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

The vertices 2x^2+y^2=50 is (0,5sqrt(2)),(0,-5sqrt(2))

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vertices 2x^2+y^2=50

vertices

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

(0, 52), (0, - 52)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 52, a = 5

(0, 0 + 52), (0, 0 - 52)

Refine

(0, 52), (0, - 52)

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

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

What is the vertices 2x^2+y^2=50 ?
The vertices 2x^2+y^2=50 is (0,5sqrt(2)),(0,-5sqrt(2))