Ex 6.3, 15 - Find equation of tangent line to y = x2 - 2x + 7 - Finding point when tangent is parallel/ perpendicular

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  1. Class 12
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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 ﷐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 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 ﷐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 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 ﷐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 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 ﷐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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.