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

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

(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})

\frac{x ^{2}}{169} + \frac{y ^{2}}{144} = 1

7 x^{2} + 16 y^{2} = 7

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0

9 x^{2} + 25 y^{2} - 225 = 0

4 x^{2} - 48 x + 9 y^{2} + 72 y = - 144

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))