Chapter 9.3, Problem 39E

(a)

To determine

To find: The standard equation of hyperbola $9y^2-x^2+2x+54y+62=0$ .

Answer to Problem 39E

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

Explanation of Solution

Given information:

The equation of the hyperbola is $9y^2-x^2+2x+54y+62=0$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}9y^2-x^2+2x+54y+62=0\\ 9\left[y^2+2\left(1\right)\left(3\right)y+3^2\right]-x^2+2x+62-81=0\\ 9\left[{\left(y+3\right)}^2\right]-\left[x^2-2x+1\right]=18\\ 9{\left(y+3\right)}^2-{\left(x-1\right)}^2=18\end{array}$

Simplify further.

  $\frac{{\left(y+3\right)}^2}{2}-\frac{{\left(x-1\right)}^2}{18}=1$

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

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

(b)

To determine

To find: The center, vertices, foci, asymptotes of hyperbola $9y^2-x^2+2x+54y+62=0$.

Answer to Problem 39E

The center of the hyperbola is $\left(1,-3\right)$ , the vertices of the hyperbola is $\left(1,-3+\sqrt{2}\right)$ , $\left(1,-3-\sqrt{2}\right)$ , the foci of the hyperbola is $\left(1,-3+2\sqrt{5}\right)$ and $\left(1,-3-2\sqrt{5}\right)$ , the asymptotes of the hyperbola is $y=\frac{x}{3}-\frac{10}{3}$ , $y=-\frac{x}{3}-\frac{8}{3}$ .

Explanation of Solution

Given information:

The equation of hyperbola is $9y^2-x^2+2x+54y+62=0$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}9y^2-x^2+2x+54y+62=0\\ 9\left[y^2+2\left(1\right)\left(3\right)y+3^2\right]-x^2+2x+62-81=0\\ 9\left[{\left(y+3\right)}^2\right]-\left[x^2-2x+1\right]=18\\ 9{\left(y+3\right)}^2-{\left(x-1\right)}^2=18\end{array}$

Simplify further.

  $\frac{{\left(y+3\right)}^2}{2}-\frac{{\left(x-1\right)}^2}{18}=1$

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

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

  $\begin{array}{l}a=\sqrt{2}\\ b=3\sqrt{2}\\ h=1\\ k=-3\end{array}$

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

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

So the vertices are, $\left(1,-3+\sqrt{2}\right)$ , $\left(1,-3-\sqrt{2}\right)$ ,

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

Calculate the value of $c$ .

  $\begin{array}{c}{c}^2={\left(\sqrt{2}\right)}^2+{\left(3\sqrt{2}\right)}^2\\ c^2=2+18\\ c^2=20\\ c=\sqrt{20}\end{array}$

Hence the foci are. $\left(1,-3+2\sqrt{5}\right)$ and $\left(1,-3-2\sqrt{5}\right)$

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

Substitute $\sqrt{2}$ for $a$ and $3\sqrt{2}$ for $b$

  $1$ for $h$ and $-3$ for $k$ to find the asymptotes.

  $y=\frac{x}{3}-\frac{10}{3}$ , $y=-\frac{x}{3}-\frac{8}{3}$

Therefore The center of the hyperbola is $\left(1,-3\right)$ , the vertices of the hyperbola is $\left(1,-3+\sqrt{2}\right)$ , $\left(1,-3-\sqrt{2}\right)$ , the foci of the hyperbola is $\left(1,-3+2\sqrt{5}\right)$ and $\left(1,-3-2\sqrt{5}\right)$ , the asymptotes of the hyperbola is $y=\frac{x}{3}-\frac{10}{3}$ , $y=-\frac{x}{3}-\frac{8}{3}$ .

(c)

To determine

To draw: The graph of the hyperbola $9y^2-x^2+2x+54y+62=0$ .

Answer to Problem 39E

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

Explanation of Solution

Given information:

The equation of the hyperbola is $9y^2-x^2+2x+54y+62=0$ .

Calculation:

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

  $\begin{array}{c}9y^2-x^2+2x+54y+62=0\\ 9\left[y^2+2\left(1\right)\left(3\right)y+3^2\right]-x^2+2x+62-81=0\\ 9\left[{\left(y+3\right)}^2\right]-\left[x^2-2x+1\right]=18\\ 9{\left(y+3\right)}^2-{\left(x-1\right)}^2=18\end{array}$

Simplify further.

  $\begin{array}{c}9{\left(y+3\right)}^2=18+{\left(x-1\right)}^2\\ {\left(y+3\right)}^2=\frac{18+{\left(x-1\right)}^2}{9}\\ \left(y+3\right)=\pm \sqrt{\frac{18+{\left(x-1\right)}^2}{9}}\\ y=\pm \sqrt{\frac{18+{\left(x-1\right)}^2}{9}}-3\end{array}$

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.