Question

Find the equation of all lines having slope 2 which are tangents to the curve y=1/x-3,x≠3.

Answer 1

it is given that y = 1/(x - 3) , x ≠ 3
differentiate y with respect to x,
dy/dx = -1/(x - 3)² , x ≠ 3
so, slope of tangent of the curve at (x,y) = dy/dx
= -1/(x - 3)²

A/C to question,
we equation of all lines having slope 2 that are tangents to the curve, y = 1/(x -3) , x ≠ 3.
so, slope of tangent = 2

so, -1/(x - 3)² = 2
=> (x - 3)² = -1/2
This is not possible since the L.H.S. is positive while the R.H.S. is negative.
Therefore, there is no tangent to the given curve having a slope 2.