Chapter 9.3, Problem 46E

To determine

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

Answer to Problem 46E

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

Explanation of Solution

Given information:

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

Calculation:

The center is the midpoint of the vertices.

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

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

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

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

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

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

  $b=3$

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

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

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