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

The vertices ((x-2)^2)/9+((y-1)^2)/4 =1 is (5,1),(-1,1)

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

vertices

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

(5, 1), (- 1, 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, 1), a = 3, b = 2

(2 + 3, 1), (2 - 3, 1)

Refine

(5, 1), (- 1, 1)

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

x^{2} + y^{2} - 10 y = 0

25 x^{2} + 4 y^{2} + 200 x - 8 y + 304 = 0

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

3 x^{2} + 7 y^{2} - 21 = 0

What is the vertices ((x-2)^2)/9+((y-1)^2)/4 =1 ?
The vertices ((x-2)^2)/9+((y-1)^2)/4 =1 is (5,1),(-1,1)