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| Question 382728:  Please help me with this: A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch at a distance of 40 ft from the center is to be 10 ft. Find the height of the arch at its center.
 Answer by ankor@dixie-net.com(22740)
      (Show Source): 
You can put this solution on YOUR website! A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch at a distance of 40 ft from the center is to be 10 ft.
 Find the height of the arch at its center.
 ;
 Find the equation using the form ax^2 + bx = y
 :
 From the information given, we can find two x,y coordinates
 x = the width of the bridge, y = height of the bridge at x
 x=100, y=0
 and it says,"height of the arch at a distance of 40 ft from the center is to be 10 ft"
 The center of the bridge = 50 ft, 40 ft to left of center: x=10
 x=10, y = 10
 and
 x=100, y=0
 100^2a + 100b = 0
 10000a + 100b = 0
 and
 x=10, y=10
 10^2a + 10b = 10
 100a + 10b = 10
 Multiply the above equation by 10 subtract from the 1st equation
 10000a + 100b = 0
 1000a + 100b = 100
 -----------------------subtraction eliminates b, find a
 9000a = -100
 a =
  a = -.011
 :
 Find b using 100a + 10b = 10
 100(-.011) + 10b = 10
 -1.11 + 10b = 10
 10b = 10 + 1.11
 b =
  b = 1.111
 :
 The equation: y =  -.011x^2 + 1.11x, graph would look like this
 
  You can see that 40' either side of the center (50') the arch is 10' high
 :
 What is the height at the center? Substitute 50' for x in the equation, find y
 y =  -.011(50^2) + 1.11(50)
 y = -27.5 + 55.5
 y = 28 ft, which is about right on the graph
 
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