# Ex 6.3,21 - Chapter 6 Class 12 Application of Derivatives

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 = 3 +2 +6 which are parallel to the line +14 +4=0. Let , be the point on the Curve at which Normal is to be taken Given Curve is = 3 +2 +6 Since point , is on the Curve , will satisfies the Equation of Curve Putting = , = = 3 +2 +6 = 3 +2 +6 We know that Slope of a tangent to the Curve is = 3 +2 +6 Differentiating w.r.t. =3 2 +2 Since tangent to be taken from , Slope of tangent at , is , =3 2 +2 We know that Slope of tangent Slope of Normal = 1 3 2 +2 Slope of Normal = 1 Slope of Normal = 1 3 2 + 2 Also, Given that Normal is parallel to the line +14 +4=0 If two lines are parallel then slopes are equal Slopes of Normal = Slope of line +14 +4=0 Now, line is +14 +4=0 14 = 4 = 4 14 = 1 14 4 14 The above equation is of the form = + where m is slope Slope of line +14 +4=0 is 1 14 Now, Slope of Normal = Slope of line +14 +4=0 1 3 2 + 2 = 1 14 1 3 2 + 2 = 1 14 14=3 2 +2 2 +2=14 3 2 =14 2 3 2 =12 2 = 12 3 2 =4 = 4 = 2 Finding equation of normal

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Chapter 6 Class 12 Application of Derivatives

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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.