What is the foci x^2+4x+4y-4=0 ?

The foci x^2+4x+4y-4=0 is (-2,1)

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foci x^2+4x+4y-4=0

foci

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

(- 2, 1)

Solution steps

Rewrite

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

in the standard form

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

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

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 + (- 1))

Refine

(- 2, 1)

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

x^{2} + 16

(y - 3)^{2} = - 20 (x + 2)

- x^{2} + 6 x

+ \frac{2 h ^{3} + 4 h ^{2} + 5 h}{h}