y^1y=(x+2)(x-1)

解答

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

(h, k) = (- \frac{1}{2}, 0), a = \frac{3}{2}, b = \frac{3}{2}

求解步骤

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

将

y^{1} y - (x + 2) (x - 1) = 0

改写为标准双曲线方程形式

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

因此双曲线的性质为:

(h, k) = (- \frac{1}{2}, 0), a = \frac{3}{2}, b = \frac{3}{2}

3 (x - 1)^{2} = 6 + 2 (y + 1)^{2}

(x - 5)^{2} - (y - 4)^{2} = 25

x^{2} - 3 y^{2} = 4

x^{2} - y^{2} = - 7

x^{2} - y^{2} = - 8