What is the vertices (x^2)/(64)+(y^2)/(16)=1 ?

The vertices (x^2)/(64)+(y^2)/(16)=1 is (8,0),(-8,0)

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vertices (x^2)/(64)+(y^2)/(16)=1

vertices

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{64} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 8, b = 4

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

Refine

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

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

y^{2} + 6 y + 9 = x + 8

x^{2} + 4 x - 4 y = 0

x^{2} + y^{2} - 4 x + 8 y - 5 = 0

x = 2 y^{2}

What is the vertices (x^2)/(64)+(y^2)/(16)=1 ?
The vertices (x^2)/(64)+(y^2)/(16)=1 is (8,0),(-8,0)