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Locus and hyperbolic functions!

  1. May 1, 2013 #1
    show that the locus of the point [itex] \left(\dfrac{a(cosh\theta + 1)}{2cosh\theta},\dfrac{b(cosh\theta - 1)}{2sinh\theta}\right) [/itex]
    has equation x(4y^2 + b^2) = ab^2

    working: http://gyazo.com/4c96af128d0293bce7f18029c2f54b0d
    where have I gone wrong :(
     
  2. jcsd
  3. May 1, 2013 #2

    haruspex

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    Your 5th and 6th lines each have the form <some expression> = 1. The second should have something other than 1 on the right hand side.
     
  4. May 1, 2013 #3

    Mark44

    Staff: Mentor

    I have to say that the image of your work is very poor quality. IMO, you are lucky that haruspex took the time to decipher your hard-to-read work. I'm sure that many others wouldn't bother.
     
  5. May 2, 2013 #4
    see below

    Apologies, I thought it was fine.

    working:

    [itex] x = \dfrac{a(cosh\theta + 1)}{2cosh\theta} [/itex]


    [itex] y = \dfrac{b(cosh\theta - a)}{2sinh\theta} [/itex]

    rearranging x = ... for cosh theta:

    [itex] cosh\theta = \dfrac{a}{2x - a} [/itex]
    subbing this into y = ...

    [itex] y = \dfrac{b(\dfrac{2a-2x}{2x-a})}{2sinh\theta} [/itex]
    rearranging for sinh:

    [itex] sinh\theta = \dfrac{b(a-x)}{y(2x-a)} [/itex]

    now using [itex] cosh^2\theta - sinh^2\theta = 1 [/itex]
    [itex] \dfrac{a^2}{(2x-a)^2} - \dfrac{b^2(a-x)^2}{y^2(2x-a)^2} = 1 [/itex]
    [itex] y^2a^2 - b^2(a-x)^2 = y^2(2x-a)^2 [/itex]
    [itex] y^2a^2 - b^2(a^2-2ax + x^2) = y^2(4x^2 - 4ax + a^2) [/itex]
    [itex] -b^2a^2 + 2ab^2x - b^2x^2 = 4y^2x^2 - 4ay^2x [/itex]

    not sure where to go from here
     
  6. May 2, 2013 #5

    haruspex

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    You're almost there. Just need to spot a factor.
     
  7. May 3, 2013 #6
    spotted it, thank you.
     
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