SOLUTION: how high is a semi elliptical arch of height 20 ft. and span 80 feet at a distance of 15 ft from the end?

To make it easy, let's change feet to "dekafeet".

1 dekafoot = 10 feet, the length of one of those measuring chains
used at a football game to see if a first down was made.  So 1
of those chains is 1 dekafoot long. 

So the problem is:

how high is a semi elliptical arch of height 2 dekafeet. and span 
8 dekafeet at a distance of 1.5 dekafeet from the end?  So we'll 
draw the arch:

We'll let 4 of the 8 dekafeet be on the left of the origin, and 4 on
the right, making the vertices (-4,0) and (4,0)



The whole ellipse looks like this:

 

Since a=4 and b=2, the equation

  1

  1

  1

Let's solve for y, so we can find the height 1.5 dekafeet
from the the end.

Clear of fractions by multiplying every term by 16

x² + 4y² = 16

     4y² = 16 - x²
      
      y² = 

       y = 

The + is the arch, the - is the dotted line lower part, below
the ground.

1.5 dekafeet from the right end is 4-1.5 = 2.5 dekafeet from the
origin, see the green line

  

So we substitute x=2.5in

       y = 

       y = 
      
       y =  = 1.5612495 dekafeet.

So the answer in feet is 15.612495.

Edwin