Chapter 11.3, Problem 9E

To determine

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

Answer to Problem 9E

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

Foci: $\left(\pm 2\sqrt{5},0\right)$

Asymptote : $y=\pm 2x$

Explanation of Solution

Given information :

The hyperbola equation is $\frac{x^2}{4}-\frac{y^2}{16}=1$ .

calculation :

First write the equation in a standard form of hyperbola,

  $\frac{x^2}{4}-\frac{y^2}{16}-1$

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

Since here

  $\begin{array}{l}{a}^2=4\\ a=2\text{ [square root]}\\ b^2=16\\ b=4\text{ [square root]}\\ c=\sqrt{a^2+b^2}\\ c=\sqrt{4+16}\text{ [substitute]}\\ c=\sqrt{20}\text{ [add]}\\ c=2\sqrt{5}\text{ [square root]}\end{array}$

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 $\text{2}$ ,

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

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

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

Now substitute $c$ by $2\sqrt{5}$ ,

So, the foci are $\left(\pm 2\sqrt{5},0\right)$ .

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

  $y=\pm \frac{b}{a}x$

Now substitute $b$ by $\text{4}$ and $a$ by $\text{2}$ ,

So asymptotes are

  $\begin{array}{l}y=\pm \frac{4}{2}x\\ y=\pm 2x\end{array}$ .

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

To sketch the graph,

x	y
$-2$	0
$-3$	17.88
$3$	17.88
$\text{2}$	0

The graph is obtained as: