Question

Find the equations of all lines having slope 2 and that are tangent to the curve $y=\frac{1}{x-3},x\ne 3$.

Answer 1

Slope of given tangent = 2
$$\begin{gathered} \text{Let}\left(x_1,y_1\right)\text{be the point where the tangent is drawn to this curve.} \\ \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},{\mathit{\text{y}}}_{\mathit{1}}\text{=}\frac{1}{x_1-3} \\ \mathrm{Now},\mathit{\text{y}}\text{=}\frac{1}{x-3} \\ \text{⇒}\frac{dy}{dx}=\frac{-1}{{\left(x-3\right)}^2} \\ \text{Slope of tangent =}\left(\frac{dy}{dx}\right)\text{=}\frac{-1}{{\left(x_1-3\right)}^2} \\ \text{Given that} \\ \text{Slope of the tangent = 2} \\ \Rightarrow \frac{-1}{{\left(x_1-3\right)}^2}=2 \\ \Rightarrow {\left(x_1-3\right)}^2=-2 \\ \Rightarrow x_1-3=\sqrt{-2},\text{which does not exist because 2 is negative.} \end{gathered}$$
So, there does not exist any such tangent.