What is the vertices ((x^2))/9+((y^2))/4 =1 ?

The vertices ((x^2))/9+((y^2))/4 =1 is (3,0),(-3,0)

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vertices ((x^2))/9+((y^2))/4 =1

vertices

\frac{( x ^{2} )}{9} + \frac{( y ^{2} )}{4} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 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}}{81} + \frac{y ^{2}}{9} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{49} = 1

9 x^{2} + 81 y^{2} - 729 = 0

16 x^{2} + 4 y^{2} = 64

y^{2} = 1 - 2 x^{2}

What is the vertices ((x^2))/9+((y^2))/4 =1 ?
The vertices ((x^2))/9+((y^2))/4 =1 is (3,0),(-3,0)