The semi-ellipse must be as above. We put the center at the origin. Since the
height is 4.78 feet, the semi-minor axis = b = 4.78.
The total width of the highway is 14.64 m, so each lane must be 7.32 m wide.
For a truck 3.56 m high to just scrape the top of the tunnel, when its
right side is at the edge of the highway, the ellipse must pass through the
point (7.32,3.56), and also the point (-7.32,3.56).
[Note: It would be more practical to make to make it a little higher, to go
through a higher point so that an 3.56-m tall truck would have a little
clearance and not scrape the top, but the problem didn't state any clearance, so
I will allow the truck to scrape the top at the edge of the highway]
The equation of an ellipse is
+ 
= 1
But since the center is (h,k) = (0,0), the equation simplifies to
+ 
= 1
We know that b = 4.78, so we have
+ 
= 1
+ 
= 1
Since it goes through the point (7.32,3.56), we substitute (x,y) = (7.32,3.56)
+ 
= 1
+ 
= 1
Solve that for "a" and get
a = 10.969227733, the semi-major axis.
The width of the opening must be the entire major axis,
so the answer to your problem is twice that or 21.93845466 m.
ii)what will be the width of each lane if the barrier between the lanes is 0.6m?
We will subtract half of 0.6 m, or 0.3 m from each lane. Then each lane will
then be 7.32-0.3 = 7.02 m wide.
Edwin
