What is the directrix (x-3)^2=-4(y+1) ?

The directrix (x-3)^2=-4(y+1) is y=0

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directrix (x-3)^2=-4(y+1)

directrix

(x - 3)^{2} = - 4 (y + 1)

y = 0

Solution steps

Rewrite

(x - 3)^{2} = - 4 (y + 1)

in the standard form:

4 (- 1) (y - (- 1)) = (x - 3)^{2}

(h, k) = (3, - 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

(3, - 1)

y-coordinate

y = - 1 - p

y = - 1 - (- 1)

Refine

y = 0

y = x^{2} - 3 x^{2}

- 2 x^{2} - 4 x + y + 70 = 0

19 - t^{2}

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

4 t^{2}