Chapter 11.3, Problem 14E

To determine

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

Answer to Problem 14E

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

Foci: $\left(\pm \sqrt{13},0\right)$

Asymptote : $y=\pm \frac{3}{2}x$

Explanation of Solution

Given information :

The hyperbola equation is $9x^2-4y^2=36$ .

calculation :

First write the equation in a standard form of hyperbola,

  $\begin{array}{l}9x^2-4y^2=36\\ \frac{x^2}{4}-\frac{y^2}{9}=1\text{ [divide by 36]}\\ \frac{x^2}{4}-\frac{y^2}{9}-1\end{array}$

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

Since here

  $\begin{array}{l}{a}^2=4\\ a=2\text{ [sqaure root]}\\ b^2=9\\ b=3\text{ [square root]}\\ c=\sqrt{a^2+b^2}\\ c=\sqrt{4+9}\text{ [substitute]}\\ c=\sqrt{13}\text{ [add]}\end{array}$

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 $\sqrt{13}$ ,

So, the foci are $\left(\pm \sqrt{13},0\right)$ .

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

  $y=\pm \frac{b}{a}x$

Now substitute $b$ by $3$ and $a$ by $2$ ,

So asymptotes are

  $y=\pm \frac{3}{2}x$ .

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

To sketch the graph,

x	y
$-2$	$0$
$-1$	$\text{2}\text{.59}$
$1$	$\text{2}\text{.59}$
$2$	$0$

The graph is obtained as: