Problem

For \((x~-~2)^2~=~8(y~+~1)\), give the parabola's vertex, \(p\text{-}\text{value}\), axis and its orientation, opening direction, focus, and directrix.

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What this problem is really about

First identify which variable is squared; that tells whether the parabola's axis is vertical or horizontal. Match the other-side coefficient to four times the signed focal distance, whose sign gives the opening. From the vertex, move that signed distance along the axis to the focus, then place the directrix the same distance on the opposite side.

Answer choices

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Preview only

The vertex is (2, -1), p = 8, the axis is x = 2 and is vertical, the parabola opens up, the focus is (2, 7), and the directrix is y = -9.

The vertex is (2, -1), p = 2, the axis is x = 2 and is vertical, the parabola opens up, the focus is (2, 1), and the directrix is y = -3.

The vertex is (2, -1), p = 2, the axis is y = -1 and is horizontal, the parabola opens right, the focus is (4, -1), and the directrix is x = 0.

The vertex is (-2, 1), p = -2, the axis is x = -2 and is vertical, the parabola opens down, the focus is (-2, -1), and the directrix is y = 3.

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For \((x~-~2)^2~=~8(y~+~1)\), give the parabola's vertex, \(p\text{-}\text{value}\), axis and its orientation, opening direction, focus, and directrix.

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