Question

If the slope of tangent of $xy+ax+by=2$ at point $(1,1)$ is $5$ then find the values of $a$ and $b$.

Answer 1

$x+ax-by=2$
Now differrentiate w.r.to $x$
$x\frac{dy}{dx}+y+a+b\frac{dy}{dx}=0$
$\therefore (x+b)\frac{dy}{dx}=-(y+a)$
$\therefore \frac{dy}{dx}=-\frac{y+a}{x+b}$
Slope of a tangent at $(1,1)$ is $5$
$\therefore {\left(\frac{dy}{dx}\right)}_{(1,1)}=\frac{a+1}{b+1}=5$
$-a-1=5b+5$
$a+5b=-\mathrm{6....}\left(1\right)$
Also point $(1,1)\in xy+ax+by=2$
$-a-5b=5+1$
$a+5b=-\mathrm{6....}\left(2\right)$
By solving eqn (1) and (2)
we get $a=\frac{11}{4},b=-\frac{7}{4}$