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

The directrix (y+1)^2=2(x-5) is x= 9/2

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

directrix

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

x = \frac{9}{2}

Solution steps

Rewrite

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

in the standard form:

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

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

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

(5, - 1)

x-coordinate

x = 5 - p

x = 5 - \frac{1}{2}

Refine

x = \frac{9}{2}

- x^{2} + 2 x - 3

y^{2} - 6 x + 6 y + 15 = 0

y = - 0.005 x^{2} + x + 5

y = x^{2} - 9 x + 3

- 2 x^{2} - 24 x - 54