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10 December, 22:06

Using a directrix of y = 5 and a focus of (4, 1), what quadratic function is created?

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  1. 10 December, 22:31
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    For the answer to the question above,

    Focus: (4, 1)

    Directrix: y = 5

    The vertex is always half of a distance between the focus and the directrix.

    The vertex must be at: (4, 3), h = 4, k = 3

    |p| is the distance between the vertex and the directrix (which is over the vertex and p is negative). p = - 2

    The equation:

    (x - h) ² = 4 p (y - k)

    (x - 4) ² = - 8 (y - 3)
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