What is the vertices (x^2)/(25)+(y^2)/(36)=1 ?

The vertices (x^2)/(25)+(y^2)/(36)=1 is (0,6),(0,-6)

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vertices (x^2)/(25)+(y^2)/(36)=1

vertices

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

(0, 6), (0, - 6)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 6, a = 5

(0, 0 + 6), (0, 0 - 6)

Refine

(0, 6), (0, - 6)

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

\frac{x ^{2}}{36} + \frac{y ^{2}}{49} = 1

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

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

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1

What is the vertices (x^2)/(25)+(y^2)/(36)=1 ?
The vertices (x^2)/(25)+(y^2)/(36)=1 is (0,6),(0,-6)