Chapter 8.3, Problem 15E

Solution

To calculate: To draw the graph of the hyperbola.

The graph of the given equation is:

  

Given information: The given equation of a hyperbola is:

  $\frac{{\left(x+3\right)}^2}{16}-\frac{{\left(y-1\right)}^2}{4}=1$ .

Calculation:

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.

  $\frac{{\left(x+3\right)}^2}{{\left(4\right)}^2}-\frac{{\left(y-1\right)}^2}{{\left(2\right)}^2}=1$

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

Center $\left(h,k\right)=\left(-3,1\right)$

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

The values of $\left(a,b\right)=\left(4,2\right)$ .

Then, the vertices will be calculated as:

  $\begin{array}{c}Vertices=\left(4-3,1\right),\left(-4-3,-1\right)\\ =\left(1,1\right),\left(-7,1\right)\end{array}$

The foci can be calculated as:

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

The value can be calculated as:

  $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{l}c=\sqrt{4+16}\\ c=\sqrt{20}\end{array}$

Thus, the value of foci is:

  $Foci=\left(-3+2\sqrt{5},1\right),\left(-3-2\sqrt{5},1\right)$

Thus, the values are:

Foci are $=\left(-3+2\sqrt{5},1\right),\left(-3-2\sqrt{5},1\right)$ .

Vertices are $\left(1,1\right),\left(-7,1\right)$ .

The graph of the given equation is: