What is the vertices x^2+1/3 y^2=x+y ?

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

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vertices x^2+1/3 y^2=x+y

vertices

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

(\frac{1}{2}, \frac{3 + 2 3}{2}), (\frac{1}{2}, \frac{3 - 2 3}{2})

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (\frac{1}{2}, \frac{3}{2}), b = 3, a = 1

(\frac{1}{2}, \frac{3}{2} + 3), (\frac{1}{2}, \frac{3}{2} - 3)

Refine

(\frac{1}{2}, \frac{3 + 2 3}{2}), (\frac{1}{2}, \frac{3 - 2 3}{2})

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

x^{2} + 8 y^{2} = 32

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