If ${\mathrm{t}}_1,{\mathrm{t}}_2,{\mathrm{t}}_3$ are the feet of normals drawn from $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ to the parabola ${\mathrm{y}}^2=4\mathrm{ax}$ then the value of ${\mathrm{t}}_1,{\mathrm{t}}_2,{\mathrm{t}}_3$

Equation of normal drawn from ${\mathrm{t}}_1, {\mathrm{t}}_2$ and ${\mathrm{t}}_3$ in the form of

$$\begin{gathered} y+xt=2\mathrm{at}+a{\mathrm{t}}^3 \\ \mathrm{it} \mathrm{is} \mathrm{pasing} \mathrm{through} ({\mathrm{x}}_1,{\mathrm{y}}_1) \mathrm{then} \\ {\mathrm{at}}^3+\mathrm{t}\left(2\mathrm{a}-{\mathrm{x}}_1\right)-{\mathrm{y}}_1=0 \\ It is a cubic equation in t ,it has three roots let it be t_1,t_2,t_3 \end{gathered}$$

product of roots$t_1{t}_2{t}_3=\frac{-(-{\mathrm{y}}_1)}{\mathrm{a}}=\raisebox{1ex}{${\mathrm{y}}_1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{a}$}\right.$