Chapter 9, Problem 54RE

To determine

Calculate the equation of the two hyperbolas that would locate the explosion.

Answer to Problem 54RE

The two parabolic equation are $\frac{{(x+5280)}^2}{10890000}-\frac{y^2}{16988400}=1$ and $\frac{{(x+5280)}^2}{19360000}-\frac{y^2}{8518400}=1$ .

Explanation of Solution

Given:

It is given in the question that there will be two friends live $4$ miles apart on the same “east-west” street, and you live halfway between them. It has to be a three − way phone conservation when you hear an explosion. Six seconds later, one of the friends to the east hears it $8$ seconds after it do. Assume that the coordinate system is measured in feet and that sound travels at $1100$ feet per second.

Concept Used:

In this, use the concept of hyperbolas and the equation $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$ , and understanding the question carefully by drawing a diagram and also using the relationship $c^2=a^2+b^2$ .

Calculation: Convert $4miles=21120ft(1mile=5280ft)$ .

Let the points be:

A $(0,0)$

  $\to$ Your house (half way between)

B $(-10560,0)$

  $\to$ friend in the east

C $(10560,0)$

  $\to$ friend in the west

Therefore, the standard equation will be:

  $\frac{{(x-h)}^2}{a^2}-\frac{{(y-k)}^2}{b^2}=1$

The two hyperbolas can be obtained by letting the foci A & B for case i & foci A & C fro case ii.

Case (i) : (Foci A & B)

The center is : $(-5280,0)$

  $\begin{array}{l}2c=10560\\ c=5280\end{array}$

Since your friend in the east heard the explosion $6$ seconds later, then it took place $6(1100)=6600ft$ farther from B than from A & on the right branch of the hyperbola. This difference also corresponds to the fixed distance $2a$ :

  $\begin{array}{l}2a=6600\\ a=3300\end{array}$

Solving $b^2$ by using the relationship $c^2=a^2+b^2$ .

  $\begin{array}{l}{c}^2=a^2+b^2\\ b^2=c^2-a^2\Rightarrow {5280}^2-{3300}^2=16988400\end{array}$

The equation will be:

  $\begin{array}{l}\frac{{(x-(-5280))}^2}{{3300}^2}-\frac{y^2}{16988400}=1\\ \frac{{(x+5280)}^2}{10890000}-\frac{y^2}{16988400}=1\end{array}$

Case (ii) : (Foci A & C)

The center is : $(5280,0)$

  $\begin{array}{l}2c=10560\\ c=5280\end{array}$

Since your friend in the east heard the explosion $8$ seconds later, then it took place $8(1100)=8800ft$ farther from C than from A & on the right branch of the hyperbola. This difference also corresponds to the fixed distance $2a$ :

  $\begin{array}{l}2a=8800\\ a=4400\end{array}$

Solving $b^2$ by using the relationship $c^2=a^2+b^2$ .

  $\begin{array}{l}{c}^2=a^2+b^2\\ b^2=c^2-a^2\Rightarrow {5280}^2-{4400}^2=8518400\end{array}$

The equation will be:

  $\begin{array}{l}\frac{{(x-5280)}^2}{{4400}^2}-\frac{{(y-0)}^2}{8518400}=1\\ \frac{{(x+5280)}^2}{19360000}-\frac{y^2}{8518400}=1\end{array}$

Conclusion:

The two parabolic equation are $\frac{{(x+5280)}^2}{10890000}-\frac{y^2}{16988400}=1$ and $\frac{{(x+5280)}^2}{19360000}-\frac{y^2}{8518400}=1$ .