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

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

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foci y=2(x+1)^2

foci

(- 1, \frac{1}{8})

Solution steps

Rewrite

y = 2 (x + 1)^{2}

in the standard form:

4 \cdot \frac{1}{8} (y - 0) = (x - (- 1))^{2}

(h, k) = (- 1, 0), p = \frac{1}{8}

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

p

from the center

(- 1, 0)

along the y-axis

(- 1, 0 + p)

= (- 1, 0 + \frac{1}{8})

Refine

(- 1, \frac{1}{8})