What is the center ((x-h)^2}{64}-\frac{(y-k)^2)/9 =1 ?

The center ((x-h)^2}{64}-\frac{(y-k)^2)/9 =1 is (h,k)

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center ((x-h)^2}{64}-\frac{(y-k)^2)/9 =1

center

\frac{( x - h ) ^{2}}{64} - \frac{( y - k ) ^{2}}{9} = 1

(h, k)

Solution steps

\frac{( x - h ) ^{2}}{64} - \frac{( y - k ) ^{2}}{9} = 1

Rewrite

\frac{( x - h ) ^{2}}{64} - \frac{( y - k ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

\frac{( x - h ) ^{2}}{8 ^{2}} - \frac{( y - k ) ^{2}}{3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (h, k), a = 8, b = 3

And the center is:

(h, k)

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

\frac{y ^{2}}{4} - \frac{x ^{2}}{25} = 1

4 x^{2} - 3 y^{2} - 8 x - 8 = 0

0 = - \frac{400}{9} x^{2} + 16 y^{2} + 400

x^{2} - y^{2} + 7 x + 8 y = 5

What is the center ((x-h)^2}{64}-\frac{(y-k)^2)/9 =1 ?
The center ((x-h)^2}{64}-\frac{(y-k)^2)/9 =1 is (h,k)