What is the foci y^2+6y+2x+1=0 ?

The foci y^2+6y+2x+1=0 is (7/2 ,-3)

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foci y^2+6y+2x+1=0

foci

y^{2} + 6 y + 2 x + 1 = 0

(\frac{7}{2}, - 3)

Solution steps

Rewrite

y^{2} + 6 y + 2 x + 1 = 0

in the standard form

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

(h, k) = (4, - 3), p = - \frac{1}{2}

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

p

from the center

(4, - 3)

along the x-axis

(4 + p, - 3)

= (4 + (- \frac{1}{2}), - 3)

Refine

(\frac{7}{2}, - 3)

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

(x + 3)^{2} = - 12 (y + 4)

y^{2} = - 32 x

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

2 y^{2} - 5 y + x - 4 = 0