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

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

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

foci

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

(8, 0)

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (8, 2), p = - 2

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

p

from the center

(8, 2)

along the y-axis

(8, 2 + p)

= (8, 2 + (- 2))

Refine

(8, 0)

(x + 3)^{2} - 8

- x^{2} + 3 x

y = 4 x^{2} + 8 x - 11

y = - x^{2} + 64 x - 292

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