If the equation of the normal to the curve $y=\frac{x-a}{\left(x+b\right)\left(x-2\right)}$ at the point $\left(1,-3\right)$ is $x-4y=13$ then the value of $a+b$ is equal to ______

Given curve is

$y=\frac{x-a}{\left(x+b\right)\left(x-2\right)} ...\left(1\right)$

At point $\left(1,-3\right)$, we have

$-3=\frac{1-a}{\left(1+b\right)\left(1-2\right)}$

$\Rightarrow 3\left(1+b\right)=1-a$

$\Rightarrow 1-a=3+3b$

Differentiating $\left(1\right)$ both sides w.r.t. $x$, we get

$\frac{dy}{dx}=\frac{(x+b)(x-2)\times \left(1\right)-(x-a)(2x+b-2)}{(x+b{)}^2(x-2{)}^2}$

At $(1,-3)$ slope of normal is $\frac{1}{4}$, hence

$\frac{dx}{dy}=-\frac{1}{4}$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(1,-3\right)}=-4$,

$\Rightarrow -4=\frac{(1+b)(-1)-(1-a)b}{(1+b{)}^2(-1{)}^2}$

$\Rightarrow -4=\frac{(1+b)(-1)-3(b+1)b}{(1+b{)}^2}$

$\Rightarrow -4=\frac{\left(-1\right)-3b}{\left(1+b\right)}$

$\left[\because b\ne -1\right]$

$\Rightarrow 4+4b=1+3b$

$\Rightarrow b=-3$

So, $\mathrm{a}=7$

Hence, $a+b=4$