What is the vertices x^2+(y^2)/(49)=1 ?

The vertices x^2+(y^2)/(49)=1 is (0,7),(0,-7)

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vertices x^2+(y^2)/(49)=1

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 7, a = 1

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

Refine

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

4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

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

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1

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