What is the foci (x+1)^2+3 ?

The foci (x+1)^2+3 is (-1, 13/4)

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foci (x+1)^2+3

foci

(x + 1)^{2} + 3

(- 1, \frac{13}{4})

Solution steps

Rewrite

y = (x + 1)^{2} + 3

in the standard form:

4 \cdot \frac{1}{4} (y - 3) = (x - (- 1))^{2}

(h, k) = (- 1, 3), p = \frac{1}{4}

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

p

from the center

(- 1, 3)

along the y-axis

(- 1, 3 + p)

= (- 1, 3 + \frac{1}{4})

Refine

(- 1, \frac{13}{4})

y^{2} + 2 y + 12 x + 25 = 0

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

- 3 x^{2} + 5 x

(x - 1)^{2} = - 2 y - 11

(x + 3)^{2} = 12 (y - 5)