3y^2-3x^2+5x+10y=5x^2+1

解答

3 y^{2} - 3 x^{2} + 5 x + 10 y = 5 x^{2} + 1

(h, k) = (\frac{5}{16}, - \frac{5}{3}), a = \frac{821}{12 2}, b = \frac{821}{16 3}

求解步骤

3 y^{2} - 3 x^{2} + 5 x + 10 y = 5 x^{2} + 1

将

3 y^{2} - 3 x^{2} + 5 x + 10 y - 5 x^{2} - 1 = 0

改写为标准双曲线方程形式

\frac{( y - ( - \frac{5}{3} ) ) ^{2}}{( \frac{821}{12 2} ) ^{2}} - \frac{( x - \frac{5}{16} ) ^{2}}{( \frac{821}{16 3} ) ^{2}} = 1

因此双曲线的性质为:

(h, k) = (\frac{5}{16}, - \frac{5}{3}), a = \frac{821}{12 2}, b = \frac{821}{16 3}

x^{2} + 8 x - y^{2} - 9 y = 6

\frac{y ^{2}}{16} - \frac{x ^{2}}{49} = 1

(\frac{3 y + 2 x}{18}) (\frac{3 y - 2 x}{2}) = 1

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

\frac{x ^{2}}{20} - \frac{( y + 1 ) ^{2}}{10} = 1