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Math Help - Equation of hyperbola

  1. #1
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    Equation of hyperbola

    write an equation of hyperbola having 3/2 as eccentricity. the point (1,-3) as focus and the line y=2 as its corresponding directrix
    Last edited by BAHADEEN; March 22nd 2011 at 12:36 PM.
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  2. #2
    Senior Member Sambit's Avatar
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    Equation of hyperbola: \frac{(y-\beta)^2}{b^2}-\frac{(x-\alpha)^2}{a^2}=1

    Eccentricity: e=\sqrt{1+\frac{b^2}{a^2}}

    Foci: (\alpha, \beta\pm ae)

    Equation of directrices: y-\beta=\pm \frac{a}{e}

    You have to solve \alpha,\beta,a,b from this.
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  3. #3
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    Quote Originally Posted by BAHADEEN View Post
    write an equation of hyperbola having 3/2 as eccentricity. the point (1,-3) as focus and the line y=2 as its corresponding directrix
    Here comes a slightly different approach:

    1. Let P denotes an arbitrary point on the hyperbola, r the distance of P to the focus F and l the distance of P to the directrix. Then

    \dfrac rl = e~\implies~r = e \cdot l

    2. Plug in the values you know:

    \sqrt{(x-1)^2+(y+3)^2} = 1.5(y+2)

    3. Simplify and re-arrange so that you'll get the equation of the hyperbola in standard form. I've got:

    -\dfrac{(x-1)^2}{\frac95}+\dfrac{\left(y+\frac65\right)^2}{\l  eft(\frac56\right)^2}} = 1
    Attached Thumbnails Attached Thumbnails Equation of hyperbola-hyperbleitlinie.png  
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  4. #4
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    Quote Originally Posted by earboth View Post
    Here comes a slightly different approach:

    1. Let P denotes an arbitrary point on the hyperbola, r the distance of P to the focus F and l the distance of P to the directrix. Then

    \dfrac rl = e~\implies~r = e \cdot l

    2. Plug in the values you know:

    \sqrt{(x-1)^2+(y+3)^2} = 1.5(y+2)

    3. Simplify and re-arrange so that you'll get the equation of the hyperbola in standard form. I've got:

    -\dfrac{(x-1)^2}{\frac95}+\dfrac{\left(y+\frac65\right)^2}{\l  eft(\frac56\right)^2}} = 1
    Thanks,but i get (y-6)^2/36 -(x-1)^2/45=1 when i used PF= e (PD) I wrong or not?
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  5. #5
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    Quote Originally Posted by BAHADEEN View Post
    Thanks,but i get (y-6)^2/36 -(x-1)^2/45=1 when i used PF= e (PD) I wrong or not?
    I've made a tiny mistake:

    \sqrt{(x-1)^2+(y+3)^2}=1.5(y-2)

    would have been the correct equation. After some transformations I would have reached your solution.

    Your result is correct. My apologies for the confusion.
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