What is the foci x^2+2x-12y+37=0 ?

The foci x^2+2x-12y+37=0 is (-1,6)

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foci x^2+2x-12y+37=0

foci

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

(- 1, 6)

Solution steps

Rewrite

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

in the standard form

4 \cdot 3 (y - 3) = (x - (- 1))^{2}

(h, k) = (- 1, 3), p = 3

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

p

from the center

(- 1, 3)

along the y-axis

(- 1, 3 + p)

= (- 1, 3 + 3)

Refine

(- 1, 6)

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

x^{2} + 6 x + 8 y = 7