What is the foci y^2-4x+6y+17=0 ?

The foci y^2-4x+6y+17=0 is (3,-3)

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foci y^2-4x+6y+17=0

foci

y^{2} - 4 x + 6 y + 17 = 0

(3, - 3)

Solution steps

Rewrite

y^{2} - 4 x + 6 y + 17 = 0

in the standard form

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

(h, k) = (2, - 3), p = 1

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

p

from the center

(2, - 3)

along the x-axis

(2 + p, - 3)

= (2 + 1, - 3)

Refine

(3, - 3)

y = x^{2} - 6 x + 5

(x - 2)^{2} = 8 (y - 1)

x^{2} - 2 x - 15

- 2 x^{2} - 4 x + y + 70 = 0

y = + 0.5 (x + 2) (x - 4)