What is the center 4x^2-12y^2-24x+24y+51=0 ?

The center 4x^2-12y^2-24x+24y+51=0 is (3,1)

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center 4x^2-12y^2-24x+24y+51=0

center

4 x^{2} - 12 y^{2} - 24 x + 24 y + 51 = 0

(3, 1)

Solution steps

4 x^{2} - 12 y^{2} - 24 x + 24 y + 51 = 0

Rewrite

4 x^{2} - 12 y^{2} - 24 x + 24 y + 51 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (3, 1), a = \frac{3}{2}, b = \frac{3 3}{2}

And the center is:

(3, 1)

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

x^{2} - y^{2} = 15

(x - 5)^{2} - (y - 4)^{2} = 1

x^{2} - 9 y^{2} = 144

\frac{( x + 10 ) ^{2}}{36} - \frac{( y - 3 ) ^{2}}{100} = 1

What is the center 4x^2-12y^2-24x+24y+51=0 ?
The center 4x^2-12y^2-24x+24y+51=0 is (3,1)