SOLUTION: A bridge is built in the shape of a semielliptical arch. The bridge has a span of 60 feet and a maximum height of 20 feet. Find the height of the arch at distances 5, 10, and 20

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A bridge is built in the shape of a semielliptical arch. The bridge has a span of 60 feet and a maximum height of 20 feet. Find the height of the arch at distances 5, 10, and 20       Log On


   

Question 711274: A bridge is built in the shape of a semielliptical arch. The bridge has a span of 60 feet and a maximum height of 20 feet. Find the height of the arch at distances 5, 10, and 20 feet from the center.
The answers are in the back of my textbook, but I have no idea how to get to the answer.
I started with finding the points on the arch (0,20) with, I'm assuming, the vertices (-30,0) and (30,0) when I drew it out.
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
A bridge is built in the shape of a semielliptical arch. The bridge has a span of 60 feet and a maximum height of 20 feet. Find the height of the arch at distances 5, 10, and 20 feet from the center.
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Equation of an ellipse with horizontal major axis with center at (0,0):
x%5E2%2Fa%5E2%2By%5E2%2Fb%5E2=1
For given problem:
b=20
b^2=400
a=30
a^2=900
equation: x%5E2%2F900%2By%5E2%2F400=1
y%5E2%2F400=1-x%5E2%2F900
y%5E2=%281-x%5E2%2F900%29%28400%29
y=√[(1-x^2/900)(400)]
plug in x-values of 5, 10, and 20 ft
f(5)=√[(1-25/900)(400)]≈19.72 ft (arch height 5 ft from center)
f(10)=√[(1-100/900)(400)]≈18.86 ft (arch height 10 ft from center)
f(20)=√[(1-400/900)(400)]≈14.91 ft (arch height 20 ft from center)