解析几何竞赛题

Eufisky posted @ 2014年7月20日 20:08 in 解析几何 with tags 解析几何竞赛题 , 865 阅读

第四届全国大学生数学竞赛决赛（数学组）解析几何试题

设 $A$ 为正整数，直线 $L$ 与双曲线 $x^2-y^2=2(x>0)$ 所围成的面积为 $A$ ，证明：

(1)上述 $L$ 被双曲线 $x^2-y^2=2(x>0)$ 所截线段的中点的轨迹为双曲线；

(2) $L$ 总是(1)中轨迹曲线的切线.

证明. (1)不妨设直线 $L$ 的方程为 $x=my+l(m^2<1)$ ，直线 $L$ 与双曲线 $x^2-y^2=2(x>0)$ 的交点 $P,Q$ 分别为 $(x_1,y_1),(x_2,y_2)$ $(\text{其中}y_1<y_2)$ ,

联立方程，有 $\left\{ \begin{array}{l}x = my + l\\{x^2} - {y^2} = 2\end{array} \right. \Rightarrow \left( {{m^2} - 1} \right){y^2} + 2mly + {l^2} - 2 = 0.$

由韦达定理，我们有： ${y_1} + {y_2} = \frac{{2ml}}{{1 - {m^2}}},{y_1}{y_2} = \frac{{{l^2} - 2}}{{{m^2} - 1}},{y_2} - {y_1} = \sqrt {{{\left( {{y_2} + {y_1}} \right)}^2} - 4{y_2}{y_1}} = \frac{{2\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}.$

由题意得：

$\begin{align}A &= \int_{{y_1}}^{{y_2}} {\left( {my + l - \sqrt {{y^2} + 2} } \right)dy} = \left[ {\frac{{m{y^2}}}{2} + ly - \left( {\frac{{y\sqrt {{y^2} + 2} }}{2} + \ln \left( {y + \sqrt {{y^2} + 2} } \right)} \right)} \right]_{{y_1}}^{{y_2}}\\&= \frac{m}{2}\left( {y_2^2 - y_1^2} \right) + l\left( {{y_2} - {y_1}} \right) - \left( {\frac{{{y_2}\sqrt {y_2^2 + 2} }}{2} - \frac{{{y_1}\sqrt {y_1^2 + 2} }}{2}} \right) - \ln \frac{{{y_2} + \sqrt {y_2^2 + 2} }}{{{y_1} + \sqrt {y_1^2 + 2} }}\\&= \frac{m}{2}\left( {y_2^2 - y_1^2} \right) + l\left( {{y_2} - {y_1}} \right) - \left( {\frac{{{x_2}{y_2}}}{2} - \frac{{{x_1}{y_1}}}{2}} \right) - \ln \frac{{{x_2} + {y_2}}}{{{x_1} + {y_1}}}\\&= \frac{m}{2}\left( {y_2^2 - y_1^2} \right) + l\left( {{y_2} - {y_1}} \right) - \left[ {\frac{m}{2}\left( {y_2^2 - y_1^2} \right) + \frac{l}{2}\left( {{y_2} - {y_1}} \right)} \right] - \ln \frac{{{x_1}{x_2} - {y_1}{y_2} + {x_1}{y_2} - {x_2}{y_1}}}{2}\\&= \frac{l}{2}\left( {{y_2} - {y_1}} \right) - \ln \frac{{\left( {{m^2} - 1} \right){y_1}{y_2} + ml\left( {{y_1} + {y_2}} \right) + l\left( {{y_2} - {y_1}} \right) + {l^2}}}{2}\end{align}$

又

$\begin{align}&\ln \frac{{\left( {{m^2} - 1} \right){y_1}{y_2} + ml\left( {{y_1} + {y_2}} \right) + l\left( {{y_2} - {y_1}} \right) + {l^2}}}{2} = \ln \frac{{{l^2} - 2 + \frac{{2{m^2}{l^2}}}{{1 - {m^2}}} + \frac{{2l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}} + {l^2}}}{2}\\&= \ln \left( {{l^2} - 1 + \frac{{{m^2}{l^2}}}{{1 - {m^2}}} + \frac{{l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}} \right) = \ln \left( {\frac{{{m^2} + {l^2} - 1}}{{1 - {m^2}}} + \frac{{l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}} \right).\end{align}$

故我们有

$A = \frac{{l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}} - \ln \frac{{{m^2} + {l^2} - 1 + l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}.$

设

$P,Q$ 中点为

$M(x,y)$ ，则有

$\left\{ \begin{array}{l}x = \frac{{{x_1} + {x_2}}}{2} = \frac{{{y_1} + {y_2}}}{2}m + l = \frac{l}{{1 - {m^2}}}\\y = \frac{{{y_1} + {y_2}}}{2} = \frac{{ml}}{{1 - {m^2}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}m = \frac{y}{x}\\l = \frac{{{x^2} - {y^2}}}{x}\end{array} \right..$

由此得

$A = \sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)} - \ln \left[ {\left( {{x^2} - {y^2}} \right) + \sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)} - 1} \right].$

令

$t = \sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)} ,$

则有

$A = t - \ln \left( {t + \sqrt {{t^2} + 1} } \right).\tag{1}$

记

$f\left( t \right) = t - \ln \left( {t + \sqrt {{t^2} + 1} } \right)$ ，注意到

$f'\left( x \right) = 1 - \frac{1}{{\sqrt {{t^2} + 1} }} > 0$ ，其值域显然为

$(0,+\infty)$ ，故方程

$\text{(1)}$ 的解唯一，记作

$t_0$ .因此我们有

$\sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)} = t_0 \Rightarrow {x^2} - {y^2} = 1 + \sqrt {1 + {t_0^2}}.$

即 $x^2-y^2=C(C>2).$ 为双曲线轨迹.

(2)再之，由 $2x - 2yy' = 0 \Rightarrow y' = \frac{x}{y} = \frac{1}{m} = {k_L}$ 及直线 $L$ 经过点 $M$ 可知，直线 $L$ 总为 $M$ 的轨迹曲线的切线.

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