Find the equation of all lines having slope 2 and that are tangent to

1.	Find the equation of all lines having slope 2 and that are tangent to the curve\(y=\frac{1}{x-3},x\neq3.\)
Answer» finding the slope of the tangent by differentiating the curve \(\frac{dy}{dx}=-\frac{1}{(x-3)^2}\) Now according to question, the slope of all tangents is equal to 2, so \(-\frac{1}{(x-3)^2}=2\) \((x-3)^2=-\frac{1}{2}\) We can see that LHS is always greater than or equal to 0, while RHS is always negative. Hence no tangent is possible.

Find the equation of all lines having slope 2 and that are tangent to the curve\(y=\frac{1}{x-3},x\neq3.\)

Answer»

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=-\frac{1}{(x-3)^2}\)

Now according to question, the slope of all tangents is equal to 2, so

\(-\frac{1}{(x-3)^2}=2\)

\((x-3)^2=-\frac{1}{2}\)

We can see that LHS is always greater than or equal to 0, while RHS is always negative.

Hence no tangent is possible.