What is the foci y^2-12y-4x+28=0 ?

The foci y^2-12y-4x+28=0 is (-1,6)

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foci y^2-12y-4x+28=0

foci

y^{2} - 12 y - 4 x + 28 = 0

(- 1, 6)

Solution steps

Rewrite

y^{2} - 12 y - 4 x + 28 = 0

in the standard form

4 \cdot 1 \cdot (x - (- 2)) = (y - 6)^{2}

(h, k) = (- 2, 6), p = 1

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

p

from the center

(- 2, 6)

along the x-axis

(- 2 + p, 6)

= (- 2 + 1, 6)

Refine

(- 1, 6)

y = x^{2} - 2 x - 3

y^{2} - 12 x = 0

x^{2} + 7 x + 12

(3 x - 4) (4 x - 3) - 2 x

0.5 x^{2} - 2 x - 2