What is the directrix 2x^2+16y=0 ?

The directrix 2x^2+16y=0 is y=2

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directrix 2x^2+16y=0

directrix

2 x^{2} + 16 y = 0

y = 2

Solution steps

Rewrite

2 x^{2} + 16 y = 0

in the standard form:

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

(h, k) = (0, 0), p = - 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

(0, 0)

y-coordinate

y = 0 - p

y = 0 - (- 2)

Refine

y = 2

y = 6 x^{2} - 29 x - 5

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

y = - 0.23 x^{2} + 1.87 x + 1.5

3 y^{2} = 24 x

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