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Ex 6.3, 10 - Find equation of all lines having slope -1 - Ex 6.3

Ex 6.3,10 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,10 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,10 - Chapter 6 Class 12 Application of Derivatives - Part 4

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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) Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑(1/(𝑥 − 1))/𝑑𝑥 (𝑑𝑦 )/𝑑𝑥 =(𝑑 )/𝑑𝑥 (𝑥−1)^(−1) (𝑑𝑦 )/𝑑𝑥 =−1〖 × (𝑥−1)〗^(−1−1) (𝑑𝑦 )/𝑑𝑥 =−(𝑥−1)^(−2) 𝑑𝑦/𝑑𝑥=(− 1)/(𝑥 − 1)^2 Given that slope = −1 Hence, 𝑑𝑦/𝑑𝑥 = −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 x = 2 x − 1 = −1 x = 0 So, x = 2 & x = 0 Finding value of y If x = 2 y = 1/(𝑥 − 1) y = 1/(2 − 1) y = 1/1 y = 1 Thus, point is (2, 1) If x = 0 y = 1/(𝑥 − 1) y = 1/(0 − 1) y = 1/(−1) y = −1 Thus, point is (0, –1) Thus, there are 2 tangents to the curve with slope 2 and passing through points (2, 1) and (0, −1) We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent through (2, 1) is 𝑦 −1 =−1 (𝑥 −2) 𝑦 −1 =−𝑥+2 𝒚+𝒙−𝟑 = 𝟎 Equation of tangent through (0, −1) is 𝑦 −(−1)=−1 (𝑥 −0) 𝑦 +1 =−𝑥 𝒚+𝒙 + 𝟏 = 𝟎

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.