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

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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