What is the foci y^2+8y+4x+20=0 ?

The foci y^2+8y+4x+20=0 is (-2,-4)

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foci y^2+8y+4x+20=0

foci

y^{2} + 8 y + 4 x + 20 = 0

(- 2, - 4)

Solution steps

Rewrite

y^{2} + 8 y + 4 x + 20 = 0

in the standard form

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

(h, k) = (- 1, - 4), p = - 1

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

p

from the center

(- 1, - 4)

along the x-axis

(- 1 + p, - 4)

= (- 1 + (- 1), - 4)

Refine

(- 2, - 4)

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

y = 4 x - x^{2}

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

- x^{2} + 4 x + 5

x = - 3 y^{2}