题目
已知抛物线 $y^{2}=2 x,$ 过点 $P(1,1)$ 分别作斜率为 $k_{1}, k_{2}$ 的抛物线的动弦 $A B, C D .$ 设 $M, N$ 分别为线段 $A B$ 和 $C D$
的中点.若 $k_{1}+k_{2}=1,$ 求证直线 $M N$ 恒过定点，并求出定点坐标.
解析
设 $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), M\left(x_{M}, y_{M}\right), N\left(x_{N}, y_{N}\right)$
由
$$
y_{1}^{2}=2 x_{1}, \quad y_{2}^{2}=2 x_{2}
$$
两式相减得
$$
y_{2}^{2}-y_{1}^{2}=2\left(x_{2}-x_{1}\right)
$$
所以
$$
\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2}{y_{2}+y_{1}}=\frac{2}{2 y_{M}}
$$
即
$$
k_{1}=\frac{1}{y_{M}}
$$
又 $k_{1}=\dfrac{y_{M}-1}{x_{M}-1},$ 所以
$$
\frac{1}{y_{M}}=\frac{y_{M}-1}{x_{M}-1}
$$
所以
$$
x_{M}=y_{M}^{2}-y_{M}+1
$$
同理可得
$$
k_{2}=\frac{1}{y_{N}}, \quad x_{N}=y_{N}^{2}-y_{N}+1
$$
由 $k_{1}+k_{2}=1$ 得
$$
\frac{1}{y_{M}}+\frac{1}{y_{N}}=1
$$
故
$$
y_{M} y_{N}=y_{M}+y_{N}
$$
直线 $MN$ 的方程为
$$
y-y_{M}=k_{M N}\left(x-x_{M}\right)
$$
令 $x=0$ 可得
$$
\begin{aligned}
y=& y_{M}-x_{M} k_{M N} \\
=& y_{M}-\frac{x_{M}\left(y_{M}-y_{N}\right)}{x_{M}-x_{N}} \\
=& \frac{x_{M} y_{N}-y_{M} x_{N}}{x_{M}-x_{N}} \\
=& \frac{\left(y_{M}^{2}-y_{M}+1\right) y_{N}-\left(y_{N}^{2}-y_{N}+1\right) y_{M}}{\left(y_{M}^{2}-y_{M}+1\right)-\left(y_{N}^{2}-y_{N}+1\right)} \\
=& \frac{y_{M}^{2} y_{N}-y_{N}^{2} y_{M}-y_{M}+y_{N}}{y_{M}^{2}-y_{N}^{2}-y_{M}+y_{N}} \\
=& \frac{y_{M}\left(y_{M}+y_{N}\right)-y_{N}\left(y_{M}+y_{N}\right)-y_{M}+y_{N}}{y_{M}^{2}-y_{N}^{2}-y_{M}+y_{N}}\left(\because y_{M} y_{N}=y_{M}+y_{N}\right) \\
=& \frac{y_{M}^{2}-y_{N}^{2}-y_{M}+y_{N}}{y_{M}^{2}-y_{N}^{2}-y_{M}+y_{N}} \\
=& 1
\end{aligned}
$$
故直线 $MN$一定经过定点 $(0, 1)$.