SOLUTION: An arch is in the form of a semiellipse. It is 48 ft wide at the base and has a height of 20ft. How wide is the arch at a height of 10 ft above the base?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: An arch is in the form of a semiellipse. It is 48 ft wide at the base and has a height of 20ft. How wide is the arch at a height of 10 ft above the base?      Log On


   

Question 1184839: An arch is in the form of a semiellipse. It is 48 ft wide at the base and has a
height of 20ft. How wide is the arch at a height of 10 ft above the base?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Here the length of the major axis is given as 2a=48 -> a=24 and the height of the semi-ellipse is given as b=20.
Now the standard equation of an ellipse with center at origin is:
x%5E2%2Fa%5E2%2By%5E2%2Fb%5E2=1
Putting the values of a and b in the above equation of an ellipse, we have
x%5E2%2F24%5E2+%2B+y%5E2%2F20%5E2+=+1
For the height of 10 feet, the equation becomes
x%5E2%2F576+%2B+10%5E2%2F400+=+1
x%5E2%2F576+%2B+100%2F400+=+1
x%5E2%2F576%2B1%2F4+=+1+
x%5E2%2F576+=+1+-1%2F4
x%5E2%2F576+=+3%2F4
x%5E2+=+%283%2F4%29576
x%5E2+=+3%2A144
x+=+sqrt%283%2A144%29
x+=+12sqrt%283%29+ ft
at a height of 10+ft above the base, the arch is +12sqrt%283%29+ ft wide