SOLUTION: The cross section of a 12 - foot -high cooling tower is in the form of a hyperbola. its top and bottom parts are 8 feet wide and the narrowest part is 4 feet. A man painting the to

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The cross section of a 12 - foot -high cooling tower is in the form of a hyperbola. its top and bottom parts are 8 feet wide and the narrowest part is 4 feet. A man painting the to      Log On


   

Question 971199: The cross section of a 12 - foot -high cooling tower is in the form of a hyperbola. its top and bottom parts are 8 feet wide and the narrowest part is 4 feet. A man painting the tower must place a horizontal plank exactly spanning the upper 6 feet width of the cross section of the tower. At what point from the ground should the plank be placed?
Pls help me with this problem
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
The cross section of a 12 - foot -high cooling tower is in the form of a hyperbola. its top and bottom parts are 8 feet wide and the narrowest part is 4 feet. A man painting the tower must place a horizontal plank exactly spanning the upper 6 feet width of the cross section of the tower. At what point from the ground should the plank be placed?
***
The cooling tower can be represented by a hyperbola with horizontal transverse axis with center at the origin.
Its standard form of equation: x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1
For given problem:
a=2 (distance from center to vertices(the narrowest part)
a^2=4
solve for b^2 using coordinates (4, 6) of the point at the top right corner of the tower
x%5E2%2Fa%5E2-y%5E2%2Fb%5E2=1
4%5E2%2F4-6%5E2%2Fb%5E2=1
16%2F4-36%2Fb%5E2=1
36/b^2=4-1=3
b^2=12
equation: x%5E2%2F4-y%5E2%2F12=1
plug in coordinates(3, y) of the point at the top right corner of the horizontal plank and solve for y
x%5E2%2F4-y%5E2%2F12=1
3%5E2%2F4-y%5E2%2F12=1
9/4-1=y^2/12
5/4=y^2/12
y^2=15
y=√15=3.87
At what point from the ground should the plank be placed? 3.87 ft