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 Minimize Angle (Posted on 2013-11-30)

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

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 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√(a²-x²)/(ax))
∠QPH=arctan(b²x/(a√(a²-x²))

We wish to minimize this sum.  The derivative (thank you WolframAlpha) is
a³b/(√(a²-x²)(a²b²+a²x²-b²x²)) + a³b/(√(a²-x²)(a^4-a²x²+b²x²))
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

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