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Mathematics Q

  1. Feb 14, 2005 #1
    Prove that the triangle formed by the asymptotes of the curve with equation x^2 - 2y^2 = 4 and any tangent to the curve is of constant area.

    Thanks. :)
     
  2. jcsd
  3. Feb 14, 2005 #2

    matt grime

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    Homework? Well, the answer is do it: find the equations of the asymptotes, and a tangent and voila do it.
     
  4. Feb 14, 2005 #3
    I'm having problems finding the asymptotes.
     
  5. Feb 14, 2005 #4
    Could anyone help me do this Q? Thanks.
     
  6. Feb 14, 2005 #5
    Please could someone show me how to do this Question? I'm having difficulties, because I've barely been taught this chapter and I want to at least see how such a question is answered. Thanks.
     
  7. Feb 14, 2005 #6
    You have to let us know what u have done?
    Post whatever working u have done(even if its wrong its fine), since that would help us to pitch the answer at the right frequency.

    -- AI
     
  8. Feb 14, 2005 #7
    I really don't know where to start, I don't know how to find the asymptotes of the curve, that's a key problem...
     
  9. Feb 14, 2005 #8
    Ok u need to run through the basics a bit then.
    First of all,
    the equation u have is a hyperbola
    Read abt hyperbolas here,
    http://colalg.math.csusb.edu/~devel/precalcdemo/conics/src/hyperbola.html

    Asymptotes are mentioned in this article and its also explained how they are obtained.
    I will leave the rest to you for now. Try to go ahead and solve your original problem. If u are getting stuck again, post your working.

    -- AI
     
  10. Feb 14, 2005 #9

    mathwonk

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    for starters, the asymptotes of xy = 1 seem to be the x and y axes y=0 and x=0, [take derivative of y = 1/x, get -1/x^2, and let x go to infinity, so the slope goes to zero, or let x go to zero, and the slope goes to infinity] so after rotating axes, the asymptotes of uv = (x-y)(x+y) = 1, are probably the lines u=0 and v=0, i.e. x = y and x = -y.

    you might look at this to be sure, as I am allergic this type of thing.
     
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