Question 1086817: Some long-range navigation systems use hyperbolas to determine a ship’s position. Suppose the system
imposes coordinates so that the location of a ship is in the first quadrant. A ship is located at the intersection of the hyperbolas with equations 25x^2 – 4y^2 = 100 and 36y^2 – x^2= 64 . Find the coordinates of the ship to the nearest hundredth of a unit.
a. (1.90, 4.28)
b. (2.07, 1.38)
c. (1.38, 2.07)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the solution is selection b. (2.07,1.38)
you can solve it algebraically as follows:
start with:
25x^2 – 4y^2 = 100
36y^2 – x^2 = 64
re-arrange the terms so that they line up like for like.
25x^2 - 4y^2 = 100
-x^2 + 36y^2 = 64
multiply both sides of the second equation by 25 to get:
25x^2 - 4y^2 = 100
-25x^2 + 900y^2 = 1600
add the equations together to get:
896y^2 = 1700
solve for y^2 to get:
y^2 = 1.897321429
solve for y to get:
y = plus or minus 1.377432913
since the intersection point has to be in the first quadrant, then you get:
y = plus 1.377432913
replace y^2 with 1.897321429 in the first equation to get:
25x^2 - 4 * 1.897321429 = 100
simplify this to get:
25x^2 - 7.589285714 = 100
add 7.589285714 to both sides of the equation to get:
25x^2 = 107.5892857
solve for x^2 to get:
x^2 = 107.5892857/25 = 4.303571429
solve for x to get:
x = plus or minus 2.074505104.
since x has to be in the first quadrant, then you get:
x = plus 2.074505104.
your intersection point in the first quadrant is:
(x,y) = (2.074505104,1.377432913)
round this to 2 decimal digits and you get:
(x,y) = (2.07,1.38)
that's selection b.
graphically, this looks like this:
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