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

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

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

foci

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

(2, - 1)

Solution steps

Rewrite

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

in the standard form:

4 (- 2) (y - 1) = (x - 2)^{2}

(h, k) = (2, 1), p = - 2

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

p

from the center

(2, 1)

along the y-axis

(2, 1 + p)

= (2, 1 + (- 2))

Refine

(2, - 1)

y = (x - 3)^{2} - 9

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

x^{2} + 5

y = x^{2} - 64

y = 5 x^{2} + 6 x - 7