What is the center ((x-2)^2)/(36)-((y+1)^2)/(64)=1 ?

The center ((x-2)^2)/(36)-((y+1)^2)/(64)=1 is (2,-1)

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center ((x-2)^2)/(36)-((y+1)^2)/(64)=1

center

\frac{( x - 2 ) ^{2}}{36} - \frac{( y + 1 ) ^{2}}{64} = 1

(2, - 1)

Solution steps

\frac{( x - 2 ) ^{2}}{36} - \frac{( y + 1 ) ^{2}}{64} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{36} - \frac{( y + 1 ) ^{2}}{64} = 1

in the form of a standard hyperbola equation

\frac{( x - 2 ) ^{2}}{6 ^{2}} - \frac{( y - ( - 1 ) ) ^{2}}{8 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (2, - 1), a = 6, b = 8

And the center is:

(2, - 1)

What is the center ((x-2)^2)/(36)-((y+1)^2)/(64)=1 ?
The center ((x-2)^2)/(36)-((y+1)^2)/(64)=1 is (2,-1)