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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise


Ex 6.3, 21 Find the equation of the normal to the curve ๐‘ฆ=๐‘ฅ^3+2๐‘ฅ+6 which are parallel to the line ๐‘ฅ+14๐‘ฆ+4=0.Let (โ„Ž , ๐‘˜) be the point on the Curve at which Normal is to be taken Given Curve is ๐‘ฆ=๐‘ฅ^3+2๐‘ฅ+6 Since point (โ„Ž , ๐‘˜) is on the Curve โˆด (โ„Ž , ๐‘˜) will satisfies the Equation of Curve Putting ๐‘ฅ=โ„Ž , ๐‘ฆ=๐‘˜ ๐‘˜=โ„Ž^3+2โ„Ž+6 We know that Slope of a tangent to the Curve is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฆ=๐‘ฅ^3+2๐‘ฅ+6 Differentiating w.r.t. ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3๐‘ฅ^2+2 Since tangent to be taken from (โ„Ž , ๐‘˜) Slope of tangent at (โ„Ž , ๐‘˜) is ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_((โ„Ž, ๐‘˜) )=3โ„Ž^2+2 We know that Slope of tangent ร— Slope of Normal =โˆ’1 (3โ„Ž^2+2) ร— Slope of Normal =โˆ’1 Slope of Normal = (โˆ’1)/(3โ„Ž^2 + 2) Also, Given that Normal is parallel to the line ๐‘ฅ+14๐‘ฆ+4=0 If two lines are parallel then slopes are equal โ‡’ Slopes of Normal = Slope of line ๐‘ฅ+14๐‘ฆ+4=0 Now, line is ๐‘ฅ+14๐‘ฆ+4=0 14๐‘ฆ=โˆ’๐‘ฅโˆ’4 ๐‘ฆ=(โˆ’ ๐‘ฅ โˆ’ 4)/14 ๐‘ฆ=((โˆ’1)/14)๐‘ฅโˆ’(4/14) The above equation is of the form ๐‘ฆ=๐‘š๐‘ฅ+๐‘ where m is slope โˆด Slope of line ๐‘ฅ+14๐‘ฆ+4=0 is (โˆ’1)/14 Now, Slope of Normal = Slope of line ๐‘ฅ+14๐‘ฆ+4=0 (โˆ’1)/(3โ„Ž^2 + 2)=(โˆ’1)/( 14) 1/(3โ„Ž^2 + 2)=1/( 14) 14=3โ„Ž^2+2 3โ„Ž^2+2=14 3โ„Ž^2=14โˆ’2 3โ„Ž^2=12 โ„Ž^2=12/3 โ„Ž^2=4 โ„Ž=ยฑโˆš4 โ„Ž=ยฑ2 When ๐’‰=๐Ÿ ๐‘˜=โ„Ž^3+2โ„Ž+6 ๐‘˜=(2)^3+2(2)+6 ๐‘˜=8+4+6 ๐‘˜=18 โˆด Point is (๐Ÿ , ๐Ÿ๐Ÿ–) When ๐’‰=โˆ’๐Ÿ ๐‘˜=โ„Ž^3+2โ„Ž+6 ๐‘˜=(โˆ’2)^3+2(โˆ’2)+6 ๐‘˜=โˆ’8โˆ’4+6 ๐‘˜=โˆ’ 6 โˆด Point is (โˆ’๐Ÿ, โˆ’๐Ÿ”) Finding equation of normal We know that Equation of line at (๐‘ฅ1 , ๐‘ฆ1)& having Slope m is ๐‘ฆโˆ’๐‘ฆ1=๐‘š(๐‘ฅโˆ’๐‘ฅ1) Equation of 1st Normal at (2 , 18) & having Slope (โˆ’1)/14 is (๐‘ฆโˆ’18)=(โˆ’1)/14 (๐‘ฅโˆ’2) 14(๐‘ฆโˆ’18)=โˆ’(๐‘ฅโˆ’2) 14๐‘ฆโˆ’252=โˆ’๐‘ฅ+2 14๐‘ฆ+๐‘ฅโˆ’252โˆ’2=0 ๐’™+๐Ÿ๐Ÿ’๐’šโˆ’๐Ÿ๐Ÿ“๐Ÿ’=๐ŸŽ Equation of 2nd Normal at (โˆ’2 , โˆ’6) & having Slope (โˆ’1)/14 is (๐‘ฆโˆ’(โˆ’6))=(โˆ’1)/14 (๐‘ฅโˆ’(โˆ’2)) ๐‘ฆ+6=(โˆ’1)/14 (๐‘ฅ+2) 14๐‘ฆ+84=โˆ’๐‘ฅโˆ’2 14๐‘ฆ+๐‘ฅ+84+2=0 ๐’™+๐Ÿ๐Ÿ’๐’š+๐Ÿ–๐Ÿ”=๐ŸŽ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.