The locus of foot of perpendicular drawn from origin upon any tangent of parabola ${\mathrm{y}}^2=8(\mathrm{x}-1)$ is

$$\begin{gathered} {\mathrm{m}}_{\mathrm{OP}}=\frac{\mathrm{k}}{\mathrm{h}}\Rightarrow {\mathrm{m}}_{\mathrm{PA}}=-\frac{\mathrm{h}}{\mathrm{k}} \\ \Rightarrow \text{ equation of }\mathrm{PA}:\mathrm{y}-\mathrm{k}=-\frac{\mathrm{h}}{\mathrm{x}}(\mathrm{x}-\mathrm{h}) \\ \Rightarrow \mathrm{hx}+\mathrm{ky}={\mathrm{h}}^2+{\mathrm{k}}^2 ....\left(1\right) \\ \text{ Also }\mathrm{PA}\text{ is tangent to }{\mathrm{y}}^2=8(\mathrm{x}-1) \\ \Rightarrow \text{ equation of }\mathrm{PA}:\mathrm{y}=\mathrm{m}(\mathrm{x}-1)+\frac{2}{\mathrm{m}}\Rightarrow \mathrm{mx}-\mathrm{y}=\mathrm{m}-\frac{2}{\mathrm{m}}\dots \left(2\right) \\ \text{ Compare (1) and (2) }\frac{\mathrm{m}}{\mathrm{h}}=\frac{-1}{\mathrm{k}}=\frac{\mathrm{m}-\frac{2}{\mathrm{m}}}{{\mathrm{h}}^2+{\mathrm{k}}^2} \\ \Rightarrow {\mathrm{x}}^3+{\mathrm{xy}}^2+2{\mathrm{y}}^2-{\mathrm{x}}^2=0 \end{gathered}$$