What is the vertices (x^2)/(81)+(y^2)/(64)=1 ?

The vertices (x^2)/(81)+(y^2)/(64)=1 is (9,0),(-9,0)

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vertices (x^2)/(81)+(y^2)/(64)=1

vertices

\frac{x ^{2}}{81} + \frac{y ^{2}}{64} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{81} + \frac{y ^{2}}{64} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 9, b = 8

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

Refine

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

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{36} = 1

x^{2} + 3 y^{2} = 84

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

\frac{( x - 2 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{16} = 1

What is the vertices (x^2)/(81)+(y^2)/(64)=1 ?
The vertices (x^2)/(81)+(y^2)/(64)=1 is (9,0),(-9,0)