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Fitting an inverse quadratic curve

  1. Apr 10, 2006 #1
    If I know 2 points in the +x,+y quadrant, and I want to find the

    Code (Text):

    f(x) =    a
    curve that passes through both points (a and b are constants).

    This is probably either really simple or impossible. :rolleyes:
  2. jcsd
  3. Apr 10, 2006 #2
    Yes... this is either really simple or impossible.

    Let [tex]k=\frac{a}{b^2}[/tex]. Since [tex]f(x) = \frac{a}{(bx)^2} = \frac{a}{b^2x^2}=\frac{\frac{a}{b^2}}{x^2}[/tex], we can say [tex]f(x) = \frac{k}{x^2}[/tex].

    If the two points are given by [tex]P_1 = (x_1,y_1)[/tex] and [tex]P_2 = (x_2,y_2)[/tex], then they both lie on such a curve if and only if [tex]y_1 = \frac{k}{x_1^2}[/tex] and [tex]y_2 = \frac{k}{x_2^2}[/tex]. That is, if [tex]x_1^2y_1=x_2^2y_2[/tex]. If this is the case, then let [tex]k=x_1^2y_1[/tex], and translate to a and b. Since we've got two variables and one equation, we're free to choose one of them. For ease, let's make b=1. Then a=k. So, [tex]f(x)=\frac{k}{x^2}[/tex].
  4. Apr 11, 2006 #3
    Though I can't think how I missed that but I do feel quite stupid now :yuck:
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