The equation of the normal to the parabola $y^2=4x$ which is perpendicular to $x+3y+1=0$ is

$\text{The equation of the parabola, }$

$y^2=4x$

$\text{The equation of the normal to the parabola, }$

$y^2=4ax\text{ at }\left(a{m}^2,-2am\right)\text{ is }y=mx-2am-a{m}^3$

$\text{Here, }a=1,\text{ the equation of the normal becomes, }$

$y=mx-2m-m^3 \dots \left(\mathrm{I}\right)$

$\text{Since the normal is perpendicular to the line, }$

$x+3 y+1=0$

$\text{Thus, }$

$m_1{m}_2=-1$

$m\left(\frac{-1}{3}\right)=-1$

$m=3$

$\text{Substitute the values in equation} \left(\mathrm{I}\right).$

$y=3x-2\left(3\right)-3^3$

$y=3x-6-27$

$3x-y=33$