Check sibling questions

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

(A)1Β 

(B) 1/3

(C) 2Β 

(D) 1/2

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.