What is the center-9x^2-32y=-16y^2+128 ?

The center-9x^2-32y=-16y^2+128 is (0,1)

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center-9x^2-32y=-16y^2+128

center

- 9 x^{2} - 32 y = - 16 y^{2} + 128

(0, 1)

Solution steps

- 9 x^{2} - 32 y = - 16 y^{2} + 128

Rewrite

- 9 x^{2} - 32 y = - 16 y^{2} + 128

in the form of a standard hyperbola equation

\frac{( y - 1 ) ^{2}}{3 ^{2}} - \frac{( x - 0 ) ^{2}}{4 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, 1), a = 3, b = 4

And the center is:

(0, 1)

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

2 x^{2} - 2 y^{2} - 20 x + 16 y + 16 = 0

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

9 y^{2} - x^{2} + 2 x + 54 y + 62 = 0

x^{2} - 4 y^{2} + 8 x + 16 y - 4 = 0

What is the center-9x^2-32y=-16y^2+128 ?
The center-9x^2-32y=-16y^2+128 is (0,1)