Problem 1171

11 (1) Let $A\left(x_{1}, y_{1}\right)$, then $y_{1}=\frac{1}{2} x_{1}^{2}$. From $y=\frac{1}{2} x^{2}$, we get $y^{\prime}=x$, so $\left.y^{\prime}\right|_{x=x_{1}}=x_{1}$. Therefore, the equation of the tangent line to the parabola $C$ at point $A$ is $y-y_{1}=x_{1}\left(x-x_{1}\right)$, which simplifies to $y=x_{1} x-y_{1}$.

Let $P\left(x_{0}, k x_{0}-1\right)$, then $k x_{0}-1=x_{0} x_{1}-y_{1}$. Let $B\left(x_{2}, y_{2}\right)$, similarly, we have $k x_{0}-1=x_{0} x_{2}-y_{2}$. Therefore, the equation of line $A B$ is $k x_{0}-1=x_{0} x-y$, which simplifies to $x_{0}(x-k)-(y-1)=0$, so line $A B$ always passes through the fixed point $Q(k, 1)$.
(2) The equation of $P Q$ is $y=\frac{k x_{0}-2}{x_{0}-k}(x-k)+1$, and combining this with the parabola equation $y=\frac{1}{2} x^{2}$, we eliminate $y$ to get
$$x^{2}-\frac{2 k x_{0}-4}{x_{0}-k} x+\frac{\left(2 k^{2}-2\right) x_{0}-2 k}{x_{0}-k}=0 .$$

Let $M\left(x_{3}, y_{3}\right)$ and $N\left(x_{4}, y_{4}\right)$, then
$$x_{3}+x_{4}=\frac{2 k x_{0}-4}{x_{0}-k}, \quad x_{3} x_{4}=\frac{\left(2 k^{2}-2\right) x_{0}-2 k}{x_{0}-k} .$$

To prove
$$\frac{|P M|}{|P N|}=\frac{|Q M|}{|Q N|},$$

it suffices to prove
$$\frac{x_{3}-x_{0}}{x_{4}-x_{0}}=\frac{k-x_{3}}{x_{4}-k},$$

which is equivalent to
$$2 x_{3} x_{4}-\left(k+x_{0}\right)\left(x_{3}+x_{4}\right)+2 k x_{0}=0 .$$

From (1), we know that the left side of (2) is
$$\frac{2\left(2 k^{2}-2\right) x_{0}-4 k}{x_{0}-k}-\left(k+x_{0}\right) \frac{2 k x_{0}-4}{x_{0}-k}+2 k x_{0}$$
$$\begin{array}{l} =\frac{2\left(2 k^{2}-2\right) x_{0}-4 k-\left(k+x_{0}\right)\left(2 k x_{0}-4\right)+2 k x_{0}\left(x_{0}-k\right)}{x_{0}-k} \\ =0 . \end{array}$$

Thus, (2) holds, and the conclusion follows.