SOLUTION: A parabolic arch 27 feet high spans a parkway. How wide is the arch if the center section of the parkway, a section that is 50 feet wide, has a minimum clearance of 15 feet?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A parabolic arch 27 feet high spans a parkway. How wide is the arch if the center section of the parkway, a section that is 50 feet wide, has a minimum clearance of 15 feet?      Log On


   



Question 1206768: A parabolic arch 27 feet high spans a parkway. How wide is the arch if the center section of the parkway, a section that is 50 feet wide, has a minimum clearance of 15 feet?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Let the parabola have vertex (0,27), and let the 50-foot parkway extend
from (-25,0) to (25,0). To have minimum clearance at the edges of 15 feet,
the parabola must pass through points (-25,15) and (25,15).



The equation of a parabola has several forms,  I'll use this one:

%28x-h%29%5E2%22%22=%22%224a%28y-k%29

where (h,k) is the vertex (0,27)

%28x-0%29%5E2%22%22=%22%224a%28y-27%29

x%5E2%22%22=%22%224a%28y-27%29

Since it passes through (25,15), we substitute

25%5E2%22%22=%22%224a%2815-27%29
625%22%22=%22%224a%28-12%29
625%22%22=%22%22-48a
625%2F%28-48%29%22%22=%22%22a
-625%2F%2848%29%22%22=%22%22a

So the equation of the parabola is

x%5E2%22%22=%22%224%28-625%2F48%29%28y-27%29

x%5E2%22%22=%22%22expr%28-625%2F12%29%28y-27%29

To find the width of the entire arch at the
bottom, we must find the x-intercepts. So we 
substitute 0 for y:

x%5E2%22%22=%22%22expr%28-625%2F12%29%280-27%29
x%5E2%22%22=%22%2216875%2F12
x%22%22=%22%22%22%22+%2B-+sqrt%281406.25%29
x%22%22=%22%22%22%22+%2B-+37.5

So the width of the arch at the bottom goes from
(-37.5,0) to {37.5,0). 

And so the width of the arch at the bottom is 

(37.5)(2) = 75 feet.

Edwin