6. Let the area of an ellipse with a major axis of length $2a$ and a minor axis of length $2b$ be $\pi a b$. Given that the hypotenuse $BC=2$ of an isosceles right $\triangle ABC$. Then, the maximum value of the area of an ellipse that is tangent to each side of $\triangle ABC$ and whose major axis is parallel to the side $BC$ is ( ).
(A) $\frac{\sqrt{3}}{4} \pi$
(B) $\frac{\sqrt{3}}{9} \pi$
(C) $\frac{\sqrt{2}}{4} \pi$
(D) $\frac{\sqrt{2}}{9} \pi$
6.(B).

As shown in Figure 3, establish a rectangular coordinate system. Let the center of the inscribed ellipse of $\triangle ABC$ be $(0, b) (b>0)$, and the semi-major axis length be $a (0<a<1)$. Then the equation of the ellipse is
$$
\frac{x^{2}}{a^{2}}+\frac{(y-b)^{2}}{b^{2}}=1.
$$

The equation of line $AC$ is
$$
x+y=1.
$$

From
$$
\left\{\begin{array}{l}
\frac{x^{2}}{a^{2}}+\frac{(y-b)^{2}}{b^{2}}=1, \\
x+y=1
\end{array}\right.
$$

we get $\left(a^{2}+b^{2}\right) y^{2}-2 b\left(a^{2}+b\right) y+b^{2}=0$.
Since $AC$ is tangent to the ellipse,
$$
\therefore \Delta=4 b^{2}\left(a^{2}+b\right)^{2}-4 b^{2}\left(a^{2}+b^{2}\right)=0.
$$

Solving for $b$ gives $b=\frac{1-a^{2}}{2}$.
$$
\begin{array}{l}
\therefore S_{\text {mew }}=\frac{\pi}{2} a\left(1-a^{2}\right) \\
\quad=\frac{\pi}{2 \sqrt{2}} \sqrt{2 a^{2}\left(1-a^{2}\right)\left(1-a^{2}\right)} \\
\quad \leqslant \frac{\pi}{2 \sqrt{2}} \sqrt{\left[\frac{2 a^{2}+\left(1-a^{2}\right)+\left(1-a^{2}\right)}{3}\right]^{3}}=\frac{\sqrt{3}}{9} \pi .
\end{array}
$$

Equality holds if and only if $2 a^{2}=1-a^{2}$, i.e., $a=\frac{\sqrt{3}}{3}$.