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

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

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


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

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


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

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