What is the directrix x=+17/4 (y+2)^2 ?

The directrix x=+17/4 (y+2)^2 is x=-1/17

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directrix x=+17/4 (y+2)^2

directrix

x = + \frac{17}{4} (y + 2)^{2}

x = - \frac{1}{17}

Solution steps

Rewrite

x = \frac{17}{4} (y + 2)^{2}

in the standard form:

4 \cdot \frac{1}{17} (x - 0) = (y - (- 2))^{2}

(h, k) = (0, - 2), p = \frac{1}{17}

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

(0, - 2)

x-coordinate

x = 0 - p

x = 0 - \frac{1}{17}

Refine

x = - \frac{1}{17}

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

(x - 2)^{2} = 12 (y - 1)

y = x^{2} - 4 x + 9

(3 x - 2) (2 x + 1)

8 - 5 x + x^{2}

What is the directrix x=+17/4 (y+2)^2 ?
The directrix x=+17/4 (y+2)^2 is x=-1/17