If the tangent to the curie xy +ax+ by= 0 at (1, 1) makes on angle tan-1 2 with x-axis, then aba+b=

We have,

$x y + a x + b y = 0$ …(i)

$\begin{array}{l} \Rightarrow x \frac{d y}{d x} + y + a + b \frac{d y}{d x} = 0 \\ \Rightarrow \frac{d y}{d x} = - (\frac{y + a}{x + b}) \end{array}$

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

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

$\Rightarrow a + 2 b = - 3$ …(ii)

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

$∴ a + b = - 1$ …(iii)

Solving (ii) and (iii), we get

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

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