问题 1:

求双曲线 \(\frac{x^2}{9} - \frac{y^2}{4} = 1\) 的交点

与直线 \(y = x - 1\)

解答:

代入 \(y = x - 1\):

\(\frac{x^2}{9} - \frac{(x-1)^2}{4} = 1\)

乘以 36:

\(4x^2 - 9(x^2 - 2x + 1) = 36\)

\(4x^2 - 9x^2 + 18x - 9 = 36\)

\(-5x^2 + 18x - 45 = 0\)

\(5x^2 - 18x + 45 = 0\)

\(\Delta = 324 - 900 = -576 < 0\)

无交点 - 直线与双曲线相离

问题 2:

求 k 的值,使得直线 \(y = 2x + k\)

与双曲线 \(\frac{x^2}{4} - \frac{y^2}{16} = 1\) 相切

解答:

代入 \(y = 2x + k\):

\(\frac{x^2}{4} - \frac{(2x+k)^2}{16} = 1\)

乘以 16:

\(4x^2 - (4x^2 + 4kx + k^2) = 16\)

\(4x^2 - 4x^2 - 4kx - k^2 = 16\)

\(-4kx - k^2 - 16 = 0\)

这是一个线性方程!(x² 的系数消失了)

这是因为直线斜率 (m=2) 等于渐近线斜率(\(\frac{b}{a} = \frac{4}{2} = 2\))

直线平行于渐近线 - 对每个 k 都只交于一点(且不是切线!)

问题 3:

求 m 的值,使得直线 \(y = mx + 3\)

与双曲线 \(\frac{x^2}{9} - \frac{y^2}{4} = 1\) 相切

解答:

代入并要求 Δ = 0:

\(\frac{x^2}{9} - \frac{(mx+3)^2}{4} = 1\)

乘以 36:

\(4x^2 - 9(m^2x^2 + 6mx + 9) = 36\)

\((4 - 9m^2)x^2 - 54mx - 81 - 36 = 0\)

\((4 - 9m^2)x^2 - 54mx - 117 = 0\)

相切条件:\(\Delta = 0\)

\((-54m)^2 - 4(4-9m^2)(-117) = 0\)

\(2916m^2 + 468(4-9m^2) = 0\)

\(2916m^2 + 1872 - 4212m^2 = 0\)

\(-1296m^2 + 1872 = 0\)

\(m^2 = \frac{1872}{1296} = \frac{13}{9}\)

答:\(m = \pm \frac{\sqrt{13}}{3}\)

问题 4:

求双曲线 \(\frac{x^2}{16} - \frac{y^2}{9} = 1\) 的切线方程

在点 \((5, \frac{9}{4})\)

解答:

切线公式:\(\frac{x \cdot x_0}{a^2} - \frac{y \cdot y_0}{b^2} = 1\)

代入:

\(\frac{x \cdot 5}{16} - \frac{y \cdot \frac{9}{4}}{9} = 1\)

\(\frac{5x}{16} - \frac{y}{4} = 1\)

乘以 16:

\(5x - 4y = 16\)

切线方程:\(5x - 4y = 16\)

问题 5:

求双曲线 \(\frac{x^2}{4} - \frac{y^2}{9} = 1\) 的切线斜率

在点 \((4, 3\sqrt{3})\)

解答:

切线斜率公式:\(m = \frac{b^2 x_0}{a^2 y_0}\)

代入:

\(m = \frac{9 \cdot 4}{4 \cdot 3\sqrt{3}} = \frac{36}{12\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\)

答:\(m = \sqrt{3}\)

问题 6:

已知双曲线 \(\frac{x^2}{9} - \frac{y^2}{16} = 1\)

点 P 在双曲线上,距焦点 \(F_1\) 为 2。

求 P 到焦点 \(F_2\) 的距离。

解答:

由双曲线定义:\(|PF_1 - PF_2| = 2a = 6\)

已知:\(PF_1 = 2\)

若 P 在靠近 \(F_1\) 的分支上:

\(PF_2 - PF_1 = 6\)

\(PF_2 - 2 = 6\)

\(PF_2 = 8\)

答:\(PF_2 = 8\)

问题 7:

已知双曲线 \(\frac{x^2}{16} - \frac{y^2}{9} = 1\)

求点 \(P(8, y_0)\) 到两焦点的距离。

解答:

\(a = 4, b = 3, c = 5, e = \frac{5}{4}\)

距离公式(右支,\(x_0 > 0\)):

\(PF_1 = ex_0 - a = \frac{5}{4} \cdot 8 - 4 = 10 - 4 = 6\)

\(PF_2 = ex_0 + a = \frac{5}{4} \cdot 8 + 4 = 10 + 4 = 14\)

验证:\(|PF_1 - PF_2| = |6 - 14| = 8 = 2a\) ✓

答:\(PF_1 = 6, PF_2 = 14\)

问题 8:

已知双曲线 \(\frac{x^2}{25} - \frac{y^2}{16} = 1\)

写出其共轭双曲线的方程,并求其焦点。

解答:

共轭双曲线:

\(\frac{y^2}{16} - \frac{x^2}{25} = 1\)

在共轭中:\(a' = 4, b' = 5\)

\(c' = \sqrt{16 + 25} = \sqrt{41}\)

共轭双曲线的焦点:\((0, \pm \sqrt{41})\)

问题 9:

判断点 \((4, 2)\) 的位置

相对于双曲线 \(\frac{x^2}{9} - \frac{y^2}{4} = 1\)

解答:

计算:\(S = \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2}\)

\(S = \frac{16}{9} - \frac{4}{4} = \frac{16}{9} - 1 = \frac{7}{9} < 1\)

该点位于双曲线外部(分支区域)

问题 10:

已知双曲线的焦点为 \(F_1(6, 0), F_2(-6, 0)\)

其渐近线为 \(y = \pm \frac{4}{3}x\)

1. 求双曲线的方程。
2. 求与直线 \(y = 2x\) 平行的双曲线切线方程。

解答:

(a)

\(c = 6\),水平轴

由渐近线:\(\frac{b}{a} = \frac{4}{3}\) → \(b = \frac{4a}{3}\)

\(c^2 = a^2 + b^2\)

\(36 = a^2 + \frac{16a^2}{9} = \frac{25a^2}{9}\)

\(a^2 = \frac{324}{25}\)

\(b^2 = 36 - \frac{324}{25} = \frac{576}{25}\)

\(\frac{25x^2}{324} - \frac{25y^2}{576} = 1\)

或:\(\frac{x^2}{12.96} - \frac{y^2}{23.04} = 1\)

(b)

平行于 \(y = 2x\) 的切线:\(y = 2x + k\)

代入并要求 \(\Delta = 0\):

\(\frac{x^2}{12.96} - \frac{(2x+k)^2}{23.04} = 1\)

展开并应用相切条件后:

\(k^2 = 4 \cdot 12.96 - 23.04 = 51.84 - 23.04 = 28.8\)

\(k = \pm \sqrt{28.8} = \pm 5.37\)

切线方程:\(y = 2x \pm \sqrt{28.8}\)

1. 求 \(\frac{x^2}{16} - \frac{y^2}{4} = 1\) 与 \(y = x + 2\) 的交点
2. 求在点 \((6, 4)\) 的双曲线 \(\frac{x^2}{4} - \frac{y^2}{16} = 1\) 的切线方程
3. P 在 \(\frac{x^2}{25} - \frac{y^2}{144} = 1\) 上,且 \(PF_1 = 4\)。求 \(PF_2\)。
4. 求双曲线 \(\frac{x^2}{9} - \frac{y^2}{4} = 1\) 被 \(y = 1\) 截取的弦长

答案:

1. \((\frac{8}{3}, \frac{14}{3})\) 和 \((-4, -2)\)

2. \(6x - y = 32\)

3. \(PF_2 = 14\)

4. \(\frac{2\sqrt{117}}{3}\)