6. Let the moving point $P(x, y)$ be on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the fixed point $M(a, 0)(0<a<3)$. If the minimum distance $|P M|$ is 1, then the value of $a$ is $\qquad$ .
6. Since point $p(x, y)$ is on the ellipse, hence
$$
\begin{array}{l}
y^{2}=4-\frac{4}{9} x^{2},|x| \leqslant 3 . \\
|P M|=\sqrt{(x-a)^{2}+y^{2}}=\sqrt{(x-a)^{2}+4-\frac{4}{9} x^{2}}=\sqrt{\frac{5}{9}\left(x-\frac{9}{5} a\right)^{2}+\frac{4(5-a)^{2}}{5}} . \\
\because a>0, \\
\therefore \text { when } x=\frac{9}{5} a, |P M|_{\text {min }}=\sqrt{\frac{4\left(5-a^{2}\right)}{5}}=1, \\
\therefore a=\frac{\sqrt{15}}{2} .
\end{array}
$$