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16 August, 06:57

Derive the equation of the parabola with a focus at (-5, 5) and a directrix of y = - 1

f (x) = - 1/12 (x - 5) 2 + 2

f (x) = 1/12 (x - 5) 2 + 2

f (x) = - 1/12 (x + 5) 2 + 2

f (x) = 1/12 (x + 5) 2 + 2

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  1. 16 August, 07:47
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    From the given information, the parabola is a regular parabola facing up with vertex at (-5, 2).

    Required equation is (x + 5) ^2 = 4p (y - 2)

    But 2 + p = 5 = > p = 5 - 2 = 3

    Therefore, required equation is (x + 5) ^2 = 4 (3) (y - 2)

    (x + 5) ^2 = 12 (y - 2)

    y - 2 = 1/12 (x + 5) ^2

    y = 1/12 (x + 5) ^2 + 2

    f (x) = 1/12 (x + 5) ^2 + 2
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