Chapter 8.5, Problem 17E

Solution

To explain : The error in graphing the equation $\frac{x^2}{4}-y^2=4$ .

The foci are at $\left(\pm \sqrt{20},0\right)$ and the asymptotes are $y=\pm \frac{1}{2}x$ .

Given information :

The graph of $\frac{x^2}{4}-y^2=4$ .

  

Calculation :

Rewrite the equation in the standard from $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ . For this, divide both the sides by $4$ .

  $\begin{array}{c}\frac{\frac{x^2}{4}}{4}-\frac{y^2}{4}=\frac{4}{4}\\ \frac{x^2}{16}-\frac{y^2}{4}=1\\ \frac{x^2}{4^2}-\frac{y^2}{2^2}=1\end{array}$

Since the $y^2$ item is negative, the transverse axis is horizontal. An equation of this form has vertices at $\left(\pm a,0\right)$ and asymptotes as $y=\pm \frac{b}{a}x$ .

Determine the value of $a\text{ and }b$ by comparing $\frac{x^2}{4^2}-\frac{y^2}{2^2}=1$ with the standard equation.

  $\begin{array}{l}{a}^2=4^2\\ b^2=2^2\end{array}$

The value of $a$ is $4$ and that of $b$ is $2$ . Thus, the vertices of the hyperbola are $\left(4,0\right)\text{ and }\left(-4,0\right)$ .

In order to find the foci, subtitute $a^2=16\text{ and }{b}^2=4$ in the formula.

  $\begin{array}{l}{c}^2=a^2+b^2\\ c^2=16+4\\ c^2=20\end{array}$

Take square root of both the sides.

  $\begin{array}{c}\sqrt{c^2}=\sqrt{20}\\ c=\sqrt{20}\end{array}$

The foci are at $\left(\pm \sqrt{20},0\right)$ .

Put $a=4\text{ and }b=2$ in $y=\pm \frac{b}{a}x$ to find the asymptotes.

  $\begin{array}{l}y=\pm \frac{2}{4}x\\ y=\pm \frac{1}{2}x\end{array}$

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

Graph of the given equation is as below: