Chapter 11.3, Problem 17E

To determine

To find : the vertices, foci, and asymptotes of the hyperbola and sketch the graph.

Answer to Problem 17E

The vertices : $\left(\pm 2\sqrt{2},0\right)$

Foci: $\left(\pm \sqrt{10},0\right)$

Asymptote : $y=\pm \frac{1}{2}x$

Explanation of Solution

Given information :

The hyperbola equation is $x^2-4y^2-8=0$ .

calculation :

First write the equation in a standard form of hyperbola,

  $\begin{array}{l}{x}^2-4y^2-8=0\\ \frac{x^2}{8}-\frac{y^2}{2}-1\text{ [divide by 8]}\end{array}$

Because the $x^2$ -term is positive, the hyperbola has ahorizontal transverse axis;

Since here

  $\begin{array}{l}{a}^2=8\\ a=2\sqrt{2}\text{ [sqaure root]}\\ b^2=2\\ b=\sqrt{2}\text{ [square root]}\\ c=\sqrt{a^2+b^2}\\ c=\sqrt{8+2}\text{ [substitute ]}\\ c=\sqrt{10}\text{ [add]}\end{array}$

Vertices: because the $x^2$ -term of hyperbola is positive so its vertices on the x -axis.

The vertices on the x -axis are $\left(\pm a,0\right)$ ,

Now substitute $a$ by $2\sqrt{2}$ ,

So vertices are $\left(\pm 2\sqrt{2},0\right)$ .

Foci: because the $x^2$ -term of hyperbola is positive,

So foci are $\left(\pm c,0\right)$ ,

Now substitute $c$ by $\sqrt{10}$ ,

So, the foci are $\left(\pm \sqrt{10},0\right)$ .

Asymptote: for the positive $x^2$ -term hyperbola the asymptote aregive as:

  $y=\pm \frac{b}{a}x$

Now substitute $b$ by $\sqrt{2}$ and $a$ by $2\sqrt{2}$ ,

So asymptotes are

  $\begin{array}{l}y=\pm \frac{\sqrt{2}}{2\sqrt{2}}x\\ y=\pm \frac{1}{2}x\end{array}$ .

Use the above information together with some additional values which is show in table below

To sketch the graph,

x	y
- $2\sqrt{2}$	$0$
$-3$	$0.5$
$3$	$0.5$
$2\sqrt{2}$	$0$

The graph is obtained as: