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

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• 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

> > 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
> >
> >
> > > 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,
> > > >
> > > > 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
> > >
> > >
> >
> >
> > ------------------------------------
> >
> >
> >
> >     Hyacinthos-fullfeatured@yahoogroups.com
> >
> >
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
• 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}

> 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
>
>
>
> > > 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
> > >
> > >
> > > > 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,
> > > > >
> > > > > 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
> > >
> > >
> >
> >
> > ------------------------------------
> >
> >
> >
> > Hyacinthos-fullfeatured@yahoogroups.com
> >
> >
>
>
> ------------------------------------
>
>
>
>
>

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

[Non-text portions of this message have been removed]
• ... 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

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
• 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