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

The center 2y^2-x^2-y+x=0 is (1/2 , 1/4)

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center 2y^2-x^2-y+x=0

center

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

(\frac{1}{2}, \frac{1}{4})

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (\frac{1}{2}, \frac{1}{4}), a = \frac{1}{2 2}, b = \frac{1}{4}

And the center is:

(\frac{1}{2}, \frac{1}{4})

\frac{( y + 5 ) ^{2}}{9} - \frac{( x - 3 ) ^{2}}{25} = 9

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

\frac{( y - 7 ) ^{2}}{9} - (x + 4)^{2} = 1

2 y^{2} - 9 x^{2} - 18 x + 20 + 5 = 0

x^{2} + 36 y - 32 x - y^{2} = 75