What is the directrix 2X^2 ?

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directrix 2X^2

directrix

y = - \frac{1}{8}

Solution steps

Rewrite

y = 2 X^{2}

in the standard form:

4 \cdot \frac{1}{8} (y - 0) = (X - 0)^{2}

(h, k) = (0, 0), p = \frac{1}{8}

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}{8}

Refine

y = - \frac{1}{8}