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

The foci (y+5)^2=-12(x-2) is (-1,-5)

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

foci

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

(- 1, - 5)

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(2, - 5)

along the x-axis

(2 + p, - 5)

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

Refine

(- 1, - 5)

y = 4 x - x^{2}

y^{2} - 12 x - 2 y + 25 = 0

- x^{2} + 4 x + 5

x = - 3 y^{2}

(y - 7)^{2} = - 12 (x + 1)