What is the foci y^2+20x+8y+56=0 ?

The foci y^2+20x+8y+56=0 is (-7,-4)

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foci y^2+20x+8y+56=0

foci

y^{2} + 20 x + 8 y + 56 = 0

(- 7, - 4)

Solution steps

Rewrite

y^{2} + 20 x + 8 y + 56 = 0

in the standard form

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

(h, k) = (- 2, - 4), p = - 5

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

p

from the center

(- 2, - 4)

along the x-axis

(- 2 + p, - 4)

= (- 2 + (- 5), - 4)

Refine

(- 7, - 4)

x^{2} = 3 y

x^{2} + 4 x - 3

(y - 4)^{2} = - 8 (x - 1)

(y - 1)^{2} = 16 (x + 3)

y = - 3 x^{2} + 12 x - 8