Let l be a line which is normal to the curve $\mathrm{y}=2{\mathrm{x}}^2+\mathrm{x}+2$ at a point P on the curve. If the point $\mathrm{Q}(6,4)$ lies on the line l, then the distance from the origin to the line l is:
 

${\mathrm{y}}^{\mathrm{\prime }}=4\mathrm{x}+1,\text{ let }\mathrm{P}(\mathrm{h},\mathrm{k})$
Normal is $\mathrm{y}-4=-\frac{1}{(4\mathrm{h}+1)}(\mathrm{x}-6)$
$\begin{array}{l}\left(2{\mathrm{h}}^2+\mathrm{h}+2\right)-4=-\frac{(\mathrm{h}-6)}{4\mathrm{h}+1}\\ \Rightarrow 8{\mathrm{h}}^3+2{\mathrm{h}}^2+4{\mathrm{h}}^2+\mathrm{h}-8\mathrm{h}-2=-\mathrm{h}+6\\ \Rightarrow 8{\mathrm{h}}^3+6{\mathrm{h}}^2-6\mathrm{h}-8=0\\ \Rightarrow \mathrm{h}=1,\mathrm{k}=5\\ \mathrm{P}(1,5),\mathrm{Q}(6,4)\end{array}$
Normal is $\mathrm{y}-4=-\frac{1}{5}(\mathrm{x}-6)$
$\begin{array}{l}\Rightarrow \mathrm{x}+5\mathrm{y}-26=0\\ \frac{\left|\mathrm{C}\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}=\frac{26}{\sqrt{26}}=\sqrt{26}.\end{array}$