**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 .

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