6. The axial section of a circular cone with vertex $P$ is an isosceles right triangle. $A$ is a point on the circumference of the base circle, $B$ is a point inside the base circle, $O$ is the center of the base circle, $A B \perp O B$, with the foot of the perpendicular at $B$, $O H \perp P B$, with the foot of the perpendicular at $H$, and $P A=4$, $C$ is the midpoint of $P A$. Then, when the volume of the tetrahedron $C-H P C$ is maximized, the length of $O B$ is
(A) $\frac{\sqrt{5}}{3}$
(B) $\frac{2 \sqrt{6}}{3}$
(C) $\frac{\sqrt{6}}{3}$
(D) $\frac{2 \sqrt{6}}{3}$
Answer $(\quad)$
(D)
6. 【Analysis and Solution】 $\because A B \perp O B, A B \perp O P, \therefore A B \perp P B$, also $O H \perp P B$
$\therefore$ plane $P A B \perp$ plane $P O B, \therefore O H \perp H C, O H \perp P A$. $C$ is the midpoint of $P A$, $\therefore O C \perp P A$ $\therefore$ when $H O=\mathrm{HC}$, $S_{\triangle H O C}$ is maximized,
which also means $V_{O-H P C}=V_{P-H C O}$ is maximized.
At this time, $H O=\sqrt{2}$, so $H O=\frac{1}{2} O P, \therefore \angle H P O=30^{\circ}$,
$$
\therefore O B=O P \cdot \tan 30^{\circ}=\frac{2 \sqrt{6}}{3} \text {. }
$$

Therefore, the answer is D.