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Question 1170541: Two stations, located at M(−1.5,0) and N(1.5,0)(units are in km), simultaneously send sound signals to a ship, with the signal traveling at the speed of 0.33 km/s. If the signal from N was received by the ship four seconds before the signal it received from M, find the equation of the curve containing the possible location of the ship.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let the ship's location be (x, y).
The distance between the ship and station M is $d_M = \sqrt{(x + 1.5)^2 + y^2}$.
The distance between the ship and station N is $d_N = \sqrt{(x - 1.5)^2 + y^2}$.
The time it takes for the signal from M to reach the ship is $t_M = \frac{d_M}{0.33}$.
The time it takes for the signal from N to reach the ship is $t_N = \frac{d_N}{0.33}$.
We are given that the signal from N was received 4 seconds before the signal from M. This means $t_M - t_N = 4$.
Substituting the expressions for $t_M$ and $t_N$, we get:
$$\frac{d_M}{0.33} - \frac{d_N}{0.33} = 4$$
$$d_M - d_N = 4(0.33) = 1.32$$
$$\sqrt{(x + 1.5)^2 + y^2} - \sqrt{(x - 1.5)^2 + y^2} = 1.32$$
Now, we need to eliminate the square roots.
$$\sqrt{(x + 1.5)^2 + y^2} = \sqrt{(x - 1.5)^2 + y^2} + 1.32$$
Square both sides:
$$(x + 1.5)^2 + y^2 = (x - 1.5)^2 + y^2 + 2(1.32)\sqrt{(x - 1.5)^2 + y^2} + (1.32)^2$$
$$x^2 + 3x + 2.25 + y^2 = x^2 - 3x + 2.25 + y^2 + 2.64\sqrt{(x - 1.5)^2 + y^2} + 1.7424$$
$$6x - 1.7424 = 2.64\sqrt{(x - 1.5)^2 + y^2}$$
Divide by 2.64:
$$\frac{6x - 1.7424}{2.64} = \sqrt{(x - 1.5)^2 + y^2}$$
$$\frac{600x - 174.24}{264} = \sqrt{(x - 1.5)^2 + y^2}$$
$$\frac{250x - 72.6}{110} = \sqrt{(x - 1.5)^2 + y^2}$$
$$\frac{125x - 36.3}{55} = \sqrt{(x - 1.5)^2 + y^2}$$
Square both sides again:
$$\left(\frac{125x - 36.3}{55}\right)^2 = (x - 1.5)^2 + y^2$$
$$\frac{(125x - 36.3)^2}{55^2} = (x - 1.5)^2 + y^2$$
$$\frac{15625x^2 - 9075x + 1317.69}{3025} = x^2 - 3x + 2.25 + y^2$$
$$15625x^2 - 9075x + 1317.69 = 3025x^2 - 9075x + 6806.25 + 3025y^2$$
$$12600x^2 - 3025y^2 = 5488.56$$
Divide by 5488.56:
$$\frac{12600x^2}{5488.56} - \frac{3025y^2}{5488.56} = 1$$
$$\frac{x^2}{0.4356} - \frac{y^2}{1.8144} = 1$$
$$\frac{x^2}{(0.66)^2} - \frac{y^2}{(1.347)^2} = 1$$
The equation of the curve is a hyperbola.
Final Answer: The final answer is $\boxed{\frac{x^2}{0.4356}-\frac{y^2}{1.8144}=1}$
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