Chapter 8.3, Problem 14E

Solution

To calculate: To draw the graph of the given equation of hyperbola.

The graph is:

  

Given information: The given equation of a hyperbola is:

  $\frac{x^2}{169}-\frac{y^2}{144}=1$ .

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.

  $\frac{x^2}{{\left(13\right)}^2}-\frac{y^2}{{\left(12\right)}^2}=1$

The vertices in given general form can be written as:

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

Here, $\left(h,k\right)=\left(0,0\right)$ .

The values of $\left(a,b\right)=\left(13,12\right)$ .

Then, the vertices will be calculated as:

  $\begin{array}{c}Vertices=\left(0+13,0\right),\left(0-13,0\right)\\ =\left(13,0\right),\left(-13,0\right)\end{array}$

The foci can be calculated as:

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

The value can be calculated as:

  $c=\sqrt{a^2+b^2}$ .

  $\begin{array}{l}c=\sqrt{169+144}\\ c=\sqrt{313}\end{array}$

Thus, the value of foci is:

  $\begin{array}{c}Foci=\left(0+\sqrt{313},0\right),\left(0-\sqrt{313},0\right)\\ =\left(\sqrt{313},0\right),\left(-\sqrt{313},0\right)\end{array}$

Thus, the values are:

Foci are $\left(\sqrt{313},0\right),\left(-\sqrt{313},0\right)$ .

Vertices are $\left(13,0\right),\left(-13,0\right)$ .

The graph of the given equation is