# Ex 6.3,15

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

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=0 We know that Slope of tangent is 𝑑𝑦𝑑𝑥 𝑦=𝑥2 −2𝑥+7 Differentiating w.r.t.𝑥 𝑑𝑦𝑑𝑥=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 the 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 𝑑𝑦𝑑𝑥=2 2𝑥−2=2 2𝑥−1=2 𝑥=2 Finding y when 𝑥=2 , 𝑦=𝑥2−2𝑥+7= 22−22+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=2𝑥−2 𝑦−7=2𝑥−4 𝑦−2𝑥−7+4=0 𝑦−2𝑥−3=0 Hence Required Equation of tangent parallel to 𝒚−𝟐𝒙−𝟑=𝟎 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=13 We know that Slope of tangent is 𝑑𝑦𝑑𝑥 𝑦=𝑥2 −2𝑥+7 Differentiating w.r.t.𝑥 𝑑𝑦𝑑𝑥=2𝑥−2 Finding Slope of line 5𝑦−15𝑥=13 5𝑦−15𝑥=13 5𝑦=15𝑥+13 𝑦=1515𝑥+13 𝑦=155𝑥+135 𝑦=3𝑥+135 Above Equation is of the 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 𝑑𝑦𝑑𝑥 ×3=−1 𝑑𝑦𝑑𝑥=−1 3 2𝑥−2=−1 3 2𝑥=−1 3+2 2𝑥=−1 + 63 2𝑥=53 𝑥=56 Finding y when 𝑥=56 𝑦=𝑥2−2𝑥+7=562−256+7=2536−106+7=21736 ∴ Point is 56 ,21736 Equation of tangent passing through 56 ,21736 & having Slope −1 3 𝑦−21736=−1 3𝑥−56 36𝑦 −21736=−1 3𝑥−56 36𝑦 −217=−36 3𝑥−5 6 36𝑦 −217=−12𝑥−5 6 36𝑦 −217=−12𝑥+12 × 56 36𝑦 −217=−12𝑥+10 36𝑦 −217=−12𝑥+10 36𝑦+12𝑥−217−10=0 𝟑𝟔𝒚+𝟏𝟐𝒙−𝟐𝟐𝟕=𝟎 is Required Equation of tangent

Ex 6.3,5
Important

Ex 6.3,7 Important

Ex 6.3,12 Important

Ex 6.3,15 Important You are here

Ex 6.3,26 Important

Example 35 Important

Example 37 Important

Example 38 Important

Example 40 Important

Ex 6.5,1 Important

Ex 6.5,5 Important

Ex 6.5,7 Important

Ex 6.5,11 Important

Ex 6.5,18 Important

Ex 6.5,20 Important

Ex 6.5,23 Important

Ex 6.5,26 Important

Ex 6.5,28 Important

Example 46 Important

Example 47 Important

Misc 6 Important

Misc 11 Important

Misc 13 Important

Misc 17 Important

Misc 22 Important

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Important Question for exams Class 12

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .