Chapter 9, Problem 4E

To determine

To find: The equation in standard form, of the hyperbola shown in graph.

Answer to Problem 4E

The equation of the hyperbola is $\frac{{\left(y+1\right)}^2}{9}-\frac{{\left(x-2\right)}^2}{16}=1$ .

Explanation of Solution

Given information:

The givenfigure is shown in figure (1).

  

Figure (1)

Calculation:

Using the graph shown in figure (1), the graph is of the vertical hyperbola with vertices at point $\left(2,2\right)$ and $\left(2,-4\right)$ . Midpoint of the vertices will be the center of the hyperbola.

  $\begin{array}{c}\left(h,k\right)=\left(\frac{2+2}{2},\frac{2+\left(-4\right)}{2}\right)\\ =\left(2,-1\right)\end{array}$

The distance between vertices will be equal to $2a.$

  $\begin{array}{c}2a=2-\left(-4\right)\\ =6\\ a=3\end{array}$

Using the equation for the asymptotes of a vertical hyperbola.

  $y-k=\pm \frac{a}{b}\left(x-h\right)$

Substitute $\left(h,k\right)=\left(2,-1\right)$ and $a=3$ in the equation of asymptotes.

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

Using the graph, one asymptote with positive slope passes through point $\left(6,2\right)$ .

  $\begin{array}{c}\left(2\right)+1=\frac{3}{b}\left(\left(6\right)-2\right)\\ 3=\frac{3}{b}\times 4\\ b=4\end{array}$

Using the equation of a vertical hyperbola in standard form.

  $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$

Substitute the values in above equation.

  $\begin{array}{c}\frac{{\left(y-\left(-1\right)\right)}^2}{3^2}-\frac{{\left(x-2\right)}^2}{4^2}=1\\ \frac{{\left(y+1\right)}^2}{9}-\frac{{\left(x-2\right)}^2}{16}=1\end{array}$

Therefore, the equation of the hyperbola is $\frac{{\left(y+1\right)}^2}{9}-\frac{{\left(x-2\right)}^2}{16}=1$ .