Let the shoreline be the x-axis, and suppose A is at (0, 0) and B at (400, 0). If the ship is at (x, y) then the distance from the ship to A is √(x^2 + y^2) and the distance to B is √((x-400)^2 + y^2), so we have
√(x^2 + y^2) - 100 = √((x-400)^2 + y^2)
<=> x^2 + y^2 + 10000 - 200√(x^2 + y^2) = x^2 - 800x + 160000 + y^2
<=> 200√(x^2 + y^2) - 800x + 150000 = 0
<=> √(x^2 + y^2) - 4x + 750 = 0
<=> x^2 + y^2 = 16x^2 - 6000x + 562500
<=> y^2 = 15x^2 - 6000x + 562500
y = 60 => 15x^2 - 6000x + 562500 - 3600 = 0
=> x^2 - 400x + 37260 = 0
=> (x - 200)^2 = 2740
=> x = 200 ± 2√685.
2007-04-10 19:59:16
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answer #1
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answered by Scarlet Manuka 7
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Let the two stations be located at the foci:
A(0,0) and B(400,0).
The foci are at a distance 2c apart.
2c = 400
c = 200
The center (h,k) of the ellipse is the midpoint of the foci.
(h,k) = (200,0)
The ship is 100 miles farther from A than B.
2a = 100
a = 50
a² = 2,500
Now solve for b².
b² = c² - a² = 200² - 50² = 40,000 - 2,500 = 37,500
The equation of the hyperbola is:
(x - h)²/a² - (y - k)²/b² = 1
(x - 200)²/2,500 - y²/37,500 = 1
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b. Find the exact coordinates of the ship if it is 60 miles from shore.
In other words solve for x when y = 60.
(x - 200)²/2,500 - y²/37,500 = 1
Multiply thru by 37,500 to clear the denominators.
15(x - 200)² - y² = 37,500
15(x - 200)² = y² + 37,500 = 60² + 37,500 = 3,600 + 37,500
15(x - 200)² = 41,100
(x - 200)² = 2,740
x² - 400x + 40,000 = 2,740
x² - 400x + 37,260 = 0
x = [400 ± â(400² - 4*1*37,260)] / (2*1)
x = (400 ± â10,960) / 2 = (400 + 4â685) / 2
x = 200 ± 2â685 â 252.34501
Since the ship is closer to B than A, the solution with the minus sign is rejected.
x = 200 + 2â685 â 252.34501
The location of the ship is (x,y) = (200 + 2â685, 60).
2007-04-13 18:23:16
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answer #2
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answered by Northstar 7
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