If two of the three feet of normals drawn from a point to the parabola ${\mathrm{y}}^2=4\mathrm{x}$ be (1, 2) (1, –2) then third foot is

Given parabola $y^2=4x$

$$\begin{gathered} \Rightarrow 4a=4 \\ \Rightarrow a=1 \end{gathered}$$

Equation of Normal to parabola is $\mathrm{y}+\mathrm{tx}=2\mathrm{t}+{\mathrm{t}}^3$

it is passing through $\left(1,2\right) {\mathrm{t}}^3+\mathrm{t}-2=0 —\overline{)1}$

it also passing through $\left(1,-2\right) {\mathrm{t}}^3+\mathrm{t}+2=0—\overline{)2}$

By solving (1) and (2) we get t=0

so third foot of the point=(0,0)