15. As shown in Figure 1, points $A$ and $A^{\prime}$ 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 points $B$ and $C$. Point $D$ is a moving point on segment $A B$, and point $E$ is on segment $A C$, satisfying $\frac{|C E|}{|C A|}=\frac{|A D|}{|A B|}$.
(1) Prove that the line $D E$ intersects the 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 $\frac{S_{1}}{S_{2}}$.
15. (1) Let $A\left(-2 a^{2}, 0\right), A^{\prime}\left(2 a^{2}, 0\right)$. Then $B\left(2 a^{2}, 2 a\right), C\left(2 a^{2},-2 a\right)$.
Let $D\left(x_{1}, y_{1}\right), \overrightarrow{A D}=\lambda \overrightarrow{A B}$. Then $\overrightarrow{C E}=\lambda \overrightarrow{C A}$.
Thus, $\left(x_{1}+2 a^{2}, y_{1}\right)=\lambda\left(4 a^{2}, 2 a\right)$.
Hence, $D\left((4 \lambda-2) a^{2}, 2 \lambda a\right)$.
Let $E\left(x_{2}, y_{2}\right)$. From $\overrightarrow{C E}=\lambda \overrightarrow{C A}$, we get $\left(x_{2}-2 a^{2}, y_{2}+2 a\right)=\lambda\left(-4 a^{2}, 2 a\right)$.
Hence, $E\left((2-4 \lambda) a^{2},(2 \lambda-2) a\right)$.
When $\lambda=\frac{1}{2}$, the line $D E$ is the $y$-axis, and the conclusion is obviously true.
When $\lambda \neq \frac{1}{2}$, the slope of line $D E$ exists, and
$$
k_{D E}=\frac{2 a}{(8 \lambda-4) a^{2}}=\frac{1}{(4 \lambda-2) a} \text {. }
$$

Thus, $l_{D E}: y-2 \lambda a=\frac{1}{(4 \lambda-2) a}\left(x-(4 \lambda-2) a^{2}\right)$
$$
\Rightarrow x=2 a(2 \lambda-1) y-2 a^{2}(2 \lambda-1)^{2} \text {. }
$$

Combining with the parabola equation, we get
$$
\begin{array}{l}
y^{2}-4 a(2 \lambda-1) y+4 a^{2}(2 \lambda-1)^{2}=0 \\
\Rightarrow y=2 a(2 \lambda-1) .
\end{array}
$$

Substituting into equation (1) gives $x=2 a^{2}(2 \lambda-1)^{2}$.
Therefore, the line $D E$ intersects the parabola at exactly one point $F\left(2 a^{2}(2 \lambda-1)^{2}, 2 a(2 \lambda-1)\right)$.
(2) Note that,
$$
S_{1}=\frac{1}{2} \cdot 4 a\left(2 a^{2}-x_{F}\right)=4 a^{3}\left(4 \lambda-4 \lambda^{2}\right) .
$$

Let the line $D E$ intersect the $x$-axis at point $G$. Setting $y=0$, substituting into equation (1) gives
$$
\begin{array}{l}
x=-2 a^{2}(2 \lambda-1)^{2} . \\
\text { Then }|A G|=2 a^{2}-2 a^{2}(2 \lambda-1)^{2} \\
=2 a^{2}\left(4 \lambda-4 \lambda^{2}\right) . \\
\text { Hence } S_{2}=\frac{1}{2}|A G|\left|y_{D}-y_{E}\right| \\
=a^{2}\left(4 \lambda-4 \lambda^{2}\right)|2 \lambda a-(2 \lambda-2) a| \\
=2 a^{3}\left(4 \lambda-4 \lambda^{2}\right) .
\end{array}
$$

Therefore, $\frac{S_{1}}{S_{2}}=2$.