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

The foci x^2+2y-6=0 is (0, 5/2)

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foci x^2+2y-6=0

foci

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

(0, \frac{5}{2})

Solution steps

Rewrite

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

in the standard form:

4 (- \frac{1}{2}) (y - 3) = (x - 0)^{2}

(h, k) = (0, 3), p = - \frac{1}{2}

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

p

from the center

(0, 3)

along the y-axis

(0, 3 + p)

= (0, 3 + (- \frac{1}{2}))

Refine

(0, \frac{5}{2})

x^{2} - 4 x + 16

x^{2} - 6 x + 8

y^{2} = - 6 x

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

y = - 12 x^{2} - 29 x + 230