What is the foci (x+1)^2=12(y-3) ?

The foci (x+1)^2=12(y-3) is (-1,6)

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

foci

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

(- 1, 6)

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(- 1, 3)

along the y-axis

(- 1, 3 + p)

= (- 1, 3 + 3)

Refine

(- 1, 6)

6 y^{2} - 12 x = 0

x^{2} - 16

(x + 1)^{2} - 2

y = - 4 x^{2}

y = \frac{x ^{2} - 4}{2}