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

The vertices 3x^2+y^2+2x-6y=3 is (-1/3 ,3+sqrt(37/3)),(-1/3 ,3-sqrt(37/3))

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

vertices

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

(- \frac{1}{3}, 3 + \frac{37}{3}), (- \frac{1}{3}, 3 - \frac{37}{3})

Solution steps

(h, k + b), (h, k - b)

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 + \frac{37}{3}), (- \frac{1}{3}, 3 - \frac{37}{3})

x^{2} + \frac{1}{3} y^{2} = x + y

25 x^{2} + 9 y^{2} - 100 x + 18 y - 116 = 0

10 x^{2} + 4 y^{2} + 2 x + 16 y = 144

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

9 x^{2} + 16 y^{2} - 54 x + 64 y + 1 = 0

What is the vertices 3x^2+y^2+2x-6y=3 ?
The vertices 3x^2+y^2+2x-6y=3 is (-1/3 ,3+sqrt(37/3)),(-1/3 ,3-sqrt(37/3))