What is the directrix 2x^2(sqrt(2)+1) ?

The directrix 2x^2(sqrt(2)+1) is y=-(sqrt(2)-1)/8

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directrix 2x^2(sqrt(2)+1)

directrix

2 x^{2} (2 + 1)

y = - \frac{2 - 1}{8}

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{2 - 1}{8} (y - 0) = (x - 0)^{2}

(h, k) = (0, 0), p = \frac{2 - 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

(0, 0)

y-coordinate

y = 0 - p

y = 0 - \frac{2 - 1}{8}

Refine

y = - \frac{2 - 1}{8}

y = x^{2} + 16 x + 64

(x - 6)^{2} = 8 (y + 4)

y = - 3 x^{2} - 42 x - 22

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

x^{2} - 2 x - 24

What is the directrix 2x^2(sqrt(2)+1) ?
The directrix 2x^2(sqrt(2)+1) is y=-(sqrt(2)-1)/8