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

The directrix (x-3)^2= 1/2 (y+1) is y=-9/8

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

directrix

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

y = - \frac{9}{8}

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{8} (y - (- 1)) = (x - 3)^{2}

(h, k) = (3, - 1), p = \frac{1}{8}

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 - \frac{1}{8}

Refine

y = - \frac{9}{8}

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

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

- x^{2} + 2 x - 3

y^{2} - 6 x + 6 y + 15 = 0

y = - 0.005 x^{2} + x + 5

What is the directrix (x-3)^2= 1/2 (y+1) ?
The directrix (x-3)^2= 1/2 (y+1) is y=-9/8