What is the foci y=(x+4)^2 ?

The foci y=(x+4)^2 is (-4, 1/4)

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foci y=(x+4)^2

foci

y = (x + 4)^{2}

(- 4, \frac{1}{4})

Solution steps

Rewrite

y = (x + 4)^{2}

in the standard form:

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

(h, k) = (- 4, 0), p = \frac{1}{4}

Parabola is symmetric around the y-axis and so the focus lies a distance

p

from the center

(- 4, 0)

along the y-axis

(- 4, 0 + p)

= (- 4, 0 + \frac{1}{4})

Refine

(- 4, \frac{1}{4})

(x + 1)^{2} = 12 (y - 3)

6 y^{2} - 12 x = 0

x^{2} - 16

(x + 1)^{2} - 2

y = - 4 x^{2}