Question 7 - Tangents and Normals (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

Last updated at July 20, 2026 by

Ex 6.3, 7 - Find points at which tangent is parallel to x-axis

Ex 6.3,7 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,7 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Question 7 Find points at which the tangent to the curve š‘¦=š‘„^3āˆ’3š‘„^2āˆ’9š‘„+7 is parallel to the x-axisEquation of Curve is š‘¦=š‘„^3āˆ’3š‘„^2āˆ’9š‘„+7 Differentiating w.r.t. š‘„ š‘‘š‘¦/š‘‘š‘„=š‘‘(š‘„^3āˆ’ 3š‘„^2 āˆ’ 9š‘„ + 7)/š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„=3š‘„^2āˆ’6š‘„āˆ’9+0 š‘‘š‘¦/š‘‘š‘„=3š‘„^2āˆ’6š‘„āˆ’9 š‘‘š‘¦/š‘‘š‘„=3(š‘„^2āˆ’2š‘„āˆ’3) š‘‘š‘¦/š‘‘š‘„=3(š‘„^2āˆ’3š‘„+š‘„āˆ’3) š‘‘š‘¦/š‘‘š‘„=3(š‘„(š‘„āˆ’3)+1(š‘„āˆ’3)) š‘‘š‘¦/š‘‘š‘„=3(š‘„+1)(š‘„āˆ’3) Given tangent to the curve is parallel to the š‘„āˆ’š‘Žš‘„š‘–š‘  i.e. the Slope of tangent = Slope of š‘„āˆ’š‘Žš‘„š‘–š‘  š’…š’š/š’…š’™=šŸŽ 3(š‘„+1)(š‘„āˆ’3)=0 (š‘„+1)(š‘„āˆ’3)=0 Thus š‘„=āˆ’1 & š‘„=3 When š’™=āˆ’šŸ š‘¦=š‘„^3āˆ’3š‘„^2āˆ’9š‘„+7 =(āˆ’1)^3āˆ’3(āˆ’1)^2āˆ’9(āˆ’1)+7 =āˆ’1āˆ’3+9+7 =12 Point is (āˆ’1 , 12) When š’™=šŸ‘ š‘¦=š‘„^3āˆ’3š‘„^2āˆ’9š‘„+7 =(3)^3āˆ’3(3)^2āˆ’9(3)+7 =27āˆ’27āˆ’27+7 =āˆ’ 20 Point is (3 , āˆ’20) Hence , the tangent to the Curve is parallel to the š‘„āˆ’š‘Žš‘„š‘–š‘  at (āˆ’šŸ , šŸšŸ) & (šŸ‘ , āˆ’šŸšŸŽ)

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