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

The foci (x+1)^2=(y+4) is (-1,-15/4)

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

foci

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

(- 1, - \frac{15}{4})

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(- 1, - 4)

along the y-axis

(- 1, - 4 + p)

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

Refine

(- 1, - \frac{15}{4})

y = - x^{2} + 4 x + 3

x^{2} - 4 x - 5 y + 7 = 0

(x - 2)^{2} = 12 (y - 1)

x^{2} - 2 x - 4

x^{2} - 2 x - 3