What is the directrix 2x^2-32x-y+127=0 ?

The directrix 2x^2-32x-y+127=0 is y=-9/8

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directrix 2x^2-32x-y+127=0

directrix

2 x^{2} - 32 x - y + 127 = 0

y = - \frac{9}{8}

Solution steps

Rewrite

2 x^{2} - 32 x - y + 127 = 0

in the standard form

4 \cdot \frac{1}{8} (y - (- 1)) = (x - 8)^{2}

(h, k) = (8, - 1), p = \frac{1}{8}

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

(8, - 1)

y-coordinate

y = - 1 - p

y = - 1 - \frac{1}{8}

Refine

y = - \frac{9}{8}

y^{2} + 2 x + 6 y + 17 = 0

(x - 7)^{2}

- 3 x^{2} + 75

y = \frac{95 x ^{2} - 10 x + 26}{10}

y = 64 x^{2}

What is the directrix 2x^2-32x-y+127=0 ?
The directrix 2x^2-32x-y+127=0 is y=-9/8