Chapter 8.5, Problem 39PS

Solution

To find : An equation that models the given hyperbola.

The equation of the given hyperbola can be modeled as $\frac{x^2}{\frac{1}{2}}-\frac{y^2}{\frac{1}{4}}=1$ .

Given information :

The outline of a cube spinning around an axis through a pair of opposite corners contains a portion of a hyperbola as shown in the figure below. The coordinates given represent a vertex and a focus of the hyperbola for a cube that measures $1$ unit on each edge.

  

Explanation :

The vertex lies on the $x-$ axis equidistant from the origin. This means that the transverse axis is horizontal. Also, the given vertex is of the form $\left(\pm a,0\right)$ and the focus is of the form $\left(\pm c,0\right)$ .

From the given information, the hyperbola is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , where $a$ is $\frac{\sqrt{2}}{2}$ and $c$ is $\frac{\sqrt{3}}{2}$ .

Use the formula $c^2=a^2+b^2$ to determine $b$ . For this, solve the formula for $b$ .

  $\begin{array}{c}{b}^2=c^2-a^2\\ b=\sqrt{c^2-a^2}\end{array}$

Substitute the value of $a$ and $c$ to simplify further:

  $\begin{array}{c}b=\sqrt{{\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{\sqrt{2}}{2}\right)}^2}\\ =\sqrt{\frac{3}{2}-\frac{2}{4}}\\ =\sqrt{\frac{1}{4}}\\ =\frac{1}{2}\end{array}$

Put $a=\frac{\sqrt{2}}{2}\text{ and }b=\frac{1}{2}$ in $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

  $\begin{array}{c}\frac{x^2}{{\left(\frac{\sqrt{2}}{2}\right)}^2}-\frac{y^2}{{\left(\frac{1}{2}\right)}^2}=1\\ \frac{x^2}{\frac{1}{2}}-\frac{y^2}{\frac{1}{4}}=1\end{array}$

Therefore, the equation of the given hyperbola can be modeled as $\frac{x^2}{\frac{1}{2}}-\frac{y^2}{\frac{1}{4}}=1$ .