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

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

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vertices ((x-4)^2)/3+(y^2)/9 =1

vertices

\frac{( x - 4 ) ^{2}}{3} + \frac{y ^{2}}{9} = 1

(4, 3), (4, - 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{3} + \frac{y ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (4, 0), b = 3, a = 3

(4, 0 + 3), (4, 0 - 3)

Refine

(4, 3), (4, - 3)

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

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

x^{2} + 3 y^{2} + 6 x - 6 y + 9 = 0

81 x^{2} + 16 y^{2} = 1296

25 x^{2} + 36 y^{2} - 350 x + 350 = 0

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