What is the vertices (x^2)/(64)+(y^2)/(121)=1 ?

The vertices (x^2)/(64)+(y^2)/(121)=1 is (0,11),(0,-11)

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vertices (x^2)/(64)+(y^2)/(121)=1

vertices

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1

(0, 11), (0, - 11)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 11, a = 8

(0, 0 + 11), (0, 0 - 11)

Refine

(0, 11), (0, - 11)

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

\frac{x ^{2}}{6.25} + \frac{y ^{2}}{4.6875} = 1

\frac{x ^{2}}{15} + \frac{y ^{2}}{45} = 1

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

\frac{x ^{2}}{25} + \frac{( y + 3 ) ^{2}}{49} = 1

What is the vertices (x^2)/(64)+(y^2)/(121)=1 ?
The vertices (x^2)/(64)+(y^2)/(121)=1 is (0,11),(0,-11)