Two control towers are located at points Q(500, 0) and R(500, 0), on a straight shore where the x-axis runs through (all distances are in meters). At the same moment, both towers sent a radio signal to a ship out at sea, each traveling at 300 m/µs. The ship received the signal from Q 3 µs (microseconds)before the message from R. Find the equation of the curve containing the possible location of the ship.

9514 1404 393

  x^2/81 -y^2/19 = 2500

Step-by-step explanation:

The time difference between signals is 3 µs, so the distance difference is ...

  (300 m/µs)(3 µs) = 900 m

If we assume the coordinates of Q are (-500, 0), then the distances from point X to the control towers are ...

  XQ = √((x +500)^2 +y^2)

  XR = √((x -500)^2 +y^2)

We want the difference in distances to be 900, so we have the equation ...

  |XQ -XR| = |√((x +500)^2 +y^2) -√((x -500)^2 +y^2)| = 900

Squaring both sides gives ...

  (x +500)^2 +2(x +500)y +y^2 -2√(((x +500)^2 +y^2)((x -500)^2 +y^2)) +((x -500)^2 +y^2) = 810000

Separating the root from the rest of the equation, squaring again, and simplifying the rather messy expression, we can arrive at the equation ...

  x^2/81 -y^2/19 = 2500 . . . . . a hyperbola opening horizontally

5   1/4   miles per hour

step-by-step explanation:

15   3/4 / 3 = 5   1/4