What is the directrix y=2x^2+4 ?

The directrix y=2x^2+4 is y= 31/8

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directrix y=2x^2+4

directrix

y = 2 x^{2} + 4

y = \frac{31}{8}

Solution steps

Rewrite

y = 2 x^{2} + 4

in the standard form:

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

(h, k) = (0, 4), 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, 4)

y-coordinate

y = 4 - p

y = 4 - \frac{1}{8}

Refine

y = \frac{31}{8}

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

(x - h)^{2} = 4 p (y - k)

y^{2} + 8 x = 0

y = 2 x^{2} - 4 x + 15

x^{2} + 6 x + 8 y = 7