Question:
Find the equation of all lines having slope 2 which are tangents to the curve $y=\frac{1}{x-3}, x \neq 3$.
Solution:
The equation of the given curve is $y=\frac{1}{x-3}, x \neq 3$.
The slope of the tangent to the given curve at any point (x, y) is given by,
$\frac{d y}{d x}=\frac{-1}{(x-3)^{2}}$
If the slope of the tangent is 2, then we have:
$\frac{-1}{(x-3)^{2}}=2$
$\Rightarrow 2(x-3)^{2}=-1$
$\Rightarrow(x-3)^{2}=\frac{-1}{2}$
This is not possible since the L.H.S. is positive while the R.H.S. is negative.
Hence, there is no tangent to the given curve having slope 2 .