Since the parabola $y^2 = 4ax$ passes through the point $(1, -2)$, $\therefore \left(-2\right)^{2} = 4a\left(I\right) \Rightarrow a = 1$ Equation of tangent to the parabola at $\left(1,-2\right)$ $yy_{1} =2a\left(x + x_{1}\right)$ or $y\left(-2\right) = 2\left(1\right)\left(x+1\right)$ or $x + y+1 = 0$