What is the directrix (y+1)^2=16(x-3) ?

The directrix (y+1)^2=16(x-3) is x=-1

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directrix (y+1)^2=16(x-3)

directrix

(y + 1)^{2} = 16 (x - 3)

x = - 1

Solution steps

Rewrite

(y + 1)^{2} = 16 (x - 3)

in the standard form:

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

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

(3, - 1)

x-coordinate

x = 3 - p

x = 3 - 4

Refine

x = - 1

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

y = 2 x^{2} - 7 x - 3

x^{2} + 5 x - 7

y^{2} - 6 y - 8 x - 23 = 0

t^{2}