What is the foci (y-2)^2=(x-1)^2+(y+4)^2 ?

The foci (y-2)^2=(x-1)^2+(y+4)^2 is (1,-3)

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foci (y-2)^2=(x-1)^2+(y+4)^2

foci

(y - 2)^{2} = (x - 1)^{2} + (y + 4)^{2}

(1, - 3)

Solution steps

Rewrite

(y - 2)^{2} = (x - 1)^{2} + (y + 4)^{2}

in the standard form

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

(h, k) = (1, 0), p = - 3

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

p

from the center

(1, 0)

along the y-axis

(1, 0 + p)

= (1, 0 + (- 3))

Refine

(1, - 3)

x^{(2)} - 6 x + 5 y = - 34

x (2 x + 3)

6 x^{2} - 49

x = + (y - 4)^{2} - 1

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

What is the foci (y-2)^2=(x-1)^2+(y+4)^2 ?
The foci (y-2)^2=(x-1)^2+(y+4)^2 is (1,-3)