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

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

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

foci

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

(3, 2)

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (- 1, 2), p = 4

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

p

from the center

(- 1, 2)

along the x-axis

(- 1 + p, 2)

= (- 1 + 4, 2)

Refine

(3, 2)

6 x - 8 y + 15 - x^{2} = 0

4 x^{2} - 8 x + 1

x^{2} - 8 x + 2 y + 7 = 0

x^{2} - 4 x - 7

y = 2 x^{2} + 28 x + 105