x^2-3y^2+5y=7

解

x^{2} - 3 y^{2} + 5 y = 7

(h, k) = (0, \frac{5}{6}), a = \frac{59}{2 3}, b = \frac{59}{6}

解答ステップ

x^{2} - 3 y^{2} + 5 y = 7

x^{2} - 3 y^{2} + 5 y = 7

を標準的な双曲線のequationで書き換える

\frac{( x - 0 ) ^{2}}{( \frac{59}{2 3} ) ^{2}} - \frac{( y - \frac{5}{6} ) ^{2}}{( \frac{59}{6} ) ^{2}} = 1

ゆえに双曲線の特性は:

(h, k) = (0, \frac{5}{6}), a = \frac{59}{2 3}, b = \frac{59}{6}