Equation of tangent at P is $$t y=x+a t^2\quad \text{.... (i)}$$

Equation of tangent at Q is $$\frac{-y}{t}=x+\frac{a}{t^2}\quad \text{... (ii)}$$

On solving equations (i) and (ii)

$$\Rightarrow \mathrm{R}=\left(-a, a\left(t-\frac{1}{t}\right)\right)$$

Given, the tangents of parabola $$y^2=4 a x$$ at P and Q meet at a point lying on the line $$y=2 x+a$$

$$\begin{aligned} & \Rightarrow \quad a\left(t-\frac{1}{t}\right)=-2 a+a \\ & \Rightarrow \quad\left(t-\frac{1}{t}\right)=-1 \quad \text{... (iii)} \end{aligned}$$

$$\Rightarrow$$ Slope of line $$\mathrm{OP}=\frac{2}{t}$$ and Slope of line OQ $$=-2 t$$

Given, $$\angle \mathrm{POQ}=\theta$$

$$\begin{aligned} & \Rightarrow \quad \tan \theta=\frac{\frac{2}{t}-(-2 t)}{1+\frac{2}{t}(-2 t)} \\ & \Rightarrow \quad \tan \theta=\frac{-2}{3}\left(t+\frac{1}{t}\right) \\ & \Rightarrow \quad \tan \theta=\frac{-2}{3} \sqrt{\left(t-\frac{1}{t}\right)^2+4} \\ & \Rightarrow \quad \tan \theta=\frac{-2 \sqrt{5}}{3} \end{aligned}$$

Hints:

(i) If slope of two lines are $$m_1$$ and $$m_2$$, then the angle of intersection between them is equal to $$\tan ^{-1}\left(\frac{m_1-m_2}{1+m_1 m_2}\right)$$

(ii) If PQ is a focal chord of parabola $$y^2=4 a x$$, then $$\mathrm{P}=\left(a t^2, 2 a t\right)$$ and $$\mathrm{Q}=\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$$.