# Ex 6.3,22 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 14, 2021 by Teachoo

Ex 6.3

Ex 6.3, 1
Deleted for CBSE Board 2023 Exams

Ex 6.3,2 Deleted for CBSE Board 2023 Exams

Ex 6.3,3 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,4 Deleted for CBSE Board 2023 Exams

Ex 6.3, 5 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,6 Deleted for CBSE Board 2023 Exams

Ex 6.3,7 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,8 Deleted for CBSE Board 2023 Exams

Ex 6.3,9 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,10 Deleted for CBSE Board 2023 Exams

Ex 6.3,11 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,12 Deleted for CBSE Board 2023 Exams

Ex 6.3,13 Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (i) Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iii) Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (iv) Important Deleted for CBSE Board 2023 Exams

Ex 6.3, 14 (v) Deleted for CBSE Board 2023 Exams

Ex 6.3,15 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,16 Deleted for CBSE Board 2023 Exams

Ex 6.3,17 Deleted for CBSE Board 2023 Exams

Ex 6.3,18 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,19 Deleted for CBSE Board 2023 Exams

Ex 6.3,20 Deleted for CBSE Board 2023 Exams

Ex 6.3,21 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,22 Deleted for CBSE Board 2023 Exams You are here

Ex 6.3,23 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,24 Important Deleted for CBSE Board 2023 Exams

Ex 6.3,25 Deleted for CBSE Board 2023 Exams

Ex 6.3,26 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.3,27 (MCQ) Deleted for CBSE Board 2023 Exams

Last updated at April 14, 2021 by Teachoo

Ex 6.3, 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