6. Let $F$ be the right focus of the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$. A tangent line is drawn from a point $P$ outside the ellipse $C$, touching the ellipse at point $M$. If $\angle P F M=90^{\circ}$, then the equation of the locus of point $P$ is $\qquad$
6. $x=4$.

Let point $P\left(x_{0}, y_{0}\right), M\left(x_{1}, y_{1}\right)$. Then the equation of the tangent line $P M$ of the ellipse $C$ is $\frac{x_{1} x}{4}+\frac{y_{1} y}{3}=1$.
$$
\text { Hence } \frac{x_{1} x_{0}}{4}+\frac{y_{1} y_{0}}{3}=1 \text {. }
$$

Given point $F(1,0)$, then
$$
\overrightarrow{F P}=\left(x_{0}-1, y_{0}\right), \overrightarrow{F M}=\left(x_{1}-1, y_{1}\right) \text {. }
$$

Since $\angle P F M=90^{\circ}$, we have $\overrightarrow{F P} \cdot \overrightarrow{F M}=0$, which means
$$
\left(x_{0}-1\right)\left(x_{1}-1\right)+y_{0} y_{1}=0 \text {. }
$$

From equations (1) and (2), eliminating $y_{0} y_{1}$ and rearranging, we get
$$
\left(x_{0}-4\right)\left(x_{1}-4\right)=0 \text {. }
$$

Since $x_{1}<4$, it follows that $x_{0}=4$.
Therefore, the equation of the locus of point $P$ is $x=4$.