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Math Help - area of the region bounded by the hyperbola?

  1. #1
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    area of the region bounded by the hyperbola?

    Please help me with this problem. I have been trying 2 do it 4 half an hour and still can figure out. Thank You!


    Find the area of the region bounded by the hyperbola 25x2 - 4y2 = 100 and the line x = 3.
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  2. #2
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    So the area looks a little like a slice of a circle right?

    the easiest way to do it is turn y in terms of x.

    25x^2-4y^2=100
    25x^2-100-4y^2=0
    25x^2-100=-4y^2
    \frac{25}{4}x^2-25=y^2
    \sqrt{\frac{25}{4}x^2-25}=y

    Since you need both the positive and negative portion instead of doing 2 integrals you can just take the integral to get the region above the x axis and multiply it's area by 2.

    can you take it from there?
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  3. #3
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    Thanks! But do i substitute x = 5 sec d(theta), If yes then when I continue, I get stuck on the way.
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  4. #4
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    \sqrt{\frac{25}{4}x^2-25}=y

    \int\sqrt{\frac{25}{4}x^2-25}dx


    Substituting when you take the integral

    x = 5sec(\theta)

    well what's x^2?

    x^2=25sec^2(\theta) right?

    what if you let u = 2x?


    and dx=5sec\theta tan\theta \text{ }d\theta right?


    when you substitute that in it gets messy.

    \sqrt{\frac{25}{4}x^2-25}=y
    =\sqrt{(25)*(\frac{1}{4}x^2-1)}
    =5\sqrt{\frac{x^2}{4}-1}

    what if you let u = 2x?

    Seems like you should do a trig substitution after that, though your initial idea to substitute secant isn't wrong, just needs somewhat different coefficients.
    Last edited by seld; September 9th 2009 at 09:33 PM. Reason: wrong direction with problem
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