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

The vertices (x^2)/5+(y^2)/9 =1 is (0,3),(0,-3)

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

\frac{( x + 4 ) ^{2}}{5} + \frac{( y - 6 ) ^{2}}{9} = 1

2 x^{2} + 2 y^{2} - 28 x + 12 y + 114 = 0

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

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

x^{2} + y^{2} - 10 y = 0

What is the vertices (x^2)/5+(y^2)/9 =1 ?
The vertices (x^2)/5+(y^2)/9 =1 is (0,3),(0,-3)