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

The directrix-1/2 (x-4)^2+3 is y= 7/2

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directrix-1/2 (x-4)^2+3

directrix

- \frac{1}{2} (x - 4)^{2} + 3

y = \frac{7}{2}

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (4, 3), 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

(4, 3)

y-coordinate

y = 3 - p

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

Refine

y = \frac{7}{2}

- 4 x^{2} + 8 x + 1

y = 2 x^{2} + 3 x + 1

x - 4 y^{2} = 0

x = (- y - 1) (y + 3)

y = 4 x^{2} - 8 x + 1