What is the directrix y^2-y-x+1=0 ?

The directrix y^2-y-x+1=0 is x= 1/2

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directrix y^2-y-x+1=0

directrix

y^{2} - y - x + 1 = 0

x = \frac{1}{2}

Solution steps

Rewrite

y^{2} - y - x + 1 = 0

in the standard form

4 \cdot \frac{1}{4} (x - \frac{3}{4}) = (y - \frac{1}{2})^{2}

(h, k) = (\frac{3}{4}, \frac{1}{2}), p = \frac{1}{4}

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{3}{4}, \frac{1}{2})

x-coordinate

x = \frac{3}{4} - p

x = \frac{3}{4} - \frac{1}{4}

Refine

x = \frac{1}{2}

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

x^{2} = 6 y

\frac{( y - 3 ) ^{2}}{11} - \frac{( x + 1 ) ^{2}}{5} = 1

x^{2} + 2 x + y^{2} + 4 y = 20

x - y^{2} = 0