8. In $\triangle A B C$, $D$ is the midpoint of side $B C$. If $\overrightarrow{A D} \cdot \overrightarrow{A C}=0$, then the minimum value of $\tan C-\cot A$ is $\qquad$
8. $\sqrt{2}$.

Take the midpoint $E$ of $A B$. Then
$$
\begin{array}{l}
D E / / A C, \angle A D E=90^{\circ}, \\
\angle D A E=\angle A-90^{\circ} \text {. } \\
\text { Let } A C=b, A D=m . \text { Then } \\
\tan C=\frac{m}{b}, \tan \left(A-90^{\circ}\right)=\frac{b}{2 m}, \\
-\cot A=\frac{b}{2 m} .
\end{array}
$$

Therefore, $\tan C-\cot A=\frac{m}{b}+\frac{b}{2 m} \geqslant \sqrt{2}$.
Equality holds if and only if $m=\frac{\sqrt{2}}{2} b$.