Chapter 8.5, Problem 35E

Solution

To find : Equations of three hyperbolas with horizontal transverse axes and asymptotes $y=\pm 2x$ and compare them.

The equations of the three hyperbolas are $\frac{x^2}{1}-\frac{y^2}{4}=1$ , $\frac{x^2}{4}-\frac{y^2}{16}=1$ and $\frac{x^2}{9}-\frac{y^2}{36}=1$ .

Given information :

The hyperbolas have horizontal transverse axes and asymptotes $y=\pm 2x$

Explanation :

The equation of a hyperbola with horizontal transverse axes is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

An equation of this form has vertices at $\left(\pm a,0\right)$ and asymptotes as $y=\pm \frac{b}{a}x$ .

It is given that the asymptotes are $y=\pm 2x.$ Compare this equation with $y=\pm \frac{b}{a}x$ .

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

In order to find the equation of the hyperbolas with asymptotes $y=\pm 2x.$ , substitute some values for $a\text{ and }b$ such that the value of $b$ is double the value of $a$ .

Put $a=1\text{ and }b=2\left(1\right)=2$ in $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

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

The equation of one of the hyperbolas will be $\frac{x^2}{1}-\frac{y^2}{4}=1$ .

Put $a=2\text{ and }b=2\left(2\right)=4$ in $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

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

The equation of the other hyperbola will be $\frac{x^2}{4}-\frac{y^2}{16}=1$ .

Put $a=3\text{ and }b=2\left(3\right)=6$ in $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

  $\begin{array}{l}\frac{x^2}{3^2}-\frac{y^2}{6^2}=1\\ \frac{x^2}{9}-\frac{y^2}{36}=1\end{array}$

The equation of the third hyperbola will be $\frac{x^2}{9}-\frac{y^2}{36}=1$ .

The hyperbola stretches vertically when the values of $a$ increases and shrinks vertically when the value of $a$ decreases. The value of $a$ increases in this case, which implies that the hyperbola stretches vertically.