SOLUTION: The hyperbolic cross-section of a cooling tower is given by the equation 4x2 − y2 + 16y − 80 = 0. The center of the cooling tower is the same as the center of the hyperbola, an

Algebra ->  Finance -> SOLUTION: The hyperbolic cross-section of a cooling tower is given by the equation 4x2 − y2 + 16y − 80 = 0. The center of the cooling tower is the same as the center of the hyperbola, an      Log On


   

Question 1180218: The hyperbolic cross-section of a cooling tower is given by the equation 4x2 − y2 + 16y − 80 = 0. The center of the cooling tower is the same as the center of the hyperbola, and the x-axis represents the ground surface.
The diameter at the center of the tower is ? meters
The center of the tower is ? meters above the ground.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

4x%5E2+-+y%5E2+%2B+16y+-+80+=+0+........complete squares
4x%5E2+-+%28y%5E2+-16y%2Bb%5E2%29+-%28-b%5E2%29-+80+=+0...........b=8
4x%5E2+-+%28y%5E2+-16y%2B8%5E2%29+%2B8%5E2-+80+=+0
4x%5E2+-+%28y+-8%29%5E2%2B64-+80+=+0
4x%5E2+-+%28y+-8%29%5E2+=16...........both sides divide by 16
4x%5E2%2F16+-+%28y+-8%29%5E2%2F16+=16%2F16
x%5E2%2F4+-+%28y+-8%29%5E2%2F16+=1
=> h=0, k=8, a=2,+b=4
center is at (0,8)
The diameter at the center of the tower is 4 meters (a=2 from each side of the center)
The center of the tower is 8 meters above the ground. (y coordinates of the center)