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

The foci x=(y+1)^2-4 is (-15/4 ,-1)

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

foci

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

(- \frac{15}{4}, - 1)

Solution steps

Rewrite

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

in the standard form:

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

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

Refine

(- \frac{15}{4}, - 1)

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

x^{2} + 14 x + 48

2 x^{2} + 7 x - 4

x^{2} + 14 x + 26

5 x^{2} - 11 x - 12