Chapter 11.3, Problem 48E

a.

To determine

To find: The value of $d\left(P,A\right)-d\left(P,B\right)$ .

Answer to Problem 48E

The value of $d\left(P,A\right)-d\left(P,B\right)$ is $-490\text{mi}$ .

Explanation of Solution

Given:

The radio signals travel at $980\text{ft/μs}$ .

Calculation:

  $d\left(P,A\right)-d\left(P,B\right)$ gives the same constant.

Here,

  $\begin{array}{l}a=245\\ b=250\end{array}$

The value of $d\left(P,A\right)-d\left(P,B\right)$ is calculated as,

  $\begin{array}{c}d\left(P,A\right)-d\left(P,B\right)=980t-980\left(t+2640\right)\\ =-2587200\text{ ft}\\ \text{=}-490\text{ mi}\end{array}$

Therefore, the value of $d\left(P,A\right)-d\left(P,B\right)=-490\text{mi}$ .

b.

To determine

To find: The equation of the hyperbola indicated in red.

Answer to Problem 48E

The equation of hyperbola is $\frac{y^2}{60,025}-\frac{x^2}{2,475}=1$ .

Explanation of Solution

Given:

The given diagram is,

  

Calculation:

The equation of hyperbola is given by,

  $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

Here,

  $\begin{array}{l}{a}^2=60,025\\ b^2=2,475\end{array}$

Substitute in the given formula,

  $\frac{y^2}{60,025}-\frac{x^2}{2,475}=1$

Therefore, the equation of hyperbola is,

  $\frac{y^2}{60,025}-\frac{x^2}{2,475}=1$

c.

To determine

To find: The distance of $P$ from $A$ .

Answer to Problem 48E

The distance from the boat is $10\text{mi}$ .

Explanation of Solution

Given:

The distance of $A$ is due north of $B$ and $P$ due east of $A$ .

Calculation:

Here,

  $\frac{b^2}{a^2}=\frac{2,475}{245}$

Substitute in the given formula,

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

To demonstrate this we use two methods, By the formula and by the distance formula.

The height of the boat is $y^2={\left(250\right)}^2$ .

Then by formula method,

  $\begin{array}{c}\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\\ \frac{250}{60,025}-\frac{x^2}{2,475}=1\\ x=10.1020\end{array}$

Hence the coordinate is $P\left(x,250\right)$ .

Now by distance formula,

  $\begin{array}{c}d\left(P,A\right)-d\left(P,B\right)=490\\ d\left(P,A\right)-500.1020=490\\ d\left(P,A\right)=10.1020\\ \approx 10\text{ mi}\end{array}$

Therefore, the distance from the boat is $10\text{mi}$ .