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Locus of zeroes

  1. Dec 30, 2007 #1
    Given the real function [tex]f(x,y) = x^2 - 4xy + 3y^2[/tex], the equation f(x,y) = 0 shows in a graph as 2 straight lines, y=x and y=x/3. For pairs (x,y) between the lines, f(x,y) < 0; for (x,y) outside the lines, f(x,y) > 0.

    It is easy to prove the above, by substituting y=mx in the equation f(x,y)=0 and finding the values of m (giving either 1 or 1/3). All the behavior above follows by playing with m.

    The question is, why straight lines? What's the theory for choosing y=mx as the appropriate substitution?
  2. jcsd
  3. Dec 30, 2007 #2


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    that equation is reducible, i.e. x^2 - 4xy +3y^2 = (x-y)(x-3y), and both factors are linear, i.e.describe lines. so the zero locus of the product is the union of the zero loci of those 2 lines.

    one way to "know" it is going to consist of lines is to note that it is homogeneous of degree 2, i.e. all terms have total degree 2, so the origin is a double point, and the only degree 2 curve with a double point is the union of 2 lines.

    thus in fact any curve with equation aX^2 + bXY + c Y^2 factors into 2 linear equations (over the complex numbers).

    this is the very simplest beginning example of the algebraic geometry of plane curves. take a look at walker,
    algebraic plane curves, for an elementary and concrete, yet deep and expert discussion.
    Last edited: Dec 30, 2007
  4. Dec 30, 2007 #3
    Hey, thanks. I have to read (or google) some things to fully understand that, but it's a good start.

    I take you mean "the product of two linear factors, (ax+by)", when you say "the union of 2 lines", here,
    (so they are actually two lines having a zero intercept, i.e. both passing through the origin).

    One little follow-up: the locus of f(x,y)=c, for c a positive constant, is an ellipse, no matter how small c is. How come this degrades to two lines when =0? I'm trying to visualize horizontal slices on a 3D shape, but this beats me.
  5. Dec 30, 2007 #4
    Oh, sorry, forget this last paragraph. Since B^2 > 4AC, this is an hyperbola, not an ellipse. I was confused due to a graph I made a few days ago, but in that case the cross-term coefficient B was -3, not -4.
    Last edited: Dec 30, 2007
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