What is the directrix y^2+4x-8y+12=0 ?

The directrix y^2+4x-8y+12=0 is x=2

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directrix y^2+4x-8y+12=0

directrix

y^{2} + 4 x - 8 y + 12 = 0

x = 2

Solution steps

Rewrite

y^{2} + 4 x - 8 y + 12 = 0

in the standard form

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

(h, k) = (1, 4), p = - 1

Parabola is symmetric around the x-axis and so the directrix is a line parallel to the y-axis, a distance

- p

from the center

(1, 4)

x-coordinate

x = 1 - p

x = 1 - (- 1)

Refine

x = 2

7 x^{2} - 3

2 y^{2} - 8 x = 0

y = \frac{1}{4} x^{2}

y = 2 x^{2} - 2

y = 4 x^{2} + 8 x - 20