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

We consider the curve $$y = \frac{x - a}{(x + b)(x - 2)}$$ which passes through the point $$(1, -3)$$ and has normal line $$x - 4y = 13$$ at that point. Substituting $$(1, -3)$$ into the normal gives $$1 - 4(-3) = 1 + 12 = 13$$, confirming consistency.

Since $$(1, -3)$$ lies on the curve, we set $$y(1)=\frac{1 - a}{(1 + b)(1 - 2)}=\frac{a - 1}{1 + b}=-3\,. $$ Hence $$a - 1 = -3(1 + b)\implies a = -2 - 3b\quad\cdots(i)$$

The normal has slope $$\tfrac14$$, so the tangent has slope $$-4$$. Writing $$u=x-a\,,\quad v=(x+b)(x-2)=x^2+(b-2)x-2b\,, $$ we get by the quotient rule $$y'=\frac{v - u\,v'}{v^2}\,. $$

At $$x=1$$ we have $$u=1-a\,,\quad v=(1+b)(-1)=-(1+b)\,,\quad v'=2(1)+b-2=b\,. $$ Thus $$y'(1)=\frac{-(1+b)-(1-a)\,b}{(1+b)^2}=\frac{-1-2b+ab}{(1+b)^2}=-4\,. $$ Clearing denominators gives $$-1-2b+ab=-4(1+b)^2=-4-8b-4b^2\,, $$ or $$4b^2+6b+ab+3=0\,. $$

Substituting $$a=-2-3b$$ from (i) yields $$4b^2+6b+(-2-3b)b+3=0\,, $$ which simplifies to $$4b^2+6b-2b-3b^2+3=0\quad\Longrightarrow\quad b^2+4b+3=0\implies(b+1)(b+3)=0\,. $$

If $$b=-1$$, then $$a=-2-3(-1)=1$$ and $$y=\frac{x-1}{(x-1)(x-2)}=\frac1{x-2}\quad(x\neq1),$$ but this gives $$y(1)=-1\neq-3$$, so $$b=-1$$ is invalid (removable discontinuity at $$x=1$$).

If $$b=-3$$, then $$a=-2-3(-3)=7$$ and $$y=\frac{x-7}{(x-3)(x-2)}\,. $$ At $$x=1$$ this yields $$y=\frac{-6}{(-2)(-1)}=-3$$ and $$y'=\frac{(x-3)(x-2)-(x-7)(2x-5)}{[(x-3)(x-2)]^2},$$ so at $$x=1$$ $$y'(1)=\frac{2-(-6)(-3)}{4}=\frac{2-18}{4}=-4\,, $$ confirming the tangent slope is $$-4$$ and hence the normal slope is $$\tfrac14\,$$.

Therefore $$a+b=7+(-3)=4\,. $$