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

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

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

center

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

(2, 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, 1), a = 3, b = 2

And the center is:

(2, 1)

25 x^{2} + 144 y^{2} = 900

x^{2} + 4 \cdot y^{2} - 4 \cdot y = 0

36 x^{2} + 288 x + 64 y^{2} - 256 y = 1472

16 x^{2} + 25 y^{2} = 400

x^{2} + 4 y^{2} + 7 = 12 y - 2 x

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