What is the foci x^2+4x-8y+12=0 ?

The foci x^2+4x-8y+12=0 is (-2,3)

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foci x^2+4x-8y+12=0

foci

x^{2} + 4 x - 8 y + 12 = 0

(- 2, 3)

Solution steps

Rewrite

x^{2} + 4 x - 8 y + 12 = 0

in the standard form

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

(h, k) = (- 2, 1), p = 2

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

p

from the center

(- 2, 1)

along the y-axis

(- 2, 1 + p)

= (- 2, 1 + 2)

Refine

(- 2, 3)

y = x^{2} + x + 1

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

y = x^{2} - 2 x^{2}

y = - \frac{1}{8} (x - 2)^{2} + 1

y = (x - 3)^{2} - 9