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 𝑦^2=4𝑎𝑥 at the point (𝑎𝑡2, 2𝑎𝑡).Given Curve is 𝑦^2=4𝑎𝑥 We need to find equation of tangent & Normal at (𝑎𝑡2, 2𝑎𝑡) We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑦^2=4𝑎𝑥 Differentiating w.r.t.𝑥 𝑑(𝑦^2 )/𝑑𝑥=𝑑(4𝑎𝑥)/𝑑𝑥 𝑑(𝑦^2 )/𝑑𝑥 × 𝑑𝑦/𝑑𝑦=4𝑎 𝑑(𝑥)/𝑑𝑥 𝑑(𝑦^2 )/𝑑𝑦 × 𝑑𝑦/𝑑𝑥=4𝑎 2𝑦 ×𝑑𝑦/𝑑𝑥=4𝑎 𝑑𝑦/𝑑𝑥=4𝑎/2𝑦 Slope of tangent at (𝑎𝑡^2 , 2𝑎𝑡) is 〖𝑑𝑦/𝑑𝑥│〗_((𝑎𝑡^2 , 2𝑎𝑡) )=4𝑎/2(2𝑎𝑡) =4𝑎/4𝑎𝑡=1/𝑡 Also we know that Slope of tangent × Slope of Normal =−1 1/𝑡× Slope of Normal =−1 Slope of Normal =−𝑡 Finding equation of tangent & normal We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent at (𝒂𝒕^𝟐 , 𝟐𝒂𝒕) & having Slope 𝟏/𝒕 is (𝑦−2𝑎𝑡)=1/𝑡 (𝑥−𝑎𝑡^2 ) 𝑡(𝑦−2𝑎𝑡)=𝑥−𝑎𝑡^2 𝑡𝑦−2𝑎𝑡^2=𝑥−𝑎𝑡^2 𝑡𝑦=𝑥−𝑎𝑡^2+2𝑎𝑡^2 𝒕𝒚=𝒙+𝒂𝒕^𝟐 Equation of Normal at (𝒂𝒕^𝟐 , 𝟐𝒂𝒕) & having Slope –t is (𝑦−2𝑎𝑡)=−𝑡(𝑥−𝑎𝑡^2 ) 𝑦−2𝑎𝑡=−𝑡𝑥+𝑎𝑡^3 𝒚=−𝒕𝒙+𝟐𝒂𝒕+𝒂𝒕^𝟑

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.