Example 8 Draw a tangent line to the parabola $y=x^{2}$ at point $A(1,1)$, which intersects the $x$-axis at $D$ and the $y$-axis at $B$. Point $C$ is on the parabola, and point $E$ is on line segment $A C$, satisfying $\frac{A E}{E C}=\lambda_{1}$; point $F$ is on line segment $B C$, satisfying $\frac{B F}{F C}=\lambda_{2}$, and $\lambda_{1}+\lambda_{2}=1$. Line segment $C D$ intersects $E F$ at point $P$. Find the equation of the trajectory of point $P$ as point $C$ moves along the parabola.
(2005 National High School Competition Problem)
The slope of the tangent line through point $A$ on the parabola is: $y^{\prime}=\left.2 x\right|_{x=1}=2$, so the equation of the tangent line $A B$ is $y=2 x-1$. Thus, the coordinates of points $B$ and $D$ are $B(0,-1), D\left(\frac{1}{2}, 0\right)$, which means $D$ is the midpoint of segment $A B$. Let $P(x, y), C\left(x_{0}, x_{0}^{2}\right), E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$. From $\frac{A E}{E C}=\lambda_{1}$, we have $x_{1}=\frac{1+\lambda_{1} x_{0}}{1+\lambda_{1}}, y_{1}=\frac{1+\lambda_{1} x_{0}^{2}}{1+\lambda_{1}}$; from $\frac{B F}{F C}=\lambda_{2}$, we get $x_{2}=\frac{\lambda_{2} x_{0}}{1+\lambda_{2}}, y_{2}=\frac{-1+\lambda_{2} x_{0}^{2}}{1+\lambda_{2}}$.

By the two-point form, the equation of line $E F$ is
$$
\begin{array}{l}
{\left[\left(\lambda_{2}-\lambda_{1}\right) x_{0}-\left(1+\lambda_{2}\right)\right] y} \\
=\left[\left(\lambda_{2}-\lambda_{1}\right) x_{0}^{2}-3\right] x+1+x_{0}-\lambda_{2} x_{0}^{2} .
\end{array}
$$

When $x_{0} \neq \frac{1}{2}$, the equation of line $C D$ is
$$
y=\frac{2 x_{0}^{2} x-x_{0}^{2}}{2 x_{0}-1} \text {. }
$$

By solving (1) and (2) together, we get $x=\frac{x_{0}+1}{3}, y=\frac{x_{0}^{2}}{3}$. Eliminating $x_{0}$ from these two equations, we obtain the trajectory equation of point $P$ as
$$
y=\frac{1}{3}(3 x-1)^{2} \text {. }
$$

When $x_{0}=\frac{1}{2}$, the equation of $E F$ is $-\frac{3}{2} y=\left(\frac{1}{4} \lambda_{2}-\frac{1}{4} \lambda_{1}-3\right) x+\frac{3}{2}-\frac{1}{4} \lambda_{2}$, and the equation of $C D$ is $x=\frac{1}{2}$. Solving these equations together, we get $(x, y)=\left(\frac{1}{2}, \frac{1}{12}\right)$, which is also on the trajectory of point $P$.
Since $C$ does not coincide with $A$, $x_{0} \neq 1, x \neq \frac{2}{3}$. Therefore, the required trajectory equation is
$$
y=\frac{1}{3}(3 x-1)^{2}, x \neq \frac{2}{3} \text {. }
$$

Alternative solution for Example 8:
It is easy to know that the equation of $A B$ is $y=2 x-1, B(0,-1), D\left(\frac{1}{2}, 0\right)$, which means $D$ is the midpoint of $A B$.
$$
\text { Let } \frac{C D}{C P}=r, \frac{C A}{C E}=1+\lambda_{1}=t_{1}, \frac{C B}{C F}=1+\lambda_{2}=t_{2} \text {, }
$$

Then $t_{1}+t_{2}=3$. Since $C D$ is the median of $\triangle A B C$, we have
$S_{\triangle C A B}=2 S_{\triangle C A D}=2 S_{\triangle C B D}$,
and $\frac{1}{t_{1} t_{2}}=\frac{C E \cdot C F}{C A \cdot C B}=\frac{S_{\triangle C E F}}{S_{\triangle C A B}}=\frac{S_{\triangle C E P}}{2 S_{\triangle C A D}}+\frac{S_{\triangle C E P}}{2 S_{\triangle C B D}}=\frac{1}{2}\left(\frac{1}{t_{1} r}+\frac{1}{t_{2} r}\right)=\frac{t_{1}+t_{2}}{2 t_{1} t_{2} r}=\frac{3}{2 t_{1} t_{2} r}$,
so $r=\frac{3}{2}$. Hence, $P$ is the centroid of $\triangle A B C$.
Let $P(x, y), C\left(x_{0}, x_{0}^{2}\right)$. Since point $C$ is different from $A$, $x_{0} \neq 1$. Therefore, the coordinates of the centroid $P$ are $x=\frac{0+1+x_{0}}{3}=\frac{1+x_{0}}{3}\left(x \neq \frac{2}{3}\right), y=\frac{-1+1+x_{0}^{2}}{3}=\frac{x_{0}^{2}}{3}$,
eliminating $x_{0}$, we get $y=\frac{1}{3}(3 x-1)^{2}$.
Thus, the required trajectory equation is $y=\frac{1}{3}(3 x-1)^{2}, x \neq \frac{2}{3}$.