What is the center 9x^2-16y^2-18x+64y-199=0 ?

The center 9x^2-16y^2-18x+64y-199=0 is (1,2)

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center 9x^2-16y^2-18x+64y-199=0

center

9 x^{2} - 16 y^{2} - 18 x + 64 y - 199 = 0

(1, 2)

Solution steps

9 x^{2} - 16 y^{2} - 18 x + 64 y - 199 = 0

Rewrite

9 x^{2} - 16 y^{2} - 18 x + 64 y - 199 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (1, 2), a = 4, b = 3

And the center is:

(1, 2)

9 x^{2} - 16 y^{2} - 144 = 0

x^{2} - \frac{y ^{2}}{25} = 1

9 x^{2} - 16 y^{2} - 126 x + 160 y - 103 = 0

\frac{( x - h ) ^{2}}{64} - \frac{( y - k ) ^{2}}{9} = 1

\frac{x ^{2}}{6} - \frac{y ^{2}}{8} = 1

What is the center 9x^2-16y^2-18x+64y-199=0 ?
The center 9x^2-16y^2-18x+64y-199=0 is (1,2)