Example 45 - Find equation of normal to x2 = 4y which passes

Example 45 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 45 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 45 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 45 - Chapter 6 Class 12 Application of Derivatives - Part 5

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Example 45 Find the equation of the normal to the curve x2 = 4y which passes through the point (1, 2).Given Curve x2 = 4y Differentiating w.r.t. x 2x = 4𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = π‘₯/2 ∴ Slope of normal = (βˆ’1)/(𝑑𝑦/𝑑π‘₯) = (βˆ’1)/((π‘₯/2) ) = (βˆ’πŸ)/𝒙 Let (h, k) be the point where normal & curve intersect We need to find equation of the normal to the curve x2 = 4y which passes through the point (1, 2). But to find equation… we need to find point on curve Let (h, k) be the point where normal & curve intersect ∴ Slope of normal at (h, k) = (βˆ’πŸ)/𝒉 Equation of normal passing through (h, k) with slope (βˆ’2)/β„Ž is y – y1 = m(x – x1) y βˆ’ k = (βˆ’πŸ)/𝒉 (x βˆ’ h) Since normal passes through (1, 2), it will satisfy its equation 2 βˆ’ k = (βˆ’2)/β„Ž (1 βˆ’ h) k = 2 + 𝟐/𝒉 (1 βˆ’ h) Also, (h, k) lies on curve x2 = 4y h2 = 4k k = 𝒉^𝟐/πŸ’ From (1) and (2) 2 + 2/β„Ž (1 βˆ’ h) = β„Ž^2/4 2 + 2/β„Ž βˆ’ 2 = β„Ž^2/4 2/β„Ž = β„Ž^2/4 β„Ž^3 = 8 h = ("8" )^(1/3) h = 2 Putting h = 2 in (2) k = β„Ž^2/4 = γ€–(2)γ€—^2/4 = 4/4 = 1 Hence, h = 2 & k = 1 Putting h = 2 & k = 1 in equation of normal π‘¦βˆ’π‘˜=(βˆ’2(π‘₯ βˆ’ β„Ž))/β„Ž π‘¦βˆ’1=(βˆ’2(π‘₯ βˆ’ 2))/2 π‘¦βˆ’1=βˆ’1(π‘₯βˆ’2) π‘¦βˆ’1=βˆ’π‘₯+2 π‘₯+𝑦=2+1 𝒙+π’š=πŸ‘

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.