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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{2} + \frac{y ^{2}}{3} = 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 y^{2} - 6 x - 16 y = 36

(\frac{x + 5}{4})^{2} + (\frac{y - 4}{16})^{2} = 1

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

x^{2} + 9 y^{2} + 4 x = 5

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1

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