What is the foci x^2+3y^2+2x-12y+10=0 ?

The foci x^2+3y^2+2x-12y+10=0 is (-1+sqrt(2),2),(-1-sqrt(2),2)

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foci x^2+3y^2+2x-12y+10=0

foci

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0

(- 1 + 2, 2), (- 1 - 2, 2)

Solution steps

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

Calculate ellipse properties

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0 :

Ellipse with center

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

(- 1 + c, 2), (- 1 - c, 2)

Compute

c :

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

(- 1 + 2, 2), (- 1 - 2, 2)

\frac{x ^{2}}{36} + \frac{y ^{2}}{11} = 1

\frac{x ^{2}}{256} + \frac{y ^{2}}{400} = 1

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

25 x^{2} + 16 y^{2} = 400

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

What is the foci x^2+3y^2+2x-12y+10=0 ?
The foci x^2+3y^2+2x-12y+10=0 is (-1+sqrt(2),2),(-1-sqrt(2),2)