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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

x^{2} + \frac{( y - 6 ) ^{2}}{16} = 1

4 x^{2} + 81 y^{2} = 324

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{20} = 1

\frac{x ^{2}}{64} + \frac{y ^{2}}{9} = 1

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{4} = 1

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