We know that on the parabola $$y^2=4 a x$$, the equation of tangent at $$p=\left(a t^2, 2 a t\right)$$ is $$t y=x+ a t^2$$ and Normal at $$\mathrm{S}\left(a s^2, 2 a s\right)$$ is $$y+s x=2 a s+a s^3$$ Let tangent at P and Normal at Q intersect at $$\left(x_0, y_0\right)$$, then

$$\begin{aligned} t y_0 & =x_0+a t^2 \quad \text{.... (i)}\\ y_0+s x_0 & =2 a s+a s^3 \quad \text{... (ii)} \end{aligned}$$

From equation (i) and (ii)

$$\begin{aligned} & x_0=t y_0-a t^2=\frac{-y_0+2 a s+a s^3}{s} \\ \Rightarrow & \text { st } y_0-\text { at.t.s }=-y_0+2 a s+a s^3 \end{aligned}$$

Given, $$s t=1$$

$$\begin{aligned} & \therefore \quad y_0-a t=-y_0+\frac{2 a}{t}+\frac{a}{t^3} \\ & \Rightarrow \quad 2 y_0=a t+\frac{2 a}{t}+\frac{a}{t^3} \\ & \Rightarrow \quad y_0=\frac{a\left(t^4+2 t^2+1\right)}{2 t^3} \\ & \Rightarrow \quad y_0=\frac{a\left(t^2+1\right)^2}{2 t^3} \end{aligned}$$

Hint:

On the parabola $$y^2=4 a x$$

(i) The equation of tangent at $$\mathrm{P}\left(a t^2, 2 a t\right)$$ is $$ty=x+a t^2$$.

(ii) The equation of normal at $$\mathrm{S}\left(a s^2, 2 a s\right)$$ is $$y+s x=2 a s+a s^3$$.