What is the center 16y^2-32y-9x^2-18x=137 ?

The center 16y^2-32y-9x^2-18x=137 is (-1,1)

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center 16y^2-32y-9x^2-18x=137

center

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

(- 1, 1)

Solution steps

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

Rewrite

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(- 1, 1)

\frac{x ^{2}}{25} - \frac{y ^{2}}{49} = 1

321 + 108 x - 9 x^{2} + 8 y + 4 y^{2} = 0

2 x^{2} - y^{2} - 2 y = - 5

- 6 x^{2} + 4 y^{2} + 6 x + 16 y - 1 = 0

y^{2} - 4 x^{2} < 1

What is the center 16y^2-32y-9x^2-18x=137 ?
The center 16y^2-32y-9x^2-18x=137 is (-1,1)