 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=x�since all parabolas are similar.

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

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

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 2k�b so the equation of this line is y-k^4b�=2k�b(x-k�b)
or x = (y+k^4b�)/(2k�b)

Setting these lines equal to solve for y
(y+b�)/(2b) = (y+k^4b�)/(2k�b)
(y+b�) = (y+k^4b�)/(k�)
k�y+k�b�= y+k^4b�
y(k�-1) = k^4b� - k�b�
y(k�-1) = k�b�(k�-1)
y = k�b�
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:
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