What is the center x^2-4y^2+8x+16y-4=0 ?

The center x^2-4y^2+8x+16y-4=0 is (-4,2)

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center x^2-4y^2+8x+16y-4=0

center

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

(- 4, 2)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(- 4, 2)

\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 5 ) ^{2}}{9} = 1

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

25 y^{2} - 16 x^{2} - 400 = 0

9 x^{2} - 16 y^{2} = 144

What is the center x^2-4y^2+8x+16y-4=0 ?
The center x^2-4y^2+8x+16y-4=0 is (-4,2)