Find the equations of all lines having slope $2$ which are tangents to the curve $y = \cfrac{1}{{x - 3}},x \ne 3.$

The given curve is
$y = \cfrac{1}{{x - 3}}$ …(i)

Differentiating (i) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = \cfrac{{ - 1}}{{{{(x - 3)}^2}}}$

For tangents having slope $2$, we must have $2 = \cfrac{{ - 1}}{{{{(x - 3)}^2}}}$
$\Rightarrow {(x - 3)^2} = - \cfrac{1}{2} \Rightarrow 2{(x - 3)^2} = - 1 \Rightarrow 2{x^2} - 12x + 19 = 0$

$\Rightarrow x = \cfrac{{12 \pm \sqrt {144 - 152} }}{4} \Rightarrow x = \cfrac{{12 \pm \sqrt { - 8} }}{4}$
which is not possible as being imaginary number.

Hence, there is no such tangent possible.