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anonymous

  • one year ago

Find the standard form of the equation of the parabola with a focus at (0, -3) and a directrix at y = 3. y2 = -12x y2 = -3x y = negative 1 divided by 12x2 y = negative 1 divided by 3x2 @ganeshie8 @dan815 @solomonzelman @nurali @

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  1. anonymous
    • one year ago
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    @Nurali please help

  2. Nurali
    • one year ago
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    (x - x)^2 + (y - 3)^2 = (x - 0)^2 + (y - -3)^2 y^2 -6y + 9 = x^2 + y^2 + 6y + 9 -12 y = x^2 y = - x^2 / 12 (or (-1/12)x^2

  3. anonymous
    • one year ago
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    can u help me with aother one? @Nurali

  4. Nurali
    • one year ago
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    i will try.

  5. anonymous
    • one year ago
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    Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ±3 divided by 2.x. y squared over 36 minus x squared over 16 = 1 y squared over 4 minus x squared over 9 = 1 y squared over 36 minus x squared over 4 = 1 y squared over 16 minus x squared over 36 = 1

  6. anonymous
    • one year ago
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    @Nurali

  7. anonymous
    • one year ago
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    @Nurali please help

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