What is the vertices (x^2)/(81)+(y^2)/(49)=1 ?

The vertices (x^2)/(81)+(y^2)/(49)=1 is (9,0),(-9,0)

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vertices (x^2)/(81)+(y^2)/(49)=1

vertices

\frac{x ^{2}}{81} + \frac{y ^{2}}{49} = 1

(9, 0), (- 9, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{81} + \frac{y ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 9, b = 7

(0 + 9, 0), (0 - 9, 0)

Refine

(9, 0), (- 9, 0)

9 x^{2} + 4 y^{2} - 36 x + 40 y + 100 = 0

9 x^{2} + 16 y^{2} + 96 y = 0

49 x^{2} + 25 y^{2} - 1225 = 0

16 x^{2} + 64 y^{2} = 128

\frac{x ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

What is the vertices (x^2)/(81)+(y^2)/(49)=1 ?
The vertices (x^2)/(81)+(y^2)/(49)=1 is (9,0),(-9,0)