What is the center 3x^2-3y^2-12x-6y+3=0 ?

The center 3x^2-3y^2-12x-6y+3=0 is (2,-1)

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center 3x^2-3y^2-12x-6y+3=0

center

3 x^{2} - 3 y^{2} - 12 x - 6 y + 3 = 0

(2, - 1)

Solution steps

3 x^{2} - 3 y^{2} - 12 x - 6 y + 3 = 0

Rewrite

3 x^{2} - 3 y^{2} - 12 x - 6 y + 3 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, - 1), a = 2, b = 2

And the center is:

(2, - 1)

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

What is the center 3x^2-3y^2-12x-6y+3=0 ?
The center 3x^2-3y^2-12x-6y+3=0 is (2,-1)