What is the foci y^2+12y-4x+48=0 ?

The foci y^2+12y-4x+48=0 is (4,-6)

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foci y^2+12y-4x+48=0

foci

y^{2} + 12 y - 4 x + 48 = 0

(4, - 6)

Solution steps

Rewrite

y^{2} + 12 y - 4 x + 48 = 0

in the standard form

4 \cdot 1 \cdot (x - 3) = (y - (- 6))^{2}

(h, k) = (3, - 6), p = 1

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

p

from the center

(3, - 6)

along the x-axis

(3 + p, - 6)

= (3 + 1, - 6)

Refine

(4, - 6)

x^{2} + 1

x^{2} + 4

x^{2} + 9

3 x^{2} - x + 5

y = x^{2} + 2