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

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

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directrix 2y=(x-3)^2

directrix

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

y = - \frac{1}{2}

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{2} (y - 0) = (x - 3)^{2}

(h, k) = (3, 0), p = \frac{1}{2}

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, 0)

y-coordinate

y = 0 - p

y = 0 - \frac{1}{2}

Refine

y = - \frac{1}{2}

y = - 3 x^{2} + 5 x

x^{2} + 16 x + 8

(x - 3)^{2} - 4

(y + 1.5)^{2} = 4 (x - 2)

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