I'm having difficulty understanding this. I'd think the geometric properties would remain the same when x and y are interchanged, and 4b replaced by some other constant, such as k or 1/k.
Also, when the ordinates are in geometric sequence so are the abscissas: to multiply y by 4, x must be multiplied by 2. So let's take y=x^2; it's just a rotation and a replacement of b by 1/4.
If the first point is x=a and y=a^2, and ratio k for the x values:
x y dy/dx
a a^2 2a
ka k^2*a^2 2ka
k^2*a k^4*a^2 2k^2*a
The tangent lines are
(y  a^2) = 2a(x  a)
(y  k^4*a^2) = 2k^2*a(x  k^2*a)
The first can be written
y = 2ax  2a^2 + a^2 = 2ax  a^2
Then substituting into the second:
2ax  a^2  k^4*a^2 = 2k^2*ax  2k^4*a^2
or
x(2a  2k^2*a) = a^2 + k^4*a^2  2k^4*a^2
x = (a^2 + k^4*a^2  2k^4*a^2) / (2a  2k^2*a)
= (a^2  k^4*a^2) / (2a(1  k^2))
= a(1  k^4) / (2(1  k^2))
This program evaluates for given a (the initial x) and k (the ratio of one x to the next):
DO
INPUT a, k
x = (a * (1  k * k * k * k)) / (2 * (1  k * k))
y = 2 * a * x  a * a
x1 = k * a: y1 = k * k * a * a
cSlope = 2 * k * a
perpQ = (y1  y) / (x1  x)
PRINT cSlope, perpQ
LOOP
(x,y) being the intersection of the tangents and (x1,y1) being point Q. However y1y always comes out zero, not perpendicular to the slope of the parabola at Q.
? 1,2
4 0
? 1.5,2
6 0
? 2,3
12 0
? 3,2
12 0
The numbers just after the ? are a and k. The two numbers on the next line after them are the slope at Q and the slope of QT. They aren't negative reciprocals of each other.
Rather than perpendicular to the tangent at Q, QT seems perpendicular to the axis of the parabola.

Posted by Charlie
on 20070921 17:11:19 