The abscissa of the point on the curve 3y = 6x – 5x 3 , the normal at which passes through origin is:

(A)1 

(B) 1/3

(C) 2 

(D) 1/2

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

Transcript

Question 1 The abscissa of the point on the curve 3y = 6x – 5x3, the normal at which passes through origin is: 1 (B) 1/3 (C) 2 (D) 1/2 Given curve 3𝑦=6π‘₯βˆ’5π‘₯^3 Finding π’…π’š/𝒅𝒙 π’…π’š/𝒅𝒙=πŸβˆ’πŸ“π’™^𝟐 Now, Slope of normal Γ— Slope of tangent = βˆ’1 Slope of normal (m) = (βˆ’πŸ)/(𝑺𝒍𝒐𝒑𝒆 𝒐𝒇 π’•π’‚π’π’ˆπ’†π’π’•) = (βˆ’1)/(𝑑𝑦/𝑑π‘₯) = (βˆ’πŸ)/((𝟐 βˆ’ πŸ“ 𝒙^𝟐)) Finding equation of Normal with Slope (βˆ’πŸ)/(𝟐 βˆ’πŸ“π’™^𝟐 ) & Passing through origin (0, 0) (𝑦 βˆ’ 𝑦_1)/(π‘₯ βˆ’ 𝑦_1 ) = m (𝑦 βˆ’ 𝟎)/(π‘₯ βˆ’ 𝟎) = (βˆ’1)/(2 βˆ’5π‘₯^2 ) (π’š )/(𝒙 ) = (βˆ’πŸ)/(𝟐 βˆ’πŸ“π’™^𝟐 ) Now, let normal and the curve passes through (h, k). Since (h, k) lies on normal So, (h, k) will satisfy the equation of normal π‘˜/β„Ž = (βˆ’1)/(2 βˆ’ 5β„Ž^2 ) π’Œ/𝒉 = 𝟏/(πŸ“π’‰^𝟐 βˆ’ 𝟐 ) Since (h, k) lies on curve So, (h, k) will satisfy the equation of curve 3π‘˜=6β„Žβˆ’5β„Ž^3 3k = h (6 βˆ’ 5β„Ž^2) π‘˜/β„Ž = (β„Ž (6 βˆ’5β„Ž^2))/(3β„Ž ) π’Œ/𝒉 = (πŸ” βˆ’πŸ“π’‰^𝟐)/(πŸ‘ ) On Solving (1) and (2) 𝟏/(πŸ“π’‰^𝟐 βˆ’ 𝟐) = (πŸ” βˆ’πŸ“π’‰^𝟐)/πŸ‘ 3 = (6 βˆ’ 5h2) (5h2 βˆ’ 2) 3 = 6 (5h2 βˆ’ 2) βˆ’ 5h2 (5h2 βˆ’ 2) 3 = 30h2 βˆ’ 12 βˆ’ 25h4 + 10h2 3 + 12 = βˆ’25h4 + 30h2 + 10h2 15 = βˆ’25h4 + 40h2 25h4 βˆ’ 40h2 + 15 = 0 5h4 βˆ’ 8h2 + 3 = 0 For simplicity, let h2 = z The equation becomes, 5z2 βˆ’ 8z + 3 = 0 5z2 βˆ’ 5z βˆ’ 3z + 3 = 0 5z (z βˆ’ 1) βˆ’ 3 (z βˆ’ 1) = 0 (5z βˆ’ 3) (z βˆ’ 1) = 0 Now, Either, (5z βˆ’ 3) = 0 z = 3/5 Putting h2 = z h2 = 3/5 h = ±√(πŸ‘/πŸ“) which does not match with any option Or, (z βˆ’ 1) = 0 z = 1 Putting h2 = z h2 = 1 h = Β± 1 But h = -1 is not in option ∴ h = 1 is correct. ∴ Abscissa = x βˆ’ coordinate = h = 1 So, the correct answer is (A)

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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.