What is the directrix 4x^2 ?

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directrix 4x^2

directrix

4 x^{2}

y = - \frac{1}{16}

Solution steps

Rewrite

y = 4 x^{2}

in the standard form:

4 \cdot \frac{1}{16} (y - 0) = (x - 0)^{2}

(h, k) = (0, 0), p = \frac{1}{16}

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}{16}

Refine

y = - \frac{1}{16}

y = 4 x^{2} - 4 x^{1}

y = 3 x^{2} - x + 5

x = y^{2} - y - 42

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

(3 x - 4) (4 x - 3) - 2 x