What is the directrix y= 1/4 (x+5)^2-1 ?

The directrix y= 1/4 (x+5)^2-1 is y=-2

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directrix y= 1/4 (x+5)^2-1

directrix

y = \frac{1}{4} (x + 5)^{2} - 1

y = - 2

Solution steps

Rewrite

y = \frac{1}{4} (x + 5)^{2} - 1

in the standard form:

4 \cdot 1 \cdot (y - (- 1)) = (x - (- 5))^{2}

(h, k) = (- 5, - 1), p = 1

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

(- 5, - 1)

y-coordinate

y = - 1 - p

y = - 1 - 1

Refine

y = - 2

y = + (x + 6)^{2}

y = - 0.5 x^{2} - 3

4 x^{2} + 48 y + 12 x = 159

(x - 3)^{2} = - 8 (y + 2)

4 x^{2} - 2 x