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Ex 6.3, 15 - Find equation of tangent line to y = x2 - 2x + 7

Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives - Part 6 Ex 6.3,15 - Chapter 6 Class 12 Application of Derivatives - Part 7


Transcript

Ex 6.3, 15 Find the equation of the tangent line to the curve 𝑦=𝑥2 −2𝑥+7 which is : (a) parallel to the line 2𝑥−𝑦+9=0We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑦=𝑥2 −2𝑥+7 Differentiating w.r.t.𝑥 𝑑𝑦/𝑑𝑥=2𝑥−2 Finding Slope of line 2𝑥−𝑦+9=0 2𝑥−𝑦+9=0 𝑦=2𝑥+9 𝑦=2𝑥+9 The Above Equation is of form 𝑦=𝑚𝑥+𝑐 where m is Slope of line Hence, Slope of line 2𝑥−𝑦+9 is 2 Now, Given tangent is parallel to 2𝑥−𝑦+9=0 Slope of tangent = Slope of line 2𝑥−𝑦+9 = 0 𝑑𝑦/𝑑𝑥=2 2𝑥−2=2 2(𝑥−1)=2 𝑥=2 Finding y when 𝑥=2 , 𝑦=𝑥^2−2𝑥+7= (2)^2−2(2)+7=4−4+7=7 We need to find Equation of tangent passes through (2, 7) & Slope is 2 Equation of tangent is (𝑦−7)=2(𝑥−2) 𝑦−7=2𝑥−4 𝑦−2𝑥−7+4=0 𝑦−2𝑥−3=0 Hence Required Equation of tangent is 𝒚−𝟐𝒙−𝟑=𝟎 We know that Equation of line at (𝑥1 , 𝑦1) & having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Ex 6.3, 15 Find the equation of the tangent line to the curve 𝑦=𝑥2−2𝑥+7 which is (b) perpendicular to the line 5𝑦−15𝑥=13We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑦=𝑥2 −2𝑥+7 Differentiating w.r.t.𝑥 𝑑𝑦/𝑑𝑥=2𝑥−2 Finding Slope of line 5𝑦−15𝑥=13 5𝑦−15𝑥=13 5𝑦=15𝑥+13 𝑦=1/5 (15𝑥+13) 𝑦=15/5 𝑥+13/5 𝑦=3𝑥+13/5 Above Equation is of form 𝑦=𝑚𝑥+𝑐 , where m is Slope of a line ∴ Slope = 3 Now, Given tangent is perpendicular to 5𝑦−15𝑥=13 Slope of tangent × Slope of line = –1 𝑑𝑦/𝑑𝑥 × 3=−1 𝑑𝑦/𝑑𝑥=(−1)/( 3) 2𝑥−2=(−1)/( 3) 2𝑥=(−1)/( 3)+2 2𝑥=(−1 + 6)/3 2𝑥=5/3 𝑥=5/6 Finding y when 𝑥=5/6 𝑦=𝑥^2−2𝑥+7=(5/6)^2−2(5/6)+7=25/36−10/6+7=217/36 ∴ Point is (5/6 ,217/36) Equation of tangent passing through (5/6 ,217/36) & having Slope (−1)/( 3) (𝑦−217/36)=(−1)/( 3) (𝑥−5/6) (36𝑦 −217)/36=(−1)/( 3) (𝑥−5/6) 36𝑦 −217=(−36)/( 3) (𝑥−5/( 6)) 36𝑦 −217=−12(𝑥−5/( 6)) 36𝑦 −217=−12𝑥+(12 × 5)/6 36𝑦 −217=−12𝑥+10 36𝑦 −217=−12𝑥+10 36𝑦+12𝑥−217−10=0 𝟑𝟔𝒚+𝟏𝟐𝒙−𝟐𝟐𝟕=𝟎 is Required Equation of tangent We know that Equation of line at (𝑥1 , 𝑦1) & having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1)

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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.