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- Path: sparky!uunet!dtix!mimsy!afterlife!rlward1
- From: rlward1@afterlife.ncsc.mil (Robert Ward)
- Newsgroups: sci.math
- Subject: Re: analytic geometry
- Keywords: polynomial curve, locus of mobile point
- Message-ID: <1993Jan7.131324.3836@afterlife.ncsc.mil>
- Date: 7 Jan 93 13:13:24 GMT
- References: <1h9pgaINN97v@uniko.uni-koblenz.de>
- Sender: rlward1@afterlife.ncsc.mil (Robert L. Ward)
- Followup-To: sci.math
- Organization: National Computer Security Center
- Lines: 24
-
- One approach that works is to compute the resultant of the two polynomials
- as functions of one of the variables, yielding a polynomial in the other
- variable. The roots of this polynomial give you one coordinate of the
- intersection points of the two polynomial curves.
-
- For example, if one curve is f(x,y) = a(x) y^2 + b(x) y + c(x) = 0, and the
- other curve is F(x,y) = A(x) y^2 + B(x) y + C(x) = 0, then the resultant is
-
- | a(x) b(x) c(x) 0 |
- | |
- | 0 a(x) b(x) c(x) |
- R(f,F) = | |
- | A(x) B(x) C(x) 0 |
- | |
- | 0 A(x) B(x) C(x) |
-
- and the roots of R(f,F) = 0 as a polynomial in x will give you the x-coord-
- inates of the points of intersection. The y-coordinates may be found by
- substituting the x-coordinates back into f(x_i,y) = 0 and solving for y_i.
-
- This all reduces the problem to one of finding roots of polynomials in one
- variable.
-
- Robert L. Ward
-