What is the foci y^2-12y+6x+36=0 ?

The foci y^2-12y+6x+36=0 is (-3/2 ,6)

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foci y^2-12y+6x+36=0

foci

y^{2} - 12 y + 6 x + 36 = 0

(- \frac{3}{2}, 6)

Solution steps

Rewrite

y^{2} - 12 y + 6 x + 36 = 0

in the standard form

4 (- \frac{3}{2}) (x - 0) = (y - 6)^{2}

(h, k) = (0, 6), p = - \frac{3}{2}

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

p

from the center

(0, 6)

along the x-axis

(0 + p, 6)

= (0 + (- \frac{3}{2}), 6)

Refine

(- \frac{3}{2}, 6)

x^{2} - 6 x - 20 y - 11 = 0

y = 3 x^{2} - 36 x + 81

(y - 6)^{2}

y = (x - 2)^{2} - 4

y^{2} = - 6 (x - 3)