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Thread: Hyperbola graphing help

  1. #1
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    Hyperbola graphing help

    Need help graphing focus, vertices, and asymptotic.
    Y^2/16 - x^2/6=1. ( sorry I Havn't figured out how to wrap text). I got a=4 b= 2.4 c= 4.7 and y=4/2.4x or y=-4/2.4x. Does that sound right?
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    Re: Hyperbola graphing help

    I got vertices as (0,4) and (0,-4) and foci (0,4.7) and (0,-4.7). Sound right?
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    Re: Hyperbola graphing help

    We are given:

    $\displaystyle \frac{y^2}{16}-\frac{x^2}{6}=1$

    or:

    $\displaystyle \frac{y^2}{4^2}-\frac{x^2}{\sqrt{6}^2}=1$

    The foci will lie along the $\displaystyle y$-axis. We compute $\displaystyle c$:

    $\displaystyle c=\sqrt{16+6}=\sqrt{22}$

    Thus, the foci are at $\displaystyle (0,\pm\sqrt{22})$

    The vertices are at:

    $\displaystyle (0\pm4)$

    The asymptotes are found by solving:

    $\displaystyle \frac{y^2}{4^2}-\frac{x^2}{\sqrt{6}^2}=0$

    $\displaystyle y=\pm\frac{2\sqrt{6}}{3}x$

    Hyperbola graphing help-mhf_0003.png
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    Re: Hyperbola graphing help

    How do you find other points on the hyperbolas?
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    MHF Contributor MarkFL's Avatar
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    Re: Hyperbola graphing help

    Quote Originally Posted by Nfalconer View Post
    How do you find other points on the hyperbolas?
    Well, any point $\displaystyle (x,y)$ that satisfies the equation of the hyperbola will be on the hyperbola. In the case of this hyperbola, we could solve for $\displaystyle y$ as follows:

    $\displaystyle y=\pm4\sqrt{1+\frac{x^6}{6}}$

    And so for any arbitrary $\displaystyle x$ the points $\displaystyle \left(x,\pm4\sqrt{1+\frac{x^6}{6}}\right)$ will be on the hyperbola.
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