If the normal at one end of the latus rectum of the parabola ${\mathrm{y}}^2=16\mathrm{x}$ meets the X-aixs at the point P, then the length of the chord passing through P and perpendicular to the normal is

$$\begin{gathered} {\mathrm{y}}^2=16\mathrm{x} \\ \mathrm{one} \mathrm{end} \mathrm{of} \mathrm{latusrectum} \mathrm{L}=(\mathrm{a},2\mathrm{a})=(4, 8) \\ \mathrm{Normal} \mathrm{at} (4, 8) is \\ \mathrm{y}-{\mathrm{y}}_1=\frac{-{\mathrm{y}}_1}{2\mathrm{a}}(\mathrm{x}-{\mathrm{x}}_1) \\ \mathrm{y}-8=\frac{-8}{8}(\mathrm{x}-4) \\ \mathrm{x}+\mathrm{y}-12=0 \\ It meets x-axis at \mathrm{P}=(12, 0) \end{gathered}$$

Equation of the chord passing through P(12, 0) tr to normal is $\mathrm{y}=1(\mathrm{x}-12)$

Length of chord $=\left|{\mathrm{x}}_1-{\mathrm{x}}_2\right|\sqrt{1+{\mathrm{m}}^2}$

$$\begin{gathered} {(\mathrm{x}-12)}^2=16\mathrm{x} {\mathrm{x}}^2-40\mathrm{x}+144=0 \\ \left|{\mathrm{x}}_1-{\mathrm{x}}_2\right|=\frac{\sqrt{1}}{\left|\mathrm{a}\right|}=\sqrt{1600-576}=32 \\ \mathrm{Length} \mathrm{of} \mathrm{chord} = 32\sqrt{2} \end{gathered}$$