What is the vertices (x^2)/(16)+(y^2)/(81)=1 ?

The vertices (x^2)/(16)+(y^2)/(81)=1 is (0,9),(0,-9)

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vertices (x^2)/(16)+(y^2)/(81)=1

vertices

\frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 9, a = 4

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

Refine

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

5 x^{2} + 2 y^{2} = 10

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0

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

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

25 x^{2} + 9 y^{2} + 50 x - 108 y + 124 = 0

What is the vertices (x^2)/(16)+(y^2)/(81)=1 ?
The vertices (x^2)/(16)+(y^2)/(81)=1 is (0,9),(0,-9)