What is the foci (x+5)^2=4(y-1) ?

The foci (x+5)^2=4(y-1) is (-5,2)

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foci (x+5)^2=4(y-1)

foci

(x + 5)^{2} = 4 (y - 1)

(- 5, 2)

Solution steps

Rewrite

(x + 5)^{2} = 4 (y - 1)

in the standard form:

4 \cdot 1 \cdot (y - 1) = (x - (- 5))^{2}

(h, k) = (- 5, 1), p = 1

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

p

from the center

(- 5, 1)

along the y-axis

(- 5, 1 + p)

= (- 5, 1 + 1)

Refine

(- 5, 2)

(x - 5)^{2} - 2 (y + 2) = 0

(x + 5)^{2} = - 12 (y - 3)

y = (x - 4) (x - 7)

x^{2} = 8 (y + 4)

y^{2} + 4 x = 0