(6) As shown in the figure, the three vertices of the right triangle $\triangle ABC$ are all on the given parabola $x^{2}=2 p y(p>0)$, and the hypotenuse $AB \parallel$ $x$-axis, then the height from $C$ to the hypotenuse $|CD|=$ $\qquad$
$62 p$ Hint: Let $A\left(2 p t_{1}, 2 p t_{1}^{2}\right) 、 C\left(2 p t_{2}, 2 p t_{2}^{2}\right)$, then $B\left(-2 p t_{1}\right.$, $2 p t_{1}^{2}$ ).
Since $A C \perp B C$, we have $\overrightarrow{A C} \cdot \overrightarrow{B C}=0$, which means
$$
4 p^{2}\left(t_{2}-t_{1}\right)\left(t_{2}+t_{1}\right)+4 p^{2}\left(t_{2}^{2}-t_{1}^{2}\right)^{2}=0 .
$$

Since $\left|t_{1}\right| \neq\left|t_{2}\right|$, we have $t_{2}^{2}-t_{1}^{2}=-1$.
Thus, $|C D|=2 p\left|t_{2}^{2}-t_{1}^{2}\right|=2 p$.