Chapter 10, Problem 28SGA

To determine

To find: The coordinates of the center, the foci, and the vertices and the equation of asymptotes, and then graph the equation.

Answer to Problem 28SGA

The coordinates of the center is $\left(2,-3\right)$ , the coordinates of the vertices is $\left(2,0\right)\text{and}\left(2,-6\right)$ , the coordinates of the foci is $\left(2,2\right)\text{and}\left(2,-8\right)$ , the equation of asymptotes is $y+3=\pm \frac{3}{4}\left(x-2\right)$ , and the graph is shown in figure (1).

Explanation of Solution

Given:

The equation is $9x^2-16y^2-36x-96y+36=0$ .

Calculation:

Write the equation $9x^2-16y^2-36x-96y+36=0$ in standard form by converting it into a perfect square.

  $\begin{array}{c}9x^2-36x-16y^2-96y=-36\\ 9\left(x^2-4x+4\right)-16\left(y^2+6y+9\right)=-36+36-144\\ 9{\left(x-2\right)}^2-16{\left(y+3\right)}^2=-144\\ \frac{{\left(y+3\right)}^2}{9}-\frac{{\left(x-2\right)}^2}{16}=1\end{array}$

Compare the equation $\frac{{\left(y+3\right)}^2}{9}-\frac{{\left(x-2\right)}^2}{16}=1$ with the standard form $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ .

  $\begin{array}{c}h=2\\ k=-3\\ a=3\\ b=4\end{array}$

The coordinates of the center will be $\left(h,k\right)=\left(2,-3\right)$ .

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

Substitute $2$ for $h$ , $-3$ for $k$ , $3$ for $a$ , and $4$ for $b$ in the equation $y-k=\pm \frac{a}{b}\left(x-h\right)$ .

  $y+3=\pm \frac{3}{4}\left(x-2\right)$

Find the coordinates of the vertices as follows.

  $\begin{array}{c}\left(h,k\pm a\right)=\left(2,-3\pm 3\right)\\ =\left(2,0\right)\text{and}\left(2,-6\right)\end{array}$

Find the value of $c$ .

  $\begin{array}{c}{c}^2=a^2+b^2\\ =3^2+4^2\\ =25\\ c=5\end{array}$

Find the coordinates of the foci as follows.

  $\begin{array}{c}\left(h,k\pm c\right)=\left(2,-3\pm 5\right)\\ =\left(2,2\right)\text{and}\left(2,-8\right)\end{array}$

The graph of the hyperbola is as follows.

  

Figure (1)

Therefore, the coordinates of the center is $\left(2,-3\right)$ , the coordinates of the vertices is $\left(2,0\right)\text{and}\left(2,-6\right)$ , the coordinates of the foci is $\left(2,2\right)\text{and}\left(2,-8\right)$ , the equation of asymptotes is $y+3=\pm \frac{3}{4}\left(x-2\right)$ , and the graph is shown in figure (1).