What is the vertices (x^2)/(144)+(y^2)/(169)=1 ?

The vertices (x^2)/(144)+(y^2)/(169)=1 is (0,13),(0,-13)

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vertices (x^2)/(144)+(y^2)/(169)=1

vertices

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

(0, 13), (0, - 13)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 13, a = 12

(0, 0 + 13), (0, 0 - 13)

Refine

(0, 13), (0, - 13)

9 x^{2} + y^{2} - 18 x - 6 y + 9 = 0

\frac{( x - 0 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{49} = 1

81 x^{2} + 64 y^{2} = 5184

16 x^{2} + 25 y^{2} - 64 x + 150 y = - 279

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{5} = 1

What is the vertices (x^2)/(144)+(y^2)/(169)=1 ?
The vertices (x^2)/(144)+(y^2)/(169)=1 is (0,13),(0,-13)