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

The vertices x^2+1/3 y^2=1 is (0,sqrt(3)),(0,-sqrt(3))

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

vertices

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

(0, 3), (0, - 3)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 3, a = 1

(0, 0 + 3), (0, 0 - 3)

Refine

(0, 3), (0, - 3)

2 x^{2} + y^{2} = 50

9 x^{2} - 54 x + 4 y^{2} = - 45

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

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

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

What is the vertices x^2+1/3 y^2=1 ?
The vertices x^2+1/3 y^2=1 is (0,sqrt(3)),(0,-sqrt(3))