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

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

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foci (y+1)^2+3

foci

(y + 1)^{2} + 3

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

Solution steps

Rewrite

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

in the standard form:

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

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

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

p

from the center

(3, - 1)

along the x-axis

(3 + p, - 1)

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

Refine

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

(x - 5)^{2} = 16 (y + 3)

y = 28 x^{2} + 47 x + 15

y = x^{2} - 4 x + 4

y^{2} + 36 x = 0

\frac{x ^{2}}{2}