What is the vertices (x^2)/(25)+(y^2)/(169)=1 ?

The vertices (x^2)/(25)+(y^2)/(169)=1 is (0,13),(0,-13)

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vertices (x^2)/(25)+(y^2)/(169)=1

vertices

\frac{x ^{2}}{25} + \frac{y ^{2}}{169} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{25} + \frac{y ^{2}}{169} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 13, a = 5

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

Refine

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

6 x^{2} + 9 y^{2} - 18 y - 45 = 0

7 x^{2} + 16 y^{2} - 112 = 0

9 x^{2} + 25 y^{2} + 90 x = 0

2 x^{2} + 3 y^{2} + 4 x - 6 y - 1 = 0

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

What is the vertices (x^2)/(25)+(y^2)/(169)=1 ?
The vertices (x^2)/(25)+(y^2)/(169)=1 is (0,13),(0,-13)