Let H be the y-intercept of the horizontal line through P.
Call any arbitrary point on Γ = P = (x,y)
∠HPO=arctan(y/x) = arctan(b√(a²-x²)/(ax))
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
H is the midpoint of OQ
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