Question

Find the values of $a$ and $b$, if the slope of the tangent to the curve $xy+ax+by=2$ at $(1,1)$ is $2$.

Answer 1

$xy+ax+by=2$ ----- ( 1 )

On differentiating both sides w.r.t. $x,$ we get

$x\frac{dy}{dx}+y+a+b\frac{dy}{dx}=0$

$\Rightarrow$ $\frac{dy}{dx}(x+b)=-a-y$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-a-y}{x+b}$

Now,

The slope of the tangent $=2$

${\left(\frac{dy}{dx}\right)}_{(1,1)}=2$

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

$\Rightarrow$ $-a-1=2+2b$

$\Rightarrow$ $-a=3+2b$

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

On substituting $a=-(3+2b),x=1$ and $y=1$ in equation ( 1 ), we get

$\Rightarrow$ $1-(3+2b)+b=2$

$\Rightarrow$ $1-3-2b+b=2$

$\Rightarrow$ $b=-4$

And

$\Rightarrow$ $a=-(3+2b)=-(3-8)=5$

$\therefore$ $a=5$ and $b=-4$