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Math Help - help please

  1. #1
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    help please

    Write an quation for each parabola with vertex at orgin.

    vertex (2,1), focus (-2,-3)
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  2. #2
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    Quote Originally Posted by norwoodjay View Post
    Write an quation for each parabola with vertex at orgin.
    I don't understand this sentence

    vertex (2,1), focus (-2,-3)
    If the parabola has it's vertex at (2,1) and it's focus at (-2,-3) then it has the equation:

    p: x^2 - 2xy + y^2 + 30x + 34y = 95
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by earboth View Post
    I don't understand this sentence



    If the parabola has it's vertex at (2,1) and it's focus at (-2,-3) then it has the equation:

    p: x^2 - 2xy + y^2 + 30x + 34y = 95

    How's that a parabola? Looks a lot like a circle to me.
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  4. #4
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    Quote Originally Posted by masters View Post
    How's that a parabola? Looks a lot like a circle to me.
    I've attached the diagram of the parabola.

    From the given points you can calculate the equation of the axis, the directrix d and the coordinates of the points B and C.

    EDIT: If you have the general equation of a conic:

    Ax + 2Bxy + Cy + 2Dx + 2Ey + F = 0

    then you can determine which conic you have by calculating

    AC - B > 0 ----> ellipse

    AC - B = 0 ----> parabola

    AC - B < 0 ----> hyperbola

    You'll see that the coefficients of the equation I posted satisfy the condition indicating a parabola.
    Attached Thumbnails Attached Thumbnails help please-schiefeparabel.gif  
    Last edited by earboth; July 26th 2008 at 04:09 AM.
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by earboth View Post
    I've attached the diagram of the parabola.

    From the given points you can calculate the equation of the axis, the directrix d and the coordinates of the points B and C.

    EDIT: If you have the general equation of a conic:

    Ax + 2Bxy + Cy + 2Dx + 2Ey + F = 0

    then you can determine which conic you have by calculating

    AC - B > 0 ----> ellipse

    AC - B = 0 ----> parabola

    AC - B < 0 ----> hyperbola

    You'll see that the coefficients of the equation I posted satisfy the condition indicating a parabola.
    Thanks, Earboth

    I haven't worked that much with transformations of conics. I knew the axis of symmetry was oblique, but had forgotten about identifying the conic by using the discriminant.

    So this parabola is a rotation about the origin, and the angle of rotation is pi/2 (45 degrees) since A = C.

    I am more familiar with: Ax^2 + Bxy + Cy^2 + Dx +Ey + F = 0
    as the general equation. Then, testing the discriminant (B^2 - 4AC),

    If B^2 - 4AC < 0, and A = C, then it's a circle. If A not equal C, then it's an ellipse

    If B^2 -4AC > 0, then it's a hyperbola

    If B^2 - 4AC = 0, then it's a parabola.

    I have to think the OP made a typo on this one, but it turned out to be a great post....for me, anyway.
    Last edited by masters; July 26th 2008 at 06:25 AM.
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