What is the directrix 2x-5x^2 ?

The directrix 2x-5x^2 is y= 1/4

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directrix 2x-5x^2

directrix

2 x - 5 x^{2}

y = \frac{1}{4}

Solution steps

Rewrite

y = 2 x - 5 x^{2}

in the standard form

4 (- \frac{1}{20}) (y - \frac{1}{5}) = (x - \frac{1}{5})^{2}

(h, k) = (\frac{1}{5}, \frac{1}{5}), 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

(\frac{1}{5}, \frac{1}{5})

y-coordinate

y = \frac{1}{5} - p

y = \frac{1}{5} - (- \frac{1}{20})

Refine

y = \frac{1}{4}

y^{2} - 6 x - 6 y + 3 = 0

y = 3 x^{2} - 31 x + 56 - (x - 8)

x^{2} - 5 x + 6

y = (x - 4)^{2}

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