What is the directrix t^2+2t-48 ?

The directrix t^2+2t-48 is y=-197/4

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directrix t^2+2t-48

directrix

t^{2} + 2 t - 48

y = - \frac{197}{4}

Solution steps

Rewrite

y = t^{2} + 2 t - 48

in the standard form

4 \cdot \frac{1}{4} (y - (- 49)) = (t - (- 1))^{2}

(h, k) = (- 1, - 49), 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

(- 1, - 49)

y-coordinate

y = - 49 - p

y = - 49 - \frac{1}{4}

Refine

y = - \frac{197}{4}

x^{2} + 9

2 x^{2} - 1 - (5 x - 6)

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

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

y = - \frac{1}{12} x^{2} + x