What is the vertices (x^2}{64}+\frac{y^2)/4 =1 ?

The vertices (x^2}{64}+\frac{y^2)/4 =1 is (8,0),(-8,0)

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vertices (x^2}{64}+\frac{y^2)/4 =1

vertices

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1

(8, 0), (- 8, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 8, b = 2

(0 + 8, 0), (0 - 8, 0)

Refine

(8, 0), (- 8, 0)

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

\frac{x ^{2}}{9} + \frac{( y + 5 ) ^{2}}{25} = 1

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

x^{2} + 4 y^{2} - 6 x + 8 y - 3 = 0

100 x^{2} + 96 y^{2} - 8840 = 0

What is the vertices (x^2}{64}+\frac{y^2)/4 =1 ?
The vertices (x^2}{64}+\frac{y^2)/4 =1 is (8,0),(-8,0)