What is the foci (x+2)^2=-12(y-5) ?

The foci (x+2)^2=-12(y-5) is (-2,2)

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foci (x+2)^2=-12(y-5)

foci

(x + 2)^{2} = - 12 (y - 5)

(- 2, 2)

Solution steps

Rewrite

(x + 2)^{2} = - 12 (y - 5)

in the standard form:

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

(h, k) = (- 2, 5), p = - 3

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

p

from the center

(- 2, 5)

along the y-axis

(- 2, 5 + p)

= (- 2, 5 + (- 3))

Refine

(- 2, 2)

y = 2 x^{2} - 1 - (5 x - 6)

(x - 3)^{2} = 8 (y + 1)

2 x^{2} - 8 x + 5

7 x - x^{2}

y^{2} = 3 x