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

 Geometric Parabola (Posted on 2013-02-02)
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.)
 Calculus solution | Comment 1 of 6
The parabola may as well be just y=x²since all parabolas are similar.

Call the points E = (b,b²), F = (kb,k²b²), G = (k²b,k^4b²)

Any line through F perpendicular to the axis of symmetry (the y-axis) will have y coordinate k²b²

E is at a point where the slope of the tangent line is 2b so the equation of this line is y-b²=2b(x-b)
or x = (y+b²)/(2b)

G is at a point where the slope of the tangent line is 2k²b so the equation of this line is y-k^4b²=2k²b(x-k²b)
or x = (y+k^4b²)/(2k²b)

Setting these lines equal to solve for y
(y+b²)/(2b) = (y+k^4b²)/(2k²b)
(y+b²) = (y+k^4b²)/(k²)
k²y+k²b²= y+k^4b²
y(k²-1) = k^4b² - k²b²
y(k²-1) = k²b²(k²-1)
y = k²b²
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, k²-1≠0 otherwise we wouldn't really have a geometric series.

 Posted by Jer on 2013-02-03 01:12:43

 Search: Search body:
Forums (0)