What is the foci (x-5)^2=12(y-3) ?

The foci (x-5)^2=12(y-3) is (5,6)

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foci (x-5)^2=12(y-3)

foci

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

(5, 6)

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (5, 3), p = 3

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

p

from the center

(5, 3)

along the y-axis

(5, 3 + p)

= (5, 3 + 3)

Refine

(5, 6)

x^{2} - 6 x - 5 y - 1 = 0

y = (x + 4)^{2}

(x + 1)^{2} = 12 (y - 3)

6 y^{2} - 12 x = 0

x^{2} - 16