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

<p>Given curve : $y = {{x - a} \over {(x + b)(x - 2)}}$ at $(1, - 3)$</p> <p>$\therefore$ $- 3 = {{1 - a} \over {(1 + b)( - 1)}} \Rightarrow 3 + 3b = 1 - a$</p> <p>$\beta \Rightarrow a + 3b + 2 = 0$</p> <p>$y = {{x - a} \over {(x + b)(x - 2)}}$</p> <p>$${{dy} \over {dx}} = {{(x + b)(x - 2) - (x - a)[(x + b) + (x - 2)]} \over {{{[(x + b)(x - 2)]}^2}}}$$</p> <p>at $(1, - 3)\,{m_T} = {{ - (1 + b) - (1 - a)(b)} \over {{{(1 + b)}^2}}} = - 4$</p> <p>$\therefore$ $1 + b + b - ab = 4{(1 + b)^2}$</p> <p>$\Rightarrow 1 + 2b + b(3b + 2) = 4{b^2} + 4 + 8b$</p> <p>$\Rightarrow {b^2} + 4b + 3 = 0$</p> <p>$(b + 1)(b + 3) = 0$</p> <p>$b = - 1,a = 1$ but $1 + b \ne 0$</p> <p>$b = - 3,a = 7$ $\therefore$ $b \ne - 1$</p> <p>$\therefore$ $a + b = 04$</p>