Chapter 11.3, Problem 7E

To determine

To match : the equation with the graph.

Answer to Problem 7E

  

Explanation of Solution

Given information :

The equation is $16y^2-x^2=144$ .

First write the equation in a standard form of hyperbola,

  $\begin{array}{l}16y^2-x^2=144\\ \frac{y^2}{9}-\frac{x^2}{144}=1\text{ [divide by 144]}\\ \frac{y^2}{9}-\frac{x^2}{144}-1\end{array}$

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

Since here

  $\begin{array}{l}{a}^2=9\\ a=3\text{ [sqaure root]}\\ b^2=144\\ b=12\text{ [sqaure root]}\\ c=\sqrt{a^2+b^2}\\ c=\sqrt{9+144}\text{ [substitute]}\\ c=\sqrt{153}\text{ [add]}\end{array}$

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 $3$ ,

So vertices are $\left(0,\pm 3\right)$ .

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

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

Now substitute $c$ by $\sqrt{153}$ ,

So, the foci are $\left(0,\pm \sqrt{153}\right)$ .

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

  $y=\pm \frac{a}{b}x$

Now substitute $b$ by $12$ and $a$ by $3$ ,

So asymptotes are

  $\begin{array}{l}y=\pm \frac{3}{12}x\\ y=\pm \frac{1}{4}x\end{array}$ .

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

To sketch the graph,

x	y
$-2$	3.04
$-3$	3.09
$3$	3.09
$2$	3.04

The graph is obtained as: