What is the vertices 300x^2+150y^2=50 ?

The vertices 300x^2+150y^2=50 is (0,sqrt(1/3)),(0,-sqrt(1/3))

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vertices 300x^2+150y^2=50

vertices

(0, \frac{1}{3}), (0, - \frac{1}{3})

Solution steps

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

Calculate ellipse properties

300 x^{2} + 150 y^{2} = 50 :

Ellipse with center

(h, k) = (0, 0), b = \frac{1}{3}, a = \frac{1}{6}

(0, 0 + \frac{1}{3}), (0, 0 - \frac{1}{3})

Refine

(0, \frac{1}{3}), (0, - \frac{1}{3})

What is the vertices 300x^2+150y^2=50 ?
The vertices 300x^2+150y^2=50 is (0,sqrt(1/3)),(0,-sqrt(1/3))