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

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

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

vertices

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

(0, 0), (0, - 4)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

(0, 0), (0, - 4)

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1

4 x^{2} + 25 y^{2} - 100 = 0

16 x^{2} + 25 y^{2} - 400 = 0

\frac{x ^{2}}{9} + y^{2} = 1

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

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