SOLUTION: 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

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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