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yeasha07
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Tyro
Asked: 2021-04-08T17:51:05+00:00 In: ,

Solving Parabola

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Solve for the vertex, focus, axis of symmetry, latus rectum and the opening of the parabola

(y+2)^2 = 20(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 positive, therefore, the parabola opens to the right.

    The vertex form of the equation of a right-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+2)^2 = 20(x+3)\Rightarrow(y-(-2))^2 = 4(5)(x-(-3))\Rightarrow h=-3, k=-2, p=5. Therefore, the vertex is at (-3, -2).

    The focus of a right-opened parabola with vertex (h, k), is given by (h+p, k). Thus the focus of the given parabola is at (-3+5, -2)=(2, -2).

    The axis of symmetry of a right-opened parabola with vertex (h, k), is given by y=k. Thus the axis of symmetry of the given parabola is at y=-2.

    The length of the latus rectum of a parabola is given by 4p. Thus the length of the latus rectum of the given parabola is at 4\times5=20.

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