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

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

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

foci

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

(- 1, - 1)

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (- 1, 2), p = - 3

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

p

from the center

(- 1, 2)

along the y-axis

(- 1, 2 + p)

= (- 1, 2 + (- 3))

Refine

(- 1, - 1)

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

y \geq x^{2}

x^{2} + 4 x - 1

(x + 5)^{2} = 4 (y - 1)

(x - 5)^{2} - 2 (y + 2) = 0