What is the directrix y^2+4x+1=0 ?

The directrix y^2+4x+1=0 is x= 3/4

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directrix y^2+4x+1=0

directrix

y^{2} + 4 x + 1 = 0

x = \frac{3}{4}

Solution steps

Rewrite

y^{2} + 4 x + 1 = 0

in the standard form:

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

(h, k) = (- \frac{1}{4}, 0), 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

(- \frac{1}{4}, 0)

x-coordinate

x = - \frac{1}{4} - p

x = - \frac{1}{4} - (- 1)

Refine

x = \frac{3}{4}

5 x^{2} - 13 x - 6

x^{2} - 4 x - 5

4 x^{2} + 12 x + 9

5 x^{2}

x^{2} + x - 1