Question

Find the equation of the lines having slope -1 that are tangent to the curve $y=\frac{1}{x-1},x\ne 1$

Answer 1

The equation of the given curve is $y=\frac{1}{x-1},x\ne 1$

The slope of the tangent to the given curve at any point (x,y) is given by

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

For tangents having slope =-1, we must have

$-1=\frac{-1}{(x-1{)}^2}\Rightarrow (x-1{)}^2=1\Rightarrow x-1=\pm 1\Rightarrow x=1\pm 1=2,0$

When x=2, then from Eq. (i), we get $y=\frac{1}{2-1}=1$

When x=0, then from Eq. (i), we get $y=\frac{1}{0-1}=-1$

Thus, the points on the given curve at which slope of tangent is -1, are (2,1) and (0,-1).

$\therefore$ Equation of tangent at (2,1) is y-1=-1(x-2)or x+y-3=0

and equation of tangent at (0,-1) is y-(-1)=-(x-0) or x + y + 1 = 0

Hence, the equation of the required line are x+y-3=0 and x + y + 1 = 0