15. As shown in the figure, point $A$ and point $A^{\prime}$ are on the $x$-axis and are symmetric with respect to the $y$-axis. A line passing through point $A^{\prime}$ and perpendicular to the $x$-axis intersects the parabola $y^{2}=2 x$ at two points $B, C$. Point $D$ is a moving point on segment $A B$, and point $E$ is on segment $A C$, satisfying $\frac{|C E|}{|C A|}=\frac{|A D|}{|D B|}$.
(1) Prove that segment $D E$ intersects this parabola at exactly one point;
(2) Let the intersection point of line $D E$ and the parabola be $F$. Denote the areas of $\triangle B C F$ and $\triangle A D E$ as $S_{1}$ and $S_{2}$, respectively. Find the value of $\frac{S_{1}}{S_{2}}$.
(1) Let $A\left(-2 t^{2}, 0\right), B\left(2 t^{2}, 2 t\right), C\left(2 t^{2},-2 t\right)$,
$D\left(\frac{-2 t^{2}+2 \lambda t^{2}}{1+\lambda}, \frac{2 \lambda t}{1+\lambda}\right), E\left(\frac{2 t^{2}-2 \lambda t^{2}}{1+\lambda}, \frac{-2 t}{1+\lambda}\right)$.
Then $k_{D E}=\frac{\lambda+1}{2 t(\lambda-1)}$,
$$
D E: x=\frac{2 t(\lambda-1)}{\lambda+1}\left(y+\frac{2 t}{\lambda+1}\right)-\frac{2 t^{2}(\lambda-1)}{\lambda+1},
$$

Substituting the parabola equation $y^{2}=2 x$ we get
$$
\begin{array}{l}
y^{2}-\frac{4 t(\lambda-1)}{\lambda+1} y-\frac{8 t^{2}(\lambda-1)}{(\lambda+1)^{2}}+\frac{4 t^{2}(\lambda-1)}{\lambda+1}=0 \\
\Rightarrow y^{2}-\frac{4 t(\lambda-1)}{\lambda+1} y+\frac{4 t^{2}\left(\lambda^{2}-2 \lambda+1\right)}{(\lambda+1)^{2}}=\left[y-\frac{2 t(\lambda-1)}{\lambda+1}\right]^{2}=0,
\end{array}
$$

Thus, the line segment $D E$ intersects the parabola at exactly one point.
(2) Setting $y=0$ in the equation of line $D E$ gives $x=\frac{2 t(\lambda-1)}{\lambda+1} \cdot \frac{2 t}{\lambda+1}-\frac{2 t^{2}(\lambda-1)}{\lambda+1}$
$$
=\frac{4 t^{2}(\lambda-1)-2 t^{2}\left(\lambda^{2}-1\right)}{(\lambda+1)^{2}}=-\frac{2 t^{2}(\lambda-1)^{2}}{(\lambda+1)^{2}},
$$

Thus, $S_{2}=\left[t^{2}-\frac{t^{2}(\lambda-1)^{2}}{(\lambda+1)^{2}}\right] \cdot\left(\frac{2 \lambda t}{1+\lambda}+\frac{2 t}{1+\lambda}\right)=2 t \cdot\left[t^{2}-\frac{t^{2}(\lambda-1)^{2}}{(\lambda+1)^{2}}\right]$,
Since $x_{F}=\frac{y_{F}^{2}}{2}=\frac{2 t^{2}(\lambda-1)^{2}}{(\lambda+1)^{2}},|B C|=4 t$, we have $S_{1}=4 t \cdot\left[t^{2}-\frac{t^{2}(\lambda-1)^{2}}{(\lambda+1)^{2}}\right]$. Therefore, the value of $\frac{S_{1}}{S_{2}}$ is 2.