Question 13 - Examples - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by

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

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Question 13 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 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 and Computer Science at Teachoo