NetNews Usenet Archive 1992 #20/NN_1992_20.iso/spool/sci/math/11237

Internet Message Format | 1992-09-10 | 1.9 KB

Path: sparky!uunet!mcsun!uknet!pavo.csi.cam.ac.uk!camcus!gjm11
From: [e-mail] (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: curves of degree 2
Message-ID: <[e-mail]>
Date: 10 Sep 92 23:28:06 GMT
References: <[e-mail]>
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In article <[e-mail]>, [e-mail] (Jeffrey J. Kosowsky) writes:

> Can anyone supply a proof for the following problem:
>
> Given any 5 points in the plane, there exists a curve of degree 2
> passing thru all 5 points.
>
> ie: There exist constants a,b,c,d,e,f such that the curve
>     ax^2 + by^2 + cxy + dx +ey +f = 0
> passes thru all 5 points.

The statement that the curve passes through (x,y) is a linear relation
between the six coefficients a,b,c,d,e,f. The statement that it passes
through the five given (x,y) is five linear relations between the six
coefficients a,b,c,d,e,f. In other words it says that an appropriate
linear map from R^6 to R^5 takes (a,b,c,d,e,f) to 0.
The rank of that map is at most 5, so its nullity (dimension of kernel)
is at least 6-5=1, so there is at least a 1-dimensional subspace of
(a,b,c,d,e,f) making all those things true. In particular there is a
sextuple other than (0,0,0,0,0,0) that works. This gives the required
curve.

If the points are (in some appropriate sense) in general position
(i.e. if the relations mentioned above are linearly independent)
then there is exactly one curve of dergee [at most] 2 passing through
the given points.

Incidentally, all the above is just a slightly highbrow way of saying:
Six degrees of freedom, five conditions, therefore one degree of freedom
subject to those conditions.

--
Gareth McCaughan     Dept. of Pure Mathematics & Mathematical Statistics,
[e-mail]  Cambridge University, England.    [Research student]