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

The foci x=-1/4 (y+2)^2-2 is (-3,-2)

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

foci

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

(- 3, - 2)

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(- 2, - 2)

along the x-axis

(- 2 + p, - 2)

= (- 2 + (- 1), - 2)

Refine

(- 3, - 2)

10 - x^{2}

x^{2} + 64

- 2 x^{2} - 12 x - 23

2 y = 3 x^{2} + x

2 x^{2} + 4 x - y = 0