What is the foci y= 1/8 (x^2-4x-12) ?

The foci y= 1/8 (x^2-4x-12) is (2,0)

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foci y= 1/8 (x^2-4x-12)

foci

y = \frac{1}{8} (x^{2} - 4 x - 12)

(2, 0)

Solution steps

Rewrite

y = \frac{1}{8} (x^{2} - 4 x - 12)

in the standard form

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

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

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

p

from the center

(2, - 2)

along the y-axis

(2, - 2 + p)

= (2, - 2 + 2)

Refine

(2, 0)

(y - 2)^{2} + 16 (x - 3) = 0

z^{2} + 10 z + 5

x^{2} = 5 y

- 3 x^{2} + 3

y^{2} + 14 y + 4 x + 45 = 0