What is the directrix (xsqrt(2)+1)^2 ?

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

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directrix (xsqrt(2)+1)^2

directrix

(x 2 + 1)^{2}

y = - \frac{1}{8}

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (- \frac{1}{2}, 0), 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

(- \frac{1}{2}, 0)

y-coordinate

y = 0 - p

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

Refine

y = - \frac{1}{8}

y = (x + 2) (x - 3)

y = + 3 x^{2} + 24 x - 51

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

x^{2} + 5 x - 14

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