What is the directrix x+5+(y-3)^2=0 ?

The directrix x+5+(y-3)^2=0 is x=-19/4

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directrix x+5+(y-3)^2=0

directrix

x + 5 + (y - 3)^{2} = 0

x = - \frac{19}{4}

Solution steps

Rewrite

x + 5 + (y - 3)^{2} = 0

in the standard form:

4 (- \frac{1}{4}) (x - (- 5)) = (y - 3)^{2}

(h, k) = (- 5, 3), p = - \frac{1}{4}

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, 3)

x-coordinate

x = - 5 - p

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

Refine

x = - \frac{19}{4}

(x + 2)^{2} = 8 (y - 3)

\frac{x ^{4} - x ^{2} - 2 x - 1}{x ^{2} - x - 1}

y = 16 x^{2} + 1

y = 2 x^{2} + 7 x + 3

y = \frac{6 x}{7} \frac{4 x}{3}