What is the vertices (x^2)/9+(y^2)/(36)=1 ?

The vertices (x^2)/9+(y^2)/(36)=1 is (0,6),(0,-6)

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vertices (x^2)/9+(y^2)/(36)=1

vertices

\frac{x ^{2}}{9} + \frac{y ^{2}}{36} = 1

(0, 6), (0, - 6)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{9} + \frac{y ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 6, a = 3

(0, 0 + 6), (0, 0 - 6)

Refine

(0, 6), (0, - 6)

16 x^{2} + 25 y^{2} = 1600

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

16 x^{2} - 96 x + 9 y^{2} = 0

4 x^{2} + 25 y^{2} - 50 y = 75

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{16} = 1

What is the vertices (x^2)/9+(y^2)/(36)=1 ?
The vertices (x^2)/9+(y^2)/(36)=1 is (0,6),(0,-6)