What is the center 16x^2-9y^2+64x+54y-161=0 ?

The center 16x^2-9y^2+64x+54y-161=0 is (-2,3)

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center 16x^2-9y^2+64x+54y-161=0

center

16 x^{2} - 9 y^{2} + 64 x + 54 y - 161 = 0

(- 2, 3)

Solution steps

16 x^{2} - 9 y^{2} + 64 x + 54 y - 161 = 0

Rewrite

16 x^{2} - 9 y^{2} + 64 x + 54 y - 161 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(- 2, 3)

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

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0

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

9 x^{2} - 18 x - 24 y - 63 = 4 y^{2}

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

What is the center 16x^2-9y^2+64x+54y-161=0 ?
The center 16x^2-9y^2+64x+54y-161=0 is (-2,3)