What is the directrix 4t^2 ?

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directrix 4t^2

directrix

4 t^{2}

y = - \frac{1}{16}

Solution steps

Rewrite

y = 4 t^{2}

in the standard form:

4 \cdot \frac{1}{16} (y - 0) = (t - 0)^{2}

(h, k) = (0, 0), p = \frac{1}{16}

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}{16}

Refine

y = - \frac{1}{16}

3 x^{2} - 2

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

(y + 3)^{2} = 4 (x - 3)

25 x^{2} - 70 x + 36

y = x^{2} - 5 x - 6