What is the vertices (x^2)/4+((y+3)^2)/9 =1 ?

The vertices (x^2)/4+((y+3)^2)/9 =1 is (0,0),(0,-6)

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vertices (x^2)/4+((y+3)^2)/9 =1

vertices

\frac{x ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (0, - 3), b = 3, a = 2

(0, - 3 + 3), (0, - 3 - 3)

Refine

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

16 y^{2} + 25 x^{2} - 80 x + 36 y + 80 = 0

4 x^{2} + 9 y^{2} - 48 x + 72 y = - 144

\frac{( x - 1 ) ^{2}}{45} + \frac{( y - 2 ) ^{2}}{20} = 1

4 x^{2} + 9 y^{2} - 16 x - 18 y - 11 = 0

x^{2} + 3 y^{2} = 6

What is the vertices (x^2)/4+((y+3)^2)/9 =1 ?
The vertices (x^2)/4+((y+3)^2)/9 =1 is (0,0),(0,-6)