What is the vertices 16x^2+64y^2=128 ?

The vertices 16x^2+64y^2=128 is (2sqrt(2),0),(-2sqrt(2),0)

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vertices 16x^2+64y^2=128

vertices

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

(22, 0), (- 22, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = 22, b = 2

(0 + 22, 0), (0 - 22, 0)

Refine

(22, 0), (- 22, 0)

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

16 y^{2} + 25 x^{2} - 80 x + 36 y + 80 = 0

4 x^{2} + 9 y^{2} - 48 x + 72 y = - 144

\frac{( x - 1 ) ^{2}}{45} + \frac{( y - 2 ) ^{2}}{20} = 1

4 x^{2} + 9 y^{2} - 16 x - 18 y - 11 = 0

What is the vertices 16x^2+64y^2=128 ?
The vertices 16x^2+64y^2=128 is (2sqrt(2),0),(-2sqrt(2),0)