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

Home > Just Math > Calculus
Minimize Angle (Posted on 2013-11-30) Difficulty: 3 of 5

  
Let Γ be the ellipse ( b2x2 + a2y2 = a2b2 with a > b ). Let O
denote the origin.

Construct a line tangent to Γ ( at point P in the 1st quadrant )
and intersecting the y-axis at point Q such ∠QPO is minimized.
  

See The Solution Submitted by Bractals    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Direct calculus approach Comment 1 of 1
Let H be the y-intercept of the horizontal line through P.
Call any arbitrary point on Γ = P = (x,y)

∠QPO=∠QPH+∠HPO
∠HPO=arctan(y/x) = arctan(b√(a-x)/(ax))
∠QPH=arctan(bx/(a√(a-x))

We wish to minimize this sum.  The derivative (thank you WolframAlpha) is
ab/(√(a-x)(ab+ax-bx)) + ab/(√(a-x)(a^4-ax+bx))
set this to zero and solve for x gives
x = a/√2
This is simple enough to construct.

Which I confirmed using Geometer's Sketchpad.
Interestingly, when ∠QPO is minimized itis also the case that ∠QPH=∠HPO
so
H is the midpoint of OQ
and
P is the midpoint of the segment connecting the x and y coordinates of the tangent lines.

  Posted by Jer on 2013-11-30 14:51:24
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 (2)
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