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primo88
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Tyro
Asked: 2021-04-08T17:40:43+00:00 In: ,

Solving Parabola

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Solve the vertex, focus, and diractrix of the following given equation (y+4)^2 = -4(x-3)

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  2. acedstud
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    The variable having squared determines the opening of a parabola. If x is the squared variable, then the parabola opens up or down but if y is the squared variable, then the parabola opens left or right. Here, the y variable is squared and the coefficient is negative, therefore, the parabola opens to the left.

    The vertex form of the equation of a left-opened parabola with vertex (h, k), is given by x=-a(y-k)^2+h or (y-k)^2=-4p(x-h), where p is the distance from the vertex to the focus of the parabola. Given (y+4)^2 = -4(x-3)\Rightarrow(y-(-4))^2 = -4(1)(x-3)\Rightarrow h=3, k=-4, p=1. Therefore, the vertex is at (3, -4).

    The focus of a left-opening parabola with vertex (h, k), is given by (h-p, k). Thus the focus of the given parabola is at (3-1, -2)=(2, -2).

    The directrix of a left-opening parabola with vertex (h, k), is given by x=h+p. Thus the directrix of the given parabola is x=2+1\Rightarrow x=3.

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