Question

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

Answer 1

The equation of the given curve is y=$\frac{1}{x-3},$ x≠3.

The slope of the tangent to the given curve at any point (x, y) is given by,

$\frac{dy}{dx}$=$-\frac{1}{(x-3)^2}$

If the slope of the tangent is 2, then we have:

$-\frac{1}{(x-3)^2}$= 2

2(x-3)2 =-1

(x-3)2=$-\frac12$

This is not possible since the L.HS. is positive while the R.H.S. is negative.

Hence, there is no tangent to the given curve having slope 2.