What is the directrix x^2+8x-8y+8=0 ?

The directrix x^2+8x-8y+8=0 is y=-3

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directrix x^2+8x-8y+8=0

directrix

x^{2} + 8 x - 8 y + 8 = 0

y = - 3

Solution steps

Rewrite

x^{2} + 8 x - 8 y + 8 = 0

in the standard form

4 \cdot 2 (y - (- 1)) = (x - (- 4))^{2}

(h, k) = (- 4, - 1), p = 2

Parabola is symmetric around the y-axis and so the directrix is a line parallel to the x-axis, a distance

- p

from the center

(- 4, - 1)

y-coordinate

y = - 1 - p

y = - 1 - 2

Refine

y = - 3

y = - 3 x^{2} + 17 x - 20

92 x^{2}

y = x^{2} + 6 x + 9

8 (x - 2) = (y + 3)^{2}

- x^{2} - 4 x - 4