What is the vertices 5/81 x^2+(y^2)/9 =1 ?

The vertices 5/81 x^2+(y^2)/9 =1 is (9/(sqrt(5)),0),(-9/(sqrt(5)),0)

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vertices 5/81 x^2+(y^2)/9 =1

vertices

\frac{5}{81} x^{2} + \frac{y ^{2}}{9} = 1

(\frac{9}{5}, 0), (- \frac{9}{5}, 0)

Solution steps

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

Calculate ellipse properties

\frac{5}{81} x^{2} + \frac{y ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (0, 0), a = \frac{9}{5}, b = 3

(0 + \frac{9}{5}, 0), (0 - \frac{9}{5}, 0)

Refine

(\frac{9}{5}, 0), (- \frac{9}{5}, 0)

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x^{2} + 3 y^{2} - 4 x + 12 y + 13 = 0

16 x^{2} + 25 y^{2} + 64 x - 200 y + 64 = 0

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

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

What is the vertices 5/81 x^2+(y^2)/9 =1 ?
The vertices 5/81 x^2+(y^2)/9 =1 is (9/(sqrt(5)),0),(-9/(sqrt(5)),0)