[Solution] Draw $E M \perp A B, D N \perp A B$ through $E, D$ respectively, and let $A M=x$, $E M=y$.
Since $E M / / D N$, and $A E=D C$, then
$$
N B=A M=x .
$$

Connect $B D$, in the right triangle $\triangle A B D$,
$$
N D^{2}=A N \cdot N B=x(2 a-x) .
$$

Also, $\mathrm{Rt} \triangle A M E \backsim \mathrm{Rt} \triangle A N D$, so $\frac{N D}{A N}=\frac{E M}{A M}$,
which means $N D=\frac{A N \cdot E M}{A M}=\frac{y(2 a-x)}{x}$. From the previous equation, we have
$x(2 a-x)=\frac{y^{2}(2 a-x)^{2}}{x^{2}}$, which simplifies to $y^{2}=\frac{x^{3}}{2 a-x}$.
Therefore, the answer is $(A)$.
$20 \cdot 29$ In a circle with center $O$ and radius $a$, $AB$ is a diameter, the extended chord $AD$ intersects the tangent at $B$ at $C, E$ is a point on $AC$ such that $AE=DC$, let $x$ and $y$ be the distances from $E$ to the tangent at $A$ and to $AB$, respectively. The relationship that can be derived is
(A) $y^{2}=\frac{x^{3}}{2 a-x}$.
(B) $y^{2}=\frac{x^{3}}{2 a+x}$.
(C) $y^{4}=\frac{x^{2}}{2 a-x}$.
(D) $x^{2}=\frac{y^{2}}{2 a-x}$.
(E) $x^{2}=\frac{y^{2}}{2 a+x}$.
(17th American High School Mathematics Examination, 1966)