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

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

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

vertices

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

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

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

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

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

\frac{x ^{2}}{56} + \frac{y ^{2}}{36} = 1

16 (x - 1)^{2} + 25 (y - 4)^{2} - 400 = 0

9 x^{2} + 4 y^{2} + 36 x + 24 y + 36 = 0

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