All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Focusing Upon The Maximum (Posted on 2007-12-31) Difficulty: 3 of 5
Determine the maximum area of the circle which is entirely contained within the parabola y2 = 36x, and passes through its focus.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 2

The parabola,
   y^2 = 36x = 4px,
where p is the distance from the vertex to the focus.
The circle,
   (x - h)^2 + y^2 = r^2.
Since the circle passes throught the focus,
   (p - h)^2 + 0^2 = r^2.
Therefore,
   (x - h)^2 + y^2 = (p - h)^2.
The intersection of the circle and ellipse,
   (x - h)^2 + 4px = (p - h)^2
              or
   x^2 - 2(h - 2p)x - (p^2 - 2ph) = 0.
Solving for x,
       
   x = h - 2p +- sqrt[(h - p)(h - 5p)].
If the radicand is zero, then the circle is
tangent to the ellipse and we have the largest
circle within the ellipse.
h = p implies the radius of the circle is zero.
Therefore, h = 5p and r^2 = (4p)^2.
For our problem,
   Area of circle = PI*(36)^2.
 

  Posted by Bractals on 2007-12-31 12:26:12
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information