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

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

(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)

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

(x^{2} + 8 (y + 5)^{2}) = 20

x^{2} + 4 y^{2} = 4 x + 8

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

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 5 ) ^{2}}{81} = 1

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)