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

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

Transcript

Ex 6.3,8 Find a point on the curve =( 2)^2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4). Given Curves is =( 2)^2 Let AB be the chord joining the Point (2 , 0) & (4 ,4) & CD be the tangent to the Curve =( 2)^2 Given that tangent is Parallel to chord i.e. CD AB Slope of CD = Slope of AB Slope of tangent CD is / =( 2)^2 / =( ( 2)^2)/ / =2( 2) . ( 2) / =2( 2) . (1 0) / =2( 2) & Slope of AB As AB is chord joining Points (2 , 0) & (4 , 4) Slope of AB =(4 0)/(4 2) =4/2 =2 Now Slope of CD = Slope of AB 2( 2)=2 2=2/2 2=1 =3 Finding y when =3 =( 2)^2 =(3 2)^2 =(1)^2 =1 Hence Point is ( , ) 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.