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

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

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directrix (x-3)^2+1

directrix

(x - 3)^{2} + 1

y = \frac{3}{4}

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (3, 1), p = \frac{1}{4}

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}{4}

Refine

y = \frac{3}{4}

x^{2} = - 6 y

y = - 12 x^{2} + 4 x

x = \frac{1}{12} y^{2}

x = - 4 (y - 7)^{2} + 7

2 y^{2} - 16 x = 0