Tangents and Normals (using Differentiation)

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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### Transcript

Question 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iii) π¦=π₯3 ππ‘ (1, 1)Given Curve is π¦=π₯3 Differentiating w.r.t.π₯ ππ¦/ππ₯=3π₯^2 We know that Slope of tangent is ππ¦/ππ₯ Given point is (1 , 3) Slope of tangent ππ¦/ππ₯ at (1 , 3) is γππ¦/ππ₯βγ_((1 , 3) )=3(1)^2 =3 We know that Slope of tangent Γ Slope of Normal =β1 3 Γ Slope of Normal =β1 Slope of Normal = (β1)/( 3) Hence Slope of tangent at (1 , 1)=3 Slope of Normal at ( 1 , 1)=(β1)/( 3) Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (1, 1) & Slope 3 is (π¦ β1)=3(π₯β1) π¦ β1=3π₯β3 π¦=3π₯β3+1 π=ππβπ Equation of Normal at (1, 1) & Slope (β1)/3 is (π¦β1)=(β1)/( 3) (π₯β1) 3(π¦β1)=β(π₯β1) 3π¦β3=βπ₯+1 3π¦+π₯β3β1=0 π+ππβπ=π