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Tyro
Asked: 2021-04-04T10:13:31+00:00 In: ,

Equation of a Parabola

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Find the equation of the parabola with directrix x + 1 = 0, axis y = 2, and latus rectum 3.

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  2. acedstud

    The standard form of the equation of a parabola with axis of symmetry at y=k is given by (y-k)^2=4p(x-h), where the directrix is x=h-p, and the endpoints of latus rectum is (h+p, k\pm2p).

    Given that the axis of symettry is y=2. Thus, k=2.

    Given that the length of the latus rectum is 3, this means that |(k+2p)-(k-2p)|=3\Rightarrow|4p|=3\Rightarrow p=\dfrac{3}{4}.

    Given that the directrix is x+1=0\Rightarrow x=-1. Thus, h-p=-1\Rightarrow h-\dfrac{3}{4}=-1\Rightarrow h=-1+\dfrac{3}{4}=-\dfrac{1}{4}.

    Therefore, the equation of the directrix is (y-2)^2=3\left(x-\left(-\dfrac{1}{4}\right)\right)\Rightarrow(y-2)^2=3\left(x+\dfrac{1}{4}\right).

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