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

Home > Shapes > Geometry
Geometric Parabola (Posted on 2013-02-02) Difficulty: 2 of 5
Three points E, F and G are taken on the parabola y2= 4ax, so that their ordinates (in order) are in geometric sequence. The ordinates of each of the three points have the same sign.

Do the tangents at E and G intersect on the line through F perpendicular to the axis of the parabola?

If this is always so, prove it. Otherwise, provide a counter example.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Calculus solution | Comment 1 of 6
The parabola may as well be just y=xsince all parabolas are similar.

Call the points E = (b,b), F = (kb,kb), G = (kb,k^4b)

Any line through F perpendicular to the axis of symmetry (the y-axis) will have y coordinate kb

E is at a point where the slope of the tangent line is 2b so the equation of this line is y-b=2b(x-b)
or x = (y+b)/(2b)

G is at a point where the slope of the tangent line is 2kb so the equation of this line is y-k^4b=2kb(x-kb)
or x = (y+k^4b)/(2kb)

Setting these lines equal to solve for y
(y+b)/(2b) = (y+k^4b)/(2kb)
(y+b) = (y+k^4b)/(k)
ky+kb= y+k^4b
y(k-1) = k^4b - kb
y(k-1) = kb(k-1)
y = kb
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, k-1≠0 otherwise we wouldn't really have a geometric series.



  Posted by Jer on 2013-02-03 01:12:43
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information