15. As shown in the figure, 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 $\frac{S_{1}}{S_{2}}$.
Proof: (1) Let $A\left(-2 a^{2}, 0\right), A^{\prime}\left(2 a^{2}, 0\right)$, so we can get $B\left(2 a^{2}, 2 a\right), C\left(2 a^{2},-2 a\right)$. 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 the coordinates of point $D$ are: $\left((4 \lambda-2) a^{2}, 2 \lambda a\right)$.
Let $E\left(x_{2}, y_{2}\right)$, similarly, the coordinates of $E$ are $\left((2-4 \lambda) a^{2},(2 \lambda-2) a\right)$.
Therefore, the slope of line $D E$ is: $k_{D E}=\frac{2 a}{(8 \lambda-4) a^{2}}=\frac{1}{(4 \lambda-2) a}$, so the equation of line $D E$ is: $y-2 \lambda a=\frac{1}{(4 \lambda-2) a}\left(x-(4 \lambda-2) a^{2}\right)$, simplifying to: $(4 \lambda-2) a y-2 \lambda a(4 \lambda-2) a=x-(4 \lambda-2) a^{2}$, which is $x=2 a(2 \lambda-1) y-2 a^{2}(2 \lambda-1)^{2}$. Combining with the parabola equation, we get $y^{2}=2\left[2 a(2 \lambda-1) y-2 a^{2}(2 \lambda-1)^{2}\right]$, i.e., $y^{2}-4 a(2 \lambda-1) y+4 a^{2}(2 \lambda-1)^{2}=0$, at this point the equation has two equal roots: $y=2 a(2 \lambda-1)$, substituting into (1) we get: $x=2 a^{2}(2 \lambda-1)^{2}$.
So 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) $S_{1}=S_{\triangle B C F}=\frac{1}{2} B C \cdot h=\frac{1}{2} \cdot 4 a \cdot\left(2 a^{2}-x_{F}\right)=4 a^{3}\left(4 \lambda-4 \lambda^{2}\right)$.

Let the line $D E$ intersect the $x$-axis at point $G$, setting $y=0$, substituting into equation (1), $x=2 a(2 \lambda-1) y-2 a^{2}(2 \lambda-1)^{2}$, solving we get: $x=-2 a^{2}(2 \lambda-1)^{2}$, hence $|A G|=2 a^{2}-2 a^{2}(2 \lambda-1)^{2}=2 a^{2}\left(4 \lambda-4 \lambda^{2}\right)$, thus $S_{2}=S_{\triangle A D E}=S_{\triangle A D G}+S_{\triangle A E G}=\frac{1}{2} \cdot|A G| \cdot\left|y_{D}-y_{E}\right|=a^{2}\left(4 \lambda-4 \lambda^{2}\right) \cdot|2 \lambda a-(2 \lambda-2) a|=2 a^{3}\left(4 \lambda-4 \lambda^{2}\right)$, so $\frac{S_{1}}{S_{2}}=2$.