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

解

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}

解答ステップ

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

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

を標準的な双曲線のequationで書き換える

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

ゆえに双曲線の特性は:

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