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If we have a subset in the complex plane $\mathbb{C}^2$ that consists of the following seven points:

$(-1,6), (8,-24), (-\frac{8}{9},\frac{152}{27}), (4,4), (-8,8), (-\frac{512}{16256}, -\frac{16256}{1331}), (-\frac{8}{9}, -\frac{152}{27})$

Then how do we prove that at most $5$ points in the subset are contained in one smooth conic in $\mathbb{C}^2$

user376343
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  • I’d do it by examining the rank of a certain matrix, but that might not be what you’ve learned. What conditions do you know of for a set of points to lie on a conic? – amd Oct 08 '19 at 21:46
  • Yes, I know that if the rank of the matrix is 6, then the only solution is 0 so no such conic exists. And if the rank is 5 then the conic exists. I know I have to compute the rank of a 6x6 matrix that corresponds to a system on linear equations. But what points would I use for this all 7 or just a certain 6 or 5, @amd –  Oct 08 '19 at 23:02
  • We are also using python to compute this –  Oct 08 '19 at 23:03
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    You only need "no three collinear" to get a smooth conic from five distinct points. Calling the points $A, \ldots, G$: $A,B, C$ lie on $10x+3y=8$ and $C,D,E$ lie on $x+3y=16.$ The ten conics $ABDEF, ABDEG, ABDFG, ABEFG, ACDFG, ACEFG, ADEFG, BCDFG, BCEFG, BDEFG$ you see in the picture are the only smooth conics through the seven points since the eleven other five element subsets of the seven contain collinear points. – Jan-Magnus Økland Oct 09 '19 at 07:36
  • therefore the statement at most 5 points in the subset are contained in one smooth conic is true? –  Oct 09 '19 at 12:08

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