-3(x^2-2x+1)+(y^2+4y+2)-2=-11

解答

- 3 (x^{2} - 2 x + 1) + (y^{2} + 4 y + 2) - 2 = - 11

(h, k) = (1, - 2), a = \frac{7}{3}, b = 7

求解步骤

- 3 (x^{2} - 2 x + 1) + y^{2} + 4 y + 2 - 2 = - 11

将

- 3 (x^{2} - 2 x + 1) + y^{2} + 4 y + 2 - 2 = - 11

改写为标准双曲线方程形式

\frac{( x - 1 ) ^{2}}{( \frac{7}{3} ) ^{2}} - \frac{( y - ( - 2 ) ) ^{2}}{( 7 ) ^{2}} = 1

因此双曲线的性质为:

(h, k) = (1, - 2), a = \frac{7}{3}, b = 7