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Math Help - parabola

  1. #1
    Junior Member mattballer082's Avatar
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    this one is confusing me

    again, find the equation of the parabola with the given fetures.

    focus (0,2) directrix x=2

    please explain step by step if possible!

    thank you
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  2. #2
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    Quote Originally Posted by mattballer082
    again, find the equation of the parabola with the given fetures.

    focus (0,2) directrix x=2

    please explain step by step if possible!

    thank you
    Definitnion:A parabola is a set of point that are equidistant from a given point (focus) and a given line (directrix)

    Let (x,y) be a point on the parabola. Then it is equidistant from the focus (0,2) and the line, x=2. The distance to the focus by the distance formula is,
    s=\sqrt{x^2+(y-2)^2}
    Now, what is distance to the line x=2. This is slightly difficult. To find the distance from a point to a line you need to draw a perpendicular line, then find the distance of that line. Note, the x=2 is a vertical line, that means if you draw (and try to imagine this) a perpendicular line from the point (x,y) you intersect at (2,y). Now the distance between these two points is,
    s=\sqrt{(2-x)^2+(y-y)^2}=\sqrt{(2-x)^2}
    By, definition these two distances are equal hence,
    \sqrt{x^2+(y-2)^2}=\sqrt{(2-x)^2}
    Square both sides,
    x^2+(y-2)^2=(2-x)^2
    Thus,
    x^2+y^2-4y+4=4-4x+x^2
    Kill the unneccesarry,
    y^2-4y=-4x
    Rewrite as,
    y^2+4x-4y=0
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  3. #3
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    Hello, mattballer0821

    Find the equation of the parabola with the given features.

    Focus (0,2) directrix x=2

    You're expected to know these basic facts . . .

    The Vertex is halfway between the Focus and the Directrix.

    There are two forms:
    . . . . Vertical: (x-h)^2 = 4p(y-k) . . . opens up or down: \cup or \cap
    . . Horizontal: (y-k)^2 = 4p(x-h) . . . opens right or left: \subset or \supset

    where (h,k) is the Vertex of the parabola
    and p is the directed distance from the Vertex to the Focus.

    The parabola always "bends around" the Focus.
    The parabola always "bends away" from the Directrix.


    Sketch the given information:
    Code:
              o   |
                  |o      :
                  |  o    :
            F(0,2)*   oV  :
                  |  o    :
          - - - - +o- - - + -
              o   |       :
                  |     -x=2

    We see that the Vertex is (1,2)\quad\Rightarrow\quad (h,k) = (1,2)
    . . the parabola opens to the left: (y - k)^2 = 4p(x - h)
    . . distance from Vertex to Focus is 1 to the left: p = -1

    Therefore: . (y - 2)^2 \,= \,-4(x - 1)

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