyperbola has a horizontal transverse axis. Use the formula for the asymptotes of a hyperbola with a horizontal transverse axis.

Substitute the center and a into the formula as shown below.

  $\begin{array}{c}y=k\pm \frac{b}{a}\left(x-h\right)\\ y=2\pm \frac{b}{3}\left(x-3\right)\end{array}$

Solve for b by comparing the equation to the two given asymptotes.

  $b=2$

Substitute 3 for $h$ , $2$ for $k$ , 3 for $a$ and 2 for $b$ in the equation $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ to find the standard form of hyperbola.

  $\begin{array}{c}\frac{{\left(x-3\right)}^2}{3^2}-\frac{{\left(y-2\right)}^2}{2^2}=1\\ \frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-2\right)}^2}{4}=1\end{array}$

Therefore, the standard form of the equation of hyperbola is $\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-2\right)}^2}{4}=1$ .