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  1. #1
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    Parabola question

    Find the equation of the parabola given that the focus is (6,0) and directrix is x=0.

    No idea as we have had no class examples but I know about cartesian transformations.

    Any help would be gratefully appreciated!

    dojo
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by dojo View Post
    Find the equation of the parabola given that the focus is (6,0) and directrix is x=0.

    No idea as we have had no class examples but I know about cartesian transformations.

    Any help would be gratefully appreciated!

    dojo

    Hi Dojo,

    Here's one way to find it.

    Your parabola will be in the form: x=a(y-k)^2+h since it has a vertical directrex.

    x=0

    The dirctrex equation is x=h-\dfrac{1}{4a}

    So, we have [1] h-\dfrac{1}{4a}=0

    The focus is at \left(h+\dfrac{1}{4a}\:,\:k\right)

    F(6, 0) and \boxed{k=0}

    So, we have [2] h+\dfrac{1}{4a}=6

    Solve the system of equations [1] and [2] for h and a.

    [1] h-\dfrac{1}{4a}=0

    [2] h+\dfrac{1}{4a}=6

    2h=6

    \boxed{h=3}

    Substituting into [1],

    3-\dfrac{1}{4a}=0

    3=\dfrac{1}{4a}

    \boxed{a=\dfrac{1}{12}}

    Now, you will have everything you need to create your equation.

    x=\dfrac{1}{12}\left(y-0\right)^2+3

    x=\dfrac{1}{12}y^2+3
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