What is the directrix 4y^2-12y+56x+177=0 ?

The directrix 4y^2-12y+56x+177=0 is x= 1/2

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directrix 4y^2-12y+56x+177=0

directrix

4 y^{2} - 12 y + 56 x + 177 = 0

x = \frac{1}{2}

Solution steps

Rewrite

4 y^{2} - 12 y + 56 x + 177 = 0

in the standard form

4 (- \frac{7}{2}) (x - (- 3)) = (y - \frac{3}{2})^{2}

(h, k) = (- 3, \frac{3}{2}), p = - \frac{7}{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

(- 3, \frac{3}{2})

x-coordinate

x = - 3 - p

x = - 3 - (- \frac{7}{2})

Refine

x = \frac{1}{2}

3 x^{2} + 19 x - 40

(y - 10)^{2} = - 12 (x + 2)

x^{2} + 4 x - 8 y + 12 = 0

x^{(2)} - 6 x + 5 y = - 34

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

What is the directrix 4y^2-12y+56x+177=0 ?
The directrix 4y^2-12y+56x+177=0 is x= 1/2