If the tangent to the curve $x y+a x+b y=0$ at $(1,1)$ makes an angle $\tan ^{-1} 2$ with x-axis, then $\frac{a+b}{a b}=$

$\frac{1}{2}$

We have,

$x y+a x+b y=0$                  ……(i)

$\Rightarrow x \frac{d y}{d x}+y+a+b \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}  =-\left(\frac{y+a}{x+b}\right) $

$\Rightarrow \left(\frac{d y}{d x}\right)_{(1,1)}  =-\left(\frac{a+1}{b+1}\right) $

$\Rightarrow 2 =-\left(\frac{a+1}{b+1}\right) $               $\left[\begin{array}{r}\text { Given : Slope of the tangent } \\ \text { at }(1,1) \text { is } \tan \theta=2\end{array}\right]$

$\Rightarrow a+2 b  =-3$                …..(ii)

Also, $(1,1)$ lies on (i).

∴ $a+b=-1$                 …….(iii)

Solving (ii) and (iii), we get

$a=1, b=-2 \Rightarrow \frac{a+b}{a b}=\frac{1}{2}$