$y=(x-1)(x-2)$, this curve intersects the $x$-axis at $A(1,0)$ and $B(2,0)$
$ \therefore \quad \frac{d y}{d x}=2 x-3 ; \frac{d y}{d x_{(x-1)}}=-1 \text { and } \frac{d y}{d x_{(x-2)}}=1
$ Equation of tangent at $\mathrm{A}(1,0)$;
$ \begin{aligned} & y=-1(x-1) \\ \Rightarrow \quad & x+y=1
\end{aligned} $ and equation of tangent at $B(2,0)$
$ \begin{aligned} & y=1(x-2) \\ \Rightarrow \quad & x-y=2
\end{aligned} $ So $\quad a=1$ and $b=2$
$ \frac{a}{b}=0.5
$