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Math Help - conics

  1. #1
    Junior Member mattballer082's Avatar
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    i dont get it!! ahhh! help

    find the equation of the parabola with the given features. graph the parabola.

    1.Vertex (0,0) focus (0,3)


    and


    2. (0,0) directrix x+1=0


    Both of them confuse me. can someone explain it step by step please!!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mattballer082
    find the equation of the parabola with the given features. graph the parabola.

    1.Vertex (0,0) focus (0,3)


    and


    2. (0,0) directrix x+1=0


    Both of them confuse me. can someone explain it step by step please!!!
    Wikipedia tells us that a parabola with vertical symetry axis (which is what you
    have in your first problem) with vertex (h,k), and focus (h,k+p) has directrix
    y=k-p, and equation:

    <br />
(x-h)^2=4p(y-k)<br />

    In your first question the vertex is at (0,0) so h=0, \ k=0, and as the focus
    is at (0,3) so p=3, and the equation becomes:

    <br />
x^2=12y<br />

    In your second problem the symetry axis is horizontal, so what I will do
    is derive the equation as though it were vertical and finaly interchenge x and y.

    Flipping the axis leaves the vertex at (0,0) but the directrix becomes y+1=0
    or equivalently y=-1

    In this case h=0, \ k=0 and p=1 so the equation
    for the vertical axis parabola is:

    <br />
x^2=4y<br />
,

    and interchanging x and y gives us:

    <br />
y^2=4x<br />

    for the horizontal axis parabola.

    RonL
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  3. #3
    Super Member

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    Hello, mattballer082!

    These problems have the vertex at the origin.

    There are two types:
    . . x^2 = 4py . . . the parabola opens up or down: \cup or \cap
    . . y^2 = 4px . . . the parabola opens right or left: \subset or \supset

    p is the distance from the Vertex to the Focus
    . . and the distance from the Vertex to the Directrix.

    Note: the curve always bends "around the Focus" and "away from the Directrix".


    Find the equation of the parabola with the given features.
    Graph the parabola.

    1. Vertex (0,0), focus (0,3)

    Make a sketch.
    Code:
                |
         *     Fo(0,3) *
          *     |     *
            *   |   *
          - - - o - - -
               V|(0,0)

    Plot the Vertex and the Focus.
    We know that the parabola bends around the Focus,
    . . so we know its orientation: opens upward.

    We use the form: . x^2 = 4py

    We see that p = 3
    . . Therefore, the equation is: . x^2 = 12y



    2. Vertex (0,0), directrix x + 1 = 0

    The vertex is at the origin; the directrix is the vertical line: x = -1
    Code:
              :   |         *
              :   |   *
              :   |*
          - - + - o - - - - - - -
              :   |*
              :   |   *
              :   |         *
             x=-1

    Plot the Vertex and the Directrix.
    We the know the parabola "bends away from the directrix"
    . . so we know its orientation: opens right.

    We use the form: . y^2 = 4px

    We see that p = 1
    . . Therefore, the equation is: . y^2 = 4x

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