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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

\frac{y ^{2}}{4} - \frac{x ^{2}}{5} = 1

25 x^{2} + 9 y^{2} = 225

\frac{x ^{2}}{100} + \frac{y ^{2}}{64} = 1

4 x^{2} - 20 x - 24 y + 97 = 0

x^{2} + y^{2} = 16

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