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in the article:

http://i.cs.hku.hk/~ykchoi/quadrics/Ellipsoid_Separation.pdf

in example 1

if I set a=3; b=2; c=4; xc=-9; yc=0; zc=0; r=5; and

f(x)=-(x/a^2+1)(x/b^2+1)(x/c^2+1)(x+r^2)+x(x/c^2+1)(x/b^2+1)(xc/a)^2+ x*(x/a^2+1)(x/c^2+1)(yc/b)^2+x*(x/a^2+1)(x/b^2+1)(zc/c)^2=0;

the calculated roots are different from what mention in the article, why?

for the calculation I use this on matlab:

syms a b c r xc yc zc x

a=3; b=2; c=4; xc=9; yc=0; zc=0; r=5;

[cc]=solve(-(x/a^2+1)(x/b^2+1)(x/c^2+1)(x+r^2)+x(x/c^2+1)(x/b^2+1)(xc/a)^2+x*(x/a^2+1)(x/c^2+1)(yc/b)^2+x*(x/a^2+1)(x/b^2+1)(zc/c)^2==0,x)

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  • $\begingroup$ I've retagged your question with geometric tags, but I wasn't sure which ones were the most appropriate. Feel free to retag again. Also, could you clarify what you mean by $E^3$, and maybe tell us in what context you encountered these notions? $\endgroup$ – Arnaud D. Jun 27 '17 at 9:47
  • $\begingroup$ A title of a book would be best. Or link to an article $\endgroup$ – Will Jagy Jun 27 '17 at 17:57
  • $\begingroup$ You might find this question and this article helpful. $\endgroup$ – amd Jun 27 '17 at 18:49
  • $\begingroup$ i.cs.hku.hk/~ykchoi/quadrics/Ellipsoid_Separation.pdf $\endgroup$ – gy ab Jun 27 '17 at 20:56
  • $\begingroup$ this is the article I am trying to understand. I understand why the ellipsoid is X^TAX=0, but I don't understand how they got $\endgroup$ – gy ab Jun 27 '17 at 20:57

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