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Question 1166634: The cabels of a suspension bridge are in the shape of a parabola. The towers supporting the cable are 400 feet apart and 150 feet high. If the cables are at its lowest is 30 feet above the bridge at its midpoint, how high is the cable 50 ft away (horizontally) from either tower?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
If you graph the parabola so that the vertex is at the origin, the center of the roadway would be at the point (0,-30). The towers are 400 feet apart, so they are 200 feet either side of center. Considering the 30 foot offset, the points of attachment of the cables to the tower are at (-200,120) and (200,120).
The equation of a parabola in vertex form with the vertex at the point (h,k) is ^2\ +\ k) . Since we chose the origin for the vertex, the equation simplifies to 
Using one of the points of attachment of the cable:
^2\ =\ 120)
Making the desired function:
\ =\ 0.003x^2)
The height of the cable above the  -axis at 50 feet in from one of the towers is given by:
\ =\ 0.003(150)^2)
Then the height above the roadway is:
\ +\ 30\ =\ 0.003(150)^2\ +\ 30)
You get to do your own arithmetic.
John
My calculator said it, I believe it, that settles it
I > Ø
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