Tangents and Normals (using Differentiation)

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Question 22 Find the equations of the tangent and normal to the parabola ๐ฆ^2=4๐๐ฅ at the point (๐๐ก2, 2๐๐ก).Given Curve is ๐ฆ^2=4๐๐ฅ We need to find equation of tangent & Normal at (๐๐ก2, 2๐๐ก) We know that Slope of tangent is ๐๐ฆ/๐๐ฅ ๐ฆ^2=4๐๐ฅ Differentiating w.r.t.๐ฅ ๐(๐ฆ^2 )/๐๐ฅ=๐(4๐๐ฅ)/๐๐ฅ ๐(๐ฆ^2 )/๐๐ฅ ร ๐๐ฆ/๐๐ฆ=4๐ ๐(๐ฅ)/๐๐ฅ ๐(๐ฆ^2 )/๐๐ฆ ร ๐๐ฆ/๐๐ฅ=4๐ 2๐ฆ ร๐๐ฆ/๐๐ฅ=4๐ ๐๐ฆ/๐๐ฅ=4๐/2๐ฆ Slope of tangent at (๐๐ก^2 , 2๐๐ก) is ใ๐๐ฆ/๐๐ฅโใ_((๐๐ก^2 , 2๐๐ก) )=4๐/2(2๐๐ก) =4๐/4๐๐ก=1/๐ก Also we know that Slope of tangent ร Slope of Normal =โ1 1/๐กร Slope of Normal =โ1 Slope of Normal =โ๐ก Finding equation of tangent & normal We know that Equation of line at (๐ฅ1 , ๐ฆ1)& having Slope m is ๐ฆโ๐ฆ1=๐(๐ฅโ๐ฅ1) Equation of tangent at (๐๐^๐ , ๐๐๐) & having Slope ๐/๐ is (๐ฆโ2๐๐ก)=1/๐ก (๐ฅโ๐๐ก^2 ) ๐ก(๐ฆโ2๐๐ก)=๐ฅโ๐๐ก^2 ๐ก๐ฆโ2๐๐ก^2=๐ฅโ๐๐ก^2 ๐ก๐ฆ=๐ฅโ๐๐ก^2+2๐๐ก^2 ๐๐=๐+๐๐^๐ Equation of Normal at (๐๐^๐ , ๐๐๐) & having Slope โt is (๐ฆโ2๐๐ก)=โ๐ก(๐ฅโ๐๐ก^2 ) ๐ฆโ2๐๐ก=โ๐ก๐ฅ+๐๐ก^3 ๐=โ๐๐+๐๐๐+๐๐^๐