Chapter 11.3, Problem 18E

To determine

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

Answer to Problem 18E

The vertices : $\left(\pm \sqrt{3},0\right)$

Foci: $\left(\pm \frac{3}{\sqrt{2}},0\right)$

Asymptote : $y=\pm \frac{1}{\sqrt{2}}x$

Explanation of Solution

Given information :

The hyperbola equation is $x^2-2y^2=3$ .

calculation :

First write the equation in a standard form of hyperbola,

  $\begin{array}{l}{x}^2-2y^2=3\\ \frac{x^2}{3}-\frac{y^2}{3/2}-1\text{ [divide by 3]}\end{array}$

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

Since here

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

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 $\sqrt{3}$ ,

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

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

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

Now substitute $c$ by $\frac{3}{\sqrt{2}}$ ,

So, the foci are $\left(\pm \frac{3}{\sqrt{2}},0\right)$ .

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

  $y=\pm \frac{b}{a}x$

Now substitute $b$ by $\frac{\sqrt{3}}{\sqrt{2}}$ and $a$ by $\sqrt{3}$ ,

So asymptotes are

  $\begin{array}{l}y=\pm \frac{\sqrt{3}}{\sqrt{3}\sqrt{2}}x\\ y=\pm \frac{1}{\sqrt{2}}x\end{array}$ .

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

To sketch the graph,

x	y
- $\sqrt{3}$	$0$
$-2$	$0.707$
$2$	$0.707$
$\sqrt{3}$	$0$

The graph is obtained as: