The point of intersection of normals to the parabola ${\mathrm{y}}^2=4\mathrm{x}$ at the points whose ordinates are 4 and 6 is

$$\begin{gathered} {\mathrm{y}}^2=4\mathrm{x} \\ 4\mathrm{a}=4 \\ \mathrm{a}=1 \\ \left({{\mathrm{at}}_1}^2,2{\mathrm{at}}_1\right) \left({{\mathrm{at}}_2}^2,2{\mathrm{at}}_2\right) \\ 2{\mathrm{at}}_1=4 2{\mathrm{at}}_2=6 \\ {\mathrm{t}}_1=2 {\mathrm{t}}_2=3 \end{gathered}$$

Points are $\left(4,4\right)\left(9,6\right)$

$$\begin{gathered} 2\mathrm{y} \mathrm{dy}/\mathrm{dx}=4 \\ \mathrm{dy}/\mathrm{dx}=2/\mathrm{y}=2/4=1/2 \end{gathered}$$

Eq. of normal at $\left(4,4\right)$

$$\begin{gathered} \mathrm{y}-4=-2\left(\mathrm{x}-4\right) \\ \mathrm{y}-4=-2\mathrm{x}+8 \\ 2\mathrm{x}+\mathrm{y}-12=0 \\ {\mathrm{m}}_2=1/3 \\ \mathrm{y}-6=-3\left(\mathrm{x}-9\right) \\ \mathrm{y}-6=-3\mathrm{x}+27 \\ 3\mathrm{x}+\mathrm{y}-33=0 \\ \mathrm{x}=21, \mathrm{y}=-30 \end{gathered}$$