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Tangent And Geometric Sequence (Posted on 2007-09-21) Difficulty: 2 of 5
P, Q and R are three points on the parabola y2 = 4bx, and the ordinates of the points P, Q and R (in this order) are in geometric sequence. b is a constant.

The tangents drawn respectively at the points P and R intersect at the point T.

Is it always the case that QT is perpendicular to the tangent drawn at Q?

No Solution Yet Submitted by K Sengupta    
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Question I'm confused | Comment 1 of 2

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.


  Posted by Charlie on 2007-09-21 17:11:19
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