What is the directrix x^2+8y=0 ?

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directrix x^2+8y=0

directrix

x^{2} + 8 y = 0

y = 2

Solution steps

Rewrite

x^{2} + 8 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

\frac{x ^{2}}{16} - \frac{y ^{2}}{36} = 1

(x + 2)^{2} + \frac{( y + 4 ) ^{2}}{\frac{1}{4}} = 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

x = - 5 y^{2}