Question

Find the equation of all lines having slope $-4$ that are tangents to the curve $y=\frac{1}{x-1}$

Answer 1

$y=1\Rightarrow \frac{dy}{dx}=\frac{-1}{(x-1{)}^2}$

Now, slope is -4

$\therefore \frac{-1}{(x-1{)}^2}=-4$

$x-1=\pm \frac{1}{2}\Rightarrow x=\frac{3}{2}$ or $\frac{1}{2}$

$\therefore (x,y)\equiv (\frac{1}{2},-2)$ OR $(\frac{3}{2},2)$

$\therefore$ lines are

$y+2=-4(x-\frac{1}{2})\Rightarrow 4x+y=0$

and

$y-2=-4\left(\frac{x-3}{2}\right)\Rightarrow 4x+y=8$