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)

2 y^{2} - 5 x^{2} - 56 y - 90 x - 807 = 0

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

\frac{x ^{2}}{0.0008} - \frac{y ^{2}}{0.999} = 1

\frac{x ^{2}}{36} - \frac{y ^{2}}{16} = 1

\frac{( x + 5 ) ^{2}}{64} - \frac{( y - 1 ) ^{2}}{36} = 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)