The 15th question of the first test:
As shown in Figure 1, a tangent line is drawn through point $A(1,1)$ on the parabola $y=x^{2}$, intersecting the $x$-axis at point $D$ and the $y$-axis at point $B$. 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 $P$. When point $C$ moves along the parabola, find the equation of the trajectory of point $P$.
Solution: Since $y^{\prime}=\left.2 x\right|_{x=1}=2$, the equation of the tangent line passing through point $A$ is
$y-1=2(x-1)$, which is $y=2 x-1$.
Thus, $A(1,1), B(0,-1), D\left(-\frac{1}{2}, 0\right)$.
Therefore, $D$ is the midpoint of $A B$, and $C D$ is a median of $\triangle A B C$.
By the given conditions,
$\boldsymbol{C} \boldsymbol{E}=\frac{1}{1+\lambda_{1}} \boldsymbol{C A}, \boldsymbol{C F}=\frac{1}{1+\lambda_{2}} \boldsymbol{C B}$,
$\boldsymbol{C P}=\alpha \boldsymbol{C D}$.
Then $\boldsymbol{C P}=\frac{\alpha}{2}(\boldsymbol{C A}+\boldsymbol{C B})$,
$F P=C P-C F$
$=\frac{\alpha}{2} \boldsymbol{C A}+\left(\frac{\alpha}{2}-\frac{1}{1+\lambda_{2}}\right) \boldsymbol{C B}$,
$\boldsymbol{F} \boldsymbol{E}=\boldsymbol{C} \boldsymbol{E}-\boldsymbol{C} F=\frac{1}{1+\lambda_{1}} \boldsymbol{C} A-\frac{1}{1+\lambda_{2}} \boldsymbol{C B}$.
Since $\boldsymbol{F P}$ is collinear with $\boldsymbol{F E}$, there exists $t \in \mathbf{R}$, such that $F P=t F E$.
From equations (1), (2), and (3), we have
$\left\{\begin{array}{l}\frac{\alpha}{2}=\frac{t}{1+\lambda_{1}}, \\ \frac{\alpha}{2}-\frac{1}{1+\lambda_{2}}=\frac{-t}{1+\lambda_{2}} .\end{array}\right.$
Then $\left\{\begin{array}{l}\frac{1}{2}\left(1+\lambda_{1}\right) \alpha=t, \\ \frac{1}{2}\left(1+\lambda_{2}\right) \alpha-1=-t .\end{array}\right.$
Adding (4) and (5) gives $\alpha=\frac{2}{3}$.
Thus, $\boldsymbol{C P}=\frac{2}{3} \boldsymbol{C D}$, and $P$ is the centroid of $\triangle A B C$.
Let $C\left(x_{0}, y_{0}\right) 、 P(x, y)$, we get
$\left\{\begin{array}{l}x=\frac{x_{0}+1+0}{3}=\frac{x_{0}+1}{3}, \\ y=\frac{x_{0}^{2}+1-1}{3}=\frac{x_{0}^{2}}{3} .\end{array}\right.$
Eliminating $x_{0}$ gives $y=\frac{1}{3}(3 x-1)^{2}$.
Since point $C$ does not coincide with point $A$, we have $x_{0} \neq 1$, and thus $x \neq \frac{2}{3}$.
Therefore, the required trajectory equation is
$$
y=\frac{1}{3}(3 x-1)^{2}\left(x \neq \frac{2}{3}\right) \text {. }
$$

It is worth noting that this problem essentially presents the inverse proposition of a similar exam question. The original problem is:

Let $G$ be the centroid of $\triangle A B C$, and a line $l$ passing through point $G$ intersects $A B$ and $A C$ at points $P$ and $Q$, respectively. Given that $A P=\lambda A B$ and $A Q=\mu A C$, prove that $\frac{1}{\lambda}+\frac{1}{\mu}=3$.