From: <HIDDEN>
Newsgroups: comp.soft-sys.matlab
Subject: Re: solving a system of multi-variable polynomial equations in mupad
Date: Tue, 16 Oct 2012 22:42:21 +0000 (UTC)
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"Roger Stafford" wrote in message <k5i3g7$ie5$>...
> "Daniel" wrote in message <k5hrmo$jtt$>...
> > .....  But why can't the K be dropped? ......
> - - - - - - - - - -
>   The purpose of K lies in deriving an algebraic identity with the expression
>  A*x^2+2*B*x*y+C*y^2
> regardless of whether it is to be equal to zero or not.  For that you do need K.  Otherwise such equations as
>  -2*B = K*sin(a+b)
>  C-A = K*cos(a+b)
>  2*sqrt(B^2-A*C) = K*sin(a-b)
>  C+A = K*cos(a-b)
> would all be inherently false with K absent and with arbitrary A, B, and C.  You will notice however that K does drop out of the picture with the two 'atan2' operations and does not play a role in solving the two line angles.  So, yes, K is definitely needed in the theory behind the matlab code but is never explicitly solved for in the code because the lines' angles are independent of its value.
>   (It is misleading to say that K scales the points on the lines.)
> Roger Stafford

Hi Rodger

Thanks again for your helpful response. I see now the use of K - I forgot the important fact that sin(a+b) and cos(a+b) are in the range [-1,1]! So yes I see that -2*B = sin(a+b) will always be false if -2*B is not in the range [-1,1], and the K can be chosen to make the equality work. I think I convinced myself it must be superfluous when it dissappered from the subsequent analysis.

I have finally got on to your message number 5. You give some very nice geometrical interpretations to the solution space of the A1-B1-C1, and A2-B2-C2 points given that the six x-y points lie somewhere on the two degenerate conics defined by equations 1/4/6 and 13/16/18 (for the coefficients A1,B1,C1), and equations 7/10/12 and 19/22/24 (for the coefficients A2,B2,C2) - here the equation numbers relate to the 24 equations I give in my original post.

I hope I have got this right by now when I say that each of the two degenerate conics can either be the point ellipse, the x-axis or y-axis (if the conic is a degenerate parabola), or two crossed lines through the origin (if the conic is a degenerate hyperbola). So the scenarios you outline correspond to the following situations;

1) all six x-y points are located at the origin     -  both conics are the point ellipse
2) all six x-y points lie along one line through the origin - both conics are the same degenerate parabola
3) 3 x-y points lie along one line through the origin whilst the other three lie on another line through the origin    -  both conics are the x-axis and y-axis respectively, or are the same hyperbola. 

Your geometric interpretations of the A-B-C solution space, i.e. the whole of R^3, a plane through the origin, or points in R^3 in the direction of a cross-product, show that, given the x-y points are located as discussed above, then there are an infinite number of (A1,B1,C1), and (A2,B2,C2) points that are solutions?

I suppose this says a similar thing as your Matlab code but in reverse - i.e., for the 12 equations using (A1,B1,C1) chosen arbitrarily, we can always find two lines from which to draw an infinite number of x-y points as solutions. If the coefficient differences making up (A2,B2,C2) are chosen so that A1=A2, B1=B2, and C1=C2, then the 12 equations using (A2,B2,C2) define the same conic, and there are an infinite number of solutions.