Ex 6.3,10 - Class 12

Transcript

Ex 6.3,10 Find the equation of all lines having slope –1 that are tangents to the curve 𝑦= 1﷮𝑥 − 1﷯ , 𝑥≠1. Equation of Curve is 𝑦= 1﷮𝑥 − 1﷯ We know that Slope of tangent is 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 1﷮𝑥 − 1﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 1﷯﷮′﷯ 𝑥 −1﷯ − 𝑥 − 1﷯﷮′﷯1﷮ 𝑥 − 1﷯﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯= 0 𝑥 −1﷯ − 1 − 0﷯ 1﷮ 𝑥 − 1﷯﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯= − 1﷮ 𝑥 − 1﷯﷮2﷯﷯ So, Slope of tangent is − 1﷮ 𝑥 − 1﷯﷮2﷯﷯ But ,Given Slope of tangent =−1 ∴ − 1﷮ 𝑥 − 1﷯﷮2﷯﷯=−1 1﷮ 𝑥 − 1﷯﷮2﷯﷯=1 1= 𝑥 − 1﷯﷮2﷯ 𝑥 − 1﷯﷮2﷯=1 𝑥 − 1=± ﷮1﷯ 𝑥 − 1=±1 Hence 𝑥=2 , 0 Equation of 1st tangent at 2 , 1﷯ & having Slope –1 is 𝑦−1﷯=−1 𝑥−2﷯ 𝑦−1=−𝑥+2 𝑦+𝑥=2+1 𝒚+𝒙−𝟑=𝟎 Equation of 2nd tangent 0 , −1﷯ & having Slope –1 is 𝑦− −1﷯﷯=−1 𝑥−0﷯ 𝑦+1=−𝑥+0 𝒚+𝒙+𝟏=𝟎 Hence the Required Equation of lines are 𝑦+𝑥−3=0 & 𝑦+𝑥+1=0