Re: Geometry with parabola...

mina_world a écrit :
"Philippe 92" <nospam@xxxxxxxxxxxx> wrote in message
news:mn.92e27d8436f76798.22155@xxxxxxxxxx
mina_world a écrit :
Hello teacher~

http://board-2.blueweb.co.kr/user/math565/data/math/paramid.jpg

How do you show it ?
Do you use complex algebra to show it ?
or pure geometry ?


http://board-2.blueweb.co.kr/user/math565/data/math/paramid2.jpg

This is a extension of original problem.
I want to listen your advice about this problem.

Of course we could consider the problem with algebra, I prefer
deductive geometry, so it is :

First of all, proof that the tangent is the perpendicular
bisector. (from the main thread, as you asked for advice also)
I gave a "short proof" at cut-the-knot, and I mentioned that this
suposed the existence of tangent, here is the full proof.
Consider two points P and P' on the parabola. Fig 1. in :
<http://cjoint.com/?esrYbV3Qnl> (valid for 21 days)

By definition they are centers of circles tangent to directrix
and going through focus (equal distance to line and focus).
The radical axis of these two circles (locus of points with same
power to both circles) is perpendicular to center line, that
is perpendicular to PP'.
It intersects the directrix in I.
I is midpoint of HH' : power of I = IH^2 = IH'^2
When P -> P', H' and I both -> H, hence line FI -> FH
and line PP' allways perpendicular to FI, -> perpendicular to
FH from P, hence has a limit, hence existence of a tangent.
And because PF = PH, this perpendicular to FH from P is the
perpendicular bisector. QED.

Now the main properties. Fig 2.
Given a parabola defined by its focus F and directrix (d)
Let a point P on that parabola, with PH = PF by definition.
The tangent in P is the perpendicular bisector of segment FH
in K, midpoint of FH. (well known property now ;-)

Choose any point I on FH and draw a perpendicular to FH in I,
intersecting the parabola in A and B. AB is parallel to tangent in P.
A and B can be constructed (with compas and straightedge) as
follows :
Let C the reflection of F through line AB. I is midpoint of FC.
As AM=AF=AC, M is the center of a circle tangent to (d) and
going through F and C.
Hence HM^2 = HF.HC and M can be constructed, for instance draw
circles with diameters FC and HI, they intersect in X and
HX^2 = HF.HC, then a circle centered in H going through X
intersects (d) in M and N. Perpendiculars from M and N to (d)
intersect line AB in A and B.

Now from that construction we derive :

1) As HM = HN, hence JA = JB (the first exercice)

2) consider a second secant A'B' with corresponding
I', C', M', N', J'.
H, P, K, F fixed, and A'B' parallel to AB.
Then A'J'/AJ = HM'/HM, and PJ'/PJ = KI'/KI (Thales)
but HM^2 = HF.HC = 2.HF.KI
and HM'^2 = HF.HC' = 2.HF.KI'
Finally we get
A'J'^2/AJ^2 = HM'^2/HM^2 = KI'/KI = PJ'/PJ
that is A'J'^2/AJ^2 = PJ'/PJ, the second one.


PS. I just saw the answer by JEMebius, fine method !
That was also my first idea : algebra and change coordinates system
by an affine transform, then obvious equation of parabola in this new
coordinate system Y^2 = m*X is the exact relation to be prooved.

Regards.

--
Philippe C., mail : chephip+news@xxxxxxx
site : http://chephip.free.fr/ (recreational mathematics)


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