What is the center 9x^2-16y^2-108x+128y+212=0 ?

The center 9x^2-16y^2-108x+128y+212=0 is (6,4)

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center 9x^2-16y^2-108x+128y+212=0

center

9 x^{2} - 16 y^{2} - 108 x + 128 y + 212 = 0

(6, 4)

Solution steps

9 x^{2} - 16 y^{2} - 108 x + 128 y + 212 = 0

Rewrite

9 x^{2} - 16 y^{2} - 108 x + 128 y + 212 = 0

in the form of a standard hyperbola equation

\frac{( y - 4 ) ^{2}}{3 ^{2}} - \frac{( x - 6 ) ^{2}}{4 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (6, 4), a = 3, b = 4

And the center is:

(6, 4)

y^{2} = x^{2} - 1

x^{2} - y^{2} - 4 x - 2 y - 1 = 0

x^{2} - y^{2} = 49

9 x^{2} - 16 y^{2} + 36 x + 32 y - 124 = 0

16 y^{2} - 25 x^{2} = 400

What is the center 9x^2-16y^2-108x+128y+212=0 ?
The center 9x^2-16y^2-108x+128y+212=0 is (6,4)