All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
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


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):

  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

(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
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information