What is the center 4x^2-9y^2+32x+36y+64=0 ?

The center 4x^2-9y^2+32x+36y+64=0 is (-4,2)

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center 4x^2-9y^2+32x+36y+64=0

center

4 x^{2} - 9 y^{2} + 32 x + 36 y + 64 = 0

(- 4, 2)

Solution steps

4 x^{2} - 9 y^{2} + 32 x + 36 y + 64 = 0

Rewrite

4 x^{2} - 9 y^{2} + 32 x + 36 y + 64 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(- 4, 2)

4 x^{2} - 45 y^{2} = 180

\frac{( y - 2 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{16} = 1

\frac{x ^{2}}{36} - \frac{y ^{2}}{9} = 1

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

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

What is the center 4x^2-9y^2+32x+36y+64=0 ?
The center 4x^2-9y^2+32x+36y+64=0 is (-4,2)