What is the directrix y^2-6y-8x+17=0 ?

The directrix y^2-6y-8x+17=0 is x=-1

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directrix y^2-6y-8x+17=0

directrix

y^{2} - 6 y - 8 x + 17 = 0

x = - 1

Solution steps

Rewrite

y^{2} - 6 y - 8 x + 17 = 0

in the standard form

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

(h, k) = (1, 3), p = 2

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

(1, 3)

x-coordinate

x = 1 - p

x = 1 - 2

Refine

x = - 1

y^{2} + 6 y + 2 x + 1 = 0

y = x^{2} - 7

2 x - 5 x^{2}

y^{2} - 6 x - 6 y + 3 = 0

y = 3 x^{2} - 31 x + 56 - (x - 8)