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

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

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

center

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

(1, - 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(1, - 1)

9 x^{2} + 25 y^{2} = 225

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

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

\frac{( x + 3 ) ^{2}}{25} + \frac{( y - 8 ) ^{2}}{16} = 1

x^{2} + 4 y^{2} - 6 x + 20 y - 2 = 0

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