If I am understanding what you wrote correctly, yes, the point of intersection of the tangents at and intersect on the x-axis.In the derivation of the tangent at P the gradient is found to be 1/p.
To get the tangent at Q, given that P and Q have the same coordinate , I would expect the intersection to be on the x-axis (since is symmetrical). The answer does not confirm this, so that's my first question - why not?
Okay. But q= -p so this is the same asIf I put the coords of Q into the parabola I get:
I therefore take the gradient at Q to be -1/p, then the equation of the tangent follows:
call this equation 1
And, of course, q= -p so this is andThe tangent at P is
which is the correct answer.
No, it's not a different result. q is still -p, q+ p= 0, and 2a(0)= a(0)= 0. The only "slip" you made was not to replace q by -p as soon as you realized, as you say above, that . Of course, q= p does not give two points so we don't have two tangent to intersect and the whole question falls apart: q= -p.The thing that puzzles me is that to get the answer I am assuming the gradient at Q is symmetrical i.e. -1/p (since the gradient at P is 1/p). Also if I express this gradient as 1/q which is the same thing and substitute into equation 1, I get:
and equation 1 is:
A different result! All these ps and qs are swimming before my eyes. Have I made a slip somewhere?