SOLUTION: Ellipse help needed? A semielliptic archway has a height of 20ft and a width of 50ft. Can a truck 14 ft wide and 10 ft wide drive under the archway without going into the othe

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Ellipse help needed? A semielliptic archway has a height of 20ft and a width of 50ft. Can a truck 14 ft wide and 10 ft wide drive under the archway without going into the othe      Log On


   



Question 154260: Ellipse help needed?

A semielliptic archway has a height of 20ft and a width of 50ft. Can a truck 14 ft wide and 10 ft wide drive under the archway without going into the other lane?

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
A semielliptic archway has a height of 20ft and a width of 50ft. Can a truck 14 ft high and 10 ft wide drive under the archway without going into the other lane?

Here is the drawing of the archway with the center of the
highway at (0,0).



Since the center of the highway is at the origin. Let's 
imagine the left side of the truck being right on the center 
line of the highway, just barely in the right lane. 

Now let's draw a rectangle 10 feet wide and 14 feet high to 
represent the back of the truck. Its corners will be (0,0), 
(0,14), (10,0), and (10,14):  



So from the looks of the graph, the truck will fit nicely 
under there, and that we can also move the truck over to 
the right a foot or two, so that it won't be sitting right 
on the line in the middle of the road.

However, your teacher doesn't want you to just draw it and 
look and see without doing any calculations. We have to make 
sure that the upper right hand corner of the truck, (10,14), 
is below the semi-elliptical archway.

So let's get the equation of the whole ellipse, which is this:



Its major axis is 50, so it has semi-major axis a=25.  
Its semi-minor axis is 20, so b=20  So the equation of 
the whole ellipse, using the equation 

x%5E2%2Fa%5E2%2By%5E2%2Fb%5E2=1

is

x%5E2%2F25%5E2%2By%5E2%2F20%5E2=1

x%5E2%2F625%2By%5E2%2F400=1

Now to find out how high the archway is above the highway at 
the right edge of the truck. To do this we substitute x=10 
into the ellipse equation and solve for y to find the height of 
the archway above the highway at the right edge of the truck
when its left edge is right on the center line:

10%5E2%2F625%2By%5E2%2F400=1

100%2F625%2By%5E2%2F400=1 

4%2F25%2By%5E2%2F400=1

Multiply thru by 400:

%28400%29%284%2F25%29%2B400%28y%5E2%2F400%29=400%281%29

64%2By%5E2=400

y%5E2=336

We take the positive square root since
we are only concerned with the top
half of the ellipse:

y=sqrt%28336%29

so y = about 18.3 feet.

So since the truck is only 14 feet tall,
it will fit there with 4.3 feet between the 
top of the right edge of the back of the truck
and the top of the archway.

So the answer is "yes". And we can move the
truck right a bit so it won't be right on the 
center line.

Edwin