What is the directrix y+1=5(x-2)^2 ?

The directrix y+1=5(x-2)^2 is y=-21/20

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directrix y+1=5(x-2)^2

directrix

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

y = - \frac{21}{20}

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{20} (y - (- 1)) = (x - 2)^{2}

(h, k) = (2, - 1), p = \frac{1}{20}

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

(2, - 1)

y-coordinate

y = - 1 - p

y = - 1 - \frac{1}{20}

Refine

y = - \frac{21}{20}

y = 4 x^{2} + 5

(x + 4)^{2} - 3

x^{2} + 12 x + 6 y + 10 = 0

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

(x - 3) (x - 7)