[f(x)]^2=3x^2-x+2

解答

[f (x)]^{2} = 3 x^{2} - x + 2

(h, k) = (\frac{1}{6}, 0), a = \frac{23}{2 3}, b = \frac{23}{6}

求解步骤

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

将

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

改写为标准双曲线方程形式

\frac{( y - 0 ) ^{2}}{( \frac{23}{2 3} ) ^{2}} - \frac{( x - \frac{1}{6} ) ^{2}}{( \frac{23}{6} ) ^{2}} = 1

因此双曲线的性质为:

(h, k) = (\frac{1}{6}, 0), a = \frac{23}{2 3}, b = \frac{23}{6}

\frac{x ^{2}}{5} - \frac{y ^{2}}{4} = 1

y^{2} = 4 + z^{2} - 2 z

a^{2} - b^{2} = 51

10 y^{2} - 9 x^{2} = 144

10 x^{2} - 7 y^{2} - 30 x + 14 y = 5