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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 2, a = 1

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

Refine

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

\frac{x ^{2}}{100} + \frac{y ^{2}}{44} = 1

\frac{x ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

25 x^{2} + 144 y^{2} = 900

4 x^{2} + 16 y^{2} - 16 x - 48 = 0

25 (x + 2)^{2} + 4 (y - 3)^{2} = 100

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