6(x+1/3)^2-6(y-1/2)^2+24=0

解答

6 (x + \frac{1}{3})^{2} - 6 (y - \frac{1}{2})^{2} + 24 = 0

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

求解步骤

6 (x + \frac{1}{3})^{2} - 6 (y - \frac{1}{2})^{2} + 24 = 0

将

6 (x + \frac{1}{3})^{2} - 6 (y - \frac{1}{2})^{2} + 24 = 0

改写为标准双曲线方程形式

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

因此双曲线的性质为:

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