What is the foci (x-2)^2=12(y-1) ?

The foci (x-2)^2=12(y-1) is (2,4)

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foci (x-2)^2=12(y-1)

foci

(x - 2)^{2} = 12 (y - 1)

(2, 4)

Solution steps

Rewrite

(x - 2)^{2} = 12 (y - 1)

in the standard form:

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

(h, k) = (2, 1), p = 3

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 + 3)

Refine

(2, 4)

x^{2} - 2 x - 4

x^{2} - 2 x - 3

x^{2} + 8 x + 18

x^{2} + 8 x + 15

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