## 15353Re: [EMHL] line/conic intersection

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• Jul 2, 2007
Construct (with R&C) the intersection of a line <r>
with a conic (parabola) given by its focus <F> and
directrix <d>.

Dear Hyacinthists,

Dear François,

I will reply directly from the archives.
I have never received this message.

===
>The center of C is one of the intersections sought.
>Otherwise perform a homothecy of C of pole P.

I will suppose the "ideal" position of the data.

Let P=d/\r. With P_0 on r as center draw a circle C
tangent to d. Let P_1 be the contact point.
Draw s=(P,F) and let F_1 and F_2 be the intersections
between <s> and <C>. Draw t=(P_0,F_1) and u=(P_0,F_2).
Draw from <F> t'||t and u'||u. The intersections O=r/\t'
and O'=r/\u' are the intersections of <r> and the parabola.

Your solution seems nice but I still have to
understand it.

Friendly,
Luis

--- In Hyacinthos@yahoogroups.com, "Francois Rideau"
<francois.rideau@...> wrote:
>
> Dear Luis
> ===
> I think we have a big problem when <d> and <r> are
> near orthogonal
> ===
> I don't think so. In this case one has P=d/\r and with P
> as pole one makes a homothecy. Draw a circle C with center
> in r and tangent do d. If C goes through F fine.
>
> >
> > That's just the problem of this construction: to find circles
tangent to
> > the directrix <d> and through the focus <F> of which the centers
are on line
> > <r>. These centers are the sought points of intersection. These
circles are
> > also on point <F'> symmetric of <F> wrt line <r>. So these
circles are in a
> > pencil and their intersections with <d> are in involution,
(Desargues
> > theorem). For example, if <Gamma> is a circle in this pencil
cutting <d> in
> > <P> and <Q>, then JP.JQ = JF.JF' = power of J wrt <Gamma>,
(product of
> > oriented segments), where J = FF' /\ <d>. In case of a touching
circle in T,
> > we have JT² = JF.JF'. The circle of center J and radius JT is
orthogonal
> > to all circles of the pencil. So draw any circle of the pencil
and draw
> > segments tangents to this circle through J and so on...
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