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

The vertices (x)^2+((y-1)^2)/4 =1 is (0,3),(0,-1)

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

vertices

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

(0, 3), (0, - 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(0, 1 + 2), (0, 1 - 2)

Refine

(0, 3), (0, - 1)

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

25 x^{2} + 16 y^{2} - 50 x + 64 y - 311 = 0

16 x^{2} + 36 y^{2} - 256 x + 448 = 0

9 x^{2} + 25 y^{2} - 36 x - 50 y + 60 = 0

\frac{( x + 1 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{48} = 1

What is the vertices (x)^2+((y-1)^2)/4 =1 ?
The vertices (x)^2+((y-1)^2)/4 =1 is (0,3),(0,-1)