What is the directrix 2x^2+5 ?

The directrix 2x^2+5 is y= 39/8

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directrix 2x^2+5

directrix

2 x^{2} + 5

y = \frac{39}{8}

Solution steps

Rewrite

y = 2 x^{2} + 5

in the standard form:

4 \cdot \frac{1}{8} (y - 5) = (x - 0)^{2}

(h, k) = (0, 5), p = \frac{1}{8}

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, 5)

y-coordinate

y = 5 - p

y = 5 - \frac{1}{8}

Refine

y = \frac{39}{8}

2 y^{2} - 44 x = 0

x = - \frac{3}{8} (y - 2)^{2} - 1

(y + 12)^{2} = 9 (x - 9)

0.5 x^{2} + 2

y = - 2 + x^{2}