Ex 6.3, 8 - Find a point on y = (x-2)2, tangent is parallel - Finding point when tangent is parallel/ perpendicular


  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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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 chord i.e. CD โˆฅ AB โ‡’ Slope of CD = Slope of AB Slope of tangent CD is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฆ=(๐‘ฅโˆ’2)^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(๐‘‘(๐‘ฅโˆ’2)^2)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2(๐‘ฅโˆ’2) . (๐‘ฅโˆ’2) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2(๐‘ฅโˆ’2) . (1โˆ’0) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=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 =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) Hence the tangent is parallel to the chord at (3 ,1)

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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.