The point of intersection of the normals to the parabola ${\mathrm{y}}^2=4\mathrm{x}$ at the ends of its latus rectum is 

$$\begin{gathered} \mathrm{Given} \mathrm{parabola} {\mathrm{y}}^2=4\mathrm{x} \\ \Rightarrow \mathrm{a}=1 \\ diff w.r.to x on both sides \\ \frac{dy}{dx}=\frac{4}{2y} \end{gathered}$$

ends of latus rectum are

$$\begin{gathered} \left(\mathrm{a}, 2\mathrm{a}\right) \mathrm{and} \left(\mathrm{a}, -2\mathrm{a}\right) \\ \left(1, 2\right) \mathrm{and} \left(1, -2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Slope} \mathrm{at} \left(1, 2\right) \mathrm{is}=\frac{4}{2\mathrm{y}}=\frac{4}{4}=1 \\ \mathrm{slope} \mathrm{at} \left(1, -2\right) \mathrm{is}=\frac{4}{2\mathrm{y}}=\frac{4}{-4}=-1 \end{gathered}$$

Normal slopes  are -1 and 1

Equation  of normal at (1,2)

$$\begin{gathered} \mathrm{y}-2=-1 \left(\mathrm{x}-1\right) \\ \mathrm{y}-2=-\mathrm{x}+1 \\ \mathrm{x}+\mathrm{y}-3=0 -\overline{)1} \end{gathered}$$

Equation  of normal at $\left(1, -2\right)$

$$\begin{gathered} \mathrm{y}+2=1 \left(\mathrm{x}-1\right) \\ \mathrm{y}+2=\mathrm{x}-1 \\ \mathrm{x}-\mathrm{y}-3=0 -\overline{)2} \end{gathered}$$

solving $\overline{)1} \& \overline{)2}$

$$\begin{gathered} 2\mathrm{x}=6 and \mathrm{y}=0 \\ \Rightarrow \mathrm{x}=3 ,y=0 \\ Required point=\left(3, 0\right) \end{gathered}$$