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

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

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

center

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

(2, - 2)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(2, - 2)

\frac{( y - ( - 2 ) ) ^{2}}{9} - \frac{( x - 5 ) ^{2}}{27} = 1

\frac{( y - 2 ) ^{2}}{14} - \frac{( x + 6 ) ^{2}}{3} = 4

4 x^{2} - 8 y^{2} = 3

4 x^{2} - 6 x + 2 = y^{2} - 10 y

4 x^{2} - y^{2} + 8 x - 2 y + 7 = 0

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