Question:
Find the equation of all lines having slope 2 and that are tangent to the curve $y=\frac{1}{x-3}, x \neq 3$.
Solution:
finding the slope of the tangent by differentiating the curve
$\frac{d y}{d x}=-\frac{1}{(x-3)^{2}}$
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