If the lines x+y=a and x-y=b touch the curve

$y=x^{2}-3 x+2$

At $x$-axis $y=0=x^{2}-3 x+2$

$x=1,2$

$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}-3$

$\mathrm{A}(1,0) \mathrm{B}(2,0)$

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{x=1}=-1$ and $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{x=2}=1$

$\# x+y=a \Rightarrow \frac{d y}{d x}=-1$ So $A(1,0)$ lies on it

$\Rightarrow 1+0=a \Rightarrow a=1$

$\# x-y=b \Rightarrow \frac{d y}{d x}=1$ So $B(2,0)$ lies on it

$2-0=b \Rightarrow b=2$

$\frac{\mathrm{a}}{\mathrm{b}}=0.50$