What is the center 9x^2-16y^2+36x+32y-124=0 ?

The center 9x^2-16y^2+36x+32y-124=0 is (-2,1)

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

center

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

(- 2, 1)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 2, 1), a = 4, b = 3

And the center is:

(- 2, 1)

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

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

9 x^{2} - y^{2} - 72 x + 8 y + 119 = 0

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

x^{2} - y^{2} - 6 x = 16

What is the center 9x^2-16y^2+36x+32y-124=0 ?
The center 9x^2-16y^2+36x+32y-124=0 is (-2,1)