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Question
Find the equations of all lines having slope 2 and that are tangent to the curve \[y = \frac{1}{x - 3}, x \neq 3\] ?
Solution
\[\text { Let }\left( x_1 , y_1 \right)\text { be the point where the tangent is drawn to this curve }.\]
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence }, y_1 =\frac{1}{x_1 - 3}\]
\[\text { Now,} y=\frac{1}{x - 3}\]
\[\Rightarrow\frac{dy}{dx} = \frac{- 1}{\left( x - 3 \right)^2}\]
\[\text { Slope of tangent } =\left( \frac{dy}{dx} \right)=\frac{- 1}{\left( x_1 - 3 \right)^2}\]
\[\text { Given that }\]
\[\text { Slope of the tangent} = 2\]
\[ \Rightarrow \frac{- 1}{\left( x_1 - 3 \right)^2} = 2\]
\[ \Rightarrow \left( x_1 - 3 \right)^2 = - 2\]
\[ \Rightarrow x_1 - 3 = \sqrt{- 2}, \text { which does not exist because 2 is negative}.\]
So, there does not exist any such tangent.
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