If the lines $x+y=a$ and $x-y=b$ touch the curve $y=x^2-3x+2$ at the points where the curve intersects the $x-$axis, then $\frac{a}{b}$ is equal to

Step 1: Calculate the $x-$intersects of the given curve

The equation of the curve, $C_1:y=x^2-3x+2$.

For the $x-$intersects, $y=0$.

$$\begin{gathered} \Rightarrow 0=x^2-3x+2 \\ \Rightarrow x^2-2x-x+2=0 \\ \Rightarrow x\left(x-2\right)-1\left(x-2\right)=0 \\ \Rightarrow \left(x-2\right)\left(x-1\right)=0 \\ \Rightarrow x=1,2 \end{gathered}$$

Therefore, the $x-$intersects of the curve are $\left(1,0\right)$ and $\left(2,0\right)$.

Step 2: Calculate the values of $a$ and $b$

The equations of the lines are,

$L_1:x+y=a$

$L_2:x-y=b$

The line $L_1$ intersects the curve $C_1$ at point $\left(1,0\right)$.

$$\begin{gathered} \left(1\right)+\left(0\right)=a \\ \Rightarrow a=1 \end{gathered}$$

The line $L_2$ intersects the curve $C_1$ at point $\left(2,0\right)$.

$$\begin{gathered} \left(2\right)-\left(0\right)=b \\ \Rightarrow b=2 \end{gathered}$$

Thus, $\frac{a}{b}=\frac{1}{2}$.

Hence, the value of $\frac{a}{b}$ is equal to $\frac{1}{2}$.