Let H be the yintercept of the horizontal line through P.
Call any arbitrary point on Γ = P = (x,y)
∠QPO=∠QPH+∠HPO
∠HPO=arctan(y/x) = arctan(b√(a²x²)/(ax))
∠QPH=arctan(b²x/(a√(a²x²))
We wish to minimize this sum. The derivative (thank you WolframAlpha) is
a³b/(√(a²x²)(a²b²+a²x²b²x²)) + a³b/(√(a²x²)(a^4a²x²+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 20131130 14:51:24 