What is the foci (y+5)^2=-16(x-3) ?

The foci (y+5)^2=-16(x-3) is (-1,-5)

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foci (y+5)^2=-16(x-3)

foci

(y + 5)^{2} = - 16 (x - 3)

(- 1, - 5)

Solution steps

Rewrite

(y + 5)^{2} = - 16 (x - 3)

in the standard form:

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

(h, k) = (3, - 5), p = - 4

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

p

from the center

(3, - 5)

along the x-axis

(3 + p, - 5)

= (3 + (- 4), - 5)

Refine

(- 1, - 5)

(x + 3)^{2} = - 12 (y + 4)

y^{2} = - 32 x

y = 8 x^{2} - x + 1

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

2 y^{2} - 16 x = 0