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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

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Transcript

Ex 6.3, 15 Find the equation of the tangent line to the curve 𝑦=π‘₯2 βˆ’2π‘₯+7 which is : (a) parallel to the line 2π‘₯βˆ’π‘¦+9=0We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ 𝑦=π‘₯2 βˆ’2π‘₯+7 Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=2π‘₯βˆ’2 Finding Slope of line 2π‘₯βˆ’π‘¦+9=0 2π‘₯βˆ’π‘¦+9=0 𝑦=2π‘₯+9 𝑦=2π‘₯+9 The Above Equation is of form 𝑦=π‘šπ‘₯+𝑐 where m is Slope of line Hence, Slope of line 2π‘₯βˆ’π‘¦+9 is 2 Now, Given tangent is parallel to 2π‘₯βˆ’π‘¦+9=0 Slope of tangent = Slope of line 2π‘₯βˆ’π‘¦+9 = 0 𝑑𝑦/𝑑π‘₯=2 2π‘₯βˆ’2=2 2(π‘₯βˆ’1)=2 π‘₯=2 Finding y when π‘₯=2 , 𝑦=π‘₯^2βˆ’2π‘₯+7= (2)^2βˆ’2(2)+7=4βˆ’4+7=7 We need to find Equation of tangent passes through (2, 7) & Slope is 2 Equation of tangent is (π‘¦βˆ’7)=2(π‘₯βˆ’2) π‘¦βˆ’7=2π‘₯βˆ’4 π‘¦βˆ’2π‘₯βˆ’7+4=0 π‘¦βˆ’2π‘₯βˆ’3=0 Hence Required Equation of tangent is π’šβˆ’πŸπ’™βˆ’πŸ‘=𝟎 We know that Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Ex 6.3, 15 Find the equation of the tangent line to the curve 𝑦=π‘₯2βˆ’2π‘₯+7 which is (b) perpendicular to the line 5π‘¦βˆ’15π‘₯=13We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ 𝑦=π‘₯2 βˆ’2π‘₯+7 Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=2π‘₯βˆ’2 Finding Slope of line 5π‘¦βˆ’15π‘₯=13 5π‘¦βˆ’15π‘₯=13 5𝑦=15π‘₯+13 𝑦=1/5 (15π‘₯+13) 𝑦=15/5 π‘₯+13/5 𝑦=3π‘₯+13/5 Above Equation is of form 𝑦=π‘šπ‘₯+𝑐 , where m is Slope of a line ∴ Slope = 3 Now, Given tangent is perpendicular to 5π‘¦βˆ’15π‘₯=13 Slope of tangent Γ— Slope of line = –1 𝑑𝑦/𝑑π‘₯ Γ— 3=βˆ’1 𝑑𝑦/𝑑π‘₯=(βˆ’1)/( 3) 2π‘₯βˆ’2=(βˆ’1)/( 3) 2π‘₯=(βˆ’1)/( 3)+2 2π‘₯=(βˆ’1 + 6)/3 2π‘₯=5/3 π‘₯=5/6 Finding y when π‘₯=5/6 𝑦=π‘₯^2βˆ’2π‘₯+7=(5/6)^2βˆ’2(5/6)+7=25/36βˆ’10/6+7=217/36 ∴ Point is (5/6 ,217/36) Equation of tangent passing through (5/6 ,217/36) & having Slope (βˆ’1)/( 3) (π‘¦βˆ’217/36)=(βˆ’1)/( 3) (π‘₯βˆ’5/6) (36𝑦 βˆ’217)/36=(βˆ’1)/( 3) (π‘₯βˆ’5/6) 36𝑦 βˆ’217=(βˆ’36)/( 3) (π‘₯βˆ’5/( 6)) 36𝑦 βˆ’217=βˆ’12(π‘₯βˆ’5/( 6)) 36𝑦 βˆ’217=βˆ’12π‘₯+(12 Γ— 5)/6 36𝑦 βˆ’217=βˆ’12π‘₯+10 36𝑦 βˆ’217=βˆ’12π‘₯+10 36𝑦+12π‘₯βˆ’217βˆ’10=0 πŸ‘πŸ”π’š+πŸπŸπ’™βˆ’πŸπŸπŸ•=𝟎 is Required Equation of tangent We know that Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.