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

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

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

foci

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

(- 3, 1)

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(- 3, - 2)

along the y-axis

(- 3, - 2 + p)

= (- 3, - 2 + 3)

Refine

(- 3, 1)

x^{2} = - 6 (y - 7)

2 y^{2} = 16 x

x^{2} + 2 x

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

- 3 x^{2} + 8 x + 896