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

The center (x^2)/(64)-(y^2)/(36)=1 is (0,0)

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

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{8 ^{2}} - \frac{( y - 0 ) ^{2}}{6 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, 0), a = 8, b = 6

And the center is:

(0, 0)

3 (x + 2)^{2} - 6 (y - 6)^{2} - 18 = 0

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

y^{2} - 4 y - 4 x^{2} + 13 = 0

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1

\frac{( x - 1 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{16} = 1

What is the center (x^2)/(64)-(y^2)/(36)=1 ?
The center (x^2)/(64)-(y^2)/(36)=1 is (0,0)