Chapter 8.3, Problem 49E

Solution

To Calculate: To draw the graph and vertices, foci , eccentricity of the hyperbola.

The graph of the given equation is:

  

Given information: The given equation of a hyperbola is:

  $9x^2-4y^2-36x+8y-4=0$ .

Calculation:

The given equation can be rewritten as below,

  $\begin{array}{c}9x^2-4y^2-36x+8y-4=0\\ 9x^2-36x-4y^2+8y=4\\ \left(9x^2-36x+36\right)-\left(4y^2-8y+4-4\right)=4+36\\ 9{\left(x-2\right)}^2-4{\left(y-1\right)}^2=4+36-4\\ 9{\left(x-2\right)}^2-4{\left(y-1\right)}^2=36\end{array}$

The standard form of a hyperbola is:

  $\frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y-1\right)}^2}{9}=1$

The general form of hyperbola is:

  $\frac{{\left(x-h\right)}^2}{b^2}-\frac{{\left(y-k\right)}^2}{a^2}=1$

So, after comparing, we get.

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

The center and vertices in given general form can be written as:

Center $\left(h,k\right)=\left(2,1\right)$

Vertices $=\left(h+a,k\right),\left(h-a,k\right)$ .

The values of $\left(a,b\right)=\left(2,3\right)$ .

Then, the vertices will be calculated as:

  $\begin{array}{c}Vertices=\left(2+2,1\right),\left(2-2,1\right)\\ =\left(4,1\right),\left(0,1\right)\end{array}$

The foci can be calculated as:

Foci $=\left(h+c,k\right),\left(h-c,k\right)$ .

The value of

  $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{l}c=\sqrt{4+9}\\ c=\sqrt{13}\end{array}$

Thus, the value of foci is:

  $Foci=\left(2+\sqrt{13},1\right),\left(2-\sqrt{13},1\right)$

The value of eccentricity can be calculated as:

  $\begin{array}{l}e=\frac{c}{a}\\ e=\frac{\sqrt{13}}{2}\end{array}$

Thus, the values are:

Foci are $=\left(2+\sqrt{13},1\right),\left(2-\sqrt{13},1\right)$ .

Vertices are $\left(4,1\right),\left(0,1\right)$ .

Eccentricity is : $e=\frac{\sqrt{13}}{2}$

The graph of the given equation is: