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Math Help - Finding equation of hyperbola from asymptotes and foci

  1. #1
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    Post Finding equation of hyperbola from asymptotes and foci

    Hi I am very confused as to how this question actually works out...

    I am asked to find the equation of a hyperbola with foci at (-3,-1) and (-3,9)
    and asymptotes at y=(-3x/4)+(7/4) and y=(3x/4)+(25/4)

    Every time I try to visualize this my foci are on a different side of the x-axis than my foci which completely throws all my thoughts out

    I also don't particularly know how to get the equation when there is asymptotic shifts...
    PLEASE HELP ME
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  2. #2
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    Re: Finding equation of hyperbola from asymptotes and foci

    have a look at this and see if it helps
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  3. #3
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    Re: Finding equation of hyperbola from asymptotes and foci

    Hello, calmo11!

    Find the equation of a hyperbola with foci (\text{-}3,\text{-}1) and (\text{-}3,9)
    and asymptotes: . y\:=\:\text{-}\tfrac{3}{4}x+\tfrac{7}{4} and y\:=\:\tfrac{3}{4}x+\tfrac{25}{4}

    Since the foci are oriented vertically, the hyperbola is "vertical": . \begin{array}{c}\cup \\ \cap \end{array}

    Its general equation is: . \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} \:=\:1
    . . where (h,k) is the center and \pm\tfrac{b}{a} is the slope of the asymptotes.


    The foci are (\text{-}3,\text{-}1) and (\text{-}3,9).
    The center is the midpoint: . (h,k) \:=\:(\text{-}3,4).

    The slope of the asymptotes is: . \pm\frac{b}{a} \:=\:\pm\frac{3}{4}
    Hence: . a = 4.\;b = 3


    Got it?
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  4. #4
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    Re: Finding equation of hyperbola from asymptotes and foci

    Thank you very much guys, I used romsek's link and the content on that site helped me heaps!
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