# Ex 6.3,21 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.3,21 Find the equation of the normal to the curve 𝑦=𝑥3+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=𝑥3+2𝑥+6 Since point ℎ , 𝑘 is on the Curve ⇒ ℎ , 𝑘 will satisfies the Equation of Curve Putting 𝑥=ℎ , 𝑦=𝑘 𝑘=ℎ3+2ℎ+6 𝑘=ℎ3+2ℎ+6 We know that Slope of a tangent to the Curve is 𝑑𝑦𝑑𝑥 𝑦=𝑥3+2𝑥+6 Differentiating w.r.t.𝑥 𝑑𝑦𝑑𝑥=3𝑥2+2 Since tangent to be taken from ℎ , 𝑘 Slope of tangent at ℎ , 𝑘 is 𝑑𝑦𝑑𝑥│ℎ, 𝑘=3ℎ2+2 We know that Slope of tangent × Slope of Normal =−1 3ℎ2+2 × Slope of Normal =−1 Slope of Normal = −13ℎ2 + 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 − 414 𝑦=−114𝑥−414 The above equation is of the form 𝑦=𝑚𝑥+𝑐 where m is slope ∴ Slope of line 𝑥+14𝑦+4=0 is −114 Now, Slope of Normal = Slope of line 𝑥+14𝑦+4=0 −13ℎ2 + 2=−1 14 13ℎ2 + 2=1 14 14=3ℎ2+2 ℎ2+2=14 3ℎ2=14−2 3ℎ2=12 ℎ2=123 ℎ2=4 ℎ=±4 ℎ=±2 Finding equation of normal

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Ex 6.3,21 You are here

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Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

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.