What is the vertices (x^2)/(16)+(y^2)/(49)=1 ?

The vertices (x^2)/(16)+(y^2)/(49)=1 is (0,7),(0,-7)

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vertices (x^2)/(16)+(y^2)/(49)=1

vertices

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 7, a = 4

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

Refine

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

4 x^{2} + 9 y^{2} - 32 x - 36 y + 64 = 0

x^{2} + 16 y^{2} = 1

2 x^{2} + 7 y^{2} = 14

16 x^{2} + 25 y^{2} - 1600 = 0

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

What is the vertices (x^2)/(16)+(y^2)/(49)=1 ?
The vertices (x^2)/(16)+(y^2)/(49)=1 is (0,7),(0,-7)