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

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

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

foci

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

(- 1, 3)

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(- 1, 6)

along the y-axis

(- 1, 6 + p)

= (- 1, 6 + (- 3))

Refine

(- 1, 3)

(x - 4)^{2} = - 8 (y + 4)

(y - 5)^{2} - 4

- x^{2} = 48 y

3 x^{2} + x - 6

- \frac{1}{20} y^{2}