Chapter 8.3, Problem 45E

Solution

To calculate: To calculate the center , vertices and foci of the hyperbola.

The value of center is $\left(-3,6\right)$ .

The value of foci are $=\left(\sqrt{17}-3,6\right),\left(-\sqrt{17}-3,6\right)$

The value of vertices are $=\left(-3+\sqrt{12},6\right),\left(-3-\sqrt{12},6\right)$ .

The graph of an equation is:

  

Given information: The given equation of a hyperbola is:

  $\frac{{\left(x+3\right)}^2}{12}-\frac{{\left(y-6\right)}^2}{5}=1$ .

Calculation:

The general form of hyperbola is:

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

So, after comparing, we get.

  $\frac{{\left(x+3\right)}^2}{12}-\frac{{\left(y-6\right)}^2}{5}=1$

The vertices in given general form can be written as:

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

Here, $\left(h,k\right)=\left(-3,6\right)$ .

The values of $\left(a,b\right)=\left(\sqrt{12},\sqrt{5}\right)$ .

Then, the vertices will be calculated as:

  $Vertices=\left(-3+\sqrt{12},6\right),\left(-3-\sqrt{12},6\right)$

The foci can be calculated as :

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

The value of

  $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{l}c=\sqrt{12+5}\\ c=\sqrt{17}\end{array}$

Thus, the value of foci is:

  $Foci=\left(\sqrt{17}-3,6\right),\left(-\sqrt{17}-3,6\right)$

Thus, the values are :

Foci are $=\left(\sqrt{17}-3,6\right),\left(-\sqrt{17}-3,6\right)$ .

Vertices are $=\left(-3+\sqrt{12},6\right),\left(-3-\sqrt{12},6\right)$ .

Center is $\left(-3,6\right)$ .

The graph of an equation is: