What is the vertices (x^2)/(36)+(y^2)/(81)=1 ?

The vertices (x^2)/(36)+(y^2)/(81)=1 is (0,9),(0,-9)

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vertices (x^2)/(36)+(y^2)/(81)=1

vertices

\frac{x ^{2}}{36} + \frac{y ^{2}}{81} = 1

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

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

\frac{x ^{2}}{36} + \frac{y ^{2}}{81} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 9, a = 6

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

Refine

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

9 x^{2} + 36 x + y^{2} + 2 y + 36 = 0

27 x^{2} + y^{2} - 18 y = 0

\frac{( x - 2 ) ^{2}}{5} + \frac{( y + 3 ) ^{2}}{9} = 1

25 x^{2} + y^{2} - 4 y - 21 = 0

9 x^{2} - 72 x + y^{2} + 4 y + 112 = 0

What is the vertices (x^2)/(36)+(y^2)/(81)=1 ?
The vertices (x^2)/(36)+(y^2)/(81)=1 is (0,9),(0,-9)