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.
2022年8月16日 00:23
Uttarakhand Board Model Paper 2023 Class 6 Pdf Download with Answers for English Medium, Hindi Medium, Urdu Medium & Students for Small Answers, Long Answer, Very Long Answer Questions, and Essay Type Questions to Term1 & Term2 Exams