Chapter 8.5, Problem 43PS

Solution

a.

When two stones are dropped simultaneously into still water, they cause circular ripples that have their intersection points at the foci where the stones first contact the water.

Enter the data into the equation for the horizontal transverse hyperbola.

  $\frac{{\text{x}}^{\text{2}}}{\text{16}}\text{-}\frac{{\text{y}}^{\text{2}}}{\text{20}}\text{=1}$

Given:

Look at the diagram to determine the positions of the vertices and foci.

  $\begin{array}{c}{a}^2+b^2=c^2\\ b^2=c^2-a^2\\ =6^2-4^2\\ =20\\ a=4\\ c=6\end{array}$

Concept Used:

Find $b^2$ using the equation that looks like the Pythagorean Theorem.

Calculation:

  $\begin{array}{c}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \frac{x^2}{4^2}-\frac{y^2}{20}=1\\ \frac{x^2}{16}-\frac{y^2}{20}=1\end{array}$

Plug the numbers into the horizontal transverse hyperbola equation.

Conclusion:

  $\frac{x^2}{16}-\frac{y^2}{20}=1$

b.

Describe a hyperbola using that

The paths of their crossings will follow a hyperbola since the speed of the waves is like ${\text{d}}_{\text{2}}$ and ${\text{d}}_{\text{1}}$ .

The speed of the waves is the same. Therefore, $d_2-d_1$ is constant if $d_1$ is the distance from a focus to the close portion of the hyperbola and $d_2$ is the distance from the farther-off focus to the same portion of the hyperbola. This implies that as you walk down the hyperbola, $d_2$ and $d_1$ rise and decrease at the same pace is the distance from the further away focus to the same part on the hyperbola, $d_2-d_1$ is constant. Which means $d_2$ and $d_1$ increase/decrease at the same rate as you move along the hyperbola.

Conclusion:

The paths of their crossings will follow a hyperbola since the speed of the waves is like ${\text{d}}_{\text{2}}$ and $d_1$ .