Chapter 9.3, Problem 40E

(a)

To determine

To find: The standard equation of hyperbola $9x^2-y^2+54x+10y+55=0$ .

Answer to Problem 40E

The standard form of the hyperbola is $\frac{{\left(x+3\right)}^2}{3^2}-\frac{{\left(y-5\right)}^2}{{\left(1\right)}^2}=1$ .

Explanation of Solution

Given information:

The equation of the hyperbola is $9x^2-y^2+54x+10y+55=0$.

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}9x^2-y^2+54x+10y+55=0\\ 9\left(x^2+2\left(1\right)\left(3\right)x+9\right)-y^2+10y+55-81=0\\ 9{\left(x+3\right)}^2-\left[y^2-2\left(1\right)\left(5\right)y+25\right]=1\\ 9{\left(x+3\right)}^2-{\left(y-5\right)}^2=1\end{array}$

Simplify further.

  $\begin{array}{l}\frac{{\left(x+3\right)}^2}{9}-\frac{{\left(y-5\right)}^2}{1}=1\\ \frac{{\left(x+3\right)}^2}{3^2}-\frac{{\left(y-5\right)}^2}{{\left(1\right)}^2}=1\end{array}$

Therefore, the standard form of the hyperbola is $\frac{{\left(x+3\right)}^2}{3^2}-\frac{{\left(y-5\right)}^2}{{\left(1\right)}^2}=1$ .

(b)

To determine

To find: The center, vertices, foci, asymptotes of hyperbola $9x^2-y^2+54x+10y+55=0$ .

Answer to Problem 40E

The center of the hyperbola is $\left(-3,5\right)$ , the vertices of the hyperbola is $\left(-\frac{8}{3},5\right)$ , $\left(-\frac{10}{3},5\right)$ , the foci of the hyperbola is $\left(-3+\frac{\sqrt{10}}{3},5\right)$ and $\left(-3-\frac{\sqrt{10}}{3},5\right)$ , the asymptotes of the hyperbola is $y=3x+14$ , $y=-3x-4$ .

Explanation of Solution

Given information:

The equation of hyperbola is $9x^2-y^2+54x+10y+55=0$ .

Calculation:

Write the equation of the hyperbola.

  $\begin{array}{c}9x^2-y^2+54x+10y+55=0\\ 9\left(x^2+2\left(1\right)\left(3\right)x+9\right)-y^2+10y+55-81=0\\ 9{\left(x+3\right)}^2-\left[y^2-2\left(1\right)\left(5\right)y+25\right]=1\\ 9{\left(x+3\right)}^2-{\left(y-5\right)}^2=1\end{array}$

Simplify further.

  $\begin{array}{l}\frac{{\left(x+3\right)}^2}{9}-\frac{{\left(y-5\right)}^2}{1}=1\\ \frac{{\left(x+3\right)}^2}{3^2}-\frac{{\left(y-5\right)}^2}{{\left(1\right)}^2}=1\end{array}$

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

  $\begin{array}{l}a=3\\ b=1\\ h=-3\\ k=5\end{array}$

The center of the hyperbola is given as $\left(-3,5\right)$ .

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

So the vertices are. $\left(-\frac{8}{3},5\right)$ , $\left(-\frac{10}{3},5\right)$ ,

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 $c$ .

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

Hence the foci are. $\left(-3+\frac{\sqrt{10}}{3},5\right)$ and $\left(-3-\frac{\sqrt{10}}{3},5\right)$

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

Substitute $3$ for $a$ and $1$ for $b$ $-3$ for $h$ and $5$ for $k$ to find the asymptotes.

  $y=3x+14$ , $y=-3x-4$ .

Therefore The center of the hyperbola is $\left(-3,5\right)$ , the vertices of the hyperbola is $\left(-\frac{8}{3},5\right)$ , $\left(-\frac{10}{3},5\right)$ , the foci of the hyperbola is $\left(-3+\frac{\sqrt{10}}{3},5\right)$ and $\left(-3-\frac{\sqrt{10}}{3},5\right)$ , the asymptotes of the hyperbola is $y=3x+14$ , $y=-3x-4$ .

(c)

To determine

To draw: The graph of the hyperbola $9x^2-y^2+54x+10y+55=0$ .

Answer to Problem 40E

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

Explanation of Solution

Given information:

The equation of the hyperbola is $9x^2-y^2+54x+10y+55=0$.

Calculation:

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

  $\begin{array}{c}9x^2-y^2+54x+10y+55=0\\ 9\left(x^2+2\left(1\right)\left(3\right)x+9\right)-y^2+10y+55-81=0\\ 9{\left(x+3\right)}^2-\left[y^2-2\left(1\right)\left(5\right)y+25\right]=1\\ 9{\left(x+3\right)}^2-{\left(y-5\right)}^2=1\end{array}$

Simplify further.

  ${\left(y-5\right)}^2=9{\left(x+3\right)}^2-1$

Taking square root both the sides

  $\begin{array}{l}y-5=\pm \sqrt{9{\left(x+3\right)}^2-1}\\ y=\pm \sqrt{9{\left(x+3\right)}^2-1}+5\end{array}$

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.