Question

Find the equation of all lines having slope −1 that are tangents to the curve y=\(\frac{1}{x-1}\) , x≠1.

Answer 1

The equation of the given curve is y=\(\frac{1}{x-1}\) , x≠1. The slope of the tangents to the given curve at any point (x, y) is given by,

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

If the slope of the tangent is −1, then we have:

\(-\frac{1}{(x-1)^2}\) =-1

=(x-1)²=1

=x-1=±1

x=2, 0

When x = 0, y = −1 and when x = 2, y = 1. 
Thus, there are two tangents to the given curve having slope −1. These are passing through the points (0, −1) and (2, 1).

∴The equation of the tangent through (0, −1) is given by,

y-(-1)=-1(x-0)

y+1=-x

y+x+1=0

∴The equation of the tangent through (2, 1) is given by

y − 1 = −1 (x − 2)

⇒ y − 1 = − x + 2

⇒ y + x − 3 = 0

Hence, the equations of the required lines are y + x + 1 = 0 and y + x − 3 = 0.