What is the directrix x^2+5 ?

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directrix x^2+5

directrix

x^{2} + 5

y = \frac{19}{4}

Solution steps

Rewrite

y = x^{2} + 5

in the standard form:

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

(h, k) = (0, 5), 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

(0, 5)

y-coordinate

y = 5 - p

y = 5 - \frac{1}{4}

Refine

y = \frac{19}{4}

y^{2} = - 32 x

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

y = - 8 x^{2} + 4

3 x^{2} + 6 x - 4

x^{2} + 2 x - 12 y + 37 = 0