The two curves x 3 – 3xy 2 + 2 = 0 and 3x 2 y – y 3 = 2

(A) touch each other  

(B) cut at right angle

(C) cut at an angle π/3 

(D) cut at an angle π/4

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

Transcript

Question 2 The two curves x3 โ€“ 3xy2 + 2 = 0 and 3x2y โ€“ y3 = 2 (A) touch each other (B) cut at right angle (C) cut at an angle ๐œ‹/3 (D) cut at an angle ๐œ‹/4 Angles between two curves is same as angle between their tangents. So, first we will find slope of their tangents. Finding slope of tangent of first curve ๐‘ฅ^3โˆ’3๐‘ฅ๐‘ฆ^2+2=0 Differentiating w.r.t x ใ€–3๐‘ฅใ€—^2โˆ’ 3๐‘ฆ^2 โˆ’ 6xy ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 ใ€–3๐‘ฅใ€—^2โˆ’ 3๐‘ฆ^2 = 6xy ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (3๐‘ฅ^2 โˆ’3๐‘ฆ^2)/6๐‘ฅ๐‘ฆ ๐’…๐’š/๐’…๐’™ = (๐’™^๐Ÿ โˆ’๐’š^๐Ÿ)/๐Ÿ๐’™๐’š Let m1 = (๐‘ฅ^2 โˆ’๐‘ฆ^2)/2๐‘ฅ๐‘ฆ Finding slope of tangent of second curve 3๐‘ฅ^2 ๐‘ฆโˆ’๐‘ฆ^3โˆ’2=8 Differentiating w.r.t x 3๐‘ฅ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+6๐‘ฅ๐‘ฆ โˆ’ 3๐‘ฆ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 (3๐‘ฅ^2โˆ’3๐‘ฆ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’6xy ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’6๐‘ฅ๐‘ฆ)/(3๐‘ฅ^2 โˆ’3๐‘ฆ^2 ) ๐’…๐’š/๐’…๐’™ = (โˆ’๐Ÿ๐’™๐’š)/(๐’™^๐Ÿ โˆ’๐’š^๐Ÿ ) Let, ๐’Ž_๐Ÿ= (โˆ’2๐‘ฅ๐‘ฆ)/(๐‘ฅ^2 โˆ’๐‘ฆ^2 ) Finding Product of m1 & m2 m1 ร— m2 = (๐‘ฅ^2 โˆ’ ๐‘ฆ^2)/2๐‘ฅ๐‘ฆ ร— ((โˆ’2๐‘ฅ๐‘ฆ)/(๐‘ฅ^(2 )โˆ’ ๐‘ฆ^2 )) = โˆ’1 Since, product of the slopes is โˆ’1 โˆด Angle between tangents is 90ยฐ Thus, curves cut each other at right angle. So, the correct answer is (B)

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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.