What is the foci x^2+4x-y+1=0 ?

The foci x^2+4x-y+1=0 is (-2,-11/4)

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foci x^2+4x-y+1=0

foci

x^{2} + 4 x - y + 1 = 0

(- 2, - \frac{11}{4})

Solution steps

Rewrite

x^{2} + 4 x - y + 1 = 0

in the standard form

4 \cdot \frac{1}{4} (y - (- 3)) = (x - (- 2))^{2}

(h, k) = (- 2, - 3), p = \frac{1}{4}

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

p

from the center

(- 2, - 3)

along the y-axis

(- 2, - 3 + p)

= (- 2, - 3 + \frac{1}{4})

Refine

(- 2, - \frac{11}{4})

+ y^{2} - 4 y + 4 x = - 16

- 2 x^{2} + 8 x

y^{2} = \frac{25}{- 2} x

- x^{2} - 3

x^{2} - 4 x - 12 y - 20 = 0