Parametric Equations: path of ship, given radio signals

[sisxixon]

This is the problem:

Two radio stations are positioned 500 miles apart, at A(-250,0) and B(250,0). A ship at point P is traveling on a path so that it always hears the signal from station A 2.5 seconds before the signal from station B. If the speed of the radio signal is c miles per second, find the equation of the path of the ship. simplify the results.

It's ten at night where I am, so I'm a bit hopeless as of now. What I got out of this, assuming this is a parametric equation, was the following:

. . .c: speeed of signal

. . .c = root[(dx/dt)² + (dy/dt)²] / dt

Am I on the right track? Either way, please nudge me down the right road.

 

[soroban]

Re: Parametric Equations

Hello, sisxixon!

This problem does not involve Calculus.
We are deriving the equation of a curve.

Two radio stations are positioned 500 miles apart, at A(-250,0) and B(250,0).
A ship at point P is traveling on a path so that it always hears the signal from station A
\(\displaystyle \;\;\)2.5 seconds before the signal from station B.
If the speed of the radio signal is c miles per second,
\(\displaystyle \;\;\)find the equation of the path of the ship. simplify the results.

                P     |
           (x,y)o     |
               *    * |
              *       | *
             *        |     *
      - - - o - - - - + - - - - o - - - 
       (-250,0)       |      (250,0)
            A         |         B

Point \(\displaystyle P\) is closer to point \(\displaystyle A\) by a distance of \(\displaystyle 2.5c\) miles.

Let \(\displaystyle 2a\,=\,2.5c\)

Then:\(\displaystyle \,\overline{PB}\,-\,\overline{PA}\:=\:2a\)

Using the Distance Formula: \(\displaystyle \.\sqrt{(x-250)^2\,+\,y^2}\,-\,\sqrt{(x+250)^2\,+\,y^2}\;=\;2a\)

We have: \(\displaystyle \,\sqrt{(x-250)^2\,+\,y^2}\;=\;\sqrt{(x+250)^2\,+\,y^2}\,+\,2a\)

Square: \(\displaystyle \,x^2\,-\,500x\,+\,250^2\,+\,y^2\;=\;x^2\,+\,500x\,+\,250^2\,+\,y^2\,+\,4a\sqrt{(x+c)^2\,+\,y^2}\,+\,4a^2\)

\(\displaystyle \;\;\)which simplifies to: \(\displaystyle \,250x\,+\,a^2\;=\;-a\sqrt{(x+250)^2\,+\,y^2}\)


Square: \(\displaystyle \,250^2x^2\,+\,500a^2x\,+\,a^4\;=\;a^2(x^2\,+\,500x\,+\,250^2\,+\,y^2)\)

\(\displaystyle \;\;\)which simplifies to: \(\displaystyle \,(250^2\,-\,a^2)x^2\,-\,a^2y^2\;=\;a^2(250^2\,-\,a^2)\)


\(\displaystyle \text{Divide by }\,a^2(250^2-a^2):\L\;\;\frac{x^2}{a^2}\,-\,\frac{y^2}{250^2-a^2} \;=\;1\)


Therefore, we have a hyperbola!

\(\displaystyle \;\;\)Its major (transverse) axis is: \(\displaystyle \,a\,=\,2.5c\)

\(\displaystyle \;\;\)Its minor (conjugate) axis is: \(\displaystyle \,b\,=\,\sqrt{250^2 - (2.5c)^2}\)



If you insist on parametric equations: \(\displaystyle \L\;\begin{array}{cc}x\,=\,a\cdot\sec\theta \\ y\,=\,b\cdot\tan\theta\end{array}\)