Find the equations of all lines having slope 2 and that are tangent to the curve `y=1/(x-3),\ \ x!=3` .

Explanation

Slope of tangent = 2
Equation of curve ` y = 1/(x-3)`
`rArr (dy)/(dx) = (-1)/((x-3)^(2))`
Slope of tangent at point `(x, y) = (-1)/((x-3)^(2))`
`:. (-1)/((x-3)^(2)) = 2`
` rArr (x-3)^(2) = - 1/2`
which is impossible.
Therefore, there is no tangent of the given curve with slope 2.