What is the directrix 2y^2-5y+x-4=0 ?

The directrix 2y^2-5y+x-4=0 is x= 29/4

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directrix 2y^2-5y+x-4=0

directrix

2 y^{2} - 5 y + x - 4 = 0

x = \frac{29}{4}

Solution steps

Rewrite

2 y^{2} - 5 y + x - 4 = 0

in the standard form

4 (- \frac{1}{8}) (x - \frac{57}{8}) = (y - \frac{5}{4})^{2}

(h, k) = (\frac{57}{8}, \frac{5}{4}), p = - \frac{1}{8}

Parabola is symmetric around the x-axis and so the directrix is a line parallel to the y-axis, a distance

- p

from the center

(\frac{57}{8}, \frac{5}{4})

x-coordinate

x = \frac{57}{8} - p

x = \frac{57}{8} - (- \frac{1}{8})

Refine

x = \frac{29}{4}

x^{2} + 2 x - 8

x^{2} + 2 x - 1

y^{2} - 16 x = 0

2 x^{2} - 8 x + 9

- 2 x^{2} + 64 x - 90