What is the vertices (x^2)/4+(y^2)/2 =1 ?

The vertices (x^2)/4+(y^2)/2 =1 is (2,0),(-2,0)

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vertices (x^2)/4+(y^2)/2 =1

vertices

\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1

(2, 0), (- 2, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1 :

Ellipse with center

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

(0 + 2, 0), (0 - 2, 0)

Refine

(2, 0), (- 2, 0)

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1

3 x^{2} + 2 y^{2} + 6 x - 4 y + 5 = 18

\frac{x ^{2}}{49} + \frac{y ^{2}}{25} = 1

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{169} = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{y ^{2}}{9} = 1

What is the vertices (x^2)/4+(y^2)/2 =1 ?
The vertices (x^2)/4+(y^2)/2 =1 is (2,0),(-2,0)