What is the foci (x-4)^2=8(y+2) ?

The foci (x-4)^2=8(y+2) is (4,0)

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foci (x-4)^2=8(y+2)

foci

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

(4, 0)

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (4, - 2), p = 2

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

p

from the center

(4, - 2)

along the y-axis

(4, - 2 + p)

= (4, - 2 + 2)

Refine

(4, 0)

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

y = \frac{1}{64} s^{2}

92 x^{2}

- 30 x^{2} + 28 x - 6

(x + 5)^{2} + 4