What is the vertices (x^2)/4+(y^2)/(25)=1 ?

The vertices (x^2)/4+(y^2)/(25)=1 is (0,5),(0,-5)

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vertices (x^2)/4+(y^2)/(25)=1

vertices

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 2

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

Refine

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

x^{2} + y^{2} - 4 y = 0

9 x^{2} - 18 x + 4 y^{2} = 27

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

4 x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

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

What is the vertices (x^2)/4+(y^2)/(25)=1 ?
The vertices (x^2)/4+(y^2)/(25)=1 is (0,5),(0,-5)