Chapter 11.3, Problem 16E

To determine

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

Answer to Problem 16E

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

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

Asymptote : $y=\pm x$

Explanation of Solution

Given information :

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

calculation :

First write the equation in a standard form of hyperbola,

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

Because the $y^2$ -term is positive, the hyperbola has avertical transverse axis;

Since here

  $\begin{array}{l}{a}^2=4\\ a=2\text{ [sqaure root]}\\ b^2=4\\ b=2\text{ [sqaure root]}\\ c=\sqrt{a^2+b^2}\\ c=\sqrt{4+4}\text{ [substitute]}\\ c=2\sqrt{2}\text{ [sqaure root]}\end{array}$

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

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

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

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

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

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

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

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

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

  $y=\pm \frac{a}{b}x$

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

So asymptotes are

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

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

To sketch the graph,

x	y
$-\text{2}$	$2.82$
$-1$	$2.23$
$1$	$2.23$
$\text{2}$	$2.82$

The graph is obtained as: