<Eddie> So, you know a parabola right? y = ax² + bx + c
<John_Wolfe_> That could be intepreted as such thing.
<John_Wolfe_> please, do continue.
<Eddie> Now imagine a tangent line to that parabola on one side and a symmetrical one on the other.
<John_Wolfe_> mmhm
<Eddie> (if it's on the vertex, they're both horizontal I guess)
<John_Wolfe_> yes.
<Eddie> The theorem states that the y distance between the tangent points and the vertex is the same as the distance between the vertex and the intersection of the tangent lines.
<Nekohime> If you tilt your head 90degrees left or right, it's no longer horizontal.
<Nekohime> mind = blown
<Eddie> You must prove this now.
<Eddie> Well not 'now' but if you wish :p
<John_Wolfe_> well.
<John_Wolfe_> let P be a point belongnin to the parabola.
<John_Wolfe_> call P' the symmetrical point with respect to the parabola's axis.
<John_Wolfe_> also le V be the vertex of the parabola.
<John_Wolfe_> *let
<Eddie> Aye.
<John_Wolfe_> call t the tangent line at P, and t' the tangent line at P'
<John_Wolfe_> ffs let's say P \neq V
<John_Wolfe_> so there is only one point in the intersection t \cap t'
<John_Wolfe_> call it
<John_Wolfe_> mmmh
<John_Wolfe_> Q
<John_Wolfe_> so.
<John_Wolfe_> you have to prove that PV \equiv VQ
<Eddie> well the y distance of PV
<John_Wolfe_> nah y messes with coordinates.
<John_Wolfe_> can we solve this with analytical arguments or have we to stick to euclidean geometry?
<Eddie> I solved it by playing with the equations I think.
<Eddie> It was a direct proof.
<John_Wolfe_> because im sniffing a geometric argument out
<Eddie> Mind you it was... 8 years ago
<John_Wolfe_> but it's a hassle to get
<Eddie> Somewhat
<Eddie> the only purpose for this to me was to find a formula for the area of the triangle formed
<John_Wolfe_> So
<Eddie> if you close it off between the tangent points
<John_Wolfe_> let's say that the parabola is y = ax^2
<Eddie> okay
<John_Wolfe_> any other form relates to this one by traslation.
<Eddie> aye
<John_Wolfe_> so if we prove the argument here, it holds for any parabola.
<John_Wolfe_> let's take P to have coordinates (x_P,y_P)
<Eddie> Most likely :3
<John_Wolfe_> obviously, y_P = a(x_P)^2
<John_Wolfe_> but who cares
<John_Wolfe_> the coordinates of P' are (-x_P, y_P)
<Eddie> Yes sir
<John_Wolfe_> because the axis of the parabola is now the y axis, x=0
<John_Wolfe_> so the tangent lines t and t' have equazions
<John_Wolfe_> t: y - y_p = 2ax_P (x - x_P)
<John_Wolfe_> and t': y - y_P = -2ax_P (x + x_P)
<John_Wolfe_> these two meet at (0, y_P - 2a(x_P)^2)
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<John_Wolfe_> which is really (0, - a(x_P)^2)
<John_Wolfe_> so the distance VQ is \abs{-a(x_P)^2}
<Solonarv> John_Wolfe_, try not to use latex here
<Solonarv> the notation is not very readable
<John_Wolfe_> uff
<John_Wolfe_> |-a(x_P)^2|
<Solonarv> :3
<John_Wolfe_> whcih is really |a| (x_P)^2 cus no complex numbers are involved.
<John_Wolfe_> now we compute the distance VP
<Solonarv> You can assume a > 0 without loss of generality and drop the '|.|
<John_Wolfe_> since V = (0,0)
<John_Wolfe_> (because the parabola is y = ax^2)
<John_Wolfe_> the distance PV is just the square root of (x_P^2 + y_P^2)
<John_Wolfe_> ah
<John_Wolfe_> im a moron :D
<John_Wolfe_> i did some error :/
<John_Wolfe_> or not?
<Eddie> I'm working at the same time and I agree with Solonarv lol sorry x3
<Solonarv> I barely even said anything :P
<John_Wolfe_> So the theorem is wrong.
<John_Wolfe_> Either me or you misunderstood that thing.
<John_Wolfe_> because QV turn out to be a (x_P)^2
<John_Wolfe_> *|a|(x_P)^2
<John_Wolfe_> and PV is \sqrt{1+a^2}(x_P)
<John_Wolfe_> where \sqrt{@} denotes the square root of @
<Solonarv> mh
<John_Wolfe_> are you sure it's the vertex?
<John_Wolfe_> because if that was the _focus_ of the parabola, the theorem would hold
<Eddie> Does it?
<Eddie> Yes vertex but if you find for the focus too...
<Solonarv> It certainly won't hold for both.
<John_Wolfe_> can i send files over?
<Solonarv> there's DCC but that doesn't work very well through routers
<John_Wolfe_> anyhows.
<Solonarv> just slap it on dropbox or google drive or something and drop a link here.
<John_Wolfe_> Well
<John_Wolfe_> the proof is identical
<Solonarv> ...mostly
<John_Wolfe_> but the focus has coordinates (0, (1+4a)/(4a))
<John_Wolfe_> the equations for t and t' still hold
<John_Wolfe_> so Q is still the point (0, -a(x_P)^2)
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<John_Wolfe_> but now the distance PQ would be |a(x_P)^2 + (1+4a)/(4a)|
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<John_Wolfe_> sorry
<John_Wolfe_> the distance QF would be |a(x_P)^2 + (1+4a)/(4a)|
<John_Wolfe_> F being the focus
<John_Wolfe_> and th distance PF would also be |a(x_P)^2 + (1+4a)/(4a)|
<John_Wolfe_> so yes
<John_Wolfe_> the theorem holds if you replace vertex with focus.
<Solonarv> TL;DR the theorem does not hold for the vertex, but it does hold for the focus
<John_Wolfe_> bascially