What is the foci 2y^2+8y+x+1=0 ?

The foci 2y^2+8y+x+1=0 is (55/8 ,-2)

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foci 2y^2+8y+x+1=0

foci

2 y^{2} + 8 y + x + 1 = 0

(\frac{55}{8}, - 2)

Solution steps

Rewrite

2 y^{2} + 8 y + x + 1 = 0

in the standard form

4 (- \frac{1}{8}) (x - 7) = (y - (- 2))^{2}

(h, k) = (7, - 2), p = - \frac{1}{8}

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

p

from the center

(7, - 2)

along the x-axis

(7 + p, - 2)

= (7 + (- \frac{1}{8}), - 2)

Refine

(\frac{55}{8}, - 2)

x^{2} - 4 x - 12

+ 18500 (0.09 - x^{2})

y = x^{2} + 4 x + 3

y = x^{2} + 10 x + 25

(x + 2)^{2} = 8 (y - 3)