4. Through a point $A$ on the line $x=-4$, draw a tangent line $l$ to the parabola $C$: $y^{2}=2 p x(p>0)$, with the point of tangency being $B(1,2)$. Line $l$ intersects the $x$-axis at point $D$. $P$ is a moving point on the parabola $C$ different from point $B$. $\overrightarrow{P E}=\lambda_{1} \overrightarrow{P A}$, $\overrightarrow{P F}=\lambda_{2} \overrightarrow{P B}$, and $E F$ intersects $P D$ at point $Q$. If $\frac{2}{\lambda_{1}}+\frac{3}{\lambda_{2}}=15\left(\lambda_{1} 、 \lambda_{2}>0\right)$, then the equation of the trajectory of point $Q$ is . $\qquad$
4. $y^{2}=\frac{8}{3}\left(x+\frac{1}{3}\right)\left(y \neq \frac{4}{3}\right)$.

It is easy to know that the parabola $y^{2}=4 x$, and the tangent line $l_{A B}: y=x+1$.
Then $D(-1,0), A(-4,-3)$
$$
\Rightarrow \overrightarrow{A D}=\frac{3}{2} \overrightarrow{D B} \text {. }
$$

Thus, $\overrightarrow{P D}=\frac{2}{5} \overrightarrow{P A}+\frac{3}{5} \overrightarrow{P B}$.
Let $\overrightarrow{P Q}=\lambda \overrightarrow{P D}=\frac{2 \lambda}{5} \overrightarrow{P A}+\frac{3 \lambda}{5} \overrightarrow{P B}$
$$
=\frac{2 \lambda}{5 \lambda_{1}} \overrightarrow{P E}+\frac{3 \lambda}{5 \lambda_{2}} \overrightarrow{P F} \text {. }
$$

Since $E, Q, F$ are collinear, then
$$
\frac{2 \lambda}{5 \lambda_{1}}+\frac{3 \lambda}{5 \lambda_{2}}=1 \Rightarrow \frac{\lambda}{5} \times 15=1 \Rightarrow \lambda=\frac{1}{3} \text {. }
$$

Let $Q(x, y), P\left(x_{0}, y_{0}\right)$.
Thus $\left(x-x_{0}, y-y_{0}\right)=\frac{1}{3}\left(-1-x_{0},-y_{0}\right)$
$$
\Rightarrow\left(x_{0}, y_{0}\right)=\left(\frac{3 x+1}{2}, \frac{3 y}{2}\right) \text {. }
$$

Substitute into $y_{0}^{2}=4 x_{0}\left(y_{0} \neq 2\right)$, we get the trajectory equation of point $Q$ as
$$
y^{2}=\frac{8}{3}\left(x+\frac{1}{3}\right)\left(y \neq \frac{4}{3}\right) .
$$