What is the directrix 0=(y-1)^2-16(x+4) ?

The directrix 0=(y-1)^2-16(x+4) is x=-8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

directrix 0=(y-1)^2-16(x+4)

directrix

0 = (y - 1)^{2} - 16 (x + 4)

x = - 8

Solution steps

Rewrite

0 = (y - 1)^{2} - 16 (x + 4)

in the standard form:

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

(h, k) = (- 4, 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

(- 4, 1)

x-coordinate

x = - 4 - p

x = - 4 - 4

Refine

x = - 8

(x + 6)^{2} = 12 (y + 1)

- \frac{1}{8} (x - 8)^{2} + 2

y = 25 x^{2} - 36

- \frac{1}{2} x^{2} + 7 x - 3

y = - 3 x^{2} + 12 x - 4

What is the directrix 0=(y-1)^2-16(x+4) ?
The directrix 0=(y-1)^2-16(x+4) is x=-8