What is the center y^2-4y-4x^2+13=0 ?

The center y^2-4y-4x^2+13=0 is (0,2)

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center y^2-4y-4x^2+13=0

center

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

(0, 2)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (0, 2), a = \frac{3}{2}, b = 3

And the center is:

(0, 2)

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

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

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

4 y^{2} - 9 x^{2} + 16 y + 18 x = 29

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137