If P(h,k) be a point on the parabola x=4y² which is the nearest to the point Q(0,33) then the distance of P from the line $3\mathrm{x}+4\mathrm{y}+5=0$ is:
 

$\begin{array}{l}\mathrm{x}=4{\mathrm{y}}^2,\mathrm{P}\left(4{\mathrm{t}}^2,\mathrm{t}\right),\mathrm{Q}(0,33)\\ {\mathrm{PQ}}^2={\left(4{\mathrm{t}}^2-0\right)}^2+(\mathrm{t}-33{)}^2\\ \mathrm{f}\left(\mathrm{t}\right)=16{\mathrm{t}}^4+{\mathrm{t}}^2-66\mathrm{t}+1089\\ {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{t}\right)=64{\mathrm{t}}^3+2\mathrm{t}-66\\ {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{t}\right)=0\Rightarrow 32{\mathrm{t}}^3+\mathrm{t}-33=0\\ \Rightarrow \mathrm{t}=1,32{\mathrm{t}}^2+32\mathrm{t}+33=0\end{array}$
Imaginary roots $\therefore \mathrm{P}(4,1)$
Distance $=\frac{|12+4+5|}{5}=\frac{21}{5}.$