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

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

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

vertices

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

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

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

7 x^{2} + 16 y^{2} = 7

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0

9 x^{2} + 25 y^{2} - 225 = 0

4 x^{2} - 48 x + 9 y^{2} + 72 y = - 144

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

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