Question 77349
A bridge is built in the shape of a semielliptical arch. It has a span of 94 feet. The height of the arch 28 feet from the center is to be 6 feet. Find the height of the arch at its center.
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Draw the picture putting the width of 94 on the x-axis centered 
at the origin.
The vertices are then (-47,0) and (47,0)------
Then x^2/47^2 + y^2/b^2 = 1
The ellipse passes thru the point (28,6)
Substitute those values in the equation to solve for "b".
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28^2/47^2 + 6^2/b^2 = 1
0.3549117248 + 36/b^2 = 1
36/b^2 = 0.6450882752..
b^2/36=1.550175439
b^2=55.806
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At the center the height is b=7.47 feet 
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EQUATION of the ellipse
x^2/47^2 + y^2/55.806 = 1

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Cheers,
Stan H.