What is the center 18x^2-32y^2-108y-128x-254=0 ?

The center 18x^2-32y^2-108y-128x-254=0 is (32/9 ,-27/16)

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center 18x^2-32y^2-108y-128x-254=0

center

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0

(\frac{32}{9}, - \frac{27}{16})

Solution steps

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0

Rewrite

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0

in the form of a standard hyperbola equation

\frac{( x - \frac{32}{9} ) ^{2}}{( \frac{28111}{36} ) ^{2}} - \frac{( y - ( - \frac{27}{16} ) ) ^{2}}{( \frac{28111}{48} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (\frac{32}{9}, - \frac{27}{16}), a = \frac{28111}{36}, b = \frac{28111}{48}

And the center is:

(\frac{32}{9}, - \frac{27}{16})

\frac{( y - 3 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{16} = 1

9 x^{2} - y^{2} - 36 x - 6 y + 18 = 0

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

\frac{( x + 4 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{25} = 1

\frac{y ^{2}}{16} - \frac{x ^{2}}{4} = 1

What is the center 18x^2-32y^2-108y-128x-254=0 ?
The center 18x^2-32y^2-108y-128x-254=0 is (32/9 ,-27/16)