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Thread: 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.

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

    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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    $\displaystyle \sqrt{\frac{25}{4}x^2-25}=y$

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


    Substituting when you take the integral

    $\displaystyle x = 5sec(\theta)$

    well what's $\displaystyle x^2$?

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

    what if you let u = 2x?


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


    when you substitute that in it gets messy.

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

    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; Sep 9th 2009 at 08:33 PM. Reason: wrong direction with problem
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