What is the center 4x^2-1/2 y^2+8x+2y=2 ?

The center 4x^2-1/2 y^2+8x+2y=2 is (-1,2)

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center 4x^2-1/2 y^2+8x+2y=2

center

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

(- 1, 2)

Solution steps

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

Rewrite

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 1, 2), a = 1, b = 22

And the center is:

(- 1, 2)

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

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

\frac{x ^{2}}{4} - \frac{( y + 1 ) ^{2}}{16} = 1

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

What is the center 4x^2-1/2 y^2+8x+2y=2 ?
The center 4x^2-1/2 y^2+8x+2y=2 is (-1,2)