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 \cdot 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)

x^{2} + 2 x - 12 y + 37 = 0

x^{2} + 4 x + 6 y - 2 = 0

(x + 4)^{2} = - 8 (y - 3)

x^{2} - 2 x + 37

4 x^{2} - 3 x + 2