What is the foci-12y+10=x^2-4x+14 ?

The foci-12y+10=x^2-4x+14 is (2,-3)

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foci-12y+10=x^2-4x+14

foci

- 12 y + 10 = x^{2} - 4 x + 14

(2, - 3)

Solution steps

Rewrite

- 12 y + 10 = x^{2} - 4 x + 14

in the standard form

4 (- 3) (y - 0) = (x - 2)^{2}

(h, k) = (2, 0), p = - 3

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

p

from the center

(2, 0)

along the y-axis

(2, 0 + p)

= (2, 0 + (- 3))

Refine

(2, - 3)

(x + 4)^{2} = - 6 (y - 3)

(y + 1)^{2} = 2 (x - 5)

y = 5 x^{2} - 12 x + 7

4 x = y^{2} - 4 y

3 x^{2} - 12 x + 2 y + 15 = 0