Tangents and Normals (using Differentiation)

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Chapter 6 Class 12 Application of Derivatives
Serial order wise

Ex 6.3, 20 - Find equation of normal at (am2, am3) for ay2 = x3

Ex 6.3,20 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,20 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.3,20 - Chapter 6 Class 12 Application of Derivatives - Part 4


Transcript

Question 20 Find the equation of the normal at the point (𝑎𝑚^2,𝑎𝑚^3) for the curve 𝑎𝑦^2=𝑥^3We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 Given 𝑎𝑦^2=𝑥^3 Differentiating w.r.t.𝑥 𝑑(𝑎𝑦^2 )/𝑑𝑥=𝑑(𝑥^3 )/𝑑𝑥 𝑎 𝑑(𝑦^2 )/𝑑𝑥=𝑑(𝑥^3 )/𝑑𝑥 𝑎 . 𝑑(𝑦^2 )/𝑑𝑥× 𝑑𝑦/𝑑𝑦=3𝑥^2 𝑎 .2𝑦 ×𝑑𝑦/𝑑𝑥=3𝑥^2 𝑑𝑦/𝑑𝑥=(3𝑥^2)/2𝑎𝑦 Slope of tangent at (𝑎𝑚^2,𝑎𝑚^3 ) is 〖𝑑𝑦/𝑑𝑥│〗_((𝑎𝑚^2,𝑎𝑚^3 ) )=(3(𝑎𝑚^2 )^2)/2𝑎(𝑎𝑚^3 ) =(3𝑎^2 𝑚^4)/(2𝑎^2 𝑚^3 )=3/2 𝑚 We know that Slope of tangent × Slope of Normal =−1 3𝑚/2 × Slope of Normal =−1 Slope of Normal =(−1)/(3𝑚/2) Slope of Normal =(−2)/3𝑚 Finding equation of normal Equation of Normal at (𝑎𝑚^2, 𝑎𝑚^3 ) & having Slope (−2)/3𝑚 is (𝑦−𝑎𝑚^3 )=(−2)/3𝑚 (𝑥−𝑎𝑚^2 ) 3𝑚(𝑦−𝑎𝑚^3 )=−2(𝑥−𝑎𝑚^2 ) 3𝑚𝑦−3 𝑎𝑚^4=−2𝑥+2𝑎𝑚^2 2𝑥+3𝑚𝑦−3𝑎𝑚^4−2𝑎𝑚^2=0 2𝑥+3𝑚𝑦−𝑎𝑚^2 (3𝑚^2+2)=0 We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Required Equation of Normal is : 𝟐𝒙+𝟑𝒎𝒚−𝒂𝒎^𝟐 (𝟑𝒎^𝟐+𝟐)=𝟎

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.