What is the foci 0=(y-1)^2-16(x+4) ?

The foci 0=(y-1)^2-16(x+4) is (0,1)

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foci 0=(y-1)^2-16(x+4)

foci

0 = (y - 1)^{2} - 16 (x + 4)

(0, 1)

Solution steps

Rewrite

0 = (y - 1)^{2} - 16 (x + 4)

in the standard form:

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

(h, k) = (- 4, 1), p = 4

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

p

from the center

(- 4, 1)

along the x-axis

(- 4 + p, 1)

= (- 4 + 4, 1)

Refine

(0, 1)

y^{2} - 6 y - 8 x + 17 = 0

- 16 t^{2} + 256

2 x^{2} + 8 x - 24

10 y^{2} = 22 x

y = \frac{- 3 x ^{2} - 2 x + 6}{5}