What is the directrix 4y^2+x-32y+68=0 ?

The directrix 4y^2+x-32y+68=0 is x=-63/16

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directrix 4y^2+x-32y+68=0

directrix

4 y^{2} + x - 32 y + 68 = 0

x = - \frac{63}{16}

Solution steps

Rewrite

4 y^{2} + x - 32 y + 68 = 0

in the standard form

4 (- \frac{1}{16}) (x - (- 4)) = (y - 4)^{2}

(h, k) = (- 4, 4), p = - \frac{1}{16}

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

(- 4, 4)

x-coordinate

x = - 4 - p

x = - 4 - (- \frac{1}{16})

Refine

x = - \frac{63}{16}

y^{2} - 12 x = 0

3 - 2 x - x^{2}

x^{2} - 16

- 0.0001 x^{2} + 6.4 x + 7600

x^{2} = 6 y

What is the directrix 4y^2+x-32y+68=0 ?
The directrix 4y^2+x-32y+68=0 is x=-63/16