Chapter 9.3, Problem 27E

To determine

To find: The center, vertices, foci, asymptotes and graph of hyperbola $\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{1}=1$ . Verify the result using graphical utility.

Answer to Problem 27E

The center of the hyperbola is $\left(3,1\right)$ , the vertices of the hyperbola is $\left(6,1\right)$ , $\left(0,1\right)$ , the foci of the hyperbola is $\left(3+\sqrt{10},1\right)$ and $\left(3-\sqrt{10},1\right)$ , the asymptotes of the hyperbola is $y=\frac{x}{3}$ , $y=-\frac{x}{3}+2$ and the graph of the hyperbola is shown in figure (1).

Explanation of Solution

Given information:

Given equation of hyperbola is $\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{1}=1$ .

Calculation:

Compare the given equation with the standard equation of hyperbola $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

  $\begin{array}{l}a=3\\ b=1\\ h=3\\ k=1\end{array}$

The center of the hyperbola is given as $\left(3,1\right)$ .

The vertices of the hyperbola is given by $\left(h+a,k\right)$ and $\left(h-a,k\right)$ .

So the vertices are $\left(6,1\right)$ and $\left(0,1\right)$ .

The foci of the hyperbola is given by $\left(h+c,k\right)$ and $\left(h-c,k\right)$ , where $c^2=a^2+b^2$ . Calculate the value of $c$ .

  $\begin{array}{c}{c}^2=3^2+1^2\\ c^2=9+1\\ c^2=10\\ c=\sqrt{10}\end{array}$

Hence the foci are $\left(3+\sqrt{10},1\right)$ and $\left(3-\sqrt{10},1\right)$ .

The asymptotes of the hyperbola is given by $y=\pm \frac{b}{a}x$ .

Substitute $3$ for $a$ and $1$ for $b$ to find the asymptotes.

  $y=\frac{x}{3}$ , $y=-\frac{x}{3}+2$

To draw the graph of hyperbola, solve the equations of hyperbola in terms of $y$

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

Simplify further to find the value of y

  $y=\pm \sqrt{\frac{{\left(x-3\right)}^2-9}{9}}+1$

Use the above equation and the asymptotes to draw the graph of the hyperbola.

.

Figure (1)

Therefore The center of the hyperbola is $\left(3,1\right)$ , the vertices of the hyperbola is $\left(6,1\right)$ , $\left(0,1\right)$ , the foci of the hyperbola is $\left(3+\sqrt{10},1\right)$ and $\left(3-\sqrt{10},1\right)$ , the asymptotes of the hyperbola is $y=\frac{x}{3}$ , $y=-\frac{x}{3}+2$ and the graph of the hyperbola is shown in figure (1).

Verify the result using graphical utility:

Draw the graph for the hyperbola $\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{1}=1$ using graphical utility.

  

From the above, it is clear that both the graphs are similar.

Hence, proved.