8. In $\triangle A B C$, $A C=3, \sin C=k \sin A(k \geq 2)$, then the maximum value of the area of $\triangle A B C$ is $\qquad$
Answer 3.
Analysis: By the Law of Sines, transform $\sin C=k \sin A$ into $c=k a$, where $c, a$ are the sides opposite to $A B, B C$. Establish a Cartesian coordinate system $x o y$ with the line segment $A C$ as the $x$-axis and the midpoint $O$ of $A C$ as the origin, then $A\left(-\frac{3}{2}, 0\right), C\left(\frac{3}{2}, 0\right)$. Let $B(x, y)$, from $c=k a$ we get
$$
\sqrt{\left(x+\frac{3}{2}\right)^{2}+y^{2}}=k \sqrt{\left(x-\frac{3}{2}\right)^{2}+y^{2}}
$$

Squaring both sides and rearranging gives
$$
\left(k^{2}-1\right) x^{2}+\left(k^{2}-1\right) y^{2}-\left(3 k^{2}+3\right) x+\frac{9}{4}\left(k^{2}-1\right)=0
$$

Since $k \geq 2$, the above equation can be simplified to
$$
x^{2}+y^{2}-\frac{\left(3 k^{2}+3\right)}{k^{2}-1} x+\frac{9}{4}=0
$$

From this, we know that the trajectory of point $B$ is a circle with center $\left(\frac{3\left(k^{2}+1\right)}{2\left(k^{2}-1\right)}, 0\right)$ and radius $r=\frac{3 k}{k^{2}-1}$. Therefore, when point $B$ moves on the circle, the maximum distance from point $B$ to the $x$-axis is the radius $r=\frac{3 k}{k^{2}-1}$, so the maximum area $S$ of $\triangle A B C$ is $S=\frac{1}{2} \times 3 \times \frac{3 k}{k^{2}-1}(k \geq 2)$, and $S=\frac{1}{2} \times 3 \times \frac{3 k}{k^{2}-1}=\frac{9}{2} \times \frac{1}{k-\frac{1}{k}}$ is monotonically decreasing for $k \geq 2$, so
$$
S_{\max }=\frac{1}{2} \times 3 \times \frac{3 k}{k^{2}-1}=\frac{9}{2} \times \frac{1}{2-\frac{1}{2}}=3
$$