Ex 6.3, 21 - Find equation of normal to y = x3 + 2x + 6 which - Finding equation of tangent/normal when slope and curve are given

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.3,21 Find the equation of the normal to the curve 𝑦=﷐𝑥﷮3﷯+2𝑥+6 which are parallel to the line 𝑥+14𝑦+4=0. Let ﷐ℎ , 𝑘﷯ be the point on the Curve at which Normal is to be taken Given Curve is 𝑦=﷐𝑥﷮3﷯+2𝑥+6 Since point ﷐ℎ , 𝑘﷯ is on the Curve ⇒ ﷐ℎ , 𝑘﷯ will satisfies the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 𝑘=﷐﷐ℎ﷯﷮3﷯+2ℎ+6 𝑘=﷐ℎ﷮3﷯+2ℎ+6 We know that Slope of a tangent to the Curve is ﷐𝑑𝑦﷮𝑑𝑥﷯ 𝑦=﷐𝑥﷮3﷯+2𝑥+6 Differentiating w.r.t.𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯=3﷐𝑥﷮2﷯+2 Since tangent to be taken from ﷐ℎ , 𝑘﷯ Slope of tangent at ﷐ℎ , 𝑘﷯ is ﷐﷐𝑑𝑦﷮𝑑𝑥﷯│﷮﷐ℎ, 𝑘﷯﷯=3﷐ℎ﷮2﷯+2 We know that Slope of tangent × Slope of Normal =−1 ﷐3﷐ℎ﷮2﷯+2﷯ × Slope of Normal =−1 Slope of Normal = ﷐−1﷮3﷐ℎ﷮2﷯ + 2﷯ Also, Given that Normal is parallel to the line 𝑥+14𝑦+4=0 If two lines are parallel then slopes are equal ⇒ Slopes of Normal = Slope of line 𝑥+14𝑦+4=0 Now, line is 𝑥+14𝑦+4=0 14𝑦=−𝑥−4 𝑦=﷐− 𝑥 − 4﷮14﷯ 𝑦=﷐﷐−1﷮14﷯﷯𝑥−﷐﷐4﷮14﷯﷯ The above equation is of the form 𝑦=𝑚𝑥+𝑐 where m is slope ∴ Slope of line 𝑥+14𝑦+4=0 is ﷐−1﷮14﷯ Now, Slope of Normal = Slope of line 𝑥+14𝑦+4=0 ﷐−1﷮3﷐ℎ﷮2﷯ + 2﷯=﷐−1﷮ 14﷯ ﷐1﷮3﷐ℎ﷮2﷯ + 2﷯=﷐1﷮ 14﷯ 14=3﷐ℎ﷮2﷯+2 ﷐ℎ﷮2﷯+2=14 3﷐ℎ﷮2﷯=14−2 3﷐ℎ﷮2﷯=12 ﷐ℎ﷮2﷯=﷐12﷮3﷯ ﷐ℎ﷮2﷯=4 ℎ=±﷐﷮4﷯ ℎ=±2 Finding equation of normal

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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.