Question:
Find the values of $a$ and $b$ if the slope of the tangent to the curve $x y+a x+b y=2$ at $(1,1)$ is 2 .
Solution:
Given :
$x y+a x+b y=2 \ldots$ .....(1)
On differentiating both sides w.r.t. $x$, we get
$x \frac{d y}{d x}+y+a+b \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}(x+b)=-a-y$
$\Rightarrow \frac{d y}{d x}=\frac{-a-y}{x+b}$
Now,
$\left(\frac{d y}{d x}\right)_{(1,1)}=2$
$\Rightarrow \frac{-a-1}{1+b}=2$
$\Rightarrow-a-1=2+2 b$
$\Rightarrow-a=3+2 b$
$\Rightarrow a=-(3+2 b)$
On substituting $a=-(3+2 b), \quad x=1$ and $y=1$ in eq. $(1)$, we get
$1-(3+2 b)+b=2$
$\Rightarrow 1-3-2 b+b=2$
$\Rightarrow b=-4$
and
$a=-(3+2 b)=-(3-8)=5$
$\therefore a=5$ and $b=-4$