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

The foci y=(x+4)^2-1 is (-4,-3/4)

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

foci

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

(- 4, - \frac{3}{4})

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(- 4, - 1)

along the y-axis

(- 4, - 1 + p)

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

Refine

(- 4, - \frac{3}{4})

9 y^{2} + 12 x = 0

8 y = x^{2}

y = 2 x^{2} - 7 x - 3

12 x^{2} - 72 x + 77

y^{2} - 10 x + 4 y - 16 = 0