If the tangent to the curve $\mathrm{y}={\mathrm{x}}^3-{\mathrm{x}}^2+\mathrm{x}$ at the point (a, b) is also tangent to thecurve  $\mathrm{y}=5{\mathrm{x}}^2+2\mathrm{x}-25$ at the point $(2,-1)$ then $|2\mathrm{a}+9\mathrm{b}|$ is equal to ………. . 

$\begin{array}{l}\mathrm{y}=5{\mathrm{x}}^2+2\mathrm{x}-25\mathrm{P}(2,-1)\\ {\mathrm{y}}^{\mathrm{\prime }}=10\mathrm{x}+2\\ {\mathrm{y}}_{\mathrm{P}}^{\mathrm{\prime }}=22\end{array}$
$\therefore$tangent to curve at P
$\begin{array}{l}\mathrm{y}+1=22(\mathrm{x}-2)\\ \mathrm{y}=22\mathrm{x}-45\\ \mathrm{y}={\mathrm{x}}^3-{\mathrm{x}}^2+\mathrm{x}\end{array}$

$\begin{array}{l}{\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{{\mathrm{C}}_2}=3{\mathrm{x}}^2-2\mathrm{x}+1\\ {\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_0=3{\mathrm{a}}^2-2\mathrm{a}+1\end{array}$
Hence $3{\mathrm{a}}^2-2\mathrm{a}+1=22$
$\begin{array}{l}\therefore 3{\mathrm{a}}^2-2\mathrm{a}-21=0\\ 3{\mathrm{a}}^2-9\mathrm{a}+7\mathrm{a}-21=0\end{array}$
$(3\mathrm{a}+7)(\mathrm{a}-3)=0\Rightarrow a=3,\frac{-7}{3}$                       
From curve $\mathrm{b}={\mathrm{a}}^3-{\mathrm{a}}^2+\mathrm{a}$
$\begin{array}{l}\mathrm{a}=3\\ \mathrm{b}=21\\ |2\mathrm{a}+9\mathrm{b}|=195\end{array}$
at $\mathrm{a}=-\frac{7}{3}$ tangent will be parallel
Hence it is rejected.