What is the directrix y=(-x)^2 ?

The directrix y=(-x)^2 is y=-1/4

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directrix y=(-x)^2

directrix

y = (- x)^{2}

y = - \frac{1}{4}

Solution steps

Rewrite

y = (- x)^{2}

in the standard form

4 \cdot \frac{1}{4} (y - 0) = (x - 0)^{2}

(h, k) = (0, 0), p = \frac{1}{4}

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

(0, 0)

y-coordinate

y = 0 - p

y = 0 - \frac{1}{4}

Refine

y = - \frac{1}{4}

(x - 3)^{2} = 4 (y + 5)

y = 3 x^{2} + 6 x - 1

9 x^{2} + 24 x + 72 y + 16 = 0

y = x^{2} - 10 x + 5

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