Find the equation of all lines having slope 2 which are tangents to th

1.	Find the equation of all lines having slope 2 which are tangents to the curvey = \(\frac{1}{x - 3},\) x ≠ 3
Answer» slope of the line = 2 slope of the curve = \(\frac{-1}{(x - 3)^2}\) \(\frac{-1}{(x - 3)^2}\) = 2 ⇒ 2(x -3)² = -1 ⇒ 2(x² – 6x + 9) = -1 ⇒ 2x² – 12x + 19 = 0 which has no real roots b² – 4ac < 0 Hence there is no point on the curve

Find the equation of all lines having slope 2 which are tangents to the curvey = \(\frac{1}{x - 3},\) x ≠ 3

Answer»

slope of the line = 2

slope of the curve = \(\frac{-1}{(x - 3)^2}\)

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

⇒ 2(x -3)² = -1

⇒ 2(x² – 6x + 9) = -1

⇒ 2x² – 12x + 19 = 0

which has no real roots b² – 4ac < 0

Hence there is no point on the curve