What is the foci x^2-2x-4y+13=0 ?

The foci x^2-2x-4y+13=0 is (1,4)

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foci x^2-2x-4y+13=0

foci

x^{2} - 2 x - 4 y + 13 = 0

(1, 4)

Solution steps

Rewrite

x^{2} - 2 x - 4 y + 13 = 0

in the standard form

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

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

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

p

from the center

(1, 3)

along the y-axis

(1, 3 + p)

= (1, 3 + 1)

Refine

(1, 4)

2 y^{2} - 16 y + 72 = 20 x

y^{2} + 4 x - 8 y + 12 = 0

- x^{2} + 4 x - 1

+ x^{2} + 3 x - 4

- x^{2} + 4 x - 9