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15353Re: [EMHL] line/conic intersection

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  • qedtexte
    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.
      > Please, gimme exactly your construction with your homothecy!

      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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