SOLUTION: An arch in the shape of a parabola has height 630ft and width 630ft. How wide is the arch 300 ft up?

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Question 1132760: An arch in the shape of a parabola has height 630ft and width 630ft. How wide is the arch 300 ft up?
Answer by MathLover1(20850) About Me  (Show Source):
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An arch in the shape of a parabola has height 630ft and width 630ft. How wide is the arch 300ft up?
The width of the bridge is 630 feet so the parabola crosses the x-axis with x-coordinates ± 630%2F2 = ± 315
=> x-intercepts are at (-315,0) and (315,0)
Assume that it stands on the x axis, that it is symmetric with respect to the y-axis, and that it is 630 feet wide at the base and 630 feet high at the center.
Then the maximum height occurs at x+=+0; so, the vertex of the parabola is =>(0,630)

Since the curve is a parabola which opens downward its equation can be written
f%28x%29+=+a%28x-h%29%5E2+%2Bk
since h=0 and k=630
f%28x%29+=+a%28x-0%29%5E2+%2B630
f%28x%29+=+ax%5E2+%2B630
x-intercepts are at (-315,0), use x-intercept to find a
0+=+a%28-315%29%5E2+%2B630
0+=99225%2Aa+%2B630
99225+a+=-630
+a+=-630%2F99225
a+=-2%2F315

and, equation is
f%28x%29+=+-%282%2F315%29x%5E2+%2B630

to find how wide is the arch 300ft up, substitute f%28x%29=+300ft
300+=+-%282%2F315%29x%5E2+%2B630
%282%2F315%29x%5E2+=630-300
%282%2F315%29x%5E2+=330
x%5E2+=330%2F%282%2F315%29
x%5E2+=%28330%2A315%29%2F2
x%5E2+=165%2A315
x%5E2+=51975
x+=sqrt%2851975%29
x = ± 227.98
the arch 300 ft have intercepts at (-227.98,300) and (227.98,300), so width will be the distance from x=-227.98 to x=227.98 which is +2%2A227.98=455.96 ft
answer: the arch 300 ft up is +455.96 ft wide