1. **State the problem:** We need to show that the coordinates of the point common to the curve $$y^2=4ax$$ and the line $$ty - x = at^2$$ are $$(at^2, 2at)$$. 2. **Write down the given equations:** - Curve: $$y^2 = 4ax$$ - Line: $$ty - x = at^2$$ 3. **Express $$x$$ from the line equation:** $$ty - x = at^2 \implies x = ty - at^2$$ 4. **Substitute $$x$$ into the curve equation:** $$y^2 = 4a(ty - at^2) = 4a t y - 4a^2 t^2$$ 5. **Rearrange the equation:** $$y^2 - 4a t y + 4a^2 t^2 = 0$$ 6. **Recognize this as a quadratic in $$y$$:** $$y^2 - 4a t y + 4a^2 t^2 = (y - 2a t)^2 = 0$$ 7. **Solve for $$y$$:** $$y - 2a t = 0 \implies y = 2a t$$ 8. **Find $$x$$ using $$y$$ in the line equation:** $$x = t y - a t^2 = t(2a t) - a t^2 = 2a t^2 - a t^2 = a t^2$$ 9. **Conclusion:** The common point has coordinates $$(a t^2, 2 a t)$$ as required.