What is the center (x^2)/4-(y^2)/(12)=1 ?

The center (x^2)/4-(y^2)/(12)=1 is (0,0)

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center (x^2)/4-(y^2)/(12)=1

center

\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1

(0, 0)

Solution steps

\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1

Rewrite

\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} - \frac{( y - 0 ) ^{2}}{( 2 3 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, 0), a = 2, b = 23

And the center is:

(0, 0)

\frac{( x - 2 ) ^{2}}{4} - \frac{y ^{2}}{9} = 1

- 321 + 108 x - 9 x^{2} + 8 y + 4 y^{2} = 0

\frac{( y - 5 ) ^{2}}{81} - \frac{( x + 8 ) ^{2}}{9} = 1

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

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

What is the center (x^2)/4-(y^2)/(12)=1 ?
The center (x^2)/4-(y^2)/(12)=1 is (0,0)