What is the directrix y=5x^2+1 ?

The directrix y=5x^2+1 is y= 19/20

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directrix y=5x^2+1

directrix

y = 5 x^{2} + 1

y = \frac{19}{20}

Solution steps

Rewrite

y = 5 x^{2} + 1

in the standard form:

4 \cdot \frac{1}{20} (y - 1) = (x - 0)^{2}

(h, k) = (0, 1), p = \frac{1}{20}

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}{20}

Refine

y = \frac{19}{20}

y = x^{2} + 9 x + 4

(x + 1)^{2} = 2 (y + 2)

3 x^{2} + 4

3 x^{2} + 2

5 x^{2} + 7 y = 3