Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12


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


Ex 6.3, 8 Find a point on the curve ๐‘ฆ=(๐‘ฅโˆ’2)^2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).Given Curves is ๐‘ฆ=(๐‘ฅโˆ’2)^2 Let AB be the chord joining the Point (2 , 0) & (4 ,4) & CD be the tangent to the Curve ๐‘ฆ=(๐‘ฅโˆ’2)^2 Given that tangent is Parallel to the chord i.e. CD โˆฅ AB โˆด Slope of CD = Slope of AB If two lines are parallel, then their slopes are equal Slope of tangent CD Slope of tangent CD = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ =(๐‘‘(๐‘ฅ โˆ’ 2)^2)/๐‘‘๐‘ฅ = 2(๐‘ฅโˆ’2) (๐‘‘ (๐‘ฅ โˆ’ 2))/๐‘‘๐‘ฅ = 2(๐‘ฅโˆ’2) (1โˆ’0) = 2(๐‘ฅโˆ’2) Slope of AB As AB is chord joining Points (2 , 0) & (4 , 4) Slope of AB =(4 โˆ’ 0)/(4 โˆ’ 2) =4/2 As slope of line joining point (๐‘ฅ , ๐‘ฆ) & (๐‘ฅ2 , ๐‘ฆ2) ๐‘–๐‘  (๐‘ฆ2 โˆ’ ๐‘ฆ1)/(๐‘ฅ2 โˆ’ ๐‘ฅ1) =2 Now, Slope of CD = Slope of AB 2(๐‘ฅโˆ’2)=2 ๐‘ฅโˆ’2=2/2 ๐‘ฅโˆ’2=1 ๐‘ฅ=3 Finding y when ๐‘ฅ=3 ๐‘ฆ=(๐‘ฅโˆ’2)^2 ๐‘ฆ=(3โˆ’2)^2 ๐‘ฆ=(1)^2 ๐‘ฆ=1 Hence, Point is (๐Ÿ‘ , ๐Ÿ) Thus, the tangent is parallel to the chord at (3 ,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 10 years. He provides courses for Maths and Science at Teachoo.