What is the directrix x=(2y+5)^2 ?

The directrix x=(2y+5)^2 is x=-1/16

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directrix x=(2y+5)^2

directrix

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

x = - \frac{1}{16}

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{16} (x - 0) = (y - (- \frac{5}{2}))^{2}

(h, k) = (0, - \frac{5}{2}), p = \frac{1}{16}

Parabola is symmetric around the x-axis and so the directrix is a line parallel to the y-axis, a distance

- p

from the center

(0, - \frac{5}{2})

x-coordinate

x = 0 - p

x = 0 - \frac{1}{16}

Refine

x = - \frac{1}{16}

6 x^{2} - 12 x

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

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

t^{2} + 2

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