What is the foci (y-3)^2=12(x+2) ?

The foci (y-3)^2=12(x+2) is (1,3)

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foci (y-3)^2=12(x+2)

foci

(y - 3)^{2} = 12 (x + 2)

(1, 3)

Solution steps

Rewrite

(y - 3)^{2} = 12 (x + 2)

in the standard form:

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

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

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

p

from the center

(- 2, 3)

along the x-axis

(- 2 + p, 3)

= (- 2 + 3, 3)

Refine

(1, 3)

- 4 x^{2}

y = 2 x^{2} - 4 x

- 3 x^{2} + 3 x - 13

y = - 2 x^{2} + 5 x - 6

x^{2} + 8 x - 5