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Re: polar conjugate of Lucas cubic

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  • Francisco Javier
    For polar conjugate of Thomson cubic, pole = X2052, pivot = X264 and for polar conjugate of Darboux cubic, pole = 2052, pivot = isotomic conjugate of X1073.
    Message 1 of 5 , Mar 28, 2013
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      For polar conjugate of Thomson cubic, pole = X2052, pivot = X264 and for polar conjugate of Darboux cubic, pole = 2052, pivot = isotomic conjugate of X1073.

      --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
      >
      > Thanks, Francisco Javier!
      >
      > I suppose less interesting would be the polar conjugates of the Thompson cubic (passing through X2, X4, X92, X253, X264, X273, X318, X342, X2052) or the Darboux cubic (passing through X2, X92, X459, X2052).
      >
      > Best regards,
      > Randy
      >
      > --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@> wrote:
      > >
      > > Dear Randy,
      > >
      > > It is the isogonal pivotal cubic with pivot G and pole X393.
      > >
      > > Its equation is
      > >
      > > a^8 x^2 y - 4 a^6 b^2 x^2 y + 6 a^4 b^4 x^2 y - 4 a^2 b^6 x^2 y + b^8 x^2 y - 2 a^4 c^4 x^2 y + 4 a^2 b^2 c^4 x^2 y - 2 b^4 c^4 x^2 y + c^8 x^2 y - a^8 x y^2 + 4 a^6 b^2 x y^2 - 6 a^4 b^4 x y^2 + 4 a^2 b^6 x y^2 - b^8 x y^2 + 2 a^4 c^4 x y^2 - 4 a^2 b^2 c^4 x y^2 + 2 b^4 c^4 x y^2 - c^8 x y^2 - a^8 x^2 z + 2 a^4 b^4 x^2 z - b^8 x^2 z + 4 a^6 c^2 x^2 z - 4 a^2 b^4 c^2 x^2 z - 6 a^4 c^4 x^2 z + 2 b^4 c^4 x^2 z + 4 a^2 c^6 x^2 z - c^8 x^2 z + a^8 y^2 z - 2 a^4 b^4 y^2 z + b^8 y^2 z + 4 a^4 b^2 c^2 y^2 z - 4 b^6 c^2 y^2 z - 2 a^4 c^4 y^2 z + 6 b^4 c^4 y^2 z - 4 b^2 c^6 y^2 z + c^8 y^2 z + a^8 x z^2 - 2 a^4 b^4 x z^2 + b^8 x z^2 - 4 a^6 c^2 x z^2 + 4 a^2 b^4 c^2 x z^2 + 6 a^4 c^4 x z^2 - 2 b^4 c^4 x z^2 - 4 a^2 c^6 x z^2 + c^8 x z^2 - a^8 y z^2 + 2 a^4 b^4 y z^2 - b^8 y z^2 - 4 a^4 b^2 c^2 y z^2 + 4 b^6 c^2 y z^2 + 2 a^4 c^4 y z^2 - 6 b^4 c^4 y z^2 + 4 b^2 c^6 y z^2 - c^8 y z^2 = 0.
      > >
      > > Best regards,
      > >
      > > Francisco Javier.
      > >
      > > --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
      > > >
      > > > Dear friends,
      > > >
      > > > What curve results from taking the polar conjugate of the Lucas cubic? The curve includes X(2), X(4), X(278), X(281), X(393), X(459), X(1249), X^-1(1433), the complement of X(3346), and is invariant under X(2)-Ceva conjugation.
      > > >
      > > > By 'polar conjugate', I mean the conjugation I introduced last November, defined equivalently as:
      > > >
      > > > The polar conjugate of P =
      > > > The pole, wrt the polar circle, of the trilinear polar of P;
      > > > The trilinear pole of the polar of P wrt the polar circle;
      > > > The X(48)-isoconjugate of P
      > > >
      > > > This is the isoconjugation that swaps G and H (B.Gibert, Hyacinthos #21304).
      > > >
      > > > If P = p:q:r (trilinears), then the polar conjugate of P has trilinears:
      > > >
      > > > (csc 2A)/p : (csc 2B)/q : (csc 2C)/r
      > > >
      > > > Best regards,
      > > > Randy
      > > >
      > >
      >
    • rhutson2
      Thanks again, Francisco! Interesting that the isotomic conjugate of X1073 is also the polar conjugate of X64. Randy ... for polar conjugate of Darboux cubic,
      Message 2 of 5 , Mar 28, 2013
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        Thanks again, Francisco! Interesting that the isotomic conjugate of
        X1073 is also the polar conjugate of X64.

        Randy

        --- In Hyacinthos@yahoogroups.com, "Francisco Javier"
        <garciacapitan@...> wrote:
        >
        > For polar conjugate of Thomson cubic, pole = X2052, pivot = X264 and
        for polar conjugate of Darboux cubic, pole = 2052, pivot = isotomic
        conjugate of X1073.
        >
        > --- In Hyacinthos@yahoogroups.com, "rhutson2" rhutson2@ wrote:
        > >
        > > Thanks, Francisco Javier!
        > >
        > > I suppose less interesting would be the polar conjugates of the
        Thompson cubic (passing through X2, X4, X92, X253, X264, X273, X318,
        X342, X2052) or the Darboux cubic (passing through X2, X92, X459,
        X2052).
        > >
        > > Best regards,
        > > Randy
        > >
        > > --- In Hyacinthos@yahoogroups.com, "Francisco Javier"
        <garciacapitan@> wrote:
        > > >
        > > > Dear Randy,
        > > >
        > > > It is the isogonal pivotal cubic with pivot G and pole X393.
        > > >
        > > > Its equation is
        > > >
        > > > a^8 x^2 y - 4 a^6 b^2 x^2 y + 6 a^4 b^4 x^2 y - 4 a^2 b^6 x^2 y +
        b^8 x^2 y - 2 a^4 c^4 x^2 y + 4 a^2 b^2 c^4 x^2 y - 2 b^4 c^4 x^2 y +
        c^8 x^2 y - a^8 x y^2 + 4 a^6 b^2 x y^2 - 6 a^4 b^4 x y^2 + 4 a^2 b^6 x
        y^2 - b^8 x y^2 + 2 a^4 c^4 x y^2 - 4 a^2 b^2 c^4 x y^2 + 2 b^4 c^4 x
        y^2 - c^8 x y^2 - a^8 x^2 z + 2 a^4 b^4 x^2 z - b^8 x^2 z + 4 a^6 c^2
        x^2 z - 4 a^2 b^4 c^2 x^2 z - 6 a^4 c^4 x^2 z + 2 b^4 c^4 x^2 z + 4 a^2
        c^6 x^2 z - c^8 x^2 z + a^8 y^2 z - 2 a^4 b^4 y^2 z + b^8 y^2 z + 4 a^4
        b^2 c^2 y^2 z - 4 b^6 c^2 y^2 z - 2 a^4 c^4 y^2 z + 6 b^4 c^4 y^2 z - 4
        b^2 c^6 y^2 z + c^8 y^2 z + a^8 x z^2 - 2 a^4 b^4 x z^2 + b^8 x z^2 - 4
        a^6 c^2 x z^2 + 4 a^2 b^4 c^2 x z^2 + 6 a^4 c^4 x z^2 - 2 b^4 c^4 x z^2
        - 4 a^2 c^6 x z^2 + c^8 x z^2 - a^8 y z^2 + 2 a^4 b^4 y z^2 - b^8 y z^2
        - 4 a^4 b^2 c^2 y z^2 + 4 b^6 c^2 y z^2 + 2 a^4 c^4 y z^2 - 6 b^4 c^4 y
        z^2 + 4 b^2 c^6 y z^2 - c^8 y z^2 = 0.
        > > >
        > > > Best regards,
        > > >
        > > > Francisco Javier.
        > > >
        > > > --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
        > > > >
        > > > > Dear friends,
        > > > >
        > > > > What curve results from taking the polar conjugate of the Lucas
        cubic? The curve includes X(2), X(4), X(278), X(281), X(393), X(459),
        X(1249), X^-1(1433), the complement of X(3346), and is invariant under
        X(2)-Ceva conjugation.
        > > > >
        > > > > By 'polar conjugate', I mean the conjugation I introduced last
        November, defined equivalently as:
        > > > >
        > > > > The polar conjugate of P =
        > > > > The pole, wrt the polar circle, of the trilinear polar of P;
        > > > > The trilinear pole of the polar of P wrt the polar circle;
        > > > > The X(48)-isoconjugate of P
        > > > >
        > > > > This is the isoconjugation that swaps G and H (B.Gibert,
        Hyacinthos #21304).
        > > > >
        > > > > If P = p:q:r (trilinears), then the polar conjugate of P has
        trilinears:
        > > > >
        > > > > (csc 2A)/p : (csc 2B)/q : (csc 2C)/r
        > > > >
        > > > > Best regards,
        > > > > Randy
        > > > >
        > > >
        > >
        >
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