What is the center (x^2)/(100)-(y^2)/(64)=1 ?

The center (x^2)/(100)-(y^2)/(64)=1 is (0,0)

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center (x^2)/(100)-(y^2)/(64)=1

center

\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1

Rewrite

\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (0, 0), a = 10, b = 8

And the center is:

(0, 0)

9 y^{2} - 25 x^{2} - 100 x - 108 y - 1 = 0

9 x^{2} - 4 y^{2} + 36 x + 8 y + 33 = 0

\frac{y ^{2}}{5} - \frac{x ^{2}}{5} = 1

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

x^{2} - y^{2} - 2 x - 2 y - 1 = 0

What is the center (x^2)/(100)-(y^2)/(64)=1 ?
The center (x^2)/(100)-(y^2)/(64)=1 is (0,0)