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.