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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


Ex 6.3, 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.