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Ex 6.3

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

Ex 6.3,8 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,8 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.3,8 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Transcript

Ex 6.3, 8 Find a point on the curve 𝑦=(𝑥−2)^2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).Given Curves is 𝑦=(𝑥−2)^2 Let AB be the chord joining the Point (2 , 0) & (4 ,4) & CD be the tangent to the Curve 𝑦=(𝑥−2)^2 Given that tangent is Parallel to the chord i.e. CD ∥ AB ∴ Slope of CD = Slope of AB If two lines are parallel, then their slopes are equal Slope of tangent CD Slope of tangent CD = 𝑑𝑦/𝑑𝑥 =(𝑑(𝑥 − 2)^2)/𝑑𝑥 = 2(𝑥−2) (𝑑 (𝑥 − 2))/𝑑𝑥 = 2(𝑥−2) (1−0) = 2(𝑥−2) Slope of AB As AB is chord joining Points (2 , 0) & (4 , 4) Slope of AB =(4 − 0)/(4 − 2) =4/2 As slope of line joining point (𝑥 , 𝑦) & (𝑥2 , 𝑦2) 𝑖𝑠 (𝑦2 − 𝑦1)/(𝑥2 − 𝑥1) =2 Now, Slope of CD = Slope of AB 2(𝑥−2)=2 𝑥−2=2/2 𝑥−2=1 𝑥=3 Finding y when 𝑥=3 𝑦=(𝑥−2)^2 𝑦=(3−2)^2 𝑦=(1)^2 𝑦=1 Hence, Point is (𝟑 , 𝟏) Thus, the tangent is parallel to the chord at (3 ,1)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.