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Math Help - finding the equation of a hyperbola!

  1. #1
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    Angry finding the equation of a hyperbola!

    Find an equation of the hyperbola having foci at and and asymptotes at and .
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  2. #2
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    Hello, ahztin13!

    Find an equation of the hyperbola having foci at \left(5 \pm\sqrt{61},\:-4\right)
    and asymptotes y \:=\:\pm\frac{6}{5}x - 10
    The equation of a "horizonal" hyperbola is: . \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} \;=\;1

    The center of a hyperbola is the midpoint between the foci.
    . . Hence: . (h,k) \:=\:(5,\,-4)

    The slopes of the asymptotes is given by: . \pm \frac{b}{a}
    . . Hence: . \frac{b}{a} \:=\:\frac{6}{5} \quad\Rightarrow\quad a = 5k,\;b = 6k\:\text{ for some integer }k


    We know that c \,=\,\sqrt{61}\:\text{ and }\:a^2+b^2\:=\:c^2

    . . So we have: . (5k)^2 + (6k)^2 \:=\:61\quad\Rightarrow\quad k \,=\,1 \quad\Rightarrow\quad a = 5,\:b = 6


    Therefore, the equation is: . \frac{(x-5)^2}{25} - \frac{y+4)^2}{36} \;=\;1

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