Find at 10-foot intervals the distance from the arch to the roadway.?

Question

The arch of a concrete bridge is a semiellipse having a span of 60 feet and a central height of 20 feet. If the roadway is a 25 feet above the base, findat 10-foot intervals the distance from the arch to the roadway.

Answer 1

x = 0 ----> y = 20
y^2 = 400
y = 0 ---->x = 30
x^2 = 900
From the above you can write the parabola in the form:
x^2/900 + y^2/400 = 1
For 10 foot intervals, we already have (-30, 0), (0, 20), and (30, 0)
x = +/-10:
(+/-10)^2/900 + y^2/400 = 1
y^2/400 = 1 - 1/9
y = √[400(8/9)] = (20/3)√8 = (40/4)√2 = 18 856feet
x = +/-20:
(+/-20)^2/900 + y^2/400 = 1
y^2/400 = 1 - 4/9
y = √[400(5/9)] = (20/3)√5 = 14 9007feet

Answer 2

Please check your answer. You may have overlooked these things:
First, the equation it will follow is that of an ellipse, not of a parabola.
Second,√[400(8/9)]= √(3200/9)= (40√2)/3 . Therefore, adding 25 to find the distance from the arch to the roadway will yield 25 + (40√2)/3 = (75+40√2)/3 or approximately 43.86 feet.
Similarly,√[400(5/9)]= √(2000/9) = (20√5)/3 . Hence, 25 + (20√5)/3 =(75+20√5)/3 or approximately 39.91 feet.

Thanks.