What is the foci (x+2)^2=-8(y+3) ?

The foci (x+2)^2=-8(y+3) is (-2,-5)

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foci (x+2)^2=-8(y+3)

foci

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

(- 2, - 5)

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(- 2, - 3)

along the y-axis

(- 2, - 3 + p)

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

Refine

(- 2, - 5)

3 x^{2} + 6 x - 2

2 x^{2} - 4 x - 5

5 x^{2} - 40 x + 4 y + 85 = 0

x^{2} - 3 x + 10

y = + x^{2} - x