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Question 1170527: LORAN navigational transmitters A and B are located at (-130,0) and (130,0) respectively. A receiver P on a fishing boat somewhere in the first quadrant listens to the pair (A,B) of the transmissions and computes the difference of the distance from boat A and B as 240 miles. Find the equation of the hyperbola on P is located.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let A = (-130, 0) and B = (130, 0). Let P = (x, y) be the position of the fishing boat.
The distance from P to A is $d_A = \sqrt{(x - (-130))^2 + (y - 0)^2} = \sqrt{(x + 130)^2 + y^2}$.
The distance from P to B is $d_B = \sqrt{(x - 130)^2 + (y - 0)^2} = \sqrt{(x - 130)^2 + y^2}$.
We are given that the difference in distances is 240 miles, i.e., $|d_A - d_B| = 240$. Since P is in the first quadrant, we can assume $d_A > d_B$, so $d_A - d_B = 240$.
Thus, $\sqrt{(x + 130)^2 + y^2} - \sqrt{(x - 130)^2 + y^2} = 240$.
We need to find the equation of the hyperbola.
First, rearrange the equation:
$\sqrt{(x + 130)^2 + y^2} = \sqrt{(x - 130)^2 + y^2} + 240$
Square both sides:
$(x + 130)^2 + y^2 = (x - 130)^2 + y^2 + 480\sqrt{(x - 130)^2 + y^2} + 240^2$
$x^2 + 260x + 130^2 + y^2 = x^2 - 260x + 130^2 + y^2 + 480\sqrt{(x - 130)^2 + y^2} + 57600$
$520x - 57600 = 480\sqrt{(x - 130)^2 + y^2}$
Divide by 80:
$6.5x - 720 = 6\sqrt{(x - 130)^2 + y^2}$
Square both sides again:
$(6.5x - 720)^2 = 36[(x - 130)^2 + y^2]$
$42.25x^2 - 9360x + 518400 = 36(x^2 - 260x + 16900 + y^2)$
$42.25x^2 - 9360x + 518400 = 36x^2 - 9360x + 608400 + 36y^2$
$6.25x^2 - 36y^2 = 90000$
Divide by 90000:
$\frac{6.25x^2}{90000} - \frac{36y^2}{90000} = 1$
$\frac{x^2}{14400} - \frac{y^2}{2500} = 1$
$\frac{x^2}{120^2} - \frac{y^2}{50^2} = 1$
Therefore, the equation of the hyperbola is $\frac{x^2}{14400} - \frac{y^2}{2500} = 1$.
Final Answer: The final answer is $\boxed{\frac{x^2}{14400}-\frac{y^2}{2500}=1}$
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