What is the center ((x-2)^2)/9-((y-1)^2)/(16)=1 ?

The center ((x-2)^2)/9-((y-1)^2)/(16)=1 is (2,1)

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center ((x-2)^2)/9-((y-1)^2)/(16)=1

center

\frac{( x - 2 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{16} = 1

(2, 1)

Solution steps

\frac{( x - 2 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, 1), a = 3, b = 4

And the center is:

(2, 1)

\frac{( y - 5 ) ^{2}}{16} - \frac{( x - 4 ) ^{2}}{64} = 1

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y - 2 = 0

y^{2} - x^{2} = 1

9 x^{2} - 18 x - 24 y - 63 = 4 y^{2}

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

What is the center ((x-2)^2)/9-((y-1)^2)/(16)=1 ?
The center ((x-2)^2)/9-((y-1)^2)/(16)=1 is (2,1)