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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

9 (x - 1)^{2} - 16 (y + 2)^{2} = 144

9 x^{2} + 4 y^{2} + 36 x - 24 y + 36 = 0

f (x) = 3 x^{2} - 18 x + 23

x^{2} + y^{2} \leq 25

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

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