6. Given that the tangents at two points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola $y=4 x$ are perpendicular to each other. Then the equation of the locus of the intersection point of the tangents is $\qquad$ .
6. $x=-1$.

Prove a more general conclusion: Given two points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola $y^{2}=2 p x$, if the tangents at these points are perpendicular to each other, then the intersection point of the tangents lies on the directrix of the parabola.

Let the intersection point of the tangents $A C$ and $B C$ be $C$. Then the equations of the tangents $A C$ and $B C$ are
$$
\begin{array}{l}
y y_{1}=p\left(x+x_{1}\right) \Leftrightarrow y=\frac{p}{y_{1}} x+\frac{1}{2} y_{1}, \\
y y_{2}=p\left(x+x_{2}\right) \Leftrightarrow y=\frac{p}{y_{2}} x+\frac{1}{2} y_{2} .
\end{array}
$$

Since $A C \perp B C$, we have $y_{1} y_{2}=-p^{2}$.
Solving the equations, we get
$$
\begin{array}{l}
\frac{p}{y_{1}} x+\frac{1}{2} y_{1}=\frac{p}{y_{2}} x+\frac{1}{2} y_{2} \\
\Rightarrow x \frac{p\left(y_{2}-y_{1}\right)}{y_{1} y_{2}}=\frac{1}{2}\left(y_{2}-y_{1}\right) \\
\Rightarrow x=-\frac{p}{2} .
\end{array}
$$

Therefore, the intersection point of the two perpendicular tangents lies on the directrix of the parabola $y=2 p x$.
Substituting into this problem, the required trajectory equation is $x=-1$.