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Math Help - Equation of a Hyperbola

  1. #1
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    Equation of a Hyperbola

    Hello,

    Can someone please verify if I'm correct with setting up the problem?

    I'm trying to find the equation for a hyperbola

    ((x-4)^2 / 2) - (y^2 / 2) = 1

    I've attached the graph.

    Thanks!
    Attached Thumbnails Attached Thumbnails Equation of a Hyperbola-hyperbola.png  
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  2. #2
    Senior Member apcalculus's Avatar
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    Your y-intercepts seem to be incorrect. When you plug in x=0 into your equation, you do not get 4 and -4 for the y-values.


    The general equation is:

    \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

    you can then shift this anyway you like.

    More at : Hyperbola -- from Wolfram MathWorld
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  3. #3
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    (oops need to redo)

    you're right, then the answer should be:

    (x-2)^2/4 - y^2/2 = 1
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  4. #4
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    I seem to be stuck...

    The equation I'm working with is

    (x-h)^2/a^2 - (y-k)^2/b^2=1

    now when i substitute the given, i get this:

    (x-4)^2/2 - (y-0)^2/b^2=1

    I'm trying to solve for b^2

    and I keep getting b^2=48

    Can someone please help me figure this out?
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  5. #5
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by l flipboi l View Post
    I seem to be stuck...

    The equation I'm working with is

    (x-h)^2/a^2 - (y-k)^2/b^2=1

    now when i substitute the given, i get this:

    (x-4)^2/2 - (y-0)^2/b^2=1

    I'm trying to solve for b^2

    and I keep getting b^2=48

    Can someone please help me figure this out?


    Note that

    \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

    has x-intercepts: -a and a
    has y-intercepts: -b and b
    centered at (0, 0)

    Shifting this 5 units to the right, and stretching it vertically, we get:

    \frac{(x-4)^2}{2^2} - \frac{3 y^2}{4^2} = 1
    Last edited by apcalculus; November 24th 2009 at 11:01 AM.
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  6. #6
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    Quote Originally Posted by apcalculus View Post
    Note that

    \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

    has x-intercepts: -a and a
    has y-intercepts: -b and b
    centered at (0, 0)

    Shifting this 5 units to the right, and stretching it vertically, we get:

    \frac{(x-4)^2}{2^2} - \frac{3 y^2}{4^2} = 1
    Thanks! what's the algebraic step of getting 3y^2/4^2
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