The parabola may as well be just y=x²since all parabolas are similar.
Call the points E = (b,b²), F = (kb,k²b²), G = (k²b,k^4b²)
Any line through F perpendicular to the axis of symmetry (the yaxis) will have y coordinate k²b²
E is at a point where the slope of the tangent line is 2b so the equation of this line is yb²=2b(xb)
or x = (y+b²)/(2b)
G is at a point where the slope of the tangent line is 2k²b so the equation of this line is yk^4b²=2k²b(xk²b)
or x = (y+k^4b²)/(2k²b)
Setting these lines equal to solve for y
(y+b²)/(2b) = (y+k^4b²)/(2k²b)
(y+b²) = (y+k^4b²)/(k²)
k²y+k²b²= y+k^4b²
y(k²1) = k^4b²  k²b²
y(k²1) = k²b²(k²1)
y = k²b²
Which is the same as point F
QED
To look for counterexamples we need to worry about dividing by zero. However k≠0, b≠0, k²1≠0 otherwise we wouldn't really have a geometric series.

Posted by Jer
on 20130203 01:12:43 