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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

2 x - y^{2} = 4

f (x) = x^{2} - 24 x - 12

x^{2} = - 12 y

f (x) = x^{2} + 6 x + 8

4 x^{2} + 16 y^{2} - 4 x - 32 y + 1 = 0

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