Chapter 9.3, Problem 36E

(a)

To determine

To find: The standard equation of hyperbola $x^2-9y^2+36y-72=0$ .

Answer to Problem 36E

The standard form of the hyperbola is $\frac{x^2}{6^2}-\frac{{\left(y-2\right)}^2}{2^2}=1$ .

Explanation of Solution

Given information:

The equation of the hyperbola is $x^2-9y^2+36y-72=0$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}{x}^2-9y^2+36y-72=0\\ x^2-9\left(y^2-4y+8\right)=0\\ x^2-9\left[{\left(y-2\right)}^2+4\right]=0\\ x^2-9{\left(y-2\right)}^2=36\end{array}$

Simplify further.

  $\begin{array}{l}\frac{x^2}{36}-\frac{{\left(y-2\right)}^2}{4}=1\\ \frac{x^2}{6^2}-\frac{{\left(y-2\right)}^2}{2^2}=1\end{array}$

Therefore, the standard form of the hyperbola is $\frac{x^2}{6^2}-\frac{{\left(y-2\right)}^2}{2^2}=1$ .

(b)

To determine

To find: The center, vertices, foci, asymptotes of hyperbola $x^2-9y^2+36y-72=0$ .

Answer to Problem 36E

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

Explanation of Solution

Given information:

The equation of hyperbola is $x^2-9y^2+36y-72=0$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}{x}^2-9y^2+36y-72=0\\ x^2-9\left(y^2-4y+8\right)=0\\ x^2-9\left[{\left(y-2\right)}^2+4\right]=0\\ x^2-9{\left(y-2\right)}^2=36\end{array}$

Simplify further.

  $\begin{array}{l}\frac{x^2}{36}-\frac{{\left(y-2\right)}^2}{4}=1\\ \frac{x^2}{6^2}-\frac{{\left(y-2\right)}^2}{2^2}=1\end{array}$

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

  $\begin{array}{l}a=6\\ b=2\\ h=0\\ k=2\end{array}$

The center of the hyperbola is given as $\left(0,2\right)$ .

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

So the vertices are $\left(6,2\right)$ and $\left(-6,2\right)$ .

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 $c$ .

  $\begin{array}{c}{c}^2=6^2+2^2\\ c^2=36+4\\ c^2=40\\ c=\sqrt{40}\end{array}$

Simplify further

  $c=2\sqrt{10}$

Hence the foci are $\left(2\sqrt{10},2\right)$ and $\left(-2\sqrt{10},2\right)$ .

The asymptotes of the hyperbola is given by $y=\pm \frac{b}{a}\left(x-h\right)+k$ .

Substitute $6$ for $a$ and $2$ for $b$

  $0$ for $h$ and $2$ for $k$ to find the asymptotes.

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

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

(c)

To determine

To draw: The graph of the hyperbola $x^2-9y^2+36y-72=0$ .

Answer to Problem 36E

The graph of the hyperbola is shown in figure (1).

Explanation of Solution

Given information:

The equation of the hyperbola is $x^2-9y^2+36y-72=0$ .

Calculation:

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

  $\begin{array}{c}{x}^2-9y^2+36y-72=0\\ x^2-9\left(y^2-4y+8\right)=0\\ x^2-9\left[{\left(y-2\right)}^2+4\right]=0\\ x^2-9{\left(y-2\right)}^2=36\end{array}$

Simplify further.

  $\begin{array}{c}\frac{x^2}{36}-\frac{{\left(y-2\right)}^2}{4}=1\\ \frac{x^2}{6^2}-\frac{{\left(y-2\right)}^2}{2^2}=1\\ \frac{{\left(y-2\right)}^2}{4}=\frac{x^2}{36}-1\\ {\left(y-2\right)}^2=4\left(\frac{x^2}{36}-1\right)\end{array}$

Taking square root both the sides

  $\begin{array}{c}\left(y-2\right)=\pm 2\sqrt{\frac{x^2}{36}-1}\\ y=\pm \left(2\sqrt{\frac{x^2}{36}-1}\right)+2\end{array}$

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

  

Figure $\left(1\right)$

Therefore, the graph of the hyperbola is shown in figure (1).

Verify the result using the graphical utility:

Draw the graph for the hyperbola using the graphical utility.

  

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

Hence, proved.