What is the directrix (5t^2)/2 ?

The directrix (5t^2)/2 is y=-1/10

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directrix (5t^2)/2

directrix

\frac{5 t ^{2}}{2}

y = - \frac{1}{10}

Solution steps

Rewrite

y = \frac{5 t ^{2}}{2}

in the standard form:

4 \cdot \frac{1}{10} (y - 0) = (t - 0)^{2}

(h, k) = (0, 0), p = \frac{1}{10}

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}{10}

Refine

y = - \frac{1}{10}

y = x^{2} - 2 x^{1}

x^{2} - 2 x

20 y = (x - 12)^{2} - 20

- 3 x^{2}

x^{2} - 8 x + 24