15. A tangent line is drawn through a 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 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 $P$. When point $C$ moves along the parabola, find the equation of the trajectory of point $P$.
The tangent line equation at point $A$ is $y=2 x-1$, then $D\left(\frac{1}{2}, 0\right)$, $B(0,-1) \Rightarrow C D$ is the median of side $A B$.
Since $\frac{C A}{C E}=\lambda_{1}+1, \frac{C B}{C F}=\lambda_{2}+1$ and $\lambda_{1}+\lambda_{2}=1$ $\Rightarrow \overrightarrow{C A}+\overrightarrow{C B}=\left(\lambda_{1}+1\right) \overrightarrow{C E}+\left(\lambda_{2}+1\right) \overrightarrow{C F}=2 \overrightarrow{C D}$ $\Rightarrow \frac{\lambda_{1}+1}{3} \overrightarrow{C E}+\frac{\lambda_{2}+1}{3} \overrightarrow{C F}=\frac{2}{3} \overrightarrow{C D}$.
Noting that $\frac{\lambda_{1}+1}{3}+\frac{\lambda_{2}+1}{3}=1$, then $E, F$ and the endpoint of $\frac{2}{3} \overrightarrow{C D}$, which is the centroid of $\triangle A B C$, are collinear, so $P$ is the centroid of $\triangle A B C$. Thus, let $P(x, y) \Rightarrow C(3 x-1,3 y)(3 x-1 \neq 1)$, substituting into the parabola equation gives $3 y=(3 x-1)^{2} \Rightarrow y=3 x^{2}-2 x+\frac{1}{3}\left(x \neq \frac{2}{3}\right)$. In summary, the trajectory equation of point $P$ is $y=3 x^{2}-2 x+\frac{1}{3}\left(x \neq \frac{2}{3}\right)$.