What is the vertices (x^2)/4+((y+2)^2)/(16)=1 ?

The vertices (x^2)/4+((y+2)^2)/(16)=1 is (0,2),(0,-6)

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vertices (x^2)/4+((y+2)^2)/(16)=1

vertices

\frac{x ^{2}}{4} + \frac{( y + 2 ) ^{2}}{16} = 1

(0, 2), (0, - 6)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{( y + 2 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, - 2), b = 4, a = 2

(0, - 2 + 4), (0, - 2 - 4)

Refine

(0, 2), (0, - 6)

\frac{( x + 6 ) ^{2}}{36} + \frac{( y - 7 ) ^{2}}{32} = 1

16 x^{2} + 4 y^{2} + 96 x - 24 y + 116 = 0

\frac{( x - 4 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{64} = 1

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

\frac{( x - 2 ) ^{2}}{16} + (y - 5)^{2} = 1

What is the vertices (x^2)/4+((y+2)^2)/(16)=1 ?
The vertices (x^2)/4+((y+2)^2)/(16)=1 is (0,2),(0,-6)