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Re: [EMHL] Re: A line and a conic

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  • Nikolaos Dergiades
    Sorry I repeat my correction Sorry, I wrote ... the correct is  -Det. I forgot to say that if the point (x : y : z) is on the conic then the second
    Message 1 of 12 , Feb 18, 2013
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      Sorry I repeat my correction


      Sorry,
      I wrote
      > By the way this T you used is the Sqrt of the Det
      the correct is  -Det.
      I forgot to say that if the point (x : y : z)
      is on the conic then the second intersection of the line
      with
      the conic is
      ( yz(g(xv-yu)^2 + h(zu-xw)^2) : zx(h(yw-zv)^2 + f(xv-yu)^2)
      : . . .)

      where u = Q - R, v = R - P, w = P - Q.

      Best regards
      Nikos Dergiades



      > > Dear Francisco,
      > > thank you very much.
      > > I will study your solution.
      > > I was expecting to have some cyclicity.
      > > I mean if in (f,g,h), (p,q,r), (P,Q,R)
      > > there is a cyclicity relative to a,b,c
      > > then I wanted to have as point of
      > > intersection a point ( F(a,b,c):F(b,c,a):F(c,a,b) ).
      > > Is this expectation false?
      > > For example I found the following solution:
      > > Let
      > > -(h P Q + h Q^2 - p P R - 2 p Q R - q Q R + g R^2 + r
      > R^2)/R
      > > = F(a,b,c)
      > > If
      > > (h P^2 + h P Q - p P R - 2 P q R - q Q R + f R^2 + r
      > R^2)/R
      > > = F(b,c,a)
      > > then the intersection points are
      > > ( F(a,b,c) + T : F(b,c,a) + T : F(c,a,b) + T )
      > > ( F(a,b,c) - T : F(b,c,a) - T : F(c,a,b) - T )
      > > But this fine result is under condition and does not
      > hold
      > > in all cases.
      > > By the way this T you used is the Sqrt of the Det
      > > 0 P Q R
      > > P f r q
      > > Q r g p
      > > R q p h
      > >
      > > Best regards
      > > Nikos Dergiades
      > >
      > >
      > > > They are the points:
      > > >
      > > > {h Q^2 - 2 p Q R + g R^2,
      > > >  -h P Q + p P R + q Q R - r R^2 - R T,
      > > >   p P Q - q Q^2 - g P R + Q r R + Q T},
      > > >
      > > > {h Q^2 - 2 p Q R + g R^2,
      > > >  -h P Q + p P R + q Q R - r R^2 + R T,
      > > >   p P Q - q Q^2 - g P R + Q r R - Q T}}
      > > >
      > > > where T is the square root of
      > > >
      > > > (p^2 - g h) P^2 + (q^2 - f h) Q^2 + (r^2 - f g)
      > R^2 +
      > > (2 f p
      > > > -
      > > >     2 q r) Q R + (2 h r - 2 p q) P Q + (2 g q -
      > 2
      > > > p r) P R
      > > >
      > > > Mathematica says.
      > > >
      > > > Best regards,
      > > >
      > > > Francisco Javier.
      > > >
      > > >
      > > >
      > > > --- In Hyacinthos@yahoogroups.com,
      > > > Nikolaos Dergiades <ndergiades@...> wrote:
      > > > >
      > > > > Dear friends
      > > > > if we know the line
      > > > > Px + Qy + Rz = 0
      > > > > and the conic
      > > > > fxx + gyy + hzz + 2pyz + 2qzx + 2rxy = 0
      > > > > is there a formula giving the barycentric
      > > > > coordinates of their intersection points?
      > > > > Best regards
      > > > > Nikos Dergiades
      > > > >
      > > >
      > > >
      > > >
      > > >
      > > > ------------------------------------
      > > >
      > > > Yahoo! Groups Links
      > > >
      > > >
      > > >     Hyacinthos-fullfeatured@yahoogroups.com
      > > >
      > > >
      > >
      > >
      > > ------------------------------------
      > >
      > > Yahoo! Groups Links
      > >
      > >
      > >     Hyacinthos-fullfeatured@yahoogroups.com
      > >
      > >
      >
      >
      > ------------------------------------
      >
      > Yahoo! Groups Links
      >
      >
      >     Hyacinthos-fullfeatured@yahoogroups.com
      >
      >
    • Francisco Javier García Capitán
      Dear Nikolaos, my solution is non symmetric because I use Mathematica s Solve in this way: Simplify[Solve[{line=0,conic=0},{y,z}], x 0] As a result we get the
      Message 2 of 12 , Feb 18, 2013
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        Dear Nikolaos, my solution is non symmetric because I use Mathematica's
        Solve in this way:

        Simplify[Solve[{line=0,conic=0},{y,z}], x>0]

        As a result we get the points

        ptSa= {h Q^2 - 2 p Q R + g R^2, -h P Q + p P R + q Q R - r R^2 - R T,
        p P Q - q Q^2 - g P R + Q r R + Q T}

        ptTa= {h Q^2 - 2 p Q R + g R^2, -h P Q + p P R + q Q R - r R^2 + R T,
        p P Q - q Q^2 - g P R + Q r R - Q T}

        where T is the square root of

        -g h P^2 + p^2 P^2 - 2 p P q Q - f h Q^2 + q^2 Q^2 + 2 h P Q r +
        2 (g P q + f p Q - (p P + q Q) r) R + (-f g + r^2) R^2 .

        This gives a result easier to simplify.

        In order to symetrize these expressions we can use the sum of the
        permutations of this solution

        You can use this function included in Baricentricas.nb:

        Simetrizar[ptP_]:=Simplificar[Total[Table[Nest[PermutarTerna,ptP,n],{n,0,2}]]]

        In order to that this works properly you have to change ReglaDePermutacion
        definition because the default value doesn't include p, q, r, P, Q, R , f,
        g, h:

        ReglaDePermutacion = {p -> q, q -> r, r -> p, P -> Q, Q -> R, R -> P,
        f -> g, g -> h, h -> f};

        For example,

        ptS = Factor[Simetrizar[ptSa]]

        gives

        {-h P Q + p P Q + h Q^2 - q Q^2 - g P R + p P R - 2 p Q R + q Q R +
        Q r R + g R^2 - r R^2 - Q T + R T,
        h P^2 - p P^2 - h P Q + P q Q + p P R - 2 P q R - f Q R + q Q R +
        P r R + f R^2 - r R^2 + P T - R T,
        g P^2 - p P^2 + p P Q + P q Q + f Q^2 - q Q^2 - 2 P Q r - g P R -
        f Q R + P r R + Q r R - P T + Q T}





        2013/2/18 Nikolaos Dergiades <ndergiades@...>

        > Sorry I repeat my correction
        >
        >
        > Sorry,
        > I wrote
        > > By the way this T you used is the Sqrt of the Det
        > the correct is -Det.
        > I forgot to say that if the point (x : y : z)
        > is on the conic then the second intersection of the line
        > with
        > the conic is
        > ( yz(g(xv-yu)^2 + h(zu-xw)^2) : zx(h(yw-zv)^2 + f(xv-yu)^2)
        > : . . .)
        >
        > where u = Q - R, v = R - P, w = P - Q.
        >
        > Best regards
        > Nikos Dergiades
        >
        >
        >
        > > > Dear Francisco,
        > > > thank you very much.
        > > > I will study your solution.
        > > > I was expecting to have some cyclicity.
        > > > I mean if in (f,g,h), (p,q,r), (P,Q,R)
        > > > there is a cyclicity relative to a,b,c
        > > > then I wanted to have as point of
        > > > intersection a point ( F(a,b,c):F(b,c,a):F(c,a,b) ).
        > > > Is this expectation false?
        > > > For example I found the following solution:
        > > > Let
        > > > -(h P Q + h Q^2 - p P R - 2 p Q R - q Q R + g R^2 + r
        > > R^2)/R
        > > > = F(a,b,c)
        > > > If
        > > > (h P^2 + h P Q - p P R - 2 P q R - q Q R + f R^2 + r
        > > R^2)/R
        > > > = F(b,c,a)
        > > > then the intersection points are
        > > > ( F(a,b,c) + T : F(b,c,a) + T : F(c,a,b) + T )
        > > > ( F(a,b,c) - T : F(b,c,a) - T : F(c,a,b) - T )
        > > > But this fine result is under condition and does not
        > > hold
        > > > in all cases.
        > > > By the way this T you used is the Sqrt of the Det
        > > > 0 P Q R
        > > > P f r q
        > > > Q r g p
        > > > R q p h
        > > >
        > > > Best regards
        > > > Nikos Dergiades
        > > >
        > > >
        > > > > They are the points:
        > > > >
        > > > > {h Q^2 - 2 p Q R + g R^2,
        > > > > -h P Q + p P R + q Q R - r R^2 - R T,
        > > > > p P Q - q Q^2 - g P R + Q r R + Q T},
        > > > >
        > > > > {h Q^2 - 2 p Q R + g R^2,
        > > > > -h P Q + p P R + q Q R - r R^2 + R T,
        > > > > p P Q - q Q^2 - g P R + Q r R - Q T}}
        > > > >
        > > > > where T is the square root of
        > > > >
        > > > > (p^2 - g h) P^2 + (q^2 - f h) Q^2 + (r^2 - f g)
        > > R^2 +
        > > > (2 f p
        > > > > -
        > > > > 2 q r) Q R + (2 h r - 2 p q) P Q + (2 g q -
        > > 2
        > > > > p r) P R
        > > > >
        > > > > Mathematica says.
        > > > >
        > > > > Best regards,
        > > > >
        > > > > Francisco Javier.
        > > > >
        > > > >
        > > > >
        > > > > --- In Hyacinthos@yahoogroups.com,
        > > > > Nikolaos Dergiades <ndergiades@...> wrote:
        > > > > >
        > > > > > Dear friends
        > > > > > if we know the line
        > > > > > Px + Qy + Rz = 0
        > > > > > and the conic
        > > > > > fxx + gyy + hzz + 2pyz + 2qzx + 2rxy = 0
        > > > > > is there a formula giving the barycentric
        > > > > > coordinates of their intersection points?
        > > > > > Best regards
        > > > > > Nikos Dergiades
        > > > > >
        > > > >
        > > > >
        > > > >
        > > > >
        > > > > ------------------------------------
        > > > >
        > > > > Yahoo! Groups Links
        > > > >
        > > > >
        > > > > Hyacinthos-fullfeatured@yahoogroups.com
        > > > >
        > > > >
        > > >
        > > >
        > > > ------------------------------------
        > > >
        > > > Yahoo! Groups Links
        > > >
        > > >
        > > > Hyacinthos-fullfeatured@yahoogroups.com
        > > >
        > > >
        > >
        > >
        > > ------------------------------------
        > >
        > > Yahoo! Groups Links
        > >
        > >
        > > Hyacinthos-fullfeatured@yahoogroups.com
        > >
        > >
        >
        >
        > ------------------------------------
        >
        > Yahoo! Groups Links
        >
        >
        >
        >


        --
        ---
        Francisco Javier García Capitán
        http://garciacapitan.99on.com


        [Non-text portions of this message have been removed]
      • Barry Wolk
        ... There were several replies, including asking for cyclically symmetric coordinates for those intersection points.   The points on the line Px+Qy+Rz=0 can
        Message 3 of 12 , Feb 24, 2013
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          > Dear friends
          > if we know the line
          > Px + Qy + Rz = 0
          > and the conic
          > fxx + gyy + hzz + 2pyz + 2qzx + 2rxy = 0
          > is there a formula giving the barycentric
          > coordinates of their intersection points?
          > Best regards
          > Nikos Dergiades

          There were several replies, including asking for cyclically symmetric coordinates for those intersection points.
           
          The points on the line Px+Qy+Rz=0 can be parametrized as
          (x, y, z)=((Q-R)(QR+t), (R-P)(RP+t), (P-Q)(PQ+t))
           
          Substitute these formulas for x,y,z into the equation of the conic, to get a cyclically symmetric quadratic equation for the parameter t. Its two roots will give the intersection points in the requested form.
          --
          Barry Wolk
        • Chris Van Tienhoven
          Dear friends, Thanks Barry! This brings me to a new problem. I never got into this. Given an equation f(x,y,z). How to convert it to a point in time (t):
          Message 4 of 12 , Feb 24, 2013
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            Dear friends,

            Thanks Barry!
            This brings me to a new problem.
            I never got into this.
            Given an equation f(x,y,z).
            How to convert it to a point in time (t): (x(t) : y(t) : z(t))?

            To make it visual and more appealing. When we have a conical orbit of a planet or comet.
            How can we describe its location as a function of time in barycentric coordinates ?

            Any ideas?

            Chris van Tienhoven


            --- In Hyacinthos@yahoogroups.com, Barry Wolk <wolkbarry@...> wrote:
            >
            > > Dear friends
            > > if we know the line
            > > Px + Qy + Rz = 0
            > > and the conic
            > > fxx + gyy + hzz + 2pyz + 2qzx + 2rxy = 0
            > > is there a formula giving the barycentric
            > > coordinates of their intersection points?
            > > Best regards
            > > Nikos Dergiades
            >
            > There were several replies, including asking for cyclically symmetric coordinates for those intersection points.
            >  
            > The points on the line Px+Qy+Rz=0 can be parametrized as
            > (x, y, z)=((Q-R)(QR+t), (R-P)(RP+t), (P-Q)(PQ+t))
            >  
            > Substitute these formulas for x,y,z into the equation of the conic, to get a cyclically symmetric quadratic equation for the parameter t. Its two roots will give the intersection points in the requested form.
            > --
            > Barry Wolk
            >
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