What is the directrix x^2-2x-4y+17=0 ?

The directrix x^2-2x-4y+17=0 is y=3

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directrix x^2-2x-4y+17=0

directrix

x^{2} - 2 x - 4 y + 17 = 0

y = 3

Solution steps

Rewrite

x^{2} - 2 x - 4 y + 17 = 0

in the standard form

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

(h, k) = (1, 4), p = 1

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

(1, 4)

y-coordinate

y = 4 - p

y = 4 - 1

Refine

y = 3

x^{2} - 12 y - 6 x + 5 = 0

x^{2} + 4 x - y + 1 = 0

3 x^{2} - 4 x - 2

x^{2} - 5 x - 6

y^{2} + 8 x - 6 y + 25 = 0