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


Ex 6.3, 24 Find the equations of the tangent and normal to the hyperbola ๐‘ฅ^2/๐‘Ž^2 โ€“ ๐‘ฆ^2/๐‘^2 = 1 at the point (๐‘ฅ0 , ๐‘ฆ0) We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Finding ๐’…๐’š/๐’…๐’™ ๐‘ฅ^2/๐‘Ž^2 โˆ’๐‘ฆ^2/๐‘^2 =1 โˆ’๐‘ฆ^2/๐‘^2 =1โˆ’๐‘ฅ^2/๐‘Ž^2 ๐‘ฆ^2/๐‘^2 =๐‘ฅ^2/๐‘Ž^2 โˆ’1 Differentiating w.r.t.๐‘ฅ ๐‘‘(๐‘ฆ^2/๐‘^2 )/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^2/๐‘Ž^2 โˆ’1)/๐‘‘๐‘ฅ 1/๐‘^2 ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฅ=๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^2/๐‘Ž^2 )โˆ’๐‘‘(1)/๐‘‘๐‘ฅ 1/๐‘^2 ร— ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ=1/๐‘Ž^2 ๐‘‘(๐‘ฅ^2 )/๐‘‘๐‘ฅโˆ’0 1/๐‘^2 ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=1/๐‘Ž^2 . 2๐‘ฅ 1/๐‘^2 ร—2๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=1/๐‘Ž^2 2๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2๐‘ฅ/๐‘Ž^2 ร— ๐‘^2/2๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(๐‘^2 ๐‘ฅ)/(๐‘Ž^2 ๐‘ฆ) Slope of tangent at (๐‘ฅ0 , ๐‘ฆ0) is ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_((๐‘ฅ0 , ๐‘ฆ0) )=(๐‘^2 ๐‘ฅ0)/(๐‘Ž^2 ๐‘ฆ0) We know that Slope of tangent ร— Slope of Normal =โˆ’1 (๐‘^2 ๐‘ฅ0)/(๐‘Ž^2 ๐‘ฆ0) ร— Slope of Normal =โˆ’1 Slope of Normal = (โˆ’ใ€– ๐‘Žใ€—^2 ๐‘ฆ0)/(๐‘^2 ๐‘ฅ0) 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 tangent at (๐‘ฅ0 , ๐‘ฆ0) & having Slope (๐‘^2 ๐‘ฅ0)/(๐‘Ž^2 ๐‘ฆ0) is (๐‘ฆโˆ’๐‘ฆ0)=(๐‘^2 ๐‘ฅ0)/(๐‘Ž^2 ๐‘ฆ0) (๐‘ฅโˆ’๐‘ฅ0) ๐‘Ž^2 ๐‘ฆ0(๐‘ฆโˆ’๐‘ฆ0)=๐‘^2 ๐‘ฅ0 (๐‘ฅโˆ’ ๐‘ฅ0) ๐‘Ž^2 (๐‘ฆ0๐‘ฆโˆ’๐‘ฆ0^2 )=๐‘^2 (๐‘ฅ0 ๐‘ฅโˆ’๐‘ฅ0^2 ) (๐‘ฆ0 ๐‘ฆ โˆ’ ๐‘ฆ0^2)/๐‘^2 =((๐‘ฅ0 ๐‘ฅ โˆ’ ๐‘ฅ0^2 ))/๐‘Ž^2 (๐‘ฆ0 ๐‘ฆ )/๐‘^2 โˆ’(๐‘ฆ0^2)/๐‘^2 =(๐‘ฅ0 ๐‘ฅ )/๐‘Ž^2 โˆ’(๐‘ฅ0^2)/๐‘Ž^2 Equation of Normal at (๐‘ฅ0 ,๐‘ฆ0) & having Slope(โˆ’ใ€– ๐‘Žใ€—^2 ๐‘ฆ0)/(๐‘^2 ๐‘ฅ0) is (๐‘ฆโˆ’๐‘ฆ0)=(โˆ’ใ€– ๐‘Žใ€—^2 ๐‘ฆ0)/(๐‘^(2 ) ๐‘ฅ0) (๐‘ฅโˆ’๐‘ฅ0) ((๐‘ฆ โˆ’ ๐‘ฆ0))/(ใ€– ๐‘Žใ€—^(2 ) ๐‘ฆ0)=(โˆ’ 1)/(๐‘^(2 ) ๐‘ฅ0) (๐‘ฅโˆ’๐‘ฅ0) (๐‘ฆ โˆ’ ๐‘ฆ0)/(ใ€– ๐‘Žใ€—^(2 ) ๐‘ฆ0)=(โˆ’ (๐‘ฅ โˆ’ ๐‘ฅ0))/(๐‘^(2 ) ๐‘ฅ0) (๐’š โˆ’ ๐’š๐ŸŽ)/(ใ€– ๐’‚ใ€—^(๐Ÿ ) ๐’š๐ŸŽ)+(๐’™ โˆ’๐’™๐ŸŽ)/(ใ€– ๐’ƒใ€—^(๐Ÿ ) ๐’™๐ŸŽ)=๐ŸŽ (๐‘ฆ0 ๐‘ฆ )/๐‘^2 โˆ’(๐‘ฅ0 ๐‘ฅ )/๐‘Ž^2 =โˆ’(๐‘ฅ0^2)/๐‘Ž^2 +(๐‘ฆ0^2)/๐‘^2 ((๐‘ฆ0 ๐‘ฆ )/๐‘^2 โˆ’(๐‘ฅ0 ๐‘ฅ )/๐‘Ž^2 )=โˆ’((๐‘ฅ0^2)/๐‘Ž^2 โˆ’(๐‘ฆ0^2)/๐‘^2 ) ((๐‘ฆ0 ๐‘ฆ )/๐‘^2 โˆ’(๐‘ฅ0 ๐‘ฅ )/๐‘Ž^2 )=โˆ’1 (๐’™๐ŸŽ ๐’™ )/๐’‚^๐Ÿ โˆ’(๐’š๐ŸŽ ๐’š )/๐’ƒ^๐Ÿ =๐Ÿ Since point (๐‘ฅ_0 ,๐‘ฆ_0 ) lie on the Curve โˆด It will satisfy the Equation of Curve โˆด (๐‘ฅ0^2)/๐‘Ž^2 โˆ’(๐‘ฆ0^2)/๐‘^2 =1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.