What is the directrix 19-t^2 ?

The directrix 19-t^2 is y= 77/4

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directrix 19-t^2

directrix

19 - t^{2}

y = \frac{77}{4}

Solution steps

Rewrite

y = 19 - t^{2}

in the standard form:

4 (- \frac{1}{4}) (y - 19) = (t - 0)^{2}

(h, k) = (0, 19), 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, 19)

y-coordinate

y = 19 - p

y = 19 - (- \frac{1}{4})

Refine

y = \frac{77}{4}

x^{2} - 4 x + 8 y - 4 = 0

4 t^{2}

3 x^{2} - 2

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

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