What is the directrix y^2-5y-x+7=0 ?

The directrix y^2-5y-x+7=0 is x= 1/2

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directrix y^2-5y-x+7=0

directrix

y^{2} - 5 y - x + 7 = 0

x = \frac{1}{2}

Solution steps

Rewrite

y^{2} - 5 y - x + 7 = 0

in the standard form

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

(h, k) = (\frac{3}{4}, \frac{5}{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{5}{2})

x-coordinate

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

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

Refine

x = \frac{1}{2}

\frac{x ^{2}}{4} - \frac{y ^{2}}{16} = 1

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

4 x^{2} + y^{2} - 8 x + 4 y - 8 = 0

(x - 1)^{2} + (y - 6)^{2} = 25

\frac{x ^{2}}{64} + \frac{y ^{2}}{36} = 1