What is the vertices (x^2)/(81)+(y^2)/(225)=1 ?

The vertices (x^2)/(81)+(y^2)/(225)=1 is (0,15),(0,-15)

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vertices (x^2)/(81)+(y^2)/(225)=1

vertices

\frac{x ^{2}}{81} + \frac{y ^{2}}{225} = 1

(0, 15), (0, - 15)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{81} + \frac{y ^{2}}{225} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 15, a = 9

(0, 0 + 15), (0, 0 - 15)

Refine

(0, 15), (0, - 15)

25 x^{2} + 16 y^{2} + 100 x - 96 y = 156

300 x^{2} + 150 y^{2} = 50

\frac{x ^{2}}{169} + \frac{y ^{2}}{144} = 1

7 x^{2} + 16 y^{2} = 7

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0

What is the vertices (x^2)/(81)+(y^2)/(225)=1 ?
The vertices (x^2)/(81)+(y^2)/(225)=1 is (0,15),(0,-15)