What is the center 4y^2=4x^2+12y-69+32x ?

The center 4y^2=4x^2+12y-69+32x is (-4, 3/2)

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center 4y^2=4x^2+12y-69+32x

center

4 y^{2} = 4 x^{2} + 12 y - 69 + 32 x

(- 4, \frac{3}{2})

Solution steps

4 y^{2} = 4 x^{2} + 12 y - 69 + 32 x

Rewrite

4 y^{2} = 4 x^{2} + 12 y - 69 + 32 x

in the form of a standard hyperbola equation

\frac{( x - ( - 4 ) ) ^{2}}{( 31 ) ^{2}} - \frac{( y - \frac{3}{2} ) ^{2}}{( 31 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 4, \frac{3}{2}), a = 31, b = 31

And the center is:

(- 4, \frac{3}{2})

4 x^{2} - 12 y^{2} - 24 x + 24 y + 51 = 0

y^{2} - x^{2} = \frac{4}{25}

x^{2} - y^{2} = 15

(x - 5)^{2} - (y - 4)^{2} = 1

x^{2} - 9 y^{2} = 144

What is the center 4y^2=4x^2+12y-69+32x ?
The center 4y^2=4x^2+12y-69+32x is (-4, 3/2)