What is the directrix (x-4)^2-2(y+8)=0 ?

The directrix (x-4)^2-2(y+8)=0 is y=-17/2

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directrix (x-4)^2-2(y+8)=0

directrix

(x - 4)^{2} - 2 (y + 8) = 0

y = - \frac{17}{2}

Solution steps

Rewrite

(x - 4)^{2} - 2 (y + 8) = 0

in the standard form:

4 \cdot \frac{1}{2} (y - (- 8)) = (x - 4)^{2}

(h, k) = (4, - 8), p = \frac{1}{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, - 8)

y-coordinate

y = - 8 - p

y = - 8 - \frac{1}{2}

Refine

y = - \frac{17}{2}

+ \frac{2 h ^{3} + 4 h ^{2} + 5 h}{h}

x^{2} + 6 x - 8

y = x^{2} + 4 x - 3

y = 18 x^{2} - 6 x + 4

7 x^{2} - 7 x

What is the directrix (x-4)^2-2(y+8)=0 ?
The directrix (x-4)^2-2(y+8)=0 is y=-17/2