What is the foci x^2+3y^2-4x+12y+13=0 ?

The foci x^2+3y^2-4x+12y+13=0 is (2+sqrt(2),-2),(2-sqrt(2),-2)

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foci x^2+3y^2-4x+12y+13=0

foci

x^{2} + 3 y^{2} - 4 x + 12 y + 13 = 0

(2 + 2, - 2), (2 - 2, - 2)

Solution steps

(h + c, k), (h - c, k)

Calculate ellipse properties

x^{2} + 3 y^{2} - 4 x + 12 y + 13 = 0 :

Ellipse with center

(h, k) = (2, - 2), a = 3, b = 1

(2 + c, - 2), (2 - c, - 2)

Compute

c :

c = 3^{2} - 1^{2} : 2

(2 + 2, - 2), (2 - 2, - 2)

49 x^{2} + 25 y^{2} = 1225

\frac{x ^{2}}{64} + \frac{y ^{2}}{49} = 1

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0

16 x^{2} - 64 x + 9 y^{2} = 80

4 x^{2} + 9 y^{2} - 8 x - 32 = 0

What is the foci x^2+3y^2-4x+12y+13=0 ?
The foci x^2+3y^2-4x+12y+13=0 is (2+sqrt(2),-2),(2-sqrt(2),-2)