The locus of middle points of normal chords of the parabola ${\mathrm{y}}^2=4\mathrm{ax}$ is

Let $\mathrm{A}\left({{\mathrm{at}}_1}^2, 2{\mathrm{at}}_1\right) \mathrm{and} \mathrm{B}=\left({{\mathrm{at}}_2}^2,2{\mathrm{at}}_2\right)$ be end points of normal chord

then we have ${\mathrm{t}}_2=-{\mathrm{t}}_1-\frac{2}{{\mathrm{t}}_1}$

$\Rightarrow t_1+t_2=\frac{-2}{t_1}and t_1{t}_2=-{{t_1}^2}_{ }-2\to \left(1\right)$

Let (h, k) is the mid point of normal chord 

$\mathrm{h}=\frac{{{\mathrm{at}}_1}^2+{{\mathrm{at}}_2}^2}{2}, \mathrm{k}=\frac{2{\mathrm{at}}_1+2{\mathrm{at}}_2}{2}, \mathrm{k}=\mathrm{a}\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)$

$$\begin{gathered} 2\mathrm{h}=\mathrm{a}\left({{\mathrm{t}}_1}^2+{{\mathrm{t}}_2}^2\right) \\ 2\mathrm{h}=\mathrm{a}\left({\left(\frac{-2}{{\mathrm{t}}_1}\right)}^2-2\left(-{{\mathrm{t}}_1}^2-2\right)\right) \\ \cancel{2}\mathrm{h}=\cancel{2}\mathrm{a}\left(\frac{2}{{{\mathrm{t}}_1}^2}+{{\mathrm{t}}_1}^2+2\right) \\ \mathrm{h}=\mathrm{a}\left(\frac{2}{{{\mathrm{t}}_1}^2}+{{\mathrm{t}}_1}^2+2\right) \\ (\mathrm{from} (1\left)\right) \\ \frac{-2}{t_1}=\frac{k}{a} \\ \therefore {\mathrm{t}}_1=-2\mathrm{a}/\mathrm{k} \end{gathered}$$

$$\begin{gathered} \mathrm{Now} \mathrm{h}=\mathrm{a}\left(\frac{{\displaystyle \frac{2}{4{\mathrm{a}}^2}}}{{\mathrm{k}}^2}+\frac{4{\mathrm{a}}^2}{{\mathrm{k}}^2}+2\right) \\ \mathrm{Replace} \mathrm{h}, \mathrm{k} \mathrm{woth} \mathrm{x}, \mathrm{y} \\ \mathrm{x}=\frac{{\mathrm{y}}^2}{2\mathrm{a}}+\frac{4{\mathrm{a}}^3}{{\mathrm{y}}^2}+2\mathrm{a} \\ \mathrm{x}-2\mathrm{a}=\frac{{\mathrm{y}}^2}{2\mathrm{a}}+\frac{4{\mathrm{a}}^3}{{\mathrm{y}}^2} \end{gathered}$$