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Tangents and Normals (using Differentiation)

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Chapter 6 Class 12 Application of Derivatives
Serial order wise

Ex 6.3, 22 - Find equations of tangent and normal to parabola

Ex 6.3,22 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,22 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Transcript

Question 22 Find the equations of the tangent and normal to the parabola ๐‘ฆ^2=4๐‘Ž๐‘ฅ at the point (๐‘Ž๐‘ก2, 2๐‘Ž๐‘ก).Given Curve is ๐‘ฆ^2=4๐‘Ž๐‘ฅ We need to find equation of tangent & Normal at (๐‘Ž๐‘ก2, 2๐‘Ž๐‘ก) We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฆ^2=4๐‘Ž๐‘ฅ Differentiating w.r.t.๐‘ฅ ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฅ=๐‘‘(4๐‘Ž๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ=4๐‘Ž ๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘Ž 2๐‘ฆ ร—๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘Ž ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘Ž/2๐‘ฆ Slope of tangent at (๐‘Ž๐‘ก^2 , 2๐‘Ž๐‘ก) is ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_((๐‘Ž๐‘ก^2 , 2๐‘Ž๐‘ก) )=4๐‘Ž/2(2๐‘Ž๐‘ก) =4๐‘Ž/4๐‘Ž๐‘ก=1/๐‘ก Also we know that Slope of tangent ร— Slope of Normal =โˆ’1 1/๐‘กร— Slope of Normal =โˆ’1 Slope of Normal =โˆ’๐‘ก Finding equation of tangent & normal We know that Equation of line at (๐‘ฅ1 , ๐‘ฆ1)& having Slope m is ๐‘ฆโˆ’๐‘ฆ1=๐‘š(๐‘ฅโˆ’๐‘ฅ1) Equation of tangent at (๐’‚๐’•^๐Ÿ , ๐Ÿ๐’‚๐’•) & having Slope ๐Ÿ/๐’• is (๐‘ฆโˆ’2๐‘Ž๐‘ก)=1/๐‘ก (๐‘ฅโˆ’๐‘Ž๐‘ก^2 ) ๐‘ก(๐‘ฆโˆ’2๐‘Ž๐‘ก)=๐‘ฅโˆ’๐‘Ž๐‘ก^2 ๐‘ก๐‘ฆโˆ’2๐‘Ž๐‘ก^2=๐‘ฅโˆ’๐‘Ž๐‘ก^2 ๐‘ก๐‘ฆ=๐‘ฅโˆ’๐‘Ž๐‘ก^2+2๐‘Ž๐‘ก^2 ๐’•๐’š=๐’™+๐’‚๐’•^๐Ÿ Equation of Normal at (๐’‚๐’•^๐Ÿ , ๐Ÿ๐’‚๐’•) & having Slope โ€“t is (๐‘ฆโˆ’2๐‘Ž๐‘ก)=โˆ’๐‘ก(๐‘ฅโˆ’๐‘Ž๐‘ก^2 ) ๐‘ฆโˆ’2๐‘Ž๐‘ก=โˆ’๐‘ก๐‘ฅ+๐‘Ž๐‘ก^3 ๐’š=โˆ’๐’•๐’™+๐Ÿ๐’‚๐’•+๐’‚๐’•^๐Ÿ‘

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.