What is the vertices (x^2}{49}+\frac{y^2)/4 =1 ?

The vertices (x^2}{49}+\frac{y^2)/4 =1 is (7,0),(-7,0)

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vertices (x^2}{49}+\frac{y^2)/4 =1

vertices

(7, 0), (- 7, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{49} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 7, b = 2

(0 + 7, 0), (0 - 7, 0)

Refine

(7, 0), (- 7, 0)

What is the vertices (x^2}{49}+\frac{y^2)/4 =1 ?
The vertices (x^2}{49}+\frac{y^2)/4 =1 is (7,0),(-7,0)