What is the foci 6x^2+12y-18=0 ?

The foci 6x^2+12y-18=0 is (0,1)

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foci 6x^2+12y-18=0

foci

6 x^{2} + 12 y - 18 = 0

(0, 1)

Solution steps

Rewrite

6 x^{2} + 12 y - 18 = 0

in the standard form:

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

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

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

p

from the center

(0, \frac{3}{2})

along the y-axis

(0, \frac{3}{2} + p)

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

Refine

(0, 1)

y = x^{2} - 3 x^{2}

y = x^{2} - 3 x^{1}

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

x^{2} + 2 x + 3

y^{2} - 12 y - 4 x + 28 = 0