SOLUTION: A bridge is built in the shape of a parabolic arch. The bridge has a span of 180 feet and a maximum height of 40 feet above the water at the center. Can a sailboat that is 39 feet

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A bridge is built in the shape of a parabolic arch. The bridge has a span of 180 feet and a maximum height of 40 feet above the water at the center. Can a sailboat that is 39 feet       Log On


   

Question 1160972: A bridge is built in the shape of a parabolic arch. The bridge has a span of 180 feet and a maximum height of 40 feet above the water at the center. Can a sailboat that is 39 feet tall fit under the bridge 10 feet from the​ center?
Found 2 solutions by Edwin McCravy, Alan3354:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Place the vertex at (0,40) and the ends of the bridge at (±90,0).



The equation is

%28x-h%29%5E2=4p%28y-k%29

Where the vertex (h,k) = (0,40)

%28x-0%29%5E2=4p%28y-40%29

x%5E2=4p%28y-40%29

Substituting the point (90,0)

90%5E2=4p%28-40%29

8100=-160p

-405%2F8=p

So the equation is now

x%5E2=4%28-405%2F8%29%28y-40%29

x%5E2=%28-405%2F2%29%28y-40%29

So when x=10

10%5E2=%28-405%2F2%29%28y-40%29

100=%28-405%2F2%29%28y-40%29

200=-405%28y-40%29

200=-405y%2B16200

405y=16200

y=16200%2F405

y=40

Yes a 39 foot tall sailboat can fit under there with 1 foot to spare.

Edwin

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A bridge is built in the shape of a parabolic arch. The bridge has a span of 180 feet and a maximum height of 40 feet above the water at the center. Can a sailboat that is 39 feet tall fit under the bridge 10 feet from the​ center?
----------------
Find the equation of the parabola.
With x-axis thru the 2 points, (-90,0) and (90,0), the Origin at (0,0), 3 points are given.
-------
y = ax^2 + bx + c
For the point (0,40):
40 = a*0 + b*0 + c
----> c = 40
====================
0 = a*8100 + 90b + 40 for the point (90,0)
0 = a*8100 - 90b + 40 for the point (-90,0)
--------------------------------------------------- Add
16200a = -80
a = -1/202.5
-------
0 = a*8100 + 90b + 40 for the point (90,0)
0 = -40 + 90b + 40
b = 0
------
---> y = -x^2/202.5 + 40
==========================================
Find y at x = 10
y = 40 - 100/202.5
y =~ 39.5 feet above the water.
---> clearance for 39 feet