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Rally easy question (but I still need help)

  1. Feb 21, 2006 #1
    I am trying to draw the contour plot of the following function:

    [tex] f(x,y) = \frac {a(x+y)} {x^2 + y^2 + a^2} [/tex]

    by setting h(x,y) = 1 and completing the square I recieved the following:

    [tex] (x-(a/2))^2 - (y-a/2)^2 = a^2 [/tex]

    surely, this is not a circle, and not a ellipse...but what is it then.

    Also, is there a possibility to draw contour plots rather fast (e.g. with Mathcad)?
     
  2. jcsd
  3. Feb 21, 2006 #2

    quasar987

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    Hyperboleas are caracterised by the implicit relation

    [tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
     
  4. Feb 21, 2006 #3
    but I thought there would be a + in between the square brackets in order for it to be a circle, i.e.

    [tex] (x-(a/2))^2 + (y-a/2)^2 = a^2 [/tex]

    or does it simply not matter?
     
  5. Feb 21, 2006 #4

    quasar987

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    You read my post before I could delete it. I had missed the - sign and though you were confused about the a/2 translation.

    See my new post.
     
  6. Feb 21, 2006 #5
    You can draw contour plots in Mathematica.
     
  7. Feb 21, 2006 #6
    Thank you very much! The entire thing does make so much more sense now! :-)

    So, is a/2 the focal point then (i.e. where the asymptotes cross)?
     
  8. Feb 21, 2006 #7

    quasar987

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    a/2 is still a translation. the assymptotes are y=±bx/a
     
  9. Feb 21, 2006 #8
    Thanks a lot! It's much appreciated!
     
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