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15355Re: [EMHL] line/conic intersection

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  • Luís Lopes
    Jul 2 10:15 AM
      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 the message below.

      To help me (you) in the future: look for msgs
      15338-342, 15344, 15352.
      I am still experiecing problems sending/receiving msgs.
      Sorry for double postings.

      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.


      --- In Hyacinthos@yahoogroups.com, "Francois Rideau" <francois.rideau@...>

      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...

      MSN Busca: fácil, rápido, direto ao ponto. http://search.msn.com.br
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