Chapter 9.3, Problem 53E

(a)

To determine

To write: The equation of the cross section of the base.

Answer to Problem 53E

The equation of the cross section of the base is $\frac{x^2}{1}-\frac{y^2}{27}=1$ .

Explanation of Solution

Given information:

The given figure is as follows.

  

Calculation:

Use the diagram to determine $a=1$ . The center of the hyperbola is at the origin.

Write the equation for the parabola with a horizontal transverse axis using any known information as shown below.

  $\frac{x^2}{1}-\frac{y^2}{b^2}=1$

Substitute 2 for $x$ and $9$ for $y$ in the above equation to find $b$ .

  $\begin{array}{c}\frac{2^2}{1}-\frac{9^2}{b^2}=1\\ 4-1=\frac{81}{b^2}\\ \frac{81}{b^2}=3\\ b^2=27\end{array}$

Write the equation of the cross section of the base.

  $\frac{x^2}{1}-\frac{y^2}{27}=1$

Therefore, the equation of the cross section of the base is $\frac{x^2}{1}-\frac{y^2}{27}=1$ .

(b)

To determine

To find: The width of the base, 4 inches from the bottom.

Answer to Problem 53E

The width of the base is $1.89\text{feet}$ .

Explanation of Solution

Given information:

It is given that, each unit in the coordinate plane represents $\frac{1}{2}\text{foot}$ .

Calculation:

Consider the equation $\frac{x^2}{1}-\frac{y^2}{27}=1$ .

Since each unit represents $\frac{1}{2}\text{foot}$ , 4 inches is equivalent to $\frac{2}{3}$ of a unit.

Let,

  $\begin{array}{c}y=-9+\frac{2}{3}\\ =-\frac{25}{3}\end{array}$

Substitute $-\frac{25}{3}$ for $y$ in the equation $\frac{x^2}{1}-\frac{y^2}{27}=1$ and solve for $x$ .

  $\begin{array}{c}\frac{x^2}{1}-\frac{{\left(-\frac{25}{3}\right)}^2}{27}=1\\ x^2-\frac{625}{243}=1\\ x=\sqrt{1+\frac{625}{243}}\\ \approx 1.89\text{feet}\end{array}$

Since, $2x$ is the width of the base, but each unit is equivalent to $\frac{1}{2}\text{foot}$ . So, the width of the base is $x$ .

Therefore, the width of the base is $1.89\text{feet}$ .