13 Draw the tangent line to the parabola $y=x^{2}$ at a point $A(1,1)$. The tangent line intersects the $x$-axis at $D$ and the $y$-axis at $B$. Point $C$ is on the parabola, and point $E$ is on the line segment $A C$, satisfying $\frac{A E}{E C}=\lambda_{1}$; point $F$ is on the line segment $B C$, satisfying $\frac{B F}{F C}=\lambda_{2}$, and $\lambda_{1}+\lambda_{2}=1$. The 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.
Solution 1 The slope of the tangent line at 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 $B$ and $D$ are $B(0,-1), D\left(\frac{1}{2}, 0\right)$, so $D$ is the midpoint of line segment $A B$. $\qquad$
Let $P(x, y), C\left(x_{0}, x_{0}^{2}\right)$, $E\left(x_{1}, y_{1}\right), F\left(x_{2}, y_{2}\right)$, then by $\frac{A E}{E C}=$ $\lambda_{1}$ we have,
$$
x_{1}=\frac{1+\lambda_{1} x_{0}}{1+\lambda_{1}}, y_{1}=\frac{1+\lambda_{1} x_{0}^{2}}{1+\lambda_{1}} ;
$$

By $\frac{B F}{F C}=\lambda_{2}$, we get
$$
x_{2}=\frac{\lambda_{2} x_{0}}{1+\lambda_{2}}, y_{2}=\frac{-1+\lambda_{2} x_{0}^{2}}{1+\lambda_{2}} \text {. }
$$

Therefore, the equation of line $E F$ is:
$$
\frac{y-\frac{1+\lambda_{1} x_{0}^{2}}{1+\lambda_{1}}}{\frac{-1+\lambda_{2} x_{0}^{2}}{1+\lambda_{2}}-\frac{1+\lambda_{1} x_{0}^{2}}{1+\lambda_{1}}}=\frac{x-\frac{1+\lambda_{1} x_{0}}{1+\lambda_{1}}}{\frac{\lambda_{2} x_{0}}{1+\lambda_{2}}-\frac{1+\lambda_{1} x_{0}}{1+\lambda_{1}}},
$$

Simplifying, we get
$$
\begin{aligned}
& {\left[\left(\lambda_{2}-\lambda_{1}\right) x_{0}-\left(1+\lambda_{2}\right)\right] y } \\
= & {\left[\left(\lambda_{2}-\lambda_{1}\right) x_{0}^{2}-3\right] x+1+x_{0}-\lambda_{2} x_{0}^{2} . }
\end{aligned}
$$
10 points
When $x_{0} \neq \frac{1}{2}$, the equation of line $C D$ is:
$$
y=\frac{2 x_{0}^{2} x-x_{0}^{2}}{2 x_{0}-1} .
$$

Solving (1) and (2) together, we get $\left\{\begin{array}{l}x=\frac{x_{0}+1}{3}, \\ y=\frac{x_{0}^{2}}{3},\end{array}\right.$ Eliminating $x_{0}$, we get the trajectory equation of point $P$:
$$
y=\frac{1}{3}(3 x-1)^{2} .
$$

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

Solution 2 From Solution 1, we know that the equation of $A B$ is $y=2 x-1, B(0,-1)$, $D\left(\frac{1}{2}, 0\right)$, so $D$ is the midpoint of $A B$. $\qquad$ 5 points
$$
\gamma=\frac{C D}{C P}, t_{1}=\frac{C A}{C E}=1+\lambda_{1}, t_{2}=\frac{C B}{C F}=1+\lambda_{2},
$$

Then $t_{1}+t_{2}=3$. Since $C D$ is the median of $\triangle A B C$, we have
$$
\begin{array}{l} 
S_{\triangle C A B}=2 S_{\triangle C A D}=2 S_{\triangle C B D} . \text { And } \\
\frac{1}{t_{1} t_{2}}= \frac{C E \cdot C F}{C A \cdot C B}=\frac{S_{\triangle C E F}}{S_{\triangle C A B}}=\frac{S_{\triangle C E P}}{2 S_{\triangle C A D}} \\
+\frac{S_{\triangle C F P}}{2 S_{\triangle C B D}}=\frac{1}{2}\left(\frac{1}{t_{1} \gamma}+\frac{1}{t_{2} \gamma}\right) \\
= \frac{t_{1}+t_{2}}{2 t_{1} t_{2} \gamma}=\frac{3}{2 t_{1} t_{2} \gamma},
\end{array}
$$

So $\gamma=\frac{3}{2}$, 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 $A$, then $x_{0} \neq 1$, so the coordinates of the centroid $P$ are
$$
\begin{array}{c}
x=\frac{0+1+x_{0}}{3}=\frac{1+x_{0}}{3}\left(x \neq \frac{2}{3}\right), \\
y=\frac{-1+1+x_{0}^{2}}{3}=\frac{x_{0}^{2}}{3},
\end{array}
$$

Eliminating $x_{0}$, we get $y=\frac{1}{3}(3 x-1)^{2}$.
Therefore, the required trajectory equation is
$$
y=\frac{1}{3}(3 x-1)^{2}\left(x \neq \frac{2}{3}\right) .
$$
. 20 points