What is the center x^2-2y^2+3x+4y=2 ?

The center x^2-2y^2+3x+4y=2 is (-3/2 ,1)

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center x^2-2y^2+3x+4y=2

center

x^{2} - 2 y^{2} + 3 x + 4 y = 2

(- \frac{3}{2}, 1)

Solution steps

x^{2} - 2 y^{2} + 3 x + 4 y = 2

Rewrite

x^{2} - 2 y^{2} + 3 x + 4 y = 2

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- \frac{3}{2}, 1), a = \frac{3}{2}, b = \frac{3}{2 2}

And the center is:

(- \frac{3}{2}, 1)

\frac{( x - 5 ) ^{2}}{4} - \frac{( y - 3 ) ^{2}}{9} = 1

4 y^{2} - 9 x^{2} + 8 y - 54 x - 113 = 0

y^{2} - x^{2} = \frac{9}{25}

9 x^{2} - 16 y^{2} - 108 x + 128 y + 212 = 0

y^{2} = x^{2} - 1