Chapter 11.3, Problem 5E

To determine

To match : the equation with the graph.

Answer to Problem 5E

  

Explanation of Solution

Given information :

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

First write the equation in a standard form of hyperbola,

  $\frac{x^2}{4}-y^2-1$

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

Since here

  $\begin{array}{l}{a}^2=4\\ a=2\text{ [square root]}\\ b^2=1\\ b=1\text{ [square root]}\\ c=\sqrt{a^2+b^2}\\ c=\sqrt{4+1}\text{ [substitute]}\\ c=\sqrt{5}\text{ [add]}\end{array}$

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 $2$ ,

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 $\sqrt{5}$ ,

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

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

  $y=\pm \frac{b}{a}x$

Now substitute $b$ by $1$ and $a$ by $2$ ,

So asymptotes are $y=\pm \frac{1}{2}x$ .

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

To sketch the graph,

x	y
$-2$	$0$
$-3$	$1.11$
$3$	$1.11$
$2$	$0$

The graph is obtained as: