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

The directrix-(x-1)^2+5 is y= 21/4

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directrix-(x-1)^2+5

directrix

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

y = \frac{21}{4}

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (1, 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

(1, 5)

y-coordinate

y = 5 - p

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

Refine

y = \frac{21}{4}

1.5 + 17.67 t - 4.905 t^{2}

x^{2} = 32 y

y = (x + 6) (x - 2)

6 x^{2} + 7 x - 3

x = - 4 y^{2}