What is the directrix (x-4)^2=-8(y+3) ?

The directrix (x-4)^2=-8(y+3) is y=-1

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directrix (x-4)^2=-8(y+3)

directrix

(x - 4)^{2} = - 8 (y + 3)

y = - 1

Solution steps

Rewrite

(x - 4)^{2} = - 8 (y + 3)

in the standard form:

4 (- 2) (y - (- 3)) = (x - 4)^{2}

(h, k) = (4, - 3), 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, - 3)

y-coordinate

y = - 3 - p

y = - 3 - (- 2)

Refine

y = - 1

y^{2} + 33 = - 8 x - 23

x^{2} + 7 x + 12

- x^{2} + 1

- x^{2} + 7

y = - 16 x^{2} + 16 x + 480