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 y1-y 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.
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Posted by Charlie
on 2007-09-21 17:11:19 |