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# Find the equation of all lines

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