What is the vertices 5x^2+2y^2=10 ?

The vertices 5x^2+2y^2=10 is (0,sqrt(5)),(0,-sqrt(5))

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vertices 5x^2+2y^2=10

vertices

5 x^{2} + 2 y^{2} = 10

(0, 5), (0, - 5)

Solution steps

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

Calculate ellipse properties

5 x^{2} + 2 y^{2} = 10 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 2

(0, 0 + 5), (0, 0 - 5)

Refine

(0, 5), (0, - 5)

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0

4 x^{2} + 16 y^{2} - 64 = 0

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{4} = 1

25 x^{2} + 9 y^{2} + 50 x - 108 y + 124 = 0

25 x^{2} + 4 y^{2} - 250 x - 16 y + 541 = 0

What is the vertices 5x^2+2y^2=10 ?
The vertices 5x^2+2y^2=10 is (0,sqrt(5)),(0,-sqrt(5))