What is the vertices (x^2)/(2^2)+(y^2)/(3^2)=1 ?

The vertices (x^2)/(2^2)+(y^2)/(3^2)=1 is (0,3),(0,-3)

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vertices (x^2)/(2^2)+(y^2)/(3^2)=1

vertices

\frac{x ^{2}}{2 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

(0, 3), (0, - 3)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{2 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 3, a = 2

(0, 0 + 3), (0, 0 - 3)

Refine

(0, 3), (0, - 3)

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

\frac{( x + 1 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{64} = 1

4 x^{2} + y^{2} - 48 x - 4 y + 48 = 0

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

7 x^{2} + 9 y^{2} - 63 = 0

What is the vertices (x^2)/(2^2)+(y^2)/(3^2)=1 ?
The vertices (x^2)/(2^2)+(y^2)/(3^2)=1 is (0,3),(0,-3)