Ex 6.3,8 - Class 12

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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 chord i.e. CD ∥ AB ⇒ Slope of CD = Slope of AB Slope of tangent CD is 𝑑𝑦/𝑑𝑥 𝑦=(𝑥−2)^2 𝑑𝑦/𝑑𝑥=(𝑑(𝑥−2)^2)/𝑑𝑥 𝑑𝑦/𝑑𝑥=2(𝑥−2) . (𝑥−2) 𝑑𝑦/𝑑𝑥=2(𝑥−2) . (1−0) 𝑑𝑦/𝑑𝑥=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 =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) Hence the tangent is parallel to the chord at (3 ,1)