Consider five points in general position (every tripple of them non collinear). There is a unique proper conic passing through all of them.
Select four out of the five {A,B,C,D} and consider the corresponding projective base, with D=A+B+C. The equations of the lines in these coordinates are: x = 0, line BC, y = 0, line CA, z = 0, line AB, y - z = 0, line AD, z - x = 0, line BD, x - y = 0, line CD. Besides pairs of lines (AB, CD) and (AC, BD) define degenerate conics (product of lines) with equations: z*(x-y) = 0 and y*(z-x) = 0, both conics passing through the four points {A,B,C,D}. Hence all the conics passing through these four points are of the form (see ProjectiveCoordinates.html ): k*z*(x-y) + k'*y*(z-x) = 0. Should the conic pass also through E(x0,y0,z0), then k*z0*(x0-y0) + k'*y0*(z0-x0) = 0, and by assumption E is not on the lines participating, hence the coefficients of k, k' are non zero, thus completely determining their ratio and the conic. Besides the conic being then -k'*x*y + (k'-k)*y*z + k*z*x = 0, corresponds to the matrix:
whose determinant is -(1/4)k'*(k'-k)*k, which is non-zero. Thus the conic is proper. In particular the coefficient k'-k is non-zero since otherwise E would satisfy x*(z-y) = 0 i.e. would be on a line from the other four. Remark The fact that any conic passing through {A,B,C,D} is a linear combination of z*(x-y) and y*(z-x) is equivalent to the fact that the four equations resulting from the general equation of the conic ax2+by2+cz2+2dxy+2eyz+2fzx=0, by substituting the coordinates of {A,B,C,D} build a homogeneous linear system (in a,b,c, ...) with the matrix of rank four:
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Conics through five points ![[0_0]](ConicsThroughFivePts_files/ConicsThroughFivePts_0_0_0.png)
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