Equation of $P Q$ is


$$
\begin{aligned}
y-2 a t_1 & =\frac{2 a t_2-2 a t_1}{a t_2^2-a t_1^2}\left(x-a t_1^2\right) \\
\Rightarrow \quad\left(t_2+t_1\right)\left(y-2 a t_1\right) & =2\left(x-a t_1^2\right) \\
\Rightarrow \quad\left(t_1+t_2\right) y-2 x & =2 a t_1 t_2
\end{aligned}
$$
This line is passing through $(a, 0)$
$$
\begin{aligned}
\Rightarrow & & \left(t_1+t_2\right)\left(-2 a t_1\right)=\frac{2}{t_1+t_2}\left(a-a t_1^2\right) \\
\Rightarrow & & t_1 t_2=-1
\end{aligned}
$$
Let $P\left(x_1, y_1\right)$ be the pole of (i) w.r.t. $y^2=4 a x$
Its polar is $y y_1=2 a\left(x+x_1\right)$
From equaiton (i) and (iii), we get
$$
\frac{t_1+t_2}{y_1}=\frac{1}{a}=\frac{2 a t_1 t_2}{2 a x_1}
$$
From last two relations, we get
$$
\begin{array}{ll}
& x_1=a t_1 t_2 \\
\Rightarrow & x_1=-a \\
\therefore \text { locus is } x=-a &
\end{array}
$$