Question 17 - Tangents and Normals (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

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Ex 6.3, 17 - Find points on y = x3 at which slope of tangent

Ex 6.3,17 - Chapter 6 Class 12 Application of Derivatives - Part 2
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Question 17 Find the points on the curve š‘¦=š‘„3 at which the slope of the tangent is equal to the y-coordinate of the pointLet the Point be (ā„Ž , š‘˜) on the Curve š‘¦=š‘„3 Where Slope of tangent at (ā„Ž , š‘˜)=š‘¦āˆ’š‘š‘œš‘œš‘Ÿš‘‘š‘–š‘›š‘Žš‘”š‘’ š‘œš‘“ (ā„Ž, š‘˜) i.e. 怖š‘‘š‘¦/š‘‘š‘„ā”‚怗_((ā„Ž, š‘˜) )=š‘˜ Given š‘¦=š‘„^3 Differentiating w.r.t.š‘„ š‘‘š‘¦/š‘‘š‘„=3š‘„^2 āˆ“ Slope of tangent at (ā„Ž , š‘˜) is 怖š‘‘š‘¦/š‘‘š‘„ā”‚怗_((ā„Ž, š‘˜) )=3ā„Ž^2 From (1) 怖š‘‘š‘¦/š‘‘š‘„ā”‚怗_((ā„Ž, š‘˜) )=š‘˜ 3ā„Ž^2=š‘˜ Also Point (ā„Ž , š‘˜) is on the Curve š‘¦=š‘„^3 Point (ā„Ž , š‘˜) must Satisfy the Equation of Curve i.e. š‘˜=ā„Ž^3 Now our equations are 3ā„Ž^2=š‘˜ ā€¦(1) & š‘˜=ā„Ž^3 ā€¦(2) Putting Value of š‘˜=3ā„Ž^2 in (3) 3ā„Ž^2=ā„Ž^3 ā„Ž^3āˆ’3ā„Ž^2=0 ā„Ž^2 (ā„Žāˆ’3)=0 ā„Ž^2=0 ā„Ž=0 ā„Žāˆ’3=0 ā„Ž=3 When š’‰=šŸŽ 3ā„Ž^2=š‘˜ 3(0)=š‘˜ š‘˜=0 Hence, point is (0, 0) When š’‰=šŸ‘ 3ā„Ž^2=š‘˜ 3(3)^2=š‘˜ š‘˜=27 Hence, point is (3 , 27)

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