What is the vertices (x^2)/9+((y+5)^2)/(25)=1 ?

The vertices (x^2)/9+((y+5)^2)/(25)=1 is (0,0),(0,-10)

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vertices (x^2)/9+((y+5)^2)/(25)=1

vertices

\frac{x ^{2}}{9} + \frac{( y + 5 ) ^{2}}{25} = 1

(0, 0), (0, - 10)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{9} + \frac{( y + 5 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, - 5), b = 5, a = 3

(0, - 5 + 5), (0, - 5 - 5)

Refine

(0, 0), (0, - 10)

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

x^{2} + 4 y^{2} - 6 x + 8 y - 3 = 0

100 x^{2} + 96 y^{2} - 8840 = 0

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{25} = 1

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

What is the vertices (x^2)/9+((y+5)^2)/(25)=1 ?
The vertices (x^2)/9+((y+5)^2)/(25)=1 is (0,0),(0,-10)