If slope of a tangent to the curve \(x y+a x+b y=0\) at the point \((1,1)\) on it is 2 , then \(a-b\) is

Given curve is
\(\begin{aligned}
& x y+\mathrm{a} x+\mathrm{b} y=0 \\
\therefore \quad & \text { Slope }=2=\frac{\mathrm{d} y}{\mathrm{~d} x} \\
\therefore \quad & x y+\mathrm{a} x+\mathrm{b} y=0
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& x \frac{\mathrm{d} y}{\mathrm{~d} x}+y+\mathrm{a}+\mathrm{b} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& \therefore \quad(x+\mathrm{b}) \frac{\mathrm{d} y}{\mathrm{~d} x}=-(y+\mathrm{a}) \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-(y+\mathrm{a})}{x+\mathrm{b}} \\
& \therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(1,1)}=2 \\
& \therefore \quad 2=\frac{-(1+\mathrm{a})}{1+\mathrm{b}} \\
& \therefore \quad \mathrm{a}+2 \mathrm{~b}=-3 ... (i)
\end{aligned}\)
Since \((1,1)\) lies on \(x y+a x+b y=0\), we get \(\mathbf{a}+\mathbf{b}=-1\) ... (ii)
Solving (i), (ii), we get
\(\begin{aligned}
& a=1, b=-2 \\
\therefore \quad & a-b=1-(-2)=3
\end{aligned}\)