What is the foci (y-2)^2=12(x+4) ?

The foci (y-2)^2=12(x+4) is (-1,2)

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foci (y-2)^2=12(x+4)

foci

(y - 2)^{2} = 12 (x + 4)

(- 1, 2)

Solution steps

Rewrite

(y - 2)^{2} = 12 (x + 4)

in the standard form:

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

(h, k) = (- 4, 2), p = 3

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

p

from the center

(- 4, 2)

along the x-axis

(- 4 + p, 2)

= (- 4 + 3, 2)

Refine

(- 1, 2)

10 x^{2} - 19 x + 5

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

x^{2} + x - 5

x^{2} + x - 4

(x + 6)^{2} = 9 (y - 1)