What is the directrix+t^2 ?

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directrix+t^2

directrix

+ t^{2}

y = - \frac{1}{4}

Solution steps

Rewrite

y = t^{2}

in the standard form:

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

(h, k) = (0, 0), p = \frac{1}{4}

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{1}{4}

Refine

y = - \frac{1}{4}

y^{2} - 4 x = 0

x^{2} - 6

4 x^{2} + 3 x - 1

x^{2} + a

y = - \frac{1}{4} (x^{2} - 14 x + 13)