What is the foci (y+6)^2=12(x+1) ?

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

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foci (y+6)^2=12(x+1)

foci

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

(2, - 6)

Solution steps

Rewrite

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

in the standard form:

4 \cdot 3 (x - (- 1)) = (y - (- 6))^{2}

(h, k) = (- 1, - 6), p = 3

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

p

from the center

(- 1, - 6)

along the x-axis

(- 1 + p, - 6)

= (- 1 + 3, - 6)

Refine

(2, - 6)

(x + 7)^{2} = 5 (y + 3)

x^{2} - 8 y = 0

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

- \frac{x ^{2}}{8} + \frac{x}{4} + \frac{23}{8}

3 x^{2} + 8