What is the directrix 4t^2+1 ?

The directrix 4t^2+1 is y= 15/16

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directrix 4t^2+1

directrix

4 t^{2} + 1

y = \frac{15}{16}

Solution steps

Rewrite

y = 4 t^{2} + 1

in the standard form:

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

(h, k) = (0, 1), 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, 1)

y-coordinate

y = 1 - p

y = 1 - \frac{1}{16}

Refine

y = \frac{15}{16}

y = (x + 5)^{2}

y = 3 x^{2} - 2

y = 3 x^{2} - 5

x^{2} + 4 x + 4 y - 4 = 0

y^{2} - 6 y - 8 x + 17 = 0