What is the foci 3x^2+y^2+2x-6y=3 ?

The foci 3x^2+y^2+2x-6y=3 is (-1/3 ,(9+sqrt(74))/3),(-1/3 ,(9-sqrt(74))/3)

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foci 3x^2+y^2+2x-6y=3

foci

3 x^{2} + y^{2} + 2 x - 6 y = 3

(- \frac{1}{3}, \frac{9 + 74}{3}), (- \frac{1}{3}, \frac{9 - 74}{3})

Solution steps

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

Calculate ellipse properties

3 x^{2} + y^{2} + 2 x - 6 y = 3 :

Ellipse with center

(h, k) = (- \frac{1}{3}, 3), b = \frac{37}{3}, a = \frac{37}{3}

(- \frac{1}{3}, 3 + c), (- \frac{1}{3}, 3 - c)

Compute

c :

c = \frac{37}{3}^{2} - (\frac{37}{3})^{2} : \frac{74}{3}

(- \frac{1}{3}, 3 + \frac{74}{3}), (- \frac{1}{3}, 3 - \frac{74}{3})

Refine

(- \frac{1}{3}, \frac{9 + 74}{3}), (- \frac{1}{3}, \frac{9 - 74}{3})

9 x^{2} + 25 y^{2} + 18 x + 150 y + 9 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{49} = 1

\frac{( x + 5 ) ^{2}}{9} + (y - 1)^{2} = 1

x^{2} + y^{2} + 2 x = 3 y

\frac{x ^{2}}{49} + \frac{y ^{2}}{9} = 1

What is the foci 3x^2+y^2+2x-6y=3 ?
The foci 3x^2+y^2+2x-6y=3 is (-1/3 ,(9+sqrt(74))/3),(-1/3 ,(9-sqrt(74))/3)