Question

Find equations of all lines having slope -1 that are tangent to the curve 𝑦=1/𝑥−1

Answer 1

Given the equation $y=\frac{1}{x-1},$

$$m=\dfrac{dy}{dx}=-\frac{1}{(x-1)^2}=-1 \qquad \text{(given)}$$

$$\implies \frac{1}{(x-1)^2} = 1\\ 1= (x-1)^2\\ 1= x^2-2x+1\\ 1-1=x^2-2x\\ 0=x(x-2)\\ \implies x=0 \text{ and } x=2$$

For $x=0,$

$$y=\frac{1}{0-1}=\frac{1}{-1}=-1\\ \therefore \text{the tangent line touches the curve at the point (0,-1)}$$

The equation of the line is:

$$y-y_1=m(x-x_1)\\ y-(-1)=-1(x-0)\\ y+1=-x\\ \implies x+y=-1 \qquad \cdots (i)$$

at $x=2,$

$$y=\frac{1}{2-1}=\frac{1}{1}=1\\ \therefore \text{the tangent line touches the curve at the point (2,1)}$$

$$y-y_1=m(x-x_1)\\ y-(-1)=-1(x-2)\\ y-1=-x+2\\ \implies x+y=3 \qquad \cdots (ii)$$

Thus, eqn (i) and (ii) are the required equations of the line that are tangential to the curve.