What is the center 9x^2-16y^2-54x+64y-127=0 ?

The center 9x^2-16y^2-54x+64y-127=0 is (3,2)

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center 9x^2-16y^2-54x+64y-127=0

center

9 x^{2} - 16 y^{2} - 54 x + 64 y - 127 = 0

(3, 2)

Solution steps

9 x^{2} - 16 y^{2} - 54 x + 64 y - 127 = 0

Rewrite

9 x^{2} - 16 y^{2} - 54 x + 64 y - 127 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(3, 2)

- \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

9 x^{2} - 36 x - 4 y^{2} - 24 y - 36 = 0

4 x^{2} - y^{2} = 16

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

9 x^{2} - 4 y^{2} + 54 x + 8 y + 78 = 0

What is the center 9x^2-16y^2-54x+64y-127=0 ?
The center 9x^2-16y^2-54x+64y-127=0 is (3,2)