Re: Geometry with parabola...

On Apr 24, 4:25 am, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
bill a écrit :



On Apr 18, 2:17 am, Tim Little <t...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
On 2008-04-18, mina_world <mina_wo...@xxxxxxxxxxx> wrote:

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 ?

I'd use coordinate geometry and a little calculus to show it, since
theparabolaand relvant lines have a very simple representation in
that case. And indeed, such a representation for theparabolaappears
to be specified in the problem.

It can also be done in pure geometry, with a suitable definition of
aparabola. I can't find such a proof that is anywhere near as easy as the
previous one, though.

- Tim

Can you be more specific about the proof to which you
refer?

I gave some proofs in my other posts.

Tim L referred to an 'easier' proof. My apologizes, but I don't
consider
your "proof" as 'easy'!

I am still hoping to hear from Tim.


Sure, calculating with algebra (especially doing the job with a CAS ;-)
is easy, but much less interesting, that's why there is so few interest
nowadays in geometry : it is reduced to calculation.

Also the proof depends on the skill and knowledge.
With projective geometry it is "obvious" :
direction of L is conjugate to axis ofparabola.

L can have any direction from nearly horizontal to vertical. The axis
of the
parabola is horizontal? Its not "obvious" to me!

Midpoints of all secants parallel to L are then on a line conjugate
of L, that is parallel to axis.


More explicit : consider a pencil of lines intersecting on a point U
at infinity (that is parallel lines in an affine map).
The line of all points at infinity is tangent toparabola(definition
ofparabola= conic section touching the line at infinity).
So we have a pencil of lines from point U (at infinity).
One of them is the line at infinity, touchingparabolaat point x, at
infinity in direction of axis.
One of these lines is tangent toparabolain P (L on mina's drawing)
Others are any secant AB (on mina's drawing).

The set of all points conjugate of U, that is of points M with
(A,B,U,M) harmonic, is a line : the polar of U.
Two points of this polar are P and x.
Hence the polar is the parallel to axis from P.
The conjugate M of U from AB is midpoint of AB when U at infinity.
QED.

Without general knowledge about conic sections in projective plane,
all proofs seem hard to find from scratch.

Scratch - yes! But a relatively simple proof based on analytical
geometry
was easy for me to find. There is a modicum of "calculation" involved;
but I
did not find that depressingly boring.

Just for my own comfort, is y = sqrt(x) actually the equation of a
parabola?

The construction of intersection points with any line andparabola
that I gave in my previous posts seems natural to get properties of
these intersection points (and their midpoint).

I feel compelled to critique your "construction". Yet, I am too lazy
to attempt
to find it myself. Would you please accommodate me and identify
appropriate
posting; hopefully by date and time?

Bill J

Regards.

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

.

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