2x^2-3y^2-x+y=0

解答

2 x^{2} - 3 y^{2} - x + y = 0

(h, k) = (\frac{1}{4}, \frac{1}{6}), a = \frac{1}{4 3}, b = \frac{1}{6 2}

求解步骤

2 x^{2} - 3 y^{2} - x + y = 0

将

2 x^{2} - 3 y^{2} - x + y = 0

改写为标准双曲线方程形式

\frac{( x - \frac{1}{4} ) ^{2}}{( \frac{1}{4 3} ) ^{2}} - \frac{( y - \frac{1}{6} ) ^{2}}{( \frac{1}{6 2} ) ^{2}} = 1

因此双曲线的性质为:

(h, k) = (\frac{1}{4}, \frac{1}{6}), a = \frac{1}{4 3}, b = \frac{1}{6 2}

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

2 a - 3 b^{2} = 4 a^{2} - 9 b^{2}

\frac{( y - 4 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1

9 x^{2} - 36 x - y^{2} - 6 y + 18 = 0

x^{2} - y^{2} - y + 9 = 0